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College Physics 



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OpenStax College 

Rice University 

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ISBN-10 1938168003 

ISBN-13 978-1-938168-00-0 

Revision CP-1-000-DW 



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RICE 



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Table of Contents 

Preface 7 

1 Introduction: The Nature of Science and Physics 11 

Physics: An Introduction 12 

Physical Quantities and Units 18 

Accuracy, Precision, and Significant Figures 25 

Approximation 29 

2 Kinematics 35 

Displacement 36 

Vectors, Scalars, and Coordinate Systems 38 

Time, Velocity, and Speed 39 

Acceleration 43 

Motion Equations for Constant Acceleration in One Dimension 51 

Problem-Solving Basics for One-Dimensional Kinematics 60 

Falling Objects 62 

Graphical Analysis of One-Dimensional Motion 68 

3 Two-Dimensional Kinematics 85 

Kinematics in Two Dimensions: An Introduction 86 

Vector Addition and Subtraction: Graphical Methods 88 

Vector Addition and Subtraction: Analytical Methods 95 

Projectile Motion 101 

Addition of Velocities 107 

4 Dynamics: Force and Newton's Laws of Motion 123 

Development of Force Concept 124 

Newton's First Law of Motion: Inertia 125 

Newton's Second Law of Motion: Concept of a System 126 

Newton's Third Law of Motion: Symmetry in Forces 132 

Normal, Tension, and Other Examples of Forces 134 

Problem-Solving Strategies 142 

Further Applications of Newton's Laws of Motion 143 

Extended Topic: The Four Basic Forces — An Introduction 148 

5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity 161 

Friction 162 

Drag Forces 166 

Elasticity: Stress and Strain 170 

6 Uniform Circular Motion and Gravitation 185 

Rotation Angle and Angular Velocity 186 

Centripetal Acceleration 189 

Centripetal Force 192 

Fictitious Forces and Non-inertial Frames: The Coriolis Force 196 

Newton's Universal Law of Gravitation 199 

Satellites and Kepler's Laws: An Argument for Simplicity 205 

7 Work, Energy, and Energy Resources 219 

Work: The Scientific Definition 221 

Kinetic Energy and the Work-Energy Theorem 223 

Gravitational Potential Energy 227 

Conservative Forces and Potential Energy 231 

Nonconservative Forces 235 

Conservation of Energy 239 

Power 242 

Work, Energy, and Power in Humans 246 

World Energy Use 248 

8 Linear Momentum and Collisions 261 

Linear Momentum and Force 262 

Impulse 264 

Conservation of Momentum 266 

Elastic Collisions in One Dimension 269 

Inelastic Collisions in One Dimension 271 

Collisions of Point Masses in Two Dimensions 274 

Introduction to Rocket Propulsion 277 

9 Statics and Torque 289 

The First Condition for Equilibrium 290 

The Second Condition for Equilibrium 291 

Stability 295 

Applications of Statics, Including Problem-Solving Strategies 298 

Simple Machines 301 

Forces and Torques in Muscles and Joints 304 

10 Rotational Motion and Angular Momentum 317 

Angular Acceleration 318 

Kinematics of Rotational Motion 322 

Dynamics of Rotational Motion: Rotational Inertia 326 

Rotational Kinetic Energy: Work and Energy Revisited 329 



Angular Momentum and Its Conservation 336 

Collisions of Extended Bodies in Two Dimensions 341 

Gyroscopic Effects: Vector Aspects of Angular Momentum 344 

11 Fluid Statics 357 

What Is a Fluid? 358 

Density 359 

Pressure 361 

Variation of Pressure with Depth in a Fluid 363 

Pascal's Principle 366 

Gauge Pressure, Absolute Pressure, and Pressure Measurement 368 

Archimedes' Principle 371 

Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action 377 

Pressures in the Body 385 

12 Fluid Dynamics and Its Biological and Medical Applications 397 

Flow Rate and Its Relation to Velocity 398 

Bernoulli's Equation 400 

The Most General Applications of Bernoulli's Equation 404 

Viscosity and Laminar Flow; Poiseuille's Law 407 

The Onset of Turbulence 413 

Motion of an Object in a Viscous Fluid 414 

Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes 416 

13 Temperature, Kinetic Theory, and the Gas Laws 429 

Temperature 430 

Thermal Expansion of Solids and Liquids 436 

The Ideal Gas Law 442 

Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature 447 

Phase Changes 453 

Humidity, Evaporation, and Boiling 458 

14 Heat and Heat Transfer Methods 469 

Heat 470 

Temperature Change and Heat Capacity 472 

Phase Change and Latent Heat 476 

Heat Transfer Methods 481 

Conduction 482 

Convection 486 

Radiation 490 

15 Thermodynamics 505 

The First Law of Thermodynamics 506 

The First Law of Thermodynamics and Some Simple Processes 510 

Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency 517 

Carnot's Perfect Heat Engine: The Second Law of Thermodynamics Restated 522 

Applications of Thermodynamics: Heat Pumps and Refrigerators 526 

Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy 530 

Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation 536 

16 Oscillatory Motion and Waves 549 

Hooke's Law: Stress and Strain Revisited 551 

Period and Frequency in Oscillations 555 

Simple Harmonic Motion: A Special Periodic Motion 556 

The Simple Pendulum 560 

Energy and the Simple Harmonic Oscillator 562 

Uniform Circular Motion and Simple Harmonic Motion 564 

Damped Harmonic Motion 566 

Forced Oscillations and Resonance 569 

Waves 571 

Superposition and Interference 574 

Energy in Waves: Intensity 578 

17 Physics of Hearing 589 

Sound 590 

Speed of Sound, Frequency, and Wavelength 592 

Sound Intensity and Sound Level 595 

Doppler Effect and Sonic Booms 598 

Sound Interference and Resonance: Standing Waves in Air Columns 603 

Hearing 609 

Ultrasound 614 

18 Electric Charge and Electric Field 627 

Static Electricity and Charge: Conservation of Charge 629 

Conductors and Insulators 633 

Coulomb's Law 637 

Electric Field: Concept of a Field Revisited 638 

Electric Field Lines: Multiple Charges 640 

Electric Forces in Biology 643 

Conductors and Electric Fields in Static Equilibrium 644 

Applications of Electrostatics 648 



19 Electric Potential and Electric Field 663 

Electric Potential Energy: Potential Difference 664 

Electric Potential in a Uniform Electric Field 668 

Electrical Potential Due to a Point Charge 671 

Equipotential Lines 673 

Capacitors and Dielectrics 675 

Capacitors in Series and Parallel 681 

Energy Stored in Capacitors 684 

20 Electric Current, Resistance, and Ohm's Law 695 

Current 696 

Ohm's Law: Resistance and Simple Circuits 701 

Resistance and Resistivity 703 

Electric Power and Energy 707 

Alternating Current versus Direct Current 710 

Electric Hazards and the Human Body 714 

Nerve Conduction-Electrocardiograms 717 

21 Circuits, Bioelectricity, and DC Instruments 733 

Resistors in Series and Parallel 734 

Electromotive Force: Terminal Voltage 742 

Kirchhoff's Rules 748 

DC Voltmeters and Ammeters 752 

Null Measurements 756 

DC Circuits Containing Resistors and Capacitors 758 

22 Magnetism 773 

Magnets 774 

Ferromagnets and Electromagnets 776 

Magnetic Fields and Magnetic Field Lines 780 

Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field 781 

Force on a Moving Charge in a Magnetic Field: Examples and Applications 783 

The Hall Effect 787 

Magnetic Force on a Current-Carrying Conductor 789 

Torque on a Current Loop: Motors and Meters 791 

Magnetic Fields Produced by Currents: Ampere's Law 793 

Magnetic Force between Two Parallel Conductors 797 

More Applications of Magnetism 798 

23 Electromagnetic Induction, AC Circuits, and Electrical Technologies 813 

Induced Emf and Magnetic Flux 815 

Faraday's Law of Induction: Lenz's Law 816 

Motional Emf 819 

Eddy Currents and Magnetic Damping 822 

Electric Generators 825 

Back Emf 828 

Transformers 828 

Electrical Safety: Systems and Devices 832 

Inductance 836 

RL Circuits 839 

Reactance, Inductive and Capacitive 841 

RLC Series AC Circuits 844 

24 Electromagnetic Waves 861 

Maxwell's Equations: Electromagnetic Waves Predicted and Observed 862 

Production of Electromagnetic Waves 864 

The Electromagnetic Spectrum 866 

Energy in Electromagnetic Waves 878 

25 Geometric Optics 887 

The Ray Aspect of Light 889 

The Law of Reflection 889 

The Law of Refraction 891 

Total Internal Reflection 896 

Dispersion: The Rainbow and Prisms 901 

Image Formation by Lenses 905 

Image Formation by Mirrors 916 

26 Vision and Optical Instruments 931 

Physics of the Eye 932 

Vision Correction 935 

Color and Color Vision 938 

Microscopes 941 

Telescopes 946 

Aberrations 949 

27 Wave Optics 957 

The Wave Aspect of Light: Interference 958 

Huygens's Principle: Diffraction 959 

Young's Double Slit Experiment 961 

Multiple Slit Diffraction 965 



Single Slit Diffraction 969 

Limits of Resolution: The Rayleigh Criterion 972 

Thin Film Interference 976 

Polarization 980 

*Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light 987 

28 Special Relativity 999 

Einstein's Postulates 1000 

Simultaneity And Time Dilation 1002 

Length Contraction 1008 

Relativistic Addition of Velocities 1011 

Relativistic Momentum 1016 

Relativistic Energy 1017 

29 Introduction to Quantum Physics 1031 

Quantization of Energy 1033 

The Photoelectric Effect 1035 

Photon Energies and the Electromagnetic Spectrum 1037 

Photon Momentum 1043 

The Particle-Wave Duality 1047 

The Wave Nature of Matter 1048 

Probability: The Heisenberg Uncertainty Principle 1051 

The Particle-Wave Duality Reviewed 1054 

30 Atomic Physics 1063 

Discovery of the Atom 1064 

Discovery of the Parts of the Atom: Electrons and Nuclei 1065 

Bohr's Theory of the Hydrogen Atom 1071 

X Rays: Atomic Origins and Applications 1077 

Applications of Atomic Excitations and De-Excitations 1081 

The Wave Nature of Matter Causes Quantization 1088 

Patterns in Spectra Reveal More Quantization 1090 

Quantum Numbers and Rules 1092 

The Pauli Exclusion Principle 1096 

31 Radioactivity and Nuclear Physics 1111 

Nuclear Radioactivity 1112 

Radiation Detection and Detectors 1115 

Substructure of the Nucleus 1117 

Nuclear Decay and Conservation Laws 1121 

Half-Life and Activity 1127 

Binding Energy 1132 

Tunneling 1136 

32 Medical Applications of Nuclear Physics 1147 

Medical Imaging and Diagnostics 1149 

Biological Effects of Ionizing Radiation 1152 

Therapeutic Uses of Ionizing Radiation 1157 

Food Irradiation 1159 

Fusion 1160 

Fission 1165 

Nuclear Weapons 1169 

33 Particle Physics 1181 

The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited 1182 

The Four Basic Forces 1183 

Accelerators Create Matter from Energy 1185 

Particles, Patterns, and Conservation Laws 1188 

Quarks: Is That All There Is? 1192 

GUTs: The Unification of Forces 1199 

34 Frontiers of Physics 1209 

Cosmology and Particle Physics 1210 

General Relativity and Quantum Gravity 1216 

Superstrings 1221 

Dark Matter and Closure 1221 

Complexity and Chaos 1224 

High-temperature Superconductors 1225 

Some Questions We Know to Ask 1226 

A Atomic Masses 1235 

B Selected Radioactive Isotopes 1241 

C Useful Information 1245 

D Glossary of Key Symbols and Notation 1251 

Index 1262 



PREFACE 7 



PREFACE 



About OpenStax College 

OpenStax College is a non-profit organization committed to improving student access to quality learning materials. Our free textbooks are developed 
and peer-reviewed by educators to ensure they are readable, accurate, and meet the scope and sequence requirements of modern college courses. 
Unlike traditional textbooks, OpenStax College resources live online and are owned by the community of educators using them. Through our 
partnerships with companies and foundations committed to reducing costs for students, OpenStax College is working to improve access to higher 
education for all. OpenStax College is an initiative of Rice University and is made possible through the generous support of several philanthropic 
foundations. 

About This Book 

Welcome to College Physics, an OpenStax College resource created with several goals in mind: accessibility, afford ability, customization, and student 
engagement — all while encouraging learners toward high levels of learning. Instructors and students alike will find that this textbook offers a strong 
foundation in introductory physics, with algebra as a prerequisite. It is available for free online and in low-cost print and e-book editions. 

To broaden access and encourage community curation, College Physics is "open source" licensed under a Creative Commons Attribution (CC-BY) 
license. Everyone is invited to submit examples, emerging research, and other feedback to enhance and strengthen the material and keep it current 
and relevant for today's students. You can make suggestions by contacting us at info@openstaxcollege.org. You can find the status of the project, as 
well as alternate versions, corrections, etc., on the StaxDash at http://openstaxcollege.org (http://openstaxcollege.org) . 

To the Student 

This book is written for you. It is based on the teaching and research experience of numerous physicists and influenced by a strong recollection of 
their own struggles as students. After reading this book, we hope you see that physics is visible everywhere. Applications range from driving a car to 
launching a rocket, from a skater whirling on ice to a neutron star spinning in space, and from taking your temperature to taking a chest X-ray. 

To the Instructor 

This text is intended for one-year introductory courses requiring algebra and some trigonometry, but no calculus. OpenStax College provides the 
essential supplemental resources at http://openstaxcollege.org ; however, we have pared down the number of supplements to keep costs low. 
College Physics can be easily customized for your course using Connexions (http://cnx.org/content/colll406). Simply select the content most 
relevant to your curriculum and create a textbook that speaks directly to the needs of your class. 

General Approach 

College Physics is organized such that topics are introduced conceptually with a steady progression to precise definitions and analytical applications. 
The analytical aspect (problem solving) is tied back to the conceptual before moving on to another topic. Each introductory chapter, for example, 
opens with an engaging photograph relevant to the subject of the chapter and interesting applications that are easy for most students to visualize. 

Organization, Level, and Content 

There is considerable latitude on the part of the instructor regarding the use, organization, level, and content of this book. By choosing the types of 
problems assigned, the instructor can determine the level of sophistication required of the student. 

Concepts and Calculations 

The ability to calculate does not guarantee conceptual understanding. In order to unify conceptual, analytical, and calculation skills within the learning 
process, we have integrated Strategies and Discussions throughout the text. 

Modern Perspective 

The chapters on modern physics are more complete than many other texts on the market, with an entire chapter devoted to medical applications of 
nuclear physics and another to particle physics. The final chapter of the text, "Frontiers of Physics," is devoted to the most exciting endeavors in 
physics. It ends with a module titled "Some Questions We Know to Ask." 

Supplements 

Accompanying the main text are a Student Solutions Manual and an Instructor Solutions Manual (http://openstaxcollege.org/textbooks/ 

college-physics) . The Student Solutions Manual provides worked-out solutions to select end-of-module Problems and Exercises. The Instructor 
Solutions Manual provides worked-out solutions to all Exercises. 

Features of OpenStax College Physics 

The following briefly describes the special features of this text. 

Modularity 

This textbook is organized on Connexions (http://cnx.org) as a collection of modules that can be rearranged and modified to suit the needs of a 
particular professor or class. That being said, modules often contain references to content in other modules, as most topics in physics cannot be 
discussed in isolation. 



8 PREFACE 

Learning Objectives 

Every module begins with a set of learning objectives. These objectives are designed to guide the instructor in deciding what content to include or 
assign, and to guide the student with respect to what he or she can expect to learn. After completing the module and end-of-module exercises, 
students should be able to demonstrate mastery of the learning objectives. 

Call-Outs 

Key definitions, concepts, and equations are called out with a special design treatment. Call-outs are designed to catch readers' attention, to make it 
clear that a specific term, concept, or equation is particularly important, and to provide easy reference for a student reviewing content. 

Key Terms 

Key terms are in bold and are followed by a definition in context. Definitions of key terms are also listed in the Glossary, which appears at the end of 
the module. 

Worked Examples 

Worked examples have four distinct parts to promote both analytical and conceptual skills. Worked examples are introduced in words, always using 
some application that should be of interest. This is followed by a Strategy section that emphasizes the concepts involved and how solving the 
problem relates to those concepts. This is followed by the mathematical Solution and Discussion. 

Many worked examples contain multiple-part problems to help the students learn how to approach normal situations, in which problems tend to have 
multiple parts. Finally, worked examples employ the techniques of the problem-solving strategies so that students can see how those strategies 
succeed in practice as well as in theory. 

Problem-Solving Strategies 

Problem solving strategies are first presented in a special section and subsequently appear at crucial points in the text where students can benefit 
most from them. Problem-solving strategies have a logical structure that is reinforced in the worked examples and supported in certain places by line 
drawings that illustrate various steps. 

Misconception Alerts 

Students come to physics with preconceptions from everyday experiences and from previous courses. Some of these preconceptions are 
misconceptions, and many are very common among students and the general public. Some are inadvertently picked up through misunderstandings 
of lectures and texts. The Misconception Alerts feature is designed to point these out and correct them explicitly. 

Take-Home Investigations 

Take Home Investigations provide the opportunity for students to apply or explore what they have learned with a hands-on activity. 

Things Great and Small 

In these special topic essays, macroscopic phenomena (such as air pressure) are explained with submicroscopic phenomena (such as atoms 
bouncing off walls). These essays support the modern perspective by describing aspects of modern physics before they are formally treated in later 
chapters. Connections are also made between apparently disparate phenomena. 

Simulations 

Where applicable, students are directed to the interactive PHeT physics simulations developed by the University of Colorado 
(http://phet.colorado.edu (http://phet.colorado.edu) ). There they can further explore the physics concepts they have learned about in the module. 

Summary 

Module summaries are thorough and functional and present all important definitions and equations. Students are able to find the definitions of all 
terms and symbols as well as their physical relationships. The structure of the summary makes plain the fundamental principles of the module or 
collection and serves as a useful study guide. 

Glossary 

At the end of every module or chapter is a glossary containing definitions of all of the key terms in the module or chapter. 

End-of-Module Problems 

At the end of every chapter is a set of Conceptual Questions and/or skills-based Problems & Exercises. Conceptual Questions challenge students' 
ability to explain what they have learned conceptually, independent of the mathematical details. Problems & Exercises challenge students to apply 
both concepts and skills to solve mathematical physics problems. Online, every other problem includes an answer that students can reveal 
immediately by clicking on a "Show Solution" button. Fully worked solutions to select problems are available in the Student Solutions Manual and the 
Teacher Solutions Manual. 

In addition to traditional skills-based problems, there are three special types of end-of-module problems: Integrated Concept Problems, Unreasonable 
Results Problems, and Construct Your Own Problems. All of these problems are indicated with a subtitle preceding the problem. 

Integrated Concept Problems 

In Unreasonable Results Problems, students are challenged not only to apply concepts and skills to solve a problem, but also to analyze the answer 
with respect to how likely or realistic it really is. These problems contain a premise that produces an unreasonable answer and are designed to further 
emphasize that properly applied physics must describe nature accurately and is not simply the process of solving equations. 



PREFACE 9 

Unreasonable Results 

In Unreasonable Results Problems, students are challenged to not only apply concepts and skills to solve a problem, but also to analyze the answer 
with respect to how likely or realistic it really is. These problems contain a premise that produces an unreasonable answer and are designed to further 
emphasize that properly applied physics must describe nature accurately and is not simply the process of solving equations. 

Construct Your Own Problem 

These problems require students to construct the details of a problem, justify their starting assumptions, show specific steps in the problem's solution, 
and finally discuss the meaning of the result. These types of problems relate well to both conceptual and analytical aspects of physics, emphasizing 
that physics must describe nature. Often they involve an integration of topics from more than one chapter. Unlike other problems, solutions are not 
provided since there is no single correct answer. Instructors should feel free to direct students regarding the level and scope of their considerations. 
Whether the problem is solved and described correctly will depend on initial assumptions. 

Appendices 

Appendix A: Atomic Masses 

Appendix B: Selected Radioactive Isotopes 

Appendix C: Useful Information 

Appendix D: Glossary of Key Symbols and Notation 

Acknowledgements 

This text is based on the work completed by Dr. Paul Peter Urone in collaboration with Roger Hinrichs, Kim Dirks, and Manjula Sharma. We would 
like to thank the authors as well as the numerous professors (a partial list follows) who have contributed their time and energy to review and provide 
feedback on the manuscript. Their input has been critical in maintaining the pedagogical integrity and accuracy of the text. 

Senior Contributing Authors 

Dr. Paul Peter Urone 

Dr. Roger Hinrichs, State University of New York, College at Oswego 

Contributing Authors 

Kim Dirks, University of Auckland, New Zealand 
Dr. Manjula Sharma, University of Sydney, Australia 

Expert Reviewers 

Erik Christensen, P.E, South Florida Community College 

Dr. Eric Kincanon, Gonzaga University 

Dr. Douglas Ingram, Texas Christian University 

Lee H. LaRue, Paris Junior College 

Dr. Marc Sher, College of William and Mary 

Dr. Ulrich Zurcher, Cleveland State University 

Dr. Matthew Adams, Crafton Hills College, San Bernardino Community College District 

Dr. Chuck Pearson, Virginia Intermont College 

Our Partners 

WebAssign 

Webassign is an independent online homework and assessment system that has been available commercially since 1998. WebAssign has recently 
begun to support the Open Education Resource community by creating a high quality online homework solution for selected open-source textbooks, 
available at an affordable price to students. These question collections include randomized values and variables, immediate feedback, links to the 
open-source textbook, and a variety of text-specific resources and tools; as well as the same level of rigorous coding and accuracy-checking as any 
commercially available online homework solution supporting traditionally available textbooks. 

Sapling Learning 

Sapling Learning provides the most effective interactive homework and instruction that improve student learning outcomes for the problem-solving 
disciplines. They offer an enjoyable teaching and effective learning experience that is distinctive in three important ways: 

• Ease of Use: Sapling Learning's easy to use interface keeps students engaged in problem-solving, not struggling with the software. 

• Targeted Instructional Content: Sapling Learning increases student engagement and comprehension by delivering immediate feedback and 
targeted instructional content. 

• Unsurpassed Service and Support: Sapling Learning makes teaching more enjoyable by providing a dedicated Masters or PhD level colleague 
to service instructors' unique needs throughout the course, including content customization. 



10 PREFACE 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 11 



INTRODUCTION: THE NATURE OF SCIENCE AND 
PHYSICS 




Figure 1.1 Galaxies are as immense as atoms are small. Yet the same laws of physics describe both, and all the rest of nature — an indication of the underlying unity in the 
universe. The laws of physics are surprisingly few in number, implying an underlying simplicity to nature's apparent complexity, (credit: NASA, JPL-Caltech, P. Barmby, 
Harvard-Smithsonian Center for Astrophysics) 



Learning Objectives 



1.1. Physics: An Introduction 

• Explain the difference between a principle and a law. 

• Explain the difference between a model and a theory. 

1.2. Physical Quantities and Units 

• Perform unit conversions both in the SI and English units. 

• Explain the most common prefixes in the SI units and be able to write them in scientific notation. 

1.3. Accuracy, Precision, and Significant Figures 

• Determine the appropriate number of significant figures in both addition and subtraction, as well as multiplication and division 
calculations. 

• Calculate the percent uncertainty of a measurement. 

1.4. Approximation 

• Make reasonable approximations based on given data. 



Introduction to Science and the Realm of Physics, Physical Quantities, and Units 

What is your first reaction when you hear the word "physics"? Did you imagine working through difficult equations or memorizing formulas that seem 
to have no real use in life outside the physics classroom? Many people come to the subject of physics with a bit of fear. But as you begin your 
exploration of this broad-ranging subject, you may soon come to realize that physics plays a much larger role in your life than you first thought, no 
matter your life goals or career choice. 

For example, take a look at the image above. This image is of the Andromeda Galaxy, which contains billions of individual stars, huge clouds of gas, 
and dust. Two smaller galaxies are also visible as bright blue spots in the background. At a staggering 2.5 million light years from the Earth, this 
galaxy is the nearest one to our own galaxy (which is called the Milky Way). The stars and planets that make up Andromeda might seem to be the 
furthest thing from most people's regular, everyday lives. But Andromeda is a great starting point to think about the forces that hold together the 
universe. The forces that cause Andromeda to act as it does are the same forces we contend with here on Earth, whether we are planning to send a 
rocket into space or simply raise the walls for a new home. The same gravity that causes the stars of Andromeda to rotate and revolve also causes 
water to flow over hydroelectric dams here on Earth. Tonight, take a moment to look up at the stars. The forces out there are the same as the ones 
here on Earth. Through a study of physics, you may gain a greater understanding of the interconnectedness of everything we can see and know in 
this universe. 

Think now about all of the technological devices that you use on a regular basis. Computers, smart phones, GPS systems, MP3 players, and satellite 
radio might come to mind. Next, think about the most exciting modern technologies that you have heard about in the news, such as trains that levitate 
above tracks, "invisibility cloaks" that bend light around them, and microscopic robots that fight cancer cells in our bodies. All of these groundbreaking 
advancements, commonplace or unbelievable, rely on the principles of physics. Aside from playing a significant role in technology, professionals such 
as engineers, pilots, physicians, physical therapists, electricians, and computer programmers apply physics concepts in their daily work. For example, 
a pilot must understand how wind forces affect a flight path and a physical therapist must understand how the muscles in the body experience forces 
as they move and bend. As you will learn in this text, physics principles are propelling new, exciting technologies, and these principles are applied in 
a wide range of careers. 

In this text, you will begin to explore the history of the formal study of physics, beginning with natural philosophy and the ancient Greeks, and leading 
up through a review of Sir Isaac Newton and the laws of physics that bear his name. You will also be introduced to the standards scientists use when 
they study physical quantities and the interrelated system of measurements most of the scientific community uses to communicate in a single 



12 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 

mathematical language. Finally, you will study the limits of our ability to be accurate and precise, and the reasons scientists go to painstaking lengths 
to be as clear as possible regarding their own limitations. 

1.1 Physics: An Introduction 




Figure 1.2 The flight formations of migratory birds such as Canada geese are governed by the laws of physics, (credit: David Merrett) 

The physical universe is enormously complex in its detail. Every day, each of us observes a great variety of objects and phenomena. Over the 
centuries, the curiosity of the human race has led us collectively to explore and catalog a tremendous wealth of information. From the flight of birds to 
the colors of flowers, from lightning to gravity, from quarks to clusters of galaxies, from the flow of time to the mystery of the creation of the universe, 
we have asked questions and assembled huge arrays of facts. In the face of all these details, we have discovered that a surprisingly small and 
unified set of physical laws can explain what we observe. As humans, we make generalizations and seek order. We have found that nature is 
remarkably cooperative — it exhibits the underlying order and simplicity we so value. 

It is the underlying order of nature that makes science in general, and physics in particular, so enjoyable to study. For example, what do a bag of 
chips and a car battery have in common? Both contain energy that can be converted to other forms. The law of conservation of energy (which says 
that energy can change form but is never lost) ties together such topics as food calories, batteries, heat, light, and watch springs. Understanding this 
law makes it easier to learn about the various forms energy takes and how they relate to one another. Apparently unrelated topics are connected 
through broadly applicable physical laws, permitting an understanding beyond just the memorization of lists of facts. 

The unifying aspect of physical laws and the basic simplicity of nature form the underlying themes of this text. In learning to apply these laws, you will, 
of course, study the most important topics in physics. More importantly, you will gain analytical abilities that will enable you to apply these laws far 
beyond the scope of what can be included in a single book. These analytical skills will help you to excel academically, and they will also help you to 
think critically in any professional career you choose to pursue. This module discusses the realm of physics (to define what physics is), some 
applications of physics (to illustrate its relevance to other disciplines), and more precisely what constitutes a physical law (to illuminate the importance 
of experimentation to theory). 

Science and the Realm of Physics 

Science consists of the theories and laws that are the general truths of nature as well as the body of knowledge they encompass. Scientists are 
continually trying to expand this body of knowledge and to perfect the expression of the laws that describe it. Physics is concerned with describing 
the interactions of energy, matter, space, and time, and it is especially interested in what fundamental mechanisms underlie every phenomenon. The 
concern for describing the basic phenomena in nature essentially defines the realm of physics. 

Physics aims to describe the function of everything around us, from the movement of tiny charged particles to the motion of people, cars, and 
spaceships. In fact, almost everything around you can be described quite accurately by the laws of physics. Consider a smart phone (Figure 1.3). 
Physics describes how electricity interacts with the various circuits inside the device. This knowledge helps engineers select the appropriate materials 
and circuit layout when building the smart phone. Next, consider a GPS system. Physics describes the relationship between the speed of an object, 
the distance over which it travels, and the time it takes to travel that distance. When you use a GPS device in a vehicle, it utilizes these physics 
equations to determine the travel time from one location to another. 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 13 




Figure 1.3 The Apple "iPhone" is a common smart phone with a GPS function. Physics describes the way that electricity flows through the circuits of this device. Engineers 
use their knowledge of physics to construct an iPhone with features that consumers will enjoy. One specific feature of an iPhone is the GPS function. GPS uses physics 
equations to determine the driving time between two locations on a map. (credit: @gletham GIS, Social, Mobile Tech Images) 



Applications of Physics 

You need not be a scientist to use physics. On the contrary, knowledge of physics is useful in everyday situations as well as in nonscientific 
professions. It can help you understand how microwave ovens work, why metals should not be put into them, and why they might affect pacemakers. 
(See Figure 1.4 and Figure 1.5.) Physics allows you to understand the hazards of radiation and rationally evaluate these hazards more easily. 
Physics also explains the reason why a black car radiator helps remove heat in a car engine, and it explains why a white roof helps keep the inside of 
a house cool. Similarly, the operation of a car's ignition system as well as the transmission of electrical signals through our body's nervous system are 
much easier to understand when you think about them in terms of basic physics. 

Physics is the foundation of many important disciplines and contributes directly to others. Chemistry, for example — since it deals with the interactions 
of atoms and molecules — is rooted in atomic and molecular physics. Most branches of engineering are applied physics. In architecture, physics is at 
the heart of structural stability, and is involved in the acoustics, heating, lighting, and cooling of buildings. Parts of geology rely heavily on physics, 
such as radioactive dating of rocks, earthquake analysis, and heat transfer in the Earth. Some disciplines, such as biophysics and geophysics, are 
hybrids of physics and other disciplines. 

Physics has many applications in the biological sciences. On the microscopic level, it helps describe the properties of cell walls and cell membranes 
(Figure 1.6 and Figure 1.7). On the macroscopic level, it can explain the heat, work, and power associated with the human body. Physics is involved 
in medical diagnostics, such as x-rays, magnetic resonance imaging (MRI), and ultrasonic blood flow measurements. Medical therapy sometimes 
directly involves physics; for example, cancer radiotherapy uses ionizing radiation. Physics can also explain sensory phenomena, such as how 
musical instruments make sound, how the eye detects color, and how lasers can transmit information. 

It is not necessary to formally study all applications of physics. What is most useful is knowledge of the basic laws of physics and a skill in the 
analytical methods for applying them. The study of physics also can improve your problem-solving skills. Furthermore, physics has retained the most 
basic aspects of science, so it is used by all of the sciences, and the study of physics makes other sciences easier to understand. 




Figure 1.4 The laws of physics help us understand how common appliances work. For example, the laws of physics can help explain how microwave ovens heat up food, and 
they also help us understand why it is dangerous to place metal objects in a microwave oven, (credit: Money BlogNewz) 



14 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 




Figure 1.5 These two applications of physics have more in common than meets the eye. Microwave ovens use electromagnetic waves to heat food. Magnetic resonance 
imaging (MRI) also uses electromagnetic waves to yield an image of the brain, from which the exact location of tumors can be determined, (credit: Rashmi Chawla, Daniel 
Smith, and Paul E. Marik) 




Figure 1.6 Physics, chemistry, and biology help describe the properties of cell walls in plant cells, such as the onion cells seen here, (credit: Umberto Salvagnin) 

Extracellular Fluid 



Protein channel 

' (transport protein) 

* Globular protein 



Hydrophilic heads 




Phospholipid bilayer 



Alpha-Helix protein ' 

(Integral protein) 

Cytoplasm 

Figure 1.7 An artist's rendition of the the structure of a cell membrane. Membranes form the boundaries of animal cells and are complex in structure and function. Many of the 
most fundamental properties of life, such as the firing of nerve cells, are related to membranes. The disciplines of biology, chemistry, and physics all help us understand the 
membranes of animal cells, (credit: Mariana Ruiz) 



Models, Theories, and Laws; The Role of Experimentation 

The laws of nature are concise descriptions of the universe around us; they are human statements of the underlying laws or rules that all natural 
processes follow. Such laws are intrinsic to the universe; humans did not create them and so cannot change them. We can only discover and 
understand them. Their discovery is a very human endeavor, with all the elements of mystery, imagination, struggle, triumph, and disappointment 
inherent in any creative effort. (See Figure 1.8 and Figure 1.9.) The cornerstone of discovering natural laws is observation; science must describe 
the universe as it is, not as we may imagine it to be. 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 15 




Sir Isaac JVflrfra 

Figure 1.8 Isaac Newton (1642-1727) was very reluctant to publish his revolutionary work and had to be convinced to do so. In his later years, he stepped down from his 
academic post and became exchequer of the Royal Mint. He took this post seriously, inventing reeding (or creating ridges) on the edge of coins to prevent unscrupulous 
people from trimming the silver off of them before using them as currency, (credit: Arthur Shuster and Arthur E. Shipley: Britain's Heritage of Science. London, 1917.) 




Figure 1.9 Marie Curie (1867-1934) sacrificed monetary assets to help finance her early research and damaged her physical well-being with radiation exposure. She is the 
only person to win Nobel prizes in both physics and chemistry. One of her daughters also won a Nobel Prize, (credit: Wikimedia Commons) 

We all are curious to some extent. We look around, make generalizations, and try to understand what we see — for example, we look up and wonder 
whether one type of cloud signals an oncoming storm. As we become serious about exploring nature, we become more organized and formal in 
collecting and analyzing data. We attempt greater precision, perform controlled experiments (if we can), and write down ideas about how the data 
may be organized and unified. We then formulate models, theories, and laws based on the data we have collected and analyzed to generalize and 
communicate the results of these experiments. 

A model is a representation of something that is often too difficult (or impossible) to display directly. While a model is justified with experimental proof, 
it is only accurate under limited situations. An example is the planetary model of the atom in which electrons are pictured as orbiting the nucleus, 
analogous to the way planets orbit the Sun. (See Figure 1.10.) We cannot observe electron orbits directly, but the mental image helps explain the 
observations we can make, such as the emission of light from hot gases (atomic spectra). Physicists use models for a variety of purposes. For 
example, models can help physicists analyze a scenario and perform a calculation, or they can be used to represent a situation in the form of a 
computer simulation. A theory is an explanation for patterns in nature that is supported by scientific evidence and verified multiple times by various 
groups of researchers. Some theories include models to help visualize phenomena, whereas others do not. Newton's theory of gravity, for example, 
does not require a model or mental image, because we can observe the objects directly with our own senses. The kinetic theory of gases, on the 
other hand, is a model in which a gas is viewed as being composed of atoms and molecules. Atoms and molecules are too small to be observed 
directly with our senses — thus, we picture them mentally to understand what our instruments tell us about the behavior of gases. 

A law uses concise language to describe a generalized pattern in nature that is supported by scientific evidence and repeated experiments. Often, a 
law can be expressed in the form of a single mathematical equation. Laws and theories are similar in that they are both scientific statements that 
result from a tested hypothesis and are supported by scientific evidence. However, the designation law is reserved for a concise and very general 
statement that describes phenomena in nature, such as the law that energy is conserved during any process, or Newton's second law of motion, 
which relates force, mass, and acceleration by the simple equation F = ma . A theory, in contrast, is a less concise statement of observed 
phenomena. For example, the Theory of Evolution and the Theory of Relativity cannot be expressed concisely enough to be considered a law. The 
biggest difference between a law and a theory is that a theory is much more complex and dynamic. A law describes a single action, whereas a theory 
explains an entire group of related phenomena. And, whereas a law is a postulate that forms the foundation of the scientific method, a theory is the 
end result of that process. 

Less broadly applicable statements are usually called principles (such as Pascal's principle, which is applicable only in fluids), but the distinction 
between laws and principles often is not carefully made. 



16 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 




Figure 1.10 What is a model? This planetary model of the atom shows electrons orbiting the nucleus. It is a drawing that we use to form a mental image of the atom that we 
cannot see directly with our eyes because it is too small. 

Models, Theories, and Laws 

Models, theories, and laws are used to help scientists analyze the data they have already collected. However, often after a model, theory, or law 
has been developed, it points scientists toward new discoveries they would not otherwise have made. 

The models, theories, and laws we devise sometimes imply the existence of objects or phenomena as yet unobserved. These predictions are 
remarkable triumphs and tributes to the power of science. It is the underlying order in the universe that enables scientists to make such spectacular 
predictions. However, if experiment does not verify our predictions, then the theory or law is wrong, no matter how elegant or convenient it is. Laws 
can never be known with absolute certainty because it is impossible to perform every imaginable experiment in order to confirm a law in every 
possible scenario. Physicists operate under the assumption that all scientific laws and theories are valid until a counterexample is observed. If a 
good-quality, verifiable experiment contradicts a well-established law, then the law must be modified or overthrown completely. 

The study of science in general and physics in particular is an adventure much like the exploration of uncharted ocean. Discoveries are made; 
models, theories, and laws are formulated; and the beauty of the physical universe is made more sublime for the insights gained. 

The Scientific Method 

As scientists inquire and gather information about the world, they follow a process called the scientific method. This process typically begins 
with an observation and question that the scientist will research. Next, the scientist typically performs some research about the topic and then 
devises a hypothesis. Then, the scientist will test the hypothesis by performing an experiment. Finally, the scientist analyzes the results of the 
experiment and draws a conclusion. Note that the scientific method can be applied to many situations that are not limited to science, and this 
method can be modified to suit the situation. 

Consider an example. Let us say that you try to turn on your car, but it will not start. You undoubtedly wonder: Why will the car not start? You can 
follow a scientific method to answer this question. First off, you may perform some research to determine a variety of reasons why the car will not 
start. Next, you will state a hypothesis. For example, you may believe that the car is not starting because it has no engine oil. To test this, you 
open the hood of the car and examine the oil level. You observe that the oil is at an acceptable level, and you thus conclude that the oil level is 
not contributing to your car issue. To troubleshoot the issue further, you may devise a new hypothesis to test and then repeat the process again. 



The Evolution of Natural Philosophy into Modern Physics 

Physics was not always a separate and distinct discipline. It remains connected to other sciences to this day. The word physics comes from Greek, 
meaning nature. The study of nature came to be called "natural philosophy." From ancient times through the Renaissance, natural philosophy 
encompassed many fields, including astronomy, biology, chemistry, physics, mathematics, and medicine. Over the last few centuries, the growth of 
knowledge has resulted in ever-increasing specialization and branching of natural philosophy into separate fields, with physics retaining the most 
basic facets. (See Figure 1.11, Figure 1.12, and Figure 1.13.) Physics as it developed from the Renaissance to the end of the 19th century is called 
classical physics. It was transformed into modern physics by revolutionary discoveries made starting at the beginning of the 20th century. 




Figure 1.11 Over the centuries, natural philosophy has evolved into more specialized disciplines, as illustrated by the contributions of some of the greatest minds in history. 
The Greek philosopher Aristotle (384-322 B.C.) wrote on a broad range of topics including physics, animals, the soul, politics, and poetry, (credit: Jastrow (2006)/Ludovisi 
Collection) 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 17 




Figure 1.12 Galileo Galilei (1564-1642) laid the foundation of modern experimentation and made contributions in mathematics, physics, and astronomy, (credit: Domenico 
Tintoretto) 




Figure 1.13 Niels Bohr (1885-1962) made fundamental contributions to the development of quantum mechanics, one part of modern physics, (credit: United States Library of 
Congress Prints and Photographs Division) 

Classical physics is not an exact description of the universe, but it is an excellent approximation under the following conditions: Matter must be 
moving at speeds less than about 1% of the speed of light, the objects dealt with must be large enough to be seen with a microscope, and only weak 
gravitational fields, such as the field generated by the Earth, can be involved. Because humans live under such circumstances, classical physics 
seems intuitively reasonable, while many aspects of modern physics seem bizarre. This is why models are so useful in modern physics — they let us 
conceptualize phenomena we do not ordinarily experience. We can relate to models in human terms and visualize what happens when objects move 
at high speeds or imagine what objects too small to observe with our senses might be like. For example, we can understand an atom's properties 
because we can picture it in our minds, although we have never seen an atom with our eyes. New tools, of course, allow us to better picture 
phenomena we cannot see. In fact, new instrumentation has allowed us in recent years to actually "picture" the atom. 

Limits on the Laws of Classical Physics 

For the laws of classical physics to apply, the following criteria must be met: Matter must be moving at speeds less than about 1% of the speed 
of light, the objects dealt with must be large enough to be seen with a microscope, and only weak gravitational fields (such as the field generated 
by the Earth) can be involved. 




Figure 1.14 Using a scanning tunneling microscope (STM), scientists can see the individual atoms that compose this sheet of gold, (credit: Erwinrossen) 

Some of the most spectacular advances in science have been made in modern physics. Many of the laws of classical physics have been modified or 
rejected, and revolutionary changes in technology, society, and our view of the universe have resulted. Like science fiction, modern physics is filled 
with fascinating objects beyond our normal experiences, but it has the advantage over science fiction of being very real. Why, then, is the majority of 
this text devoted to topics of classical physics? There are two main reasons: Classical physics gives an extremely accurate description of the 
universe under a wide range of everyday circumstances, and knowledge of classical physics is necessary to understand modern physics. 



18 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 

Modern physics itself consists of the two revolutionary theories, relativity and quantum mechanics. These theories deal with the very fast and the 
very small, respectively. Relativity must be used whenever an object is traveling at greater than about 1% of the speed of light or experiences a 
strong gravitational field such as that near the Sun. Quantum mechanics must be used for objects smaller than can be seen with a microscope. The 
combination of these two theories is relativistic quantum mechanics, and it describes the behavior of small objects traveling at high speeds or 
experiencing a strong gravitational field. Relativistic quantum mechanics is the best universally applicable theory we have. Because of its 
mathematical complexity, it is used only when necessary, and the other theories are used whenever they will produce sufficiently accurate results. We 
will find, however, that we can do a great deal of modern physics with the algebra and trigonometry used in this text. 



Check Your Understandinc 



A friend tells you he has learned about a new law of nature. What can you know about the information even before your friend describes the law? 
How would the information be different if your friend told you he had learned about a scientific theory rather than a law? 

Solution 

Without knowing the details of the law, you can still infer that the information your friend has learned conforms to the requirements of all laws of 
nature: it will be a concise description of the universe around us; a statement of the underlying rules that all natural processes follow. If the 
information had been a theory, you would be able to infer that the information will be a large-scale, broadly applicable generalization. 

PhET Explorations: Equation Grapher 

Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. 
y = bx ) to see how they add to generate the polynomial curve. 



M 



***** 



,j j 



PhET Interactive Simulation 



Figure 1.15 Equation Grapher (http://cnx.Org/content/m42092/l.4/equation-grapher_en.jar) 



1.2 Physical Quantities and Units 




Figure 1.16 The distance from Earth to the Moon may seem immense, but it is just a tiny fraction of the distances from Earth to other celestial bodies, (credit: NASA) 

The range of objects and phenomena studied in physics is immense. From the incredibly short lifetime of a nucleus to the age of the Earth, from the 
tiny sizes of sub-nuclear particles to the vast distance to the edges of the known universe, from the force exerted by a jumping flea to the force 
between Earth and the Sun, there are enough factors of 10 to challenge the imagination of even the most experienced scientist. Giving numerical 
values for physical quantities and equations for physical principles allows us to understand nature much more deeply than does qualitative 
description alone. To comprehend these vast ranges, we must also have accepted units in which to express them. And we shall find that (even in the 
potentially mundane discussion of meters, kilograms, and seconds) a profound simplicity of nature appears — all physical quantities can be expressed 
as combinations of only four fundamental physical quantities: length, mass, time, and electric current. 

We define a physical quantity either by specifying how it is measured or by stating how it is calculated from other measurements. For example, we 
define distance and time by specifying methods for measuring them, whereas we define average speed by stating that it is calculated as distance 
traveled divided by time of travel. 

Measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the length of a race, which is a 
physical quantity, can be expressed in units of meters (for sprinters) or kilometers (for distance runners). Without standardized units, it would be 
extremely difficult for scientists to express and compare measured values in a meaningful way. (See Figure 1.17.) 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 19 




Figure 1.17 Distances given in unknown units are maddeningly useless. 

There are two major systems of units used in the world: SI units (also known as the metric system) and English units (also known as the customary 
or imperial system). English units were historically used in nations once ruled by the British Empire and are still widely used in the United States. 
Virtually every other country in the world now uses SI units as the standard; the metric system is also the standard system agreed upon by scientists 
and mathematicians. The acronym "SI" is derived from the French Systeme International. 

SI Units: Fundamental and Derived Units 

Table 1.1 gives the fundamental SI units that are used throughout this textbook. This text uses non-SI units in a few applications where they are in 
very common use, such as the measurement of blood pressure in millimeters of mercury (mm Hg). Whenever non-SI units are discussed, they will be 
tied to SI units through conversions. 



Table 1.1 Fundamental SI Units 






Length Mass 








ilectric Current 


meter (m) 


kilogram (kg) 


second (s) 


ampere (A) 



It is an intriguing fact that some physical quantities are more fundamental than others and that the most fundamental physical quantities can be 
defined only in terms of the procedure used to measure them. The units in which they are measured are thus called fundamental units. In this 
textbook, the fundamental physical quantities are taken to be length, mass, time, and electric current. (Note that electric current will not be introduced 
until much later in this text.) All other physical quantities, such as force and electric current, can be expressed as algebraic combinations of length, 
mass, time, and current (for example, speed is length divided by time); these units are called derived units. 

Units of Time, Length, and Mass: The Second, Meter, and Kilogram 

The Second 

The SI unit for time, the second(abbreviated s), has a long history. For many years it was defined as 1/86,400 of a mean solar day. More recently, a 
new standard was adopted to gain greater accuracy and to define the second in terms of a non-varying, or constant, physical phenomenon (because 
the solar day is getting longer due to very gradual slowing of the Earth's rotation). Cesium atoms can be made to vibrate in a very steady way, and 
these vibrations can be readily observed and counted. In 1967 the second was redefined as the time required for 9,192,631,770 of these vibrations. 
(See Figure 1.18.) Accuracy in the fundamental units is essential, because all measurements are ultimately expressed in terms of fundamental units 
and can be no more accurate than are the fundamental units themselves. 




Figure 1.18 An atomic clock such as this one uses the vibrations of cesium atoms to keep time to a precision of better than a microsecond per year. The fundamental unit of 
time, the second, is based on such clocks. This image is looking down from the top of an atomic fountain nearly 30 feet tall! (credit: Steve Jurvetson/Flickr) 



20 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 



The Meter 



The SI unit for length is the meter (abbreviated m); its definition has also changed over time to become more accurate and precise. The meter was 
first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This measurement was improved in 1889 by redefining the 
meter to be the distance between two engraved lines on a platinum-iridium bar now kept near Paris. By 1960, it had become possible to define the 
meter even more accurately in terms of the wavelength of light, so it was again redefined as 1,650,763.73 wavelengths of orange light emitted by 
krypton atoms. In 1983, the meter was given its present definition (partly for greater accuracy) as the distance light travels in a vacuum in 
1/299,792,458 of a second. (See Figure 1.19.) This change defines the speed of light to be exactly 299,792,458 meters per second. The length of 
the meter will change if the speed of light is someday measured with greater accuracy. 

The Kilogram 

The SI unit for mass is the kilogram (abbreviated kg); it is defined to be the mass of a platinum-iridium cylinder kept with the old meter standard at 
the International Bureau of Weights and Measures near Paris. Exact replicas of the standard kilogram are also kept at the United States' National 
Institute of Standards and Technology, or NIST, located in Gaithersburg, Maryland outside of Washington D.C., and at other locations around the 
world. The determination of all other masses can be ultimately traced to a comparison with the standard mass. 



I c» lb ' " ' " ' ' " A" ' " " " ' AT " ""*&"" "■■■&'"■ ' ' ' lab" " ' "'■'#■'■ ' " ' id ' ' " " ' " W Mfll 



Light travels a distance of 1 meter 
in 1/299,792,458 seconds 

Figure 1.19 The meter is defined to be the distance light travels in 1/299,792,458 of a second in a vacuum. Distance traveled is speed multiplied by time. 

Electric current and its accompanying unit, the ampere, will be introduced in Introduction to Electric Current, Resistance, and Ohm's Law when 
electricity and magnetism are covered. The initial modules in this textbook are concerned with mechanics, fluids, heat, and waves. In these subjects 
all pertinent physical quantities can be expressed in terms of the fundamental units of length, mass, and time. 

Metric Prefixes 

SI units are part of the metric system. The metric system is convenient for scientific and engineering calculations because the units are categorized 
by factors of 10. Table 1.2 gives metric prefixes and symbols used to denote various factors of 10. 

Metric systems have the advantage that conversions of units involve only powers of 10. There are 100 centimeters in a meter, 1000 meters in a 
kilometer, and so on. In nonmetric systems, such as the system of U.S. customary units, the relationships are not as simple — there are 12 inches in a 
foot, 5280 feet in a mile, and so on. Another advantage of the metric system is that the same unit can be used over extremely large ranges of values 
simply by using an appropriate metric prefix. For example, distances in meters are suitable in construction, while distances in kilometers are 
appropriate for air travel, and the tiny measure of nanometers are convenient in optical design. With the metric system there is no need to invent new 
units for particular applications. 

The term order of magnitude refers to the scale of a value expressed in the metric system. Each power of 10 in the metric system represents a 

19 3 

different order of magnitude. For example, 10 , 10 , 10 , and so forth are all different orders of magnitude. All quantities that can be expressed as 

n 

a product of a specific power of 10 are said to be of the same order of magnitude. For example, the number 800 can be written as 8x 10 , and 

9 9 

the number 450 can be written as 4.5x10 . Thus, the numbers 800 and 450 are of the same order of magnitude: 10 . Order of magnitude 
can be thought of as a ballpark estimate for the scale of a value. The diameter of an atom is on the order of 10~ m, while the diameter of the Sun 
is on the order of 10 m. 

The Quest for Microscopic Standards for Basic Units 



The fundamental units described in this chapter are those that produce the greatest accuracy and precision in measurement. There is a sense 
among physicists that, because there is an underlying microscopic substructure to matter, it would be most satisfying to base our standards of 
measurement on microscopic objects and fundamental physical phenomena such as the speed of light. A microscopic standard has been 
accomplished for the standard of time, which is based on the oscillations of the cesium atom. 

The standard for length was once based on the wavelength of light (a small-scale length) emitted by a certain type of atom, but it has been 
supplanted by the more precise measurement of the speed of light. If it becomes possible to measure the mass of atoms or a particular 
arrangement of atoms such as a silicon sphere to greater precision than the kilogram standard, it may become possible to base mass 
measurements on the small scale. There are also possibilities that electrical phenomena on the small scale may someday allow us to base a unit 
of charge on the charge of electrons and protons, but at present current and charge are related to large-scale currents and forces between wires. 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 21 

Table 1.2 Metric Prefixes for Powers of 10 and their Symbols 











exa 


E 


10 18 


exameter 


Em 


IO 18 m 


distance light travels in a century 


peta 


P 


10 15 


petasecond 


Ps 


10 15 s 


30 million years 


tera 


T 


10 12 


terawatt 


TW 


IO 12 W 


powerful laser output 


giga 


G 


10 9 


gigahertz 


GHz 


IO 9 Hz 


a microwave frequency 


mega 


M 


10 6 


megacurie 


MCi 


IO 6 Ci 


high radioactivity 


kilo 


k 


10 3 


kilometer 


km 


IO 3 m 


about 6/10 mile 


hecto 


h 


10 2 


hectoliter 


hl_ 


IO 2 L 


26 gallons 


deka 


da 


10 1 


dekagram 


dag 


IO 1 g 


teaspoon of butter 


— 


— 


10° (=1) 










deci 


d 


io- 1 


deciliter 


di- 


10 _1 L 


less than half a soda 


centi 


c 


io- 2 


centimeter 


em 


IO" 2 m 


fingertip thickness 


milli 


m 


IO -3 


millimeter 


mm 


IO" 3 m 


flea at its shoulders 


micro 


H 


IO -6 


micrometer 


u.m 


IO" 6 m 


detail in microscope 


nano 


n 


IO" 9 


nanogram 


ng 


10" 9 g 


small speck of dust 


pico 


P 


io- 12 


picofarad 


PF 


IO" 12 F 


small capacitor in radio 


femto 


f 


io- 15 


femtometer 


fm 


IO" 15 m 


size of a proton 


atto 


a 


io- 18 


attosecond 


as 


IO" 18 s 


time light crosses an atom 



Known Ranges of Length, Mass, and Time 

The vastness of the universe and the breadth over which physics applies are illustrated by the wide range of examples of known lengths, masses, 
and times in Table 1.3. Examination of this table will give you some feeling for the range of possible topics and numerical values. (See Figure 1.20 
and Figure 1.21.) 




Figure 1.20 Tiny phytoplankton swims among crystals of ice in the Antarctic Sea. They range from a few micrometers to as much as 2 millimeters in length, (credit: Prof. 
Gordon T Taylor, Stony Brook University; NOAA Corps Collections) 



1. See Appendix A for a discussion of powers of 10. 



22 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 




Figure 1.21 Galaxies collide 2.4 billion light years away from Earth. The tremendous range of observable phenomena in nature challenges the imagination, (credit: NASA/ 
CXC/UVic./A. Mahdavi et al. Optical/lensing: CFHT/UVic./H. Hoekstra et al.) 

Unit Conversion and Dimensional Analysis 

It is often necessary to convert from one type of unit to another. For example, if you are reading a European cookbook, some quantities may be 
expressed in units of liters and you need to convert them to cups. Or, perhaps you are reading walking directions from one location to another and 
you are interested in how many miles you will be walking. In this case, you will need to convert units of feet to miles. 

Let us consider a simple example of how to convert units. Let us say that we want to convert 80 meters (m) to kilometers (km). 

The first thing to do is to list the units that you have and the units that you want to convert to. In this case, we have units in meters and we want to 
convert to kilometers. 

Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio expressing how many of one unit are 
equal to another unit. For example, there are 12 inches in 1 foot, 100 centimeters in 1 meter, 60 seconds in 1 minute, and so on. In this case, we 
know that there are 1,000 meters in 1 kilometer. 

Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor so that the units cancel 
out, as shown: 

SO^X-JJ^b = 0-080 km. (11) 

1000>f 

Note that the unwanted m unit cancels, leaving only the desired km unit. You can use this method to convert between any types of unit. 
Click Section C.l for a more complete list of conversion factors. 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 23 



Table 1.3 


Approximate Values of Length, Mass, and Time 












^B^r?K>55^ffiWiTr!TMf?TTTKWnTT»nj^rTijT5WTj^H 




ffTWyjTJJJfffjFWiTTTjTvJ^RWJTJffyj^B 


Lengins in 




values in parentheses) 


values in parentheses) 


10 -18 


Present experimental limit to smallest 
observable detail 


10 -30 


Mass of an electron 
(9.11X10 -31 kg) 


io- 23 


Time for light to cross a proton 


io- 15 


Diameter of a proton 


io- 27 


Mass of a hydrogen atom 
(l.67xl0 -27 kg) 


io- 22 


Mean life of an extremely unstable 
nucleus 


10 -14 


Diameter of a uranium nucleus 


IO" 15 


Mass of a bacterium 


io- 15 


Time for one oscillation of visible 
light 


10 -io 


Diameter of a hydrogen atom 


io- 5 


Mass of a mosquito 


io- 13 


Time for one vibration of an atom 
in a solid 


10~ 8 


Thickness of membranes in cells of living 
organisms 


io- 2 


Mass of a hummingbird 


IO -8 


Time for one oscillation of an FM 
radio wave 


1(T 6 


Wavelength of visible light 


1 


Mass of a liter of water (about a 
quart) 


IO -3 


Duration of a nerve impulse 


IO -3 


Size of a grain of sand 


IO 2 


Mass of a person 


1 


Time for one heartbeat 


1 


Height of a 4-year-old child 


IO 3 


Mass of a car 


IO 5 


One day (8.64xl0 4 s) 


10 2 


Length of a football field 


IO 8 


Mass of a large ship 


IO 7 


One year (y) (3.16xl0 7 s) 


10 4 


Greatest ocean depth 


IO 12 


Mass of a large iceberg 


IO 9 


About half the life expectancy of a 
human 


IO 7 


Diameter of the Earth 


IO 15 


Mass of the nucleus of a comet 


IO 11 


Recorded history 


10 11 


Distance from the Earth to the Sun 


IO 23 


Mass of the Moon (7.35xl0 22 kg) 


IO 17 


Age of the Earth 


10 16 


Distance traveled by light in 1 year (a light 
year) 


IO 25 


Mass of the Earth (5.97X10 24 kg) 


IO 18 


Age of the universe 


10 21 


Diameter of the Milky Way galaxy 


IO 30 


Mass of the Sun (l.99xl0 30 kg) 






10 22 


Distance from the Earth to the nearest large 
galaxy (Andromeda) 


IO 42 


Mass of the Milky Way galaxy 
(current upper limit) 






10 26 


Distance from the Earth to the edges of the 
known universe 


IO 53 


Mass of the known universe (current 
upper limit) 


















Exarr 


iple 1.1 Unit Conversions: A Short 


Drive H 







Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers per hour (km/h) and 
(b) in meters per second (m/s). (Note: Average speed is distance traveled divided by time of travel.) 

Strategy 

First we calculate the average speed using the given units. Then we can get the average speed into the desired units by picking the correct 
conversion factor and multiplying by it. The correct conversion factor is the one that cancels the unwanted unit and leaves the desired unit in its 
place. 

Solution for (a) 

(1) Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for now — average 
speed and other motion concepts will be covered in a later module.) In equation form, 



average speed - distance 
time 



(1.2) 



(2) Substitute the given values for distance and time. 



average speed = }™ km = 0.500^L. (13) 

° 20.0 mm mm 

(3) Convert km/min to km/h: multiply by the conversion factor that will cancel minutes and leave hours. That conversion factor is 60 min/hr . 
Thus, 



average speed =0.500^^x^2^ = 30.0^1. 

° mm In h 



(1.4) 



24 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 

Discussion for (a) 

To check your answer, consider the following: 

(1) Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units 
will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the 
wrong units as follows: 

km x 1 hr _ 1 km • hr (1.5) 

min 60 min 60 mm 2 ' 

which are obviously not the desired units of km/h. 

(2) Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km/h and we have 
indeed obtained these units. 

(3) Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have 
three significant figures. The answer 30.0 km/hr does indeed have three significant figures, so this is appropriate. Note that the significant figures 
in the conversion factor are not relevant because an hour is defined to be 60 minutes, so the precision of the conversion factor is perfect. 

(4) Next, check whether the answer is reasonable. Let us consider some information from the problem — if you travel 10 km in a third of an hour 
(20 min), you would travel three times that far in an hour. The answer does seem reasonable. 

Solution for (b) 

There are several ways to convert the average speed into meters per second. 

(1) Start with the answer to (a) and convert km/h to m/s. Two conversion factors are needed — one to convert hours to seconds, and another to 
convert kilometers to meters. 

(2) Multiplying by these yields 

Average speed = 30.0^X T ^-X-^?^, (16) 

h 3,600 s 1 km 

Average speed = 8.33^-. C 1 - 7 ) 

Discussion for (b) 

If we had started with 0.500 km/min, we would have needed different conversion factors, but the answer would have been the same: 8.33 m/s. 

You may have noted that the answers in the worked example just covered were given to three digits. Why? When do you need to be concerned 
about the number of digits in something you calculate? Why not write down all the digits your calculator produces? The module Accuracy, 
Precision, and Significant Figures will help you answer these questions. 

Nonstandard Units 

While there are numerous types of units that we are all familiar with, there are others that are much more obscure. For example, a firkin is a unit 
of volume that was once used to measure beer. One firkin equals about 34 liters. To learn more about nonstandard units, use a dictionary or 
encyclopedia to research different "weights and measures." Take note of any unusual units, such as a barleycorn, that are not listed in the text. 
Think about how the unit is defined and state its relationship to SI units. 



Check Your Unclerstandim 



Some hummingbirds beat their wings more than 50 times per second. A scientist is measuring the time it takes for a hummingbird to beat its 
wings once. Which fundamental unit should the scientist use to describe the measurement? Which factor of 10 is the scientist likely to use to 
describe the motion precisely? Identify the metric prefix that corresponds to this factor of 10. 

Solution 

The scientist will measure the time between each movement using the fundamental unit of seconds. Because the wings beat so fast, the scientist 

will probably need to measure in milliseconds, or 10 seconds. (50 beats per second corresponds to 20 milliseconds per beat.) 



Check Your Understanding 



One cubic centimeter is equal to one milliliter. What does this tell you about the different units in the SI metric system? 

Solution 

The fundamental unit of length (meter) is probably used to create the derived unit of volume (liter). The measure of a milliliter is dependent on the 
measure of a centimeter. 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 25 



1.3 Accuracy, Precision, and Significant Figures 




Figure 1.22 A double-pan mechanical balance is used to compare different masses. Usually an object with unknown mass is placed in one pan and objects of known mass are 
placed in the other pan. When the bar that connects the two pans is horizontal, then the masses in both pans are equal. The "known masses" are typically metal cylinders of 
standard mass such as 1 gram, 10 grams, and 100 grams, (credit: Serge Melki) 




Figure 1.23 Many mechanical balances, such as double-pan balances, have been replaced by digital scales, which can typically measure the mass of an object more 
precisely. Whereas a mechanical balance may only read the mass of an object to the nearest tenth of a gram, many digital scales can measure the mass of an object up to the 
nearest thousandth of a gram, (credit: Karel Jakubec) 



Accuracy and Precision of a Measurement 

Science is based on observation and experiment — that is, on measurements. Accuracy is how close a measurement is to the correct value for that 
measurement. For example, let us say that you are measuring the length of standard computer paper. The packaging in which you purchased the 
paper states that it is 11.0 inches long. You measure the length of the paper three times and obtain the following measurements: 11.1 in., 11.2 in., and 
10.9 in. These measurements are quite accurate because they are very close to the correct value of 11.0 inches. In contrast, if you had obtained a 
measurement of 12 inches, your measurement would not be very accurate. 

The precision of a measurement system is refers to how close the agreement is between repeated measurements (which are repeated under the 
same conditions). Consider the example of the paper measurements. The precision of the measurements refers to the spread of the measured 
values. One way to analyze the precision of the measurements would be to determine the range, or difference, between the lowest and the highest 
measured values. In that case, the lowest value was 10.9 in. and the highest value was 11.2 in. Thus, the measured values deviated from each other 
by at most 0.3 in. These measurements were relatively precise because they did not vary too much in value. However, if the measured values had 
been 10.9, 11.1, and 11.9, then the measurements would not be very precise because there would be significant variation from one measurement to 
another. 

The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are 
precise but not accurate. Let us consider an example of a GPS system that is attempting to locate the position of a restaurant in a city. Think of the 
restaurant location as existing at the center of a bull's-eye target, and think of each GPS attempt to locate the restaurant as a black dot. In Figure 
1.24, you can see that the GPS measurements are spread out far apart from each other, but they are all relatively close to the actual location of the 
restaurant at the center of the target. This indicates a low precision, high accuracy measuring system. However, in Figure 1.25, the GPS 
measurements are concentrated quite closely to one another, but they are far away from the target location. This indicates a high precision, low 
accuracy measuring system. 



26 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 




Figure 1.24 A GPS system attempts to locate a restaurant at the center of the bull's-eye. The black dots represent each attempt to pinpoint the location of the restaurant. The 
dots are spread out quite far apart from one another, indicating low precision, but they are each rather close to the actual location of the restaurant, indicating high accuracy, 
(credit: Dark Evil) 




Figure 1.25 In this figure, the dots are concentrated rather closely to one another, indicating high precision, but they are rather far away from the actual location of the 
restaurant, indicating low accuracy, (credit: Dark Evil) 

Accuracy, Precision, and Uncertainty 

The degree of accuracy and precision of a measuring system are related to the uncertainty in the measurements. Uncertainty is a quantitative 
measure of how much your measured values deviate from a standard or expected value. If your measurements are not very accurate or precise, then 
the uncertainty of your values will be very high. In more general terms, uncertainty can be thought of as a disclaimer for your measured values. For 
example, if someone asked you to provide the mileage on your car, you might say that it is 45,000 miles, plus or minus 500 miles. The plus or minus 
amount is the uncertainty in your value. That is, you are indicating that the actual mileage of your car might be as low as 44,500 miles or as high as 
45,500 miles, or anywhere in between. All measurements contain some amount of uncertainty. In our example of measuring the length of the paper, 
we might say that the length of the paper is 11 in., plus or minus 0.2 in. The uncertainty in a measurement, A , is often denoted as 5A ("delta A "), 
so the measurement result would be recorded as A ± SA . In our paper example, the length of the paper could be expressed as 11 in. ± 0.2. 

The factors contributing to uncertainty in a measurement include: 

1. Limitations of the measuring device, 

2. The skill of the person making the measurement, 

3. Irregularities in the object being measured, 

4. Any other factors that affect the outcome (highly dependent on the situation). 

In our example, such factors contributing to the uncertainty could be the following: the smallest division on the ruler is 0.1 in., the person using the 
ruler has bad eyesight, or one side of the paper is slightly longer than the other. At any rate, the uncertainty in a measurement must be based on a 
careful consideration of all the factors that might contribute and their possible effects. 

Making Connections: Real-World Connections - Fevers or Chills? 



Uncertainty is a critical piece of information, both in physics and in many other real-world applications. Imagine you are caring for a sick child. 
You suspect the child has a fever, so you check his or her temperature with a thermometer. What if the uncertainty of the thermometer were 
3.0°C ? If the child's temperature reading was 37.0°C (which is normal body temperature), the "true" temperature could be anywhere from a 
hypothermic 34.0°C to a dangerously high 40.0°C . A thermometer with an uncertainty of 3.0°C would be useless. 

Percent Uncertainty 

One method of expressing uncertainty is as a percent of the measured value. If a measurement A is expressed with uncertainty, 8A , the percent 
uncertainty (%unc) is defined to be 

%unc=%<100%. (18) 

A 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 27 



Example 1.2 Calculating Percent Uncertainty: A Bag of Apples 



A grocery store sells 5-lb bags of apples. You purchase four bags over the course of a month and weigh the apples each time. You obtain the 
following measurements: 

• Week 1 weight: 4.8 lb 

• Week 2 weight: 5.3 lb 

• Week 3 weight: 4.9 lb 

• Week 4 weight: 5.41b 

You determine that the weight of the 5-lb bag has an uncertainty of ±0.4 lb . What is the percent uncertainty of the bag's weight? 
Strategy 

First, observe that the expected value of the bag's weight, A , is 5 lb. The uncertainty in this value, SA , is 0.4 lb. We can use the following 
equation to determine the percent uncertainty of the weight: 



Solution 

Plug the known values into the equation: 



%unc=%<100%. (19) 

A 



% unc =^#X 100% = 8%. (110) 

5 lb 



Discussion 

We can conclude that the weight of the apple bag is 5 lb ± 8% . Consider how this percent uncertainty would change if the bag of apples were 

half as heavy, but the uncertainty in the weight remained the same. Hint for future calculations: when calculating percent uncertainty, always 
remember that you must multiply the fraction by 100%. If you do not do this, you will have a decimal quantity, not a percent value. 

Uncertainties in Calculations 

There is an uncertainty in anything calculated from measured quantities. For example, the area of a floor calculated from measurements of its length 
and width has an uncertainty because the length and width have uncertainties. How big is the uncertainty in something you calculate by multiplication 
or division? If the measurements going into the calculation have small uncertainties (a few percent or less), then the method of adding percents can 
be used for multiplication or division. This method says that the percent uncertainty in a quantity calculated by multiplication or division is the sum of 
the percent uncertainties in the items used to make the calculation. For example, if a floor has a length of 4.00 m and a width of 3.00 m , with 

uncertainties of 2% and 1% , respectively, then the area of the floor is 12.0 m and has an uncertainty of 3% . (Expressed as an area this is 

0.36 m , which we round to 0.4 m since the area of the floor is given to a tenth of a square meter.) 



Check Your Understanding 



A high school track coach has just purchased a new stopwatch. The stopwatch manual states that the stopwatch has an uncertainty of ±0.05 s 
. Runners on the track coach's team regularly clock 100-m sprints of 11.49 s to 15.01 s . At the school's last track meet, the first-place sprinter 
came in at 12.04 s and the second-place sprinter came in at 12.07 s . Will the coach's new stopwatch be helpful in timing the sprint team? 
Why or why not? 

Solution 

No, the uncertainty in the stopwatch is too great to effectively differentiate between the sprint times. 

Precision of Measuring Tools and Significant Figures 

An important factor in the accuracy and precision of measurements involves the precision of the measuring tool. In general, a precise measuring tool 
is one that can measure values in very small increments. For example, a standard ruler can measure length to the nearest millimeter, while a caliper 
can measure length to the nearest 0.01 millimeter. The caliper is a more precise measuring tool because it can measure extremely small differences 
in length. The more precise the measuring tool, the more precise and accurate the measurements can be. 

When we express measured values, we can only list as many digits as we initially measured with our measuring tool. For example, if you use a 
standard ruler to measure the length of a stick, you may measure it to be 36.7 cm . You could not express this value as 36.71 cm because your 
measuring tool was not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been 
estimated in some way by the person performing the measurement. For example, the person measuring the length of a stick with a ruler notices that 
the stick length seems to be somewhere in between 36.6 cm and 36.7 cm , and he or she must estimate the value of the last digit. Using the 
method of significant figures, the rule is that the last digit written down in a measurement is the first digit with some uncertainty. In order to 
determine the number of significant digits in a value, start with the first measured value at the left and count the number of digits through the last digit 
written on the right. For example, the measured value 36.7 cm has three digits, or significant figures. Significant figures indicate the precision of a 
measuring tool that was used to measure a value. 



28 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 



Zeros 



Special consideration is given to zeros when counting significant figures. The zeros in 0.053 are not significant, because they are only placekeepers 
that locate the decimal point. There are two significant figures in 0.053. The zeros in 10.053 are not placekeepers but are significant — this number 
has five significant figures. The zeros in 1300 may or may not be significant depending on the style of writing numbers. They could mean the number 
is known to the last digit, or they could be placekeepers. So 1300 could have two, three, or four significant figures. (To avoid this ambiguity, write 
1300 in scientific notation.) Zeros are significant except when they serve only as placekeepers. 



Check Your Understanding 



Determine the number of significant figures in the following measurements: 

a. 0.0009 

b. 15,450.0 

c. 6xl0 3 

d. 87.990 

e. 30.42 
Solution 

(a) 1; the zeros in this number are placekeepers that indicate the decimal point 

(b) 6; here, the zeros indicate that a measurement was made to the 0.1 decimal point, so the zeros are significant 

(c) 1; the value 10 signifies the decimal place, not the number of measured values 

(d) 5; the final zero indicates that a measurement was made to the 0.001 decimal point, so it is significant 

(e) 4; any zeros located in between significant figures in a number are also significant 

Significant Figures in Calculations 

When combining measurements with different degrees of accuracy and precision, the number of significant digits in the final answer can be no 
greater than the number of significant digits in the least precise measured value. There are two different rules, one for multiplication and division and 
the other for addition and subtraction, as discussed below. 

1. For multiplication and division: The result should have the same number of significant figures as the quantity having the least significant figures 
entering into the calculation. For example, the area of a circle can be calculated from its radius using A = nr . Let us see how many significant 
figures the area has if the radius has only two — say, r = 1.2 m . Then, 

A = 7ir 2 = (3. 1415927.. .)X(1.2 m) 2 = 4.5238934 m 2 ( LU ) 

is what you would get using a calculator that has an eight-digit output. But because the radius has only two significant figures, it limits the calculated 
quantity to two significant figures or 

A=4.5 m 2 , (!■") 

even though n is good to at least eight digits. 

2. For addition and subtraction: The answer can contain no more decimal places than the least precise measurement. Suppose that you buy 
7.56-kg of potatoes in a grocery store as measured with a scale with precision 0.01 kg. Then you drop off 6.052-kg of potatoes at your laboratory as 
measured by a scale with precision 0.001 kg. Finally, you go home and add 13.7 kg of potatoes as measured by a bathroom scale with precision 0.1 
kg. How many kilograms of potatoes do you now have, and how many significant figures are appropriate in the answer? The mass is found by simple 
addition and subtraction: 

7.56 kg (1.13) 

- 6.052 kg 
+ 13.7 kg _ 

15.208 kg " 15 ' 2kg - 

Next, we identify the least precise measurement: 13.7 kg. This measurement is expressed to the 0.1 decimal place, so our final answer must also be 
expressed to the 0.1 decimal place. Thus, the answer is rounded to the tenths place, giving us 15.2 kg. 

Significant Figures in this Text 

In this text, most numbers are assumed to have three significant figures. Furthermore, consistent numbers of significant figures are used in all worked 
examples. You will note that an answer given to three digits is based on input good to at least three digits, for example. If the input has fewer 
significant figures, the answer will also have fewer significant figures. Care is also taken that the number of significant figures is reasonable for the 
situation posed. In some topics, particularly in optics, more accurate numbers are needed and more than three significant figures will be used. Finally, 
if a number is exact, such as the two in the formula for the circumference of a circle, c = 2%r , it does not affect the number of significant figures in a 
calculation. 



Check Your Understanding 



Perform the following calculations and express your answer using the correct number of significant digits. 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 29 

(a) A woman has two bags weighing 13.5 pounds and one bag with a weight of 10.2 pounds. What is the total weight of the bags? 

(b) The force F on an object is equal to its mass m multiplied by its acceleration a . If a wagon with mass 55 kg accelerates at a rate of 
0.0255 m/s , what is the force on the wagon? (The unit of force is called the newton, and it is expressed with the symbol N.) 
Solution 

(a) 37.2 pounds; Because the number of bags is an exact value, it is not considered in the significant figures. 

(b) 1.4 N; Because the value 55 kg has only two significant figures, the final value must also contain two significant figures. 

PhET Explorati ons: Estimation 

Explore size estimation in one, two, and three dimensions! Multiple levels of difficulty allow for progressive skill improvement. 

cij fp'PhET Interactive Simulation 

Figure 1.26 Estimation (http://cnx.Org/content/m42120/l.6/estimation_en.jar) 

1.4 Approximation 

On many occasions, physicists, other scientists, and engineers need to make approximations or "guesstimates" for a particular quantity. What is the 
distance to a certain destination? What is the approximate density of a given item? About how large a current will there be in a circuit? Many 
approximate numbers are based on formulae in which the input quantities are known only to a limited accuracy. As you develop problem-solving skills 
(that can be applied to a variety of fields through a study of physics), you will also develop skills at approximating. You will develop these skills 
through thinking more quantitatively, and by being willing to take risks. As with any endeavor, experience helps, as well as familiarity with units. These 
approximations allow us to rule out certain scenarios or unrealistic numbers. Approximations also allow us to challenge others and guide us in our 
approaches to our scientific world. Let us do two examples to illustrate this concept. 



Example 1.3 Approximate the Height of a Building 



Can you approximate the height of one of the buildings on your campus, or in your neighborhood? Let us make an approximation based upon 
the height of a person. In this example, we will calculate the height of a 39-story building. 

Strategy 

Think about the average height of an adult male. We can approximate the height of the building by scaling up from the height of a person. 

Solution 

Based on information in the example, we know there are 39 stories in the building. If we use the fact that the height of one story is approximately 
equal to about the length of two adult humans (each human is about 2-m tall), then we can estimate the total height of the building to be 

?m 2 person ork ^ . . cr (1.14) 

— ^^ — X-rH X39 stories = 156 m. 



1 person 1 story 

Discussion 

You can use known quantities to determine an approximate measurement of unknown quantities. If your hand measures 10 cm across, how 
many hand lengths equal the width of your desk? What other measurements can you approximate besides length? 



30 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 



Example 1.4 Approximating Vast Numbers: a Trillion Dollars 




Figure 1.27 A bank stack contains one-hundred $100 bills, and is worth $10,000. How many bank stacks make up a trillion dollars? (credit: Andrew Magill) 

The U.S. federal deficit in the 2008 fiscal year was a little greater than $10 trillion. Most of us do not have any concept of how much even one 
trillion actually is. Suppose that you were given a trillion dollars in $100 bills. If you made 100-bill stacks and used them to evenly cover a football 
field (between the end zones), make an approximation of how high the money pile would become. (We will use feet/inches rather than meters 
here because football fields are measured in yards.) One of your friends says 3 in., while another says 10 ft. What do you think? 

Strategy 

When you imagine the situation, you probably envision thousands of small stacks of 100 wrapped $100 bills, such as you might see in movies or 
at a bank. Since this is an easy-to-approximate quantity, let us start there. We can find the volume of a stack of 100 bills, find out how many 
stacks make up one trillion dollars, and then set this volume equal to the area of the football field multiplied by the unknown height. 

Solution 

(1) Calculate the volume of a stack of 100 bills. The dimensions of a single bill are approximately 3 in. by 6 in. A stack of 100 of these is about 
0.5 in. thick. So the total volume of a stack of 100 bills is: 

volume of stack = lengthxwidthx height, (115) 

volume of stack = 6 in.x3 in.x0.5 in., 

volume of stack = 9 in. . 

(2) Calculate the number of stacks. Note that a trillion dollars is equal to $1x10 , and a stack of one-hundred $100 bills is equal to 
$10,000, or $1x10 . The number of stacks you will have is: 

$lxl0 12 (a trillion dollars)/ $lxl0 4 per stack = lxlO 8 stacks. ( 1 - 16 ) 



(3) Calculate the area of a football field in square inches. The area of a football field is 100 ydx50 yd, which gives 5,000 yd . Because we 
are working in inches, we need to convert square yards to square inches: 

Area = 5,000 yd 2 X^X^X#^X^4^ = 6,480,000 in. 2 , (117) 

J 1 yd 1 yd 1 ft 1 ft 

Area«6xl0 6 in. 2 . 
This conversion gives us 6x 10 in. for the area of the field. (Note that we are using only one significant figure in these calculations.) 

(4) Calculate the total volume of the bills. The volume of all the $100 -bill stacks is 9 in. 3 /stackxlO 8 stacks = 9xl0 8 in. 3 . 

(5) Calculate the height. To determine the height of the bills, use the equation: 

volume of bills = area of fieldx height of money: 



(1.18) 



Height of money 



_ volume of bills 
area of field 



Height of money = SxlOJlk- = 1.33xl0 2 in., 
6xl0 6 in. 2 

Height of money « lx 10 2 in. = 100 in. 
The height of the money will be about 100 in. high. Converting this value to feet gives 



100 in.x 



lft 
12 in. 



= 8.33 ft « 8 ft. 



(1.19) 



Discussion 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 31 

The final approximate value is much higher than the early estimate of 3 in., but the other early estimate of 10 ft (120 in.) was roughly correct. 
How did the approximation measure up to your first guess? What can this exercise tell you in terms of rough "guesstimates" versus carefully 
calculated approximations? 



Check Your Understanding 



Using mental math and your understanding of fundamental units, approximate the area of a regulation basketball court. Describe the process 
you used to arrive at your final approximation. 

Solution 

An average male is about two meters tall. It would take approximately 15 men laid out end to end to cover the length, and about 7 to cover the 

width. That gives an approximate area of 420 m . 



Glossary 



accuracy: the degree to which a measured value agrees with correct value for that measurement 

approximation: an estimated value based on prior experience and reasoning 

classical physics: physics that was developed from the Renaissance to the end of the 19th century 

conversion factor: a ratio expressing how many of one unit are equal to another unit 

derived units: units that can be calculated using algebraic combinations of the fundamental units 

English units: system of measurement used in the United States; includes units of measurement such as feet, gallons, and pounds 

fundamental units: units that can only be expressed relative to the procedure used to measure them 

kilogram: the SI unit for mass, abbreviated (kg) 

law: a description, using concise language or a mathematical formula, a generalized pattern in nature that is supported by scientific evidence and 
repeated experiments 

meter: the SI unit for length, abbreviated (m) 

method of adding percents: the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent uncertainties in 
the items used to make the calculation 

metric system: a system in which values can be calculated in factors of 10 

model: representation of something that is often too difficult (or impossible) to display directly 

modern physics: the study of relativity, quantum mechanics, or both 

order of magnitude: refers to the size of a quantity as it relates to a power of 10 

percent uncertainty: the ratio of the uncertainty of a measurement to the measured value, expressed as a percentage 

physical quantity : a characteristic or property of an object that can be measured or calculated from other measurements 

physics: the science concerned with describing the interactions of energy, matter, space, and time; it is especially interested in what fundamental 
mechanisms underlie every phenomenon 

precision: the degree to which repeated measurements agree with each other 

quantum mechanics: the study of objects smaller than can be seen with a microscope 

relativity: the study of objects moving at speeds greater than about 1% of the speed of light, or of objects being affected by a strong gravitational 
field 

SI units : the international system of units that scientists in most countries have agreed to use; includes units such as meters, liters, and grams 

scientific method: a method that typically begins with an observation and question that the scientist will research; next, the scientist typically 
performs some research about the topic and then devises a hypothesis; then, the scientist will test the hypothesis by performing an 
experiment; finally, the scientist analyzes the results of the experiment and draws a conclusion 

second: the SI unit for time, abbreviated (s) 

significant figures: express the precision of a measuring tool used to measure a value 

theory: an explanation for patterns in nature that is supported by scientific evidence and verified multiple times by various groups of researchers 

uncertainty: a quantitative measure of how much your measured values deviate from a standard or expected value 



32 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 

units : a standard used for expressing and comparing measurements 



Section Summary 



1.1 Physics: An Introduction 

• Science seeks to discover and describe the underlying order and simplicity in nature. 

• Physics is the most basic of the sciences, concerning itself with energy, matter, space and time, and their interactions. 

• Scientific laws and theories express the general truths of nature and the body of knowledge they encompass. These laws of nature are rules 
that all natural processes appear to follow. 

1.2 Physical Quantities and Units 

• Physical quantities are a characteristic or property of an object that can be measured or calculated from other measurements. 

• Units are standards for expressing and comparing the measurement of physical quantities. All units can be expressed as combinations of four 
fundamental units. 

• The four fundamental units we will use in this text are the meter (for length), the kilogram (for mass), the second (for time), and the ampere (for 
electric current). These units are part of the metric system, which uses powers of 10 to relate quantities over the vast ranges encountered in 
nature. 

• The four fundamental units are abbreviated as follows: meter, m; kilogram, kg; second, s; and ampere, A. The metric system also uses a 
standard set of prefixes to denote each order of magnitude greater than or lesser than the fundamental unit itself. 

• Unit conversions involve changing a value expressed in one type of unit to another type of unit. This is done by using conversion factors, which 
are ratios relating equal quantities of different units. 

1.3 Accuracy, Precision, and Significant Figures 

• Accuracy of a measured value refers to how close a measurement is to the correct value. The uncertainty in a measurement is an estimate of 
the amount by which the measurement result may differ from this value. 

• Precision of measured values refers to how close the agreement is between repeated measurements. 

• The precision of a measuring tool is related to the size of its measurement increments. The smaller the measurement increment, the more 
precise the tool. 

• Significant figures express the precision of a measuring tool. 

• When multiplying or dividing measured values, the final answer can contain only as many significant figures as the least precise value. 

• When adding or subtracting measured values, the final answer cannot contain more decimal places than the least precise value. 

1.4 Approximation 

Scientists often approximate the values of quantities to perform calculations and analyze systems. 



Conceptual Questions 



1.1 Physics: An Introduction 

1. Models are particularly useful in relativity and quantum mechanics, where conditions are outside those normally encountered by humans. What is a 
model? 

2. How does a model differ from a theory? 

3. If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assuming both use 
accepted rules of logic)? 

4. What determines the validity of a theory? 

5. Certain criteria must be satisfied if a measurement or observation is to be believed. Will the criteria necessarily be as strict for an expected result 
as for an unexpected result? 

6. Can the validity of a model be limited, or must it be universally valid? How does this compare to the required validity of a theory or a law? 

7. Classical physics is a good approximation to modern physics under certain circumstances. What are they? 

8. When is it necessary to use relativistic quantum mechanics? 

9. Can classical physics be used to accurately describe a satellite moving at a speed of 7500 m/s? Explain why or why not. 

1.2 Physical Quantities and Units 

10. Identify some advantages of metric units. 

1.3 Accuracy, Precision, and Significant Figures 

11. What is the relationship between the accuracy and uncertainty of a measurement? 

12. Prescriptions for vision correction are given in units called diopters (D). Determine the meaning of that unit. Obtain information (perhaps by calling 
an optometrist or performing an internet search) on the minimum uncertainty with which corrections in diopters are determined and the accuracy with 
which corrective lenses can be produced. Discuss the sources of uncertainties in both the prescription and accuracy in the manufacture of lenses. 



CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 33 



Problems & Exercises 



1.2 Physical Quantities and Units 

1. The speed limit on some interstate highways is roughly 100 km/h. (a) 
What is this in meters per second? (b) How many miles per hour is this? 

2. A car is traveling at a speed of 33 m/s . (a) What is its speed in 
kilometers per hour? (b) Is it exceeding the 90 km/h speed limit? 

3. Show that 1.0 m/s = 3.6 km/h . Hint: Show the explicit steps 
involved in converting 1.0 m/s = 3.6 km/h. 

4. American football is played on a 100-yd-long field, excluding the end 
zones. How long is the field in meters? (Assume that 1 meter equals 
3.281 feet.) 

5. Soccer fields vary in size. A large soccer field is 115 m long and 85 m 
wide. What are its dimensions in feet and inches? (Assume that 1 meter 
equals 3.281 feet.) 

6. What is the height in meters of a person who is 6 ft 1.0 in. tall? 
(Assume that 1 meter equals 39.37 in.) 

7. Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. 
What is its height in kilometers? (Assume that 1 kilometer equals 3,281 
feet.) 

8. The speed of sound is measured to be 342 m/s on a certain day. 
What is this in km/h? 

9. Tectonic plates are large segments of the Earth's crust that move 
slowly. Suppose that one such plate has an average speed of 4.0 cm/ 
year, (a) What distance does it move in 1 s at this speed? (b) What is its 
speed in kilometers per million years? 

10. (a) Refer to Table 1.3 to determine the average distance between the 
Earth and the Sun. Then calculate the average speed of the Earth in its 
orbit in kilometers per second, (b) What is this in meters per second? 

1.3 Accuracy, Precision, and Significant Figures 

Express your answers to problems in this section to the correct 
number of significant figures and proper units. 

11. Suppose that your bathroom scale reads your mass as 65 kg with a 3 
uncertainty. What is the uncertainty in your mass (in kilograms)? 

12. A good-quality measuring tape can be off by 0.50 cm over a distance 
of 20 m. What is its percent uncertainty? 

13. (a) A car speedometer has a 5.0% uncertainty. What is the range of 
possible speeds when it reads 90 km/h ? (b) Convert this range to miles 
per hour. (1 km = 0.6214 mi) 

14. An infant's pulse rate is measured to be 130 ± 5 beats/min. What is 
the percent uncertainty in this measurement? 

15. (a) Suppose that a person has an average heart rate of 72.0 beats/ 
min. How many beats does he or she have in 2.0 y? (b) In 2.00 y? (c) In 
2.000 y? 

16. A can contains 375 ml_ of soda. How much is left after 308 ml_ is 
removed? 

17. State how many significant figures are proper in the results of the 
following calculations: (a) (106.7)(98.2)/(46.210)(1.01) (b) (18.7) 2 

(c) (l.60xl0" 19 )(3712). 

18. (a) How many significant figures are in the numbers 99 and 100? (b) 
If the uncertainty in each number is 1, what is the percent uncertainty in 
each? (c) Which is a more meaningful way to express the accuracy of 
these two numbers, significant figures or percent uncertainties? 

19. (a) If your speedometer has an uncertainty of 2.0 km/h at a speed 
of 90 km/h , what is the percent uncertainty? (b) If it has the same 
percent uncertainty when it reads 60 km/h , what is the range of speeds 
you could be going? 



20. (a) A person's blood pressure is measured to be 120 ± 2 mm Hg . 

What is its percent uncertainty? (b) Assuming the same percent 
uncertainty, what is the uncertainty in a blood pressure measurement of 
80 mm Hg ? 

21. A person measures his or her heart rate by counting the number of 
beats in 30 s . If 40 ± 1 beats are counted in 30.0 ± 0.5 s , what is 
the heart rate and its uncertainty in beats per minute? 

22. What is the area of a circle 3.102 cm in diameter? 

23. If a marathon runner averages 9.5 mi/h, how long does it take him or 
her to run a 26.22-mi marathon? 

24. A marathon runner completes a 42.188-km course in 2 h , 30 min, 
and 12 s. There is an uncertainty of 25 m in the distance traveled and 
an uncertainty of 1 s in the elapsed time, (a) Calculate the percent 
uncertainty in the distance, (b) Calculate the uncertainty in the elapsed 
time, (c) What is the average speed in meters per second? (d) What is 
the uncertainty in the average speed? 

25. The sides of a small rectangular box are measured to be 
1.80 ± 0.01 cm , 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm long. 

Calculate its volume and uncertainty in cubic centimeters. 

26. When non-metric units were used in the United Kingdom, a unit of 
mass called the pound-mass (Ibm) was employed, where 

1 lbm = 0.4539 kg . (a) If there is an uncertainty of 0.0001 kg in the 

pound-mass unit, what is its percent uncertainty? (b) Based on that 
percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg 
when converted to kilograms? 

27. The length and width of a rectangular room are measured to be 
3.955 ± 0.005 m and 3.050 ± 0.005 m . Calculate the area of the 
room and its uncertainty in square meters. 

28. A car engine moves a piston with a circular cross section of 
7.500 ± 0.002 cm diameter a distance of 3.250 ± 0.001 cm to 

compress the gas in the cylinder, (a) By what amount is the gas 
decreased in volume in cubic centimeters? (b) Find the uncertainty in this 
volume. 

1.4 Approximation 

29. How many heartbeats are there in a lifetime? 

30. A generation is about one-third of a lifetime. Approximately how many 
generations have passed since the year AD? 

31. How many times longer than the mean life of an extremely unstable 
atomic nucleus is the lifetime of a human? (Hint: The lifetime of an 

unstable atomic nucleus is on the order of 10 s .) 

32. Calculate the approximate number of atoms in a bacterium. Assume 
that the average mass of an atom in the bacterium is ten times the mass 
of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order 

— 97 

of 10 kg and the mass of a bacterium is on the order of 
10" 15 kg.) 



34 CHAPTER 1 | INTRODUCTION: THE NATURE OF SCIENCE AND PHYSICS 




Figure 1.28 This color-enhanced photo shows Salmonella typhimurium (red) 
attacking human cells. These bacteria are commonly known for causing foodborne 
illness. Can you estimate the number of atoms in each bacterium? (credit: Rocky 
Mountain Laboratories, NIAID, NIH) 

33. Approximately how many atoms thick is a cell membrane, assuming 
all atoms there average about twice the size of a hydrogen atom? 

34. (a) What fraction of Earth's diameter is the greatest ocean depth? (b) 
The greatest mountain height? 

35. (a) Calculate the number of cells in a hummingbird assuming the 
mass of an average cell is ten times the mass of a bacterium, (b) Making 
the same assumption, how many cells are there in a human? 

36. Assuming one nerve impulse must end before another can begin, 
what is the maximum firing rate of a nerve in impulses per second? 



CHAPTER 2 | KINEMATICS 35 




Figure 2.1 The motion of an American kestrel through the air can be described by the bird's displacement, speed, velocity, and acceleration. When it flies in a straight line 
without any change in direction, its motion is said to be one dimensional, (credit: Vince Maidens, Wikimedia Commons) 



Learning Objectives 



2.1. Displacement 

• Define position, displacement, distance, and distance traveled. 

• Explain the relationship between position and displacement. 

• Distinguish between displacement and distance traveled. 

• Calculate displacement and distance given initial position, final position, and the path between the two. 

2.2. Vectors, Scalars, and Coordinate Systems 

• Define and distinguish between scalar and vector quantities. 

• Assign a coordinate system for a scenario involving one-dimensional motion. 

2.3. Time, Velocity, and Speed 

• Explain the relationships between instantaneous velocity, average velocity, instantaneous speed, average speed, displacement, and time, 

• Calculate velocity and speed given initial position, initial time, final position, and final time. 

• Derive a graph of velocity vs. time given a graph of position vs. time. 

• Interpret a graph of velocity vs. time. 

2.4. Acceleration 

• Define and distinguish between instantaneous acceleration, average acceleration, and deceleration. 

• Calculate acceleration given initial time, initial velocity, final time, and final velocity. 

2.5. Motion Equations for Constant Acceleration in One Dimension 

• Calculate displacement of an object that is not accelerating, given initial position and velocity. 

• Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time. 

• Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration. 

2.6. Problem-Solving Basics for One-Dimensional Kinematics 

• Apply problem-solving steps and strategies to solve problems of one-dimensional kinematics. 

• Apply strategies to determine whether or not the result of a problem is reasonable, and if not, determine the cause. 

2.7. Falling Objects 

• Describe the effects of gravity on objects in motion. 

• Describe the motion of objects that are in free fall. 

• Calculate the position and velocity of objects in free fall. 

2.8. Graphical Analysis of One-Dimensional Motion 

• Describe a straight-line graph in terms of its slope and y-intercept. 

• Determine average velocity or instantaneous velocity from a graph of position vs. time. 

• Determine average or instantaneous acceleration from a graph of velocity vs. time. 

• Derive a graph of velocity vs. time from a graph of position vs. time. 

• Derive a graph of acceleration vs. time from a graph of velocity vs. time. 



36 CHAPTER 2 | KINEMATICS 



Introduction to One-Dimensional Kinematics 



Objects are in motion everywhere we look. Everything from a tennis game to a space-probe flyby of the planet Neptune involves motion. When you 
are resting, your heart moves blood through your veins. And even in inanimate objects, there is continuous motion in the vibrations of atoms and 
molecules. Questions about motion are interesting in and of themselves: How long will it take for a space probe to get to Mars? Where will a football 
land if it is thrown at a certain angle? But an understanding of motion is also key to understanding other concepts in physics. An understanding of 
acceleration, for example, is crucial to the study of force. 

Our formal study of physics begins with kinematics which is defined as the study of motion without considering its causes. The word "kinematics" 
comes from a Greek term meaning motion and is related to other English words such as "cinema" (movies) and "kinesiology" (the study of human 
motion). In one-dimensional kinematics and Two-Dimensional Kinematics we will study only the motion of a football, for example, without worrying 
about what forces cause or change its motion. Such considerations come in other chapters. In this chapter, we examine the simplest type of 
motion — namely, motion along a straight line, or one-dimensional motion. In Two-Dimensional Kinematics, we apply concepts developed here to 
study motion along curved paths (two- and three-dimensional motion); for example, that of a car rounding a curve. 

2.1 Displacement 




Figure 2.2 These cyclists in Vietnam can be described by their position relative to buildings and a canal. Their motion can be described by their change in position, or 
displacement, in the frame of reference, (credit: Suzan Black, Fotopedia) 

Position 

In order to describe the motion of an object, you must first be able to describe its position — where it is at any particular time. More precisely, you 
need to specify its position relative to a convenient reference frame. Earth is often used as a reference frame, and we often describe the position of 
an object as it relates to stationary objects in that reference frame. For example, a rocket launch would be described in terms of the position of the 
rocket with respect to the Earth as a whole, while a professor's position could be described in terms of where she is in relation to the nearby white 
board. (See Figure 2.3.) In other cases, we use reference frames that are not stationary but are in motion relative to the Earth. To describe the 
position of a person in an airplane, for example, we use the airplane, not the Earth, as the reference frame. (See Figure 2.4.) 

Displacement 

If an object moves relative to a reference frame (for example, if a professor moves to the right relative to a white board or a passenger moves toward 
the rear of an airplane), then the object's position changes. This change in position is known as displacement. The word "displacement" implies that 
an object has moved, or has been displaced. 



Displacement 


Displacement is the change in position of an object: 




ZaX = Xf — Xr\, 


(2.1) 


where Ax is displacement, x f is the final position, and Xq is the initial position. 





In this text the upper case Greek letter A (delta) always means "change in" whatever quantity follows it; thus, Ax means change in position. 
Always solve for displacement by subtracting initial position Xq from final position x f . 

Note that the SI unit for displacement is the meter (m) (see Physical Quantities and Units), but sometimes kilometers, miles, feet, and other units of 
length are used. Keep in mind that when units other than the meter are used in a problem, you may need to convert them into meters to complete the 
calculation. 



CHAPTER 2 | KINEMATICS 37 




&x — Xf — X(j — +2>0 m 



1 1 ^ A 

1.5 m 3.5 m 

Figure 2.3 A professor paces left and right while lecturing. Her position relative to Earth is given by X . The +2.0 m displacement of the professor relative to Earth is 
represented by an arrow pointing to the right. 




T 

2,0 m 6.0 m 

Figure 2.4 A passenger moves from his seat to the back of the plane. His location relative to the airplane is given by X . The — 4.0-m displacement of the passenger 
relative to the plane is represented by an arrow toward the rear of the plane. Notice that the arrow representing his displacement is twice as long as the arrow representing the 
displacement of the professor (he moves twice as far) in Figure 2.3. 

Note that displacement has a direction as well as a magnitude. The professor's displacement is 2.0 m to the right, and the airline passenger's 
displacement is 4.0 m toward the rear. In one-dimensional motion, direction can be specified with a plus or minus sign. When you begin a problem, 
you should select which direction is positive (usually that will be to the right or up, but you are free to select positive as being any direction). The 
professor's initial position is x = 1.5 m and her final position is Xf = 3.5 m . Thus her displacement is 



Ax = Xf—xo = 3.5 m — 1.5 m = + 2.0 m. 



(2.2) 



In this coordinate system, motion to the right is positive, whereas motion to the left is negative. Similarly, the airplane passenger's initial position is 
Xq = 6.0 m and his final position is Xf = 2.0 m , so his displacement is 



Ax = x f -x = 2.0 m - 6.0 m = -4.0 m. 
His displacement is negative because his motion is toward the rear of the plane, or in the negative x direction in our coordinate system. 



(2.3) 



Distance 

Although displacement is described in terms of direction, distance is not. Distance is defined to be the magnitude or size of displacement between 
two positions. Note that the distance between two positions is not the same as the distance traveled between them. Distance traveled is the total 
length of the path traveled between two positions. Distance has no direction and, thus, no sign. For example, the distance the professor walks is 2.0 
m. The distance the airplane passenger walks is 4.0 m. 



38 CHAPTER 2 | KINEMATICS 



Misconception Alert: Distance Traveled vs. Magnitude of Displacement 



It is important to note that the distance traveled, however, can be greater than the magnitude of the displacement (by magnitude, we mean just 
the size of the displacement without regard to its direction; that is, just a number with a unit). For example, the professor could pace back and 
forth many times, perhaps walking a distance of 150 m during a lecture, yet still end up only 2.0 m to the right of her starting point. In this case 
her displacement would be +2.0 m, the magnitude of her displacement would be 2.0 m, but the distance she traveled would be 150 m. In 
kinematics we nearly always deal with displacement and magnitude of displacement, and almost never with distance traveled. One way to think 
about this is to assume you marked the start of the motion and the end of the motion. The displacement is simply the difference in the position of 
the two marks and is independent of the path taken in traveling between the two marks. The distance traveled, however, is the total length of the 
path taken between the two marks. 



Check Your Understanding 



A cyclist rides 3 km west and then turns around and rides 2 km east, (a) What is her displacement? (b) What distance does she ride? (c) What is 
the magnitude of her displacement? 

Solution 





Ax j = ~3 km 


x 


N 
\y^ "i > rf i A 




r< 










s 




1 — , — ' 

A#2 = +2 km 





Figure 2.5 

(a) The rider's displacement is Ax = Xf — Xq = — 1 km . (The displacement is negative because we take east to be positive and west to be 
negative.) 

(b) The distance traveled is 3 km + 2 km = 5 km . 

(c) The magnitude of the displacement is 1 km . 



2.2 Vectors, Scalars, and Coordinate Systems 




Figure 2.6 The motion of this Eclipse Concept jet can be described in terms of the distance it has traveled (a scalar quantity) or its displacement in a specific direction (a vector 
quantity). In order to specify the direction of motion, its displacement must be described based on a coordinate system. In this case, it may be convenient to choose motion 
toward the left as positive motion (it is the forward direction for the plane), although in many cases, the X -coordinate runs from left to right, with motion to the right as positive 
and motion to the left as negative, (credit: Armchair Aviator, Flickr) 

What is the difference between distance and displacement? Whereas displacement is defined by both direction and magnitude, distance is defined 
only by magnitude. Displacement is an example of a vector quantity. Distance is an example of a scalar quantity. A vector is any quantity with both 
magnitude and direction. Other examples of vectors include a velocity of 90 km/h east and a force of 500 newtons straight down. 

The direction of a vector in one-dimensional motion is given simply by a plus ( + ) or minus ( — ) sign. Vectors are represented graphically by 

arrows. An arrow used to represent a vector has a length proportional to the vector's magnitude (e.g., the larger the magnitude, the longer the length 
of the vector) and points in the same direction as the vector. 

Some physical quantities, like distance, either have no direction or none is specified. A scalar is any quantity that has a magnitude, but no direction. 
For example, a 20°C temperature, the 250 kilocalories (250 Calories) of energy in a candy bar, a 90 km/h speed limit, a person's 1.8 m height, and 
a distance of 2.0 m are all scalars — quantities with no specified direction. Note, however, that a scalar can be negative, such as a — 20°C 
temperature. In this case, the minus sign indicates a point on a scale rather than a direction. Scalars are never represented by arrows. 



CHAPTER 2 | KINEMATICS 39 

Coordinate Systems for One-Dimensional Motion 

In order to describe the direction of a vector quantity, you must designate a coordinate system within the reference frame. For one-dimensional 
motion, this is a simple coordinate system consisting of a one-dimensional coordinate line. In general, when describing horizontal motion, motion to 
the right is usually considered positive, and motion to the left is considered negative. With vertical motion, motion up is usually positive and motion 
down is negative. In some cases, however, as with the jet in Figure 2.6, it can be more convenient to switch the positive and negative directions. For 
example, if you are analyzing the motion of falling objects, it can be useful to define downwards as the positive direction. If people in a race are 
running to the left, it is useful to define left as the positive direction. It does not matter as long as the system is clear and consistent. Once you assign 
a positive direction and start solving a problem, you cannot change it. 




Figure 2.7 It is usually convenient to consider motion upward or to the right as positive ( + ) and motion downward or to the left as negative ( — ) . 



Check Your Understanding 



A person's speed can stay the same as he or she rounds a corner and changes direction. Given this information, is speed a scalar or a vector 
quantity? Explain. 

Solution 

Speed is a scalar quantity. It does not change at all with direction changes; therefore, it has magnitude only. If it were a vector quantity, it would 
change as direction changes (even if its magnitude remained constant). 



2.3 Time, Velocity, and Speed 




Figure 2.8 The motion of these racing snails can be described by their speeds and their velocities, (credit: tobitasflickr, Flickr) 

There is more to motion than distance and displacement. Questions such as, "How long does a foot race take?" and "What was the runner's speed?" 
cannot be answered without an understanding of other concepts. In this section we add definitions of time, velocity, and speed to expand our 
description of motion. 



Time 

As discussed in Physical Quantities and Units, the most fundamental physical quantities are defined by how they are measured. This is the case 
with time. Every measurement of time involves measuring a change in some physical quantity. It may be a number on a digital clock, a heartbeat, or 
the position of the Sun in the sky. In physics, the definition of time is simple — time is change, or the interval over which change occurs. It is 
impossible to know that time has passed unless something changes. 

The amount of time or change is calibrated by comparison with a standard. The SI unit for time is the second, abbreviated s. We might, for example, 
observe that a certain pendulum makes one full swing every 0.75 s. We could then use the pendulum to measure time by counting its swings or, of 
course, by connecting the pendulum to a clock mechanism that registers time on a dial. This allows us to not only measure the amount of time, but 
also to determine a sequence of events. 

How does time relate to motion? We are usually interested in elapsed time for a particular motion, such as how long it takes an airplane passenger to 
get from his seat to the back of the plane. To find elapsed time, we note the time at the beginning and end of the motion and subtract the two. For 



40 CHAPTER 2 | KINEMATICS 

example, a lecture may start at 11:00 A.M. and end at 11:50 A.M., so that the elapsed time would be 50 min. Elapsed time A* is the difference 
between the ending time and beginning time, 

At = t f - 1 , (2.4) 

where At is the change in time or elapsed time, tf is the time at the end of the motion, and t^ is the time at the beginning of the motion. (As usual, 
the delta symbol, A , means the change in the quantity that follows it.) 

Life is simpler if the beginning time t$ is taken to be zero, as when we use a stopwatch. If we were using a stopwatch, it would simply read zero at 
the start of the lecture and 50 min at the end. If t$ = , then At = tf = t . 

In this text, for simplicity's sake, 

• motion starts at time equal to zero (7 = 0) 

• the symbol t is used for elapsed time unless otherwise specified (At = t f = t) 

Velocity 

Your notion of velocity is probably the same as its scientific definition. You know that if you have a large displacement in a small amount of time you 
have a large velocity, and that velocity has units of distance divided by time, such as miles per hour or kilometers per hour. 

Average Velocity 



Average velocity is displacement (change in position) divided by the time of travel, 

- _ Ax _ x f ~ X Q 
A; h - t > 



(2.5) 



where v is the average (indicated by the bar over the v ) velocity, Ax is the change in position (or displacement), and x f and Xq are the 
final and beginning positions at times t^ and t$ , respectively. If the starting time t$ is taken to be zero, then the average velocity is simply 

v = A* (2-6) 

t ' 

Notice that this definition indicates that velocity is a vector because displacement is a vector. It has both magnitude and direction. The SI unit for 
velocity is meters per second or m/s, but many other units, such as km/h, mi/h (also written as mph), and cm/s, are in common use. Suppose, for 
example, an airplane passenger took 5 seconds to move -4 m (the negative sign indicates that displacement is toward the back of the plane). His 
average velocity would be 

- = Ax = -4 m _ _ n 8 m/ c ( 2 - 7 ) 

t 5s 



■= -0.8 m/s. 



The minus sign indicates the average velocity is also toward the rear of the plane. 

The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point, however. For 
example, we cannot tell from average velocity whether the airplane passenger stops momentarily or backs up before he goes to the back of the 
plane. To get more details, we must consider smaller segments of the trip over smaller time intervals. 




A*a 



A*b 



Aa;. 



Ax, 



Ax ro 



1 1 ^ x 

Figure 2.9 A more detailed record of an airplane passenger heading toward the back of the plane, showing smaller segments of his trip. 

The smaller the time intervals considered in a motion, the more detailed the information. When we carry this process to its logical conclusion, we are 
left with an infinitesimally small interval. Over such an interval, the average velocity becomes the instantaneous velocity or the velocity at a specific 
instant. A car's speedometer, for example, shows the magnitude (but not the direction) of the instantaneous velocity of the car. (Police give tickets 
based on instantaneous velocity, but when calculating how long it will take to get from one place to another on a road trip, you need to use average 
velocity.) Instantaneous velocity v is the average velocity at a specific instant in time (or over an infinitesimally small time interval). 

Mathematically, finding instantaneous velocity, v , at a precise instant t can involve taking a limit, a calculus operation beyond the scope of this text. 
However, under many circumstances, we can find precise values for instantaneous velocity without calculus. 



CHAPTER 2 | KINEMATICS 41 

Speed 

In everyday language, most people use the terms "speed" and "velocity" interchangeably. In physics, however, they do not have the same meaning 
and they are distinct concepts. One major difference is that speed has no direction. Thus speed is a scalar. Just as we need to distinguish between 
instantaneous velocity and average velocity, we also need to distinguish between instantaneous speed and average speed. 

Instantaneous speed is the magnitude of instantaneous velocity. For example, suppose the airplane passenger at one instant had an instantaneous 
velocity of -3.0 m/s (the minus meaning toward the rear of the plane). At that same time his instantaneous speed was 3.0 m/s. Or suppose that at 
one time during a shopping trip your instantaneous velocity is 40 km/h due north. Your instantaneous speed at that instant would be 40 km/h — the 
same magnitude but without a direction. Average speed, however, is very different from average velocity. Average speed is the distance traveled 
divided by elapsed time. 

We have noted that distance traveled can be greater than displacement. So average speed can be greater than average velocity, which is 
displacement divided by time. For example, if you drive to a store and return home in half an hour, and your car's odometer shows the total distance 
traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however, was zero, because your displacement for the round trip is 
zero. (Displacement is change in position and, thus, is zero for a round trip.) Thus average speed is not simply the magnitude of average velocity. 




?S5^ 



Store 



Home 

Figure 2.10 During a 30-minute round trip to the store, the total distance traveled is 6 km. The average speed is 12 km/h. The displacement for the round trip is zero, since 
there was no net change in position. Thus the average velocity is zero. 

Another way of visualizing the motion of an object is to use a graph. A plot of position or of velocity as a function of time can be very useful. For 
example, for this trip to the store, the position, velocity, and speed-vs.-time graphs are displayed in Figure 2.11. (Note that these graphs depict a very 
simplified model of the trip. We are assuming that speed is constant during the trip, which is unrealistic given that we'll probably stop at the store. But 
for simplicity's sake, we will model it with no stops or changes in speed. We are also assuming that the route between the store and the house is a 
perfectly straight line.) 



42 CHAPTER 2 | KINEMATICS 




15- 



10- 



5 







4 -5 + 
-10-- 

-15-- 



14 -r 

12<i- 



3 10 

J 8 

<u u T 

Oh / 

y) 4-- 
2-- 





0,1 



0.2 0,3 0.4 
Time (hours) 



Velocity vs. Time 



Time (hours) 



Speed vs + Time 



H- 



H- 



-h 



0.5 0.6 



0.1 0.2 0.3 0.4 0.5 0.6 



-h 



H 



0.1 0.2 0.3 0.4 0.5 0.6 
Time (hours) 



Figure 2.11 Position vs. time, velocity vs. time, and speed vs. time on a trip. Note that the velocity for the return trip is negative. 



Making Connections: Take-Home Investigation — Getting a Sense of Speed 



If you have spent much time driving, you probably have a good sense of speeds between about 10 and 70 miles per hour. But what are these in 
meters per second? What do we mean when we say that something is moving at 10 m/s? To get a better sense of what these values really 
mean, do some observations and calculations on your own: 

• calculate typical car speeds in meters per second 

• estimate jogging and walking speed by timing yourself; convert the measurements into both m/s and mi/h 

• determine the speed of an ant, snail, or falling leaf 



Check Your Understanding 



A commuter train travels from Baltimore to Washington, DC, and back in 1 hour and 45 minutes. The distance between the two stations is 
approximately 40 miles. What is (a) the average velocity of the train, and (b) the average speed of the train in m/s? 

Solution 



CHAPTER 2 | KINEMATICS 43 



(a) The average velocity of the train is zero because Xf = Xq ; the train ends up at the same place it starts. 

(b) The average speed of the train is calculated below. Note that the train travels 40 miles one way and 40 miles back, for a total distance of 80 
miles. 



distance 
time 



80 miles 
105 minutes 



80 miles x 5280 feet x 1 meter x 1 minute _ 20 m / s 
105 minutes 1 mile 3.28 feet 60 seconds 



(2.8) 
(2.9) 



2.4 Acceleration 




Figure 2.12 A plane decelerates, or slows down, as it comes in for landing in St. Maarten. Its acceleration is opposite in direction to its velocity, (credit: Steve Conry, Flickr) 

In everyday conversation, to accelerate means to speed up. The accelerator in a car can in fact cause it to speed up. The greater the acceleration, 
the greater the change in velocity over a given time. The formal definition of acceleration is consistent with these notions, but more inclusive. 



Average Acceleration 



Average Acceleration is the rate at which velocity changes, 



_ Av _ v f ~ Vq 



(2.10) 



A; 



f o 



where a is average acceleration, v is velocity, and t is time. (The bar over the a means average acceleration.) 

Because acceleration is velocity in m/s divided by time in s, the SI units for acceleration are m/s , meters per second squared or meters per second 
per second, which literally means by how many meters per second the velocity changes every second. 

Recall that velocity is a vector — it has both magnitude and direction. This means that a change in velocity can be a change in magnitude (or speed), 
but it can also be a change in direction. For example, if a car turns a corner at constant speed, it is accelerating because its direction is changing. The 
quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in 
speed) or in direction, or both. 

Acceleration a s a Vector 

Acceleration is a vector in the same direction as the change in velocity, Av . Since velocity is a vector, it can change either in magnitude or in 
direction. Acceleration is therefore a change in either speed or direction, or both. 



Keep in mind that although acceleration is in the direction of the change in velocity, it is not always in the direction of motion. When an object slows 
down, its acceleration is opposite to the direction of its motion. This is known as deceleration. 



44 CHAPTER 2 | KINEMATICS 




Figure 2.13 A subway train in Sao Paulo, Brazil, decelerates as it comes into a station. It is accelerating in a direction opposite to its direction of motion, (credit: Yusuke 
Kawasaki, Flickr) 



Misconception Alert: Deceleration vs. Negative Acceleration 



Deceleration always refers to acceleration in the direction opposite to the direction of the velocity. Deceleration always reduces speed. Negative 
acceleration, however, is acceleration in the negative direction in the chosen coordinate system. Negative acceleration may or may not be 
deceleration, and deceleration may or may not be considered negative acceleration. For example, consider Figure 2.14. 



(a) 




y 

A 



->-X 



■*~a 



(b) 




■*► v 



a -*■ 



(c) 




»-*■ 



■>-« 



(d) 




v~4- 



a -<- 



Figure 2.14 (a) This car is speeding up as it moves toward the right. It therefore has positive acceleration in our coordinate system, (b) This car is slowing down as it 
moves toward the right. Therefore, it has negative acceleration in our coordinate system, because its acceleration is toward the left. The car is also decelerating: the 
direction of its acceleration is opposite to its direction of motion, (c) This car is moving toward the left, but slowing down over time. Therefore, its acceleration is positive in 
our coordinate system because it is toward the right. However, the car is decelerating because its acceleration is opposite to its motion, (d) This car is speeding up as it 
moves toward the left. It has negative acceleration because it is accelerating toward the left. However, because its acceleration is in the same direction as its motion, it is 
speeding up (not decelerating). 



CHAPTER 2 | KINEMATICS 45 



Example 2.1 Calculating Acceleration: A Racehorse Leaves the Gate 



A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration? 




Figure 2.15 (credit: Jon Sullivan, PD Photo.org) 

Strategy 

First we draw a sketch and assign a coordinate system to the problem. This is a simple problem, but it always helps to visualize it. Notice that we 
assign east as positive and west as negative. Thus, in this case, we have negative velocity. 



F= > 



N(+y) 
A 



v = 



W(-x) - 



Vf= -15.0 m/s 



+E(+x) 



S(-y) 



Figure 2.16 

We can solve this problem by identifying Av and At from the given information and then calculating the average acceleration directly from the 



equation a 



Av 
At 



t f - 1 

Solution 

1. Identify the knowns. v = , v f = — 15.0 m/s (the negative sign indicates direction toward the west), At = 1.80 s . 

2. Find the change in velocity. Since the horse is going from zero to — 15.0 m/s , its change in velocity equals its final velocity: 
Av = v f = -15.0 m/s. 

3. Plug in the known values ( Av and At) and solve for the unknown a . 

a=M = -15 Q 0m/s = _ 8 33 ^2 
A? 1.80 s 

Discussion 



(2.11) 



The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of 8.33 m/s due west means that the horse 

increases its velocity by 8.33 m/s due west each second, that is, 8.33 meters per second per second, which we write as 8.33 m/s . This is truly 
an average acceleration, because the ride is not smooth. We shall see later that an acceleration of this magnitude would require the rider to hang 
on with a force nearly equal to his weight. 



Instantaneous Acceleration 

Instantaneous acceleration a , or the acceleration at a specific instant in time, is obtained by the same process as discussed for instantaneous 
velocity in Time, Velocity, and Speed — that is, by considering an infinitesimally small interval of time. How do we find instantaneous acceleration 
using only algebra? The answer is that we choose an average acceleration that is representative of the motion. Figure 2.17 shows graphs of 
instantaneous acceleration versus time for two very different motions. In Figure 2.17(a), the acceleration varies slightly and the average over the 
entire interval is nearly the same as the instantaneous acceleration at any time. In this case, we should treat this motion as if it had a constant 

acceleration equal to the average (in this case about 1.8 m/s ). In Figure 2.17(b), the acceleration varies drastically over time. In such situations it 



46 CHAPTER 2 | KINEMATICS 



is best to consider smaller time intervals and choose an average acceleration for each. For example, we could consider motion over the time intervals 

9 9 

from to 1.0 s and from 1.0 to 3.0 s as separate motions with accelerations of +3.0 m/s and -2.0 m/s , respectively. 



a (m/s 2 ) 
2.0 ' 
1.0 



0.0 



a = Average 

-I 1 *- 



1.0 2.0 3.0 4.0 5.0 *(s) 



(a) 



a (m/s 2 ) 

5.0 
4.0 
3.0 
2.0 
1.0 



0.0 



-1.0 - 

-2.0 
-3.0 
-4.0 
-5.0 



_i_ 



2,0 



c 



4,0 5.0 



6.0 t(s) 



(b) 



Figure 2.17 Graphs of instantaneous acceleration versus time for two different one-dimensional motions, (a) Here acceleration varies only slightly and is always in the same 
direction, since it is positive. The average over the interval is nearly the same as the acceleration at any given time, (b) Here the acceleration varies greatly, perhaps 
representing a package on a post office conveyor belt that is accelerated forward and backward as it bumps along. It is necessary to consider small time intervals (such as 
from to 1.0 s) with constant or nearly constant acceleration in such a situation. 

The next several examples consider the motion of the subway train shown in Figure 2.18. In (a) the shuttle moves to the right, and in (b) it moves to 
the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems. 



;■■' 



{L__Jl JL__J} til 



Qoaor 



Ax = 2.00 km 



i 
i 







(a) 



-»-x 



(km) 



i xq = 4.70 km i xf = 6.70 km 



Ax' = -1,50 km 
'-< 




x (km) 



Figure 2.18 One-dimensional motion of a subway train considered in Example 2.2, Example 2.3, Example 2.4, Example 2.5, Example 2.6, and Example 2.7. Here we have 
chosen the X -axis so that + means to the right and — means to the left for displacements, velocities, and accelerations, (a) The subway train moves to the right from Xq to 

Xf . Its displacement Ax is +2.0 km. (b) The train moves to the left from x' q to x' f . Its displacement Ax r is — 1.5 km . (Note that the prime symbol (') is used simply 
to distinguish between displacement in the two different situations. The distances of travel and the size of the cars are on different scales to fit everything into the diagram.) 



Example 2.2 Calculating Displacement: A Subway Train 



What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure 2.18? 

Strategy 

A drawing with a coordinate system is already provided, so we don't need to make a sketch, but we should analyze it to make sure we 
understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation Ax = Xf — Xq . This 

is straightforward since the initial and final positions are given. 

Solution 



CHAPTER 2 | KINEMATICS 47 

1. Identify the knowns. In the figure we see that Xf = 6.70 km and x = 4.70 km for part (a), and x' f = 3.75 km and x' q = 5.25 km for 
part (b). 

2. Solve for displacement in part (a). 

Ax = x f - x = 6.70 km - 4.70 km= +2.00 km (2.12) 

3. Solve for displacement in part (b). 

Ax' = x' f - x' = 3.75 km - 5.25 km = - 1.50 km (2.13) 

Discussion 

The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus 
has a negative sign. 



Example 2.3 Comparing Distance Traveled with Displacement: A Subway Train 



What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in Figure 2.18? 

Strategy 

To answer this question, think about the definitions of distance and distance traveled, and how they are related to displacement. Distance 
between two positions is defined to be the magnitude of displacement, which was found in Example 2.2. Distance traveled is the total length of 
the path traveled between the two positions. (See Displacement.) In the case of the subway train shown in Figure 2.18, the distance traveled is 
the same as the distance between the initial and final positions of the train. 

Solution 

1. The displacement for part (a) was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km, and the distance 
traveled was 2.00 km. 

2. The displacement for part (b) was —1.5 km. Therefore, the distance between the initial and final positions was 1.50 km, and the distance 
traveled was 1.50 km. 

Discussion 

Distance is a scalar. It has magnitude but no sign to indicate direction. 



Example 2.4 Calculating Acceleration: A Subway Train Speeding Up 



Suppose the train in Figure 2.18(a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that 
time interval? 

Strategy 

It is worth it at this point to make a simple sketch: 



• 




vq — km/h j 

i 




v { = 30,0 km/h 




a= ? 



Figure 2.19 

This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we 
use these values to calculate the acceleration. 

Solution 

1. Identify the knowns. v = (the trains starts at rest), v f = 30.0 km/h , and At = 20.0 s . 

2. Calculate Av . Since the train starts from rest, its change in velocity is Av= +30.0 km/h , where the plus sign means velocity to the right. 

3. Plug in known values and solve for the unknown, a . 

- _ Av _ +30.0 km/h (2.14) 

A; 20.0 s 

4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See 
Physical Quantities and Units for more guidance.) 



48 CHAPTER 2 | KINEMATICS 



H^^^Ah) -""»»' 



(2.15) 



Discussion 

The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right 
(also positive). So acceleration is in the same direction as the change in velocity, as is always the case. 



Example 2.5 Calculate Acceleration: A Subway Train Slowing Down 



Now suppose that at the end of its trip, the train in Figure 2.18(a) slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average 
acceleration while stopping? 

Strategy 





V 

i 


i/ = 30.0 km/h ; 

• 
V£ = km/h 


^ r ' 

a= ? 



Figure 2.20 

In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the 
change in velocity and the change in time and then solve for acceleration. 

Solution 

1. Identify the knowns. v = 30.0 km/h , v f = km/h (the train is stopped, so its velocity is 0), and At = 8.00 s . 

2. Solve for the change in velocity, Av . 



Av = v. 



f- v o : 



3. Plug in the knowns, Av and At , and solve for a 



4. Convert the units to meters and seconds. 



: - 30.0 km/h = -30.0 km/h 



Av _ -30.0 km/h 



A; 



8.00 s 



^iH^SHwto- 104 "* 2 



(2.16) 



(2.17) 



(2.18) 



Discussion 

The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, 
and a negative acceleration would oppose the motion. Again, acceleration is in the same direction as the change in velocity, which is negative 
here. This acceleration can be called a deceleration because it has a direction opposite to the velocity. 



The graphs of position, velocity, and acceleration vs. time for the trains in Example 2.4 and Example 2.5 are displayed in Figure 2.21. (We have 
taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.) 



CHAPTER 2 | KINEMATICS 49 



i 



u 
O 

2 



o 

- 



Position vs. Time 




10 



20 30 40 

Time (s) 



50 



60 




10 



20 30 
Time (s) 



40 



50 



60 



Acceleration vs. Time 

-H — X 




Figure 2.21 (a) Position of the train over time. Notice that the train's position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. 
Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a 
constant rate, (b) Velocity of the train over time. The train's velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey 
(where there is no acceleration). It decreases as the train decelerates at the end of the journey, (c) The acceleration of the train over time. The train has positive acceleration 
as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down 
at the end of the journey. 



Example 2.6 Calculating Average Velocity: The Subway Train 



What is the average velocity of the train in part b of Example 2.2, and shown again below, if it takes 5.00 min to make its trip? 



50 CHAPTER 2 | KINEMATICS 




Figure 2.22 

Strategy 

Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement. 
Solution 

1. Identify the knowns. x\ = 3.75 km , x' q = 5.25 km , At = 5.00 min . 

2. Determine displacement, Ax' . We found Ax' to be — 1.5 km in Example 2.2. 

3. Solve for average velocity. 

- _ Ajk' _ -1.50 km (2.19) 



A J 5.00 min 



4. Convert units. 



v = A*l = ( -1^50km Y60mn) = _i 8 .0 km/h 
At V 5.00 mm A 1 h / 



(2.20) 



Discussion 

The negative velocity indicates motion to the left. 



Example 2.7 Calculating Deceleration: The Subway Train 



Finally, suppose the train in Figure 2.22 slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration? 

Strategy 

Once again, let's draw a sketch: 







v = —20 km/h j 
• i 
v^ — km/h 




a= ? 



Figure 2.23 

As before, we must find the change in velocity and the change in time to calculate average acceleration. 
Solution 

1. Identify the knowns. v = -20 km/h , v f = km/h , A£ = 10.0 s . 

2. Calculate Av . The change in velocity here is actually positive, since 

Av = v f - v = - (-20 km/h)=+20 km/h. 

3. Solve for a . 

- = Av = +20.0 km/h 
Af 10.0 s 



4. Convert units. 



Discussion 



"-( ±2 v^i i f^^>^^ 2 



(2.21) 



(2.22) 



(2.23) 



The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to the left) in this 
problem and a positive acceleration opposes the motion (and so it is to the right). Again, acceleration is in the same direction as the change in 



CHAPTER 2 | KINEMATICS 51 

velocity, which is positive here. As in Example 2.5, this acceleration can be called a deceleration since it is in the direction opposite to the 
velocity. 

Sign and Direction 

Perhaps the most important thing to note about these examples is the signs of the answers. In our chosen coordinate system, plus means the 
quantity is to the right and minus means it is to the left. This is easy to imagine for displacement and velocity. But it is a little less obvious for 
acceleration. Most people interpret negative acceleration as the slowing of an object. This was not the case in Example 2.7, where a positive 
acceleration slowed a negative velocity. The crucial distinction was that the acceleration was in the opposite direction from the velocity. In fact, a 
negative acceleration will increase a negative velocity. For example, the train moving to the left in Figure 2.22 is sped up by an acceleration to the 
left. In that case, both v and a are negative. The plus and minus signs give the directions of the accelerations. If acceleration has the same sign as 
the change in velocity, the object is speeding up. If acceleration has the opposite sign of the change in velocity, the object is slowing down. 



Check Your Understanding 



An airplane lands on a runway traveling east. Describe its acceleration. 

Solution 

If we take east to be positive, then the airplane has negative acceleration, as it is accelerating toward the west. It is also decelerating: its 
acceleration is opposite in direction to its velocity. 

PhET Explorations: Moving Man Simulation 



Learn about position, velocity, and acceleration graphs. Move the little man back and forth with the mouse and plot his motion. Set the position, 
velocity, or acceleration and let the simulation move the man for you. 



y 



P 

ct ^>PhET Interactive Simulation 



Figure 2.24 Moving Man (http://cnx.Org/content/m42100/l.3/moving-man_en.jar) 



2.5 Motion Equations for Constant Acceleration in One Dimension 



\ 




r *Vf | 


"4vv- 


_ J! 





Figure 2.25 Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England, (credit: Barry Skeates, 
Flickr) 

We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we 
have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for 
kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered. 

Notation: t, x, v, a 

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. 
Since elapsed time is At = tf — t$, taking t$ = means that At = tf , the final time on the stopwatch. When initial time is taken to be zero, we 

use the subscript to denote initial values of position and velocity. That is, Xq is the initial position and Vq is the initial velocity. We put no 
subscripts on the final values. That is, t is the final time, x is the final position, and v is the final velocity. This gives a simpler expression for 
elapsed time — now, At = t .\t also simplifies the expression for displacement, which is now Ax = x — Xq . Also, it simplifies the expression for 

change in velocity, which is now Av = v — Vq . To summarize, using the simplified notation, with the initial time taken to be zero, 



52 CHAPTER 2 | KINEMATICS 




(2.24) 



where the subscript denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration. 

We now make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus to find instantaneous 
acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is, 

a = a = constant, 



(2.25) 



so we use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor 
degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations 
we can accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where 
acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, the motion can be considered in separate parts, 
each of which has its own constant acceleration. 



Solving for Displacement ( Ax ) and Final Position ( x ) from Average Velocity when Acceleration ( a ) 


is Constant 




To get our first two new equations, 


we start with the definition of average velocity: 








*-& 




(2.26) 


Substituting the simplified notation for Ax and At yields 








X — Xc\ 

v = t ■ 




(2.27) 


Solving for x yields 










x = Xq + v t (constant a), 




(2.28) 


where the average velocity is 










v n + v . 
v = — *-= — (constant a). 




(2.29) 



v n + V 
The equation v = — ^ — reflects the fact that, when acceleration is constant, v is just the simple average of the initial and final velocities. For 

example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady 



increase is 45 km/h. Using the equation v 



-10 



v n + v 



to check this, we see that 



v ^Z0±I^ 3Qkm/h + 6Qkm/h =45km/h; 



(2.30) 



which seems logical. 



Example 2.8 Calculating Displacement: How Far does the Jogger Run? 



A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position 
to be zero? 



Strategy 

Draw a sketch. 






















v = 4.00 m/s 




J 

i 


























> 


ft 


x = 


= ? 









Figure 2.26 

The final position x is given by the equation 

x = Xq + vt. 

To find x , we identify the values of Xq , v , and t from the statement of the problem and substitute them into the equation. 



(2.31) 



CHAPTER 2 | KINEMATICS 53 



Solution 




1. Identify the knowns. v = 4.00 m/s , At = 2.00 min , and Xq = m . 




2. Enter the known values into the equation. 




jt = * +w = + (4.00 m/s)(120 s) = 480 m 


(2.32) 


Discussion 




Velocity and final displacement are both positive, which means they are in the same direction. 





The equation x = Xq + vt gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that 

displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on v rather than on v raised to 

some other power, such as v . When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get 
twice as far in a given time if we average 90 km/h than if we average 45 km/h. 



Displacement vs. Velocity 
for a given time, t 




Average velocity* v (m/s) 



Figure 2.27 There is a linear relationship between displacement and average velocity. For a given time t , an object moving twice as fast as another object will move twice as 
far as the other object. 



Solving for Final Velocity 


We can derive another useful equation by manipulating the definition of acceleration. 




a At 


(2.33) 


Substituting the simplified notation for Av and At gives us 




v — v n , 
a = — y~ (constant a). 


(2.34) 


Solving for v yields 




v = v + at (constant a). 


(2.35) 



Example 2.9 Calculating Final Velocity: An Airplane Slowing Down after Landing 



An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s for 40.0 s. What is its final velocity? 

Strategy 

Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating. 



54 CHAPTER 2 | KINEMATICS 





1/ 

*-# 


v = 70.0 m/s J 

< 1 


a = -1.50 m/s 2 


V f = ? 



Figure 2.28 
Solution 

1. Identify the knowns. Av = 70.0 m/s , a = -1.50 m/s 2 , t = 40.0 s . 

2. Identify the unknown. In this case, it is final velocity, v f . 

3. Determine which equation to use. We can calculate the final velocity using the equation v = v + at . 

4. Plug in the known values and solve. 

v = v + at = 70.0 m/s + (-1.50 m/s 2 )(40.0 s) = 10.0 m/s 
Discussion 



(2.36) 



The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be 
maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the 
case here. 



"'V s '' 


v = 70,0 m/s 

n 




■"V^ 


v = 10,0 m/s 


f — 


r- 


X 


a = —1.5 m/s 2 




s 


tf = -1,50 m/s 2 


%=o 






t= 40.0 s 





Figure 2.29 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before heading for the terminal. Note that the acceleration is 
negative because its direction is opposite to its velocity, which is positive. 

In addition to being useful in problem solving, the equation v = v + at gives us insight into the relationships among velocity, acceleration, and 
time. From it we can see, for example, that 

• final velocity depends on how large the acceleration is and how long it lasts 

• if the acceleration is zero, then the final velocity equals the initial velocity (v = Vq) , as expected (i.e., velocity is constant) 

• if a is negative, then the final velocity is less than the initial velocity 

(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they 
do indeed describe nature accurately.) 



Making Connections: Real-World Connection 




Figure 2.30 The Space Shuttle Endeavor blasts off from the Kennedy Space Center in February 2010. (credit: Matthew Simantov, Flickr) 

An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first 
minute or two of flight (actual ICBM burn times are classified — short-burn-time missiles are more difficult for an enemy to destroy). But the Space 



CHAPTER 2 | KINEMATICS 55 



Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle 
does this by accelerating for a longer time. 



Solving for Final Position When Velocity is Not Constant (a ^ 0) 


We can combine the equations above to find a third equation that allows us to calculate the final position of an 
acceleration. We start with 


object experiencing constant 


v = Vq + at. 








(2.37) 


Adding v to each side of this equation and dividing by 2 gives 










v n + V 1 








(2.38) 


Vn + V 

Since — *-= — = v for constant acceleration, then 










v = v + ^at. 








(2.39) 


Now we substitute this expression for v into the equation for displacement, x = 


: *0 + 


vt 


yielding 




1 2 

x = Xq + v$t + -^at (constant a). 






(2.40) 



Example 2.10 Calculating Displacement of an Accelerating Object: Dragsters 



Dragsters can achieve average accelerations of 26.0 m/s . Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does 
it travel in this time? 




Figure 2.31 U.S. Army Top Fuel pilot Tony "The Sarge" Schumacher begins a race with a controlled burnout, (credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. 
Army.) 

Strategy 

Draw a sketch. 



^ 




t= ? 


y 


W V 


% 




; 


i 


k 




a = 


26.0 m/s 2 










W A 



Figure 2.32 

We are asked to find displacement, which is x if we take Xq to be zero. (Think about it like the starting line of a race. It can be anywhere, but 

1 2 
we call it and measure all other positions relative to it.) We can use the equation x = Xq + vtf + -^-at once we identify v , a , and t from 

the statement of the problem. 
Solution 



1. Identify the knowns. Starting from rest means that v = , a is given as 26.0 m/s and t is given as 5.56 s. 



56 CHAPTER 2 | KINEMATICS 



2. Plug the known values into the equation to solve for the unknown x : 














1 2 
x = Xq + v t + ^at . 












(2.41) 


Since the initial position and velocity are both zero, this simplifies to 














x = -^at . 












(2.42) 


Substituting the identified values of a and t gives 














x = 1(26.0 m/s 2 )(5.56s) 2 , 












(2.43) 


yielding 














x = 402 m. 












(2.44) 


Discussion 














If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, 
the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters 
this. 


the standard distance for drag racing. So 
can do a quarter mile in even less time than 



1^2, 



What else can we learn by examining the equation x = Xq + vtf + ^at ? We see that: 

• displacement depends on the square of the elapsed time when acceleration is not zero. In Example 2.10, the dragster covers only one fourth of 
the total distance in the first half of the elapsed time 

1 2 

• if acceleration is zero, then the initial velocity equals average velocity ( v = v ) and x = Xq + v$t + ^at becomes x = Xq + vtf 



Solving for Final Velocity when Velocity Is Not Constant ( <z # ) 



A fourth useful equation can be obtained from another algebraic manipulation of previous equations. 
If we solve v = v + at for t , we get 

v-v 



t = ■ 



a • 



_ Vq_+V 



Substituting this and v = — ^ — into x = Xq + v t , we get 



v 2 = Vq + 2a(x — Xq) (constants). 



(2.45) 



(2.46) 



Example 2.11 Calculating Final Velocity: Dragsters 



Calculate the final velocity of the dragster in Example 2.10 without using information about time. 

Strategy 

Draw a sketch. 



1 




/= ? 


y 


ki V 


'0 




\ 


i 


i 




a = 


26.0 m/ S 2 










W X 



Figure 2.33 

The equation v = Vq + 2a(x — Xq) is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time 

information is required. 

Solution 

1. Identify the known values. We know that v = , since the dragster starts from rest. Then we note that x — Xq = 402 m (this was the 

answer in Example 2.10). Finally, the average acceleration was given to be a = 26.0 m/s . 



2 _ ,,2 



2. Plug the knowns into the equation v = v Q + 2a(x — x ) and solve for v. 



CHAPTER 2 | KINEMATICS 57 



v 2 = + 2(26.0 m/s 2 )(402 m). 












(2.47) 


Thus 














v 2 = 2.09xl0 4 m 2 /s 2 . 












(2.48) 


To get v , we take the square root: 












(2.49) 


v = V2.09xl0 4 m 2 /s 2 = 145 m/s. 


Discussion 














145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter 
root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration. 


mile. 


Also, 


note that 


a square 



An examination of the equation v = v Q + 2a(x — Xq) can produce further insights into the general relationships among physical quantities: 

• The final velocity depends on how large the acceleration is and the distance over which it acts 

• For a fixed deceleration, a car that is going twice as fast doesn't simply stop in twice the distance — it takes much further to stop. (This is why we 
have reduced speed zones near schools.) 

Putting Equations Together 

In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples 
also give insight into problem-solving techniques. The box below provides easy reference to the equations needed. 



Summary of Kinematic Equations (constant a ) 




X = Xq + vt 


(2.50) 




v + v 
V = 2 


(2.51) 




v = Vq + at 


(2.52) 




1 2 
X = Xq + Vgf + 7rdt 


(2.53) 




9 9 
V = Vq + 2a(x — Xq) 


(2.54) 



Example 2.12 Calculating Displacement: How Far Does a Car Go When Coming to a Halt? 



9 9 

On dry concrete, a car can decelerate at a rate of 7.00 m/s , whereas on wet concrete it can decelerate at only 5.00 m/s . Find the distances 
necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete, (c) Repeat both calculations, finding 
the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the 
brake. 

Strategy 

Draw a sketch. 



&X= ? 



Vq = 30,0 m/s 



pf=0 m/s 



a^y = -7,00 m/s 2 
^Vet = ~5.00 m/s 2 



Figure 2.34 

In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We 
shall do this explicitly in the next several examples, using tables to set them off. 

Solution for (a) 



58 CHAPTER 2 | KINEMATICS 

1. Identify the knowns and what we want to solve for. We know that v = 30.0 m/s ; v = ; a = —7.00 m/s ( a is negative because it is in 
a direction opposite to velocity). We take Xq to be 0. We are looking for displacement Ax , or x — Xq . 



2. Identify the equation that will help up solve the problem. The best equation to use is 

9 9 

V = Vq + 2a(x — Xq). 



(2.55) 



This equation is best because it includes only one unknown, x . We know the values of all the other variables in this equation. (There are other 
equations that would allow us to solve for x , but they require us to know the stopping time, t , which we do not know. We could use them but it 
would entail additional calculations.) 
3. Rearrange the equation to solve for x . 



X — Xq — ■ 



2a 



4. Enter known values. 



Thus, 



Q = 2 - (30.0 m/s) 2 
2(-7.00 m/s 2 ) 

i = 64.3 m on dry concrete. 
Solution for (b) 

This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is - 5.00 m/s . The result is 

r p t = 90.0 m on wet concrete. 



(2.56) 



(2.57) 



(2.58) 



(2.59) 



Solution for (c) 

Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need 
to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity 
remains constant during the driver's reaction time. 

1. Identify the knowns and what we want to solve for. We know that v = 30.0 m/s ; £ react i on = 0.500 s ; ^ react i on = . We take x _ react i on 
to be 0. We are looking for ^ react i on ■ 

2. Identify the best equation to use. 

x = Xq 4- v t works well because the only unknown value is x , which is what we want to solve for. 

3. Plug in the knowns to solve the equation. 

x = + (30.0 m/s)(0.500 s) = 15.0 m. (2.60) 

This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m 
greater than if he reacted instantly. 

4. Add the displacement during the reaction time to the displacement when braking. 

^braking + -^reaction = x total (2.61) 

a. 64.3 m + 15.0 m = 79.3 m when dry 

b. 90.0 m + 15.0 m = 105 m when wet 



643 




90.0m ' 



Reaction 
time 





105m «P 
wet 



— r~ 
50 



100 



T 



Position x (m) 



Figure 2.35 The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and 
wet pavement, as calculated in this example, for a car initially traveling at 30.0 m/s. Also shown are the total distances traveled from the point where the driver first sees a 
light turn red, assuming a 0.500 s reaction time. 



CHAPTER 2 | KINEMATICS 59 

Discussion 

The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than 
dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving 
problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way 
to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest. 



Example 2.13 Calculating Time: A Car Merges into Traffic 



Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at 2.00 m/s , how long does 
it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.) 
Strategy 
Draw a sketch. 



x = 


t= ? 


= 200 m 
= ? 


-0 x = 

= 10.0 m/s v = 




a = 2.00 m/s 2 











Figure 2.36 

We are asked to solve for the time * . As before, we identify the known quantities in order to choose a convenient physical relationship (that is, 
an equation with one unknown, *). 

Solution 

1. Identify the knowns and what we want to solve for. We know that v = 10 m/s ; a = 2.00 m/s ; and x = 200 m . 

1 2 

2. We need to solve for * . Choose the best equation, x = Xq 4- v * + -jrat works best because the only unknown in the equation is the 

variable * for which we need to solve. 

3. We will need to rearrange the equation to solve for * . In this case, it will be easier to plug in the knowns first. 

200 m = m + (10.0 m/s)* + 1(2.00 m/s 2 )* 2 (262) 

4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking 
t = t s , where * is the magnitude of time and s is the unit. Doing so leaves 



200 = 10* + t\ 
5. Use the quadratic formula to solve for * . 

(a) Rearrange the equation to get on one side of the equation. 

* 2 + 10* - 200 = 
This is a quadratic equation of the form 

at 2 + bt + c = 0, 
where the constants are a = 1.00, b = 10.0, and c = —200 . 

(b) Its solutions are given by the quadratic formula: 



* = ■ 



-b ± Jb 2 - Aac 
2a 



This yields two solutions for * , which are 



In this case, then, the time is * = * in seconds, or 



* = 10.0 and- 20.0. 



* = 10.0 s and - 20.0 s. 



(2.63) 

(2.64) 
(2.65) 

(2.66) 

(2.67) 
(2.68) 



60 CHAPTER 2 | KINEMATICS 

A negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We can discard that 
solution. Thus, 

t = 10.0 s. (2.69) 

Discussion 

Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, 
such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a typical freeway on-ramp. 

With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing 
kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical 
relationships. Problem-Solving Basics discusses problem-solving basics and outlines an approach that will help you succeed in this invaluable task. 

Making Connections: Take-Home Experiment — Breaking News 

We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and 
trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car doing a slow (and safe) stop. Recall that, 
for average acceleration, a = Av I At . While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note 
the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. 
Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in 
braking. 











A manned rocket accelerates at a rate of 20 m/s during launch. How long does it take the rocket reach a velocity of 400 m/s? 




Solution 




To answer this, choose an equation that allows you to solve for time t , given only a , v , and v . 




v = Vq + at 


(2.70) 


Rearrange to solve for t . 




t _ v — v _ 400 m/s - m/s _ oq . 
a 20 m/s 2 


(2.71) 



2.6 Problem-Solving Basics for One-Dimensional Kinematics 




Figure 2.37 Problem-solving skills are essential to your success in Physics, (credit: scui3asteveo, Flickr) 

Problem-solving skills are obviously essential to success in a quantitative course in physics. More importantly, the ability to apply broad physical 
principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a 
list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to 
contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics in everyday and 
professional life. 

Problem-Solving Steps 

While there is no simple step-by-step method that works for every problem, the following general procedures facilitate problem solving and make it 
more meaningful. A certain amount of creativity and insight is required as well. 

Stepl 

Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketch at the outset. You will also need to 
decide which direction is positive and note that on your sketch. Once you have identified the physical principles, it is much easier to find and apply the 



CHAPTER 2 | KINEMATICS 61 

equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, 
laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless. 

Step 2 

Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Many problems are stated very succinctly and require 
some inspection to determine what is known. A sketch can also be very useful at this point. Formally identifying the knowns is of particular 
importance in applying physics to real-world situations. Remember, "stopped" means velocity is zero, and we often can take initial time and position 
as zero. 

Step 3 

Identify exactly what needs to be determined in the problem (identify the unknowns). In complex problems, especially, it is not always obvious what 
needs to be found or in what sequence. Making a list can help. 

Step 4 

Find an equation or set of equations that can help you solve the problem. Your list of knowns and unknowns can help here. It is easiest if you can find 
equations that contain only one unknown — that is, all of the other variables are known, so you can easily solve for the unknown. If the equation 
contains more than one unknown, then an additional equation is needed to solve the problem. In some problems, several unknowns must be 
determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a 
sea of equations. You may have to use two (or more) different equations to get the final answer. 

Step 5 

Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units. This step produces the 
numerical answer; it also provides a check on units that can help you find errors. If the units of the answer are incorrect, then an error has been 
made. However, be warned that correct units do not guarantee that the numerical part of the answer is also correct. 

Step 6 

Check the answer to see if it is reasonable: Does it make sense? This final step is extremely important — the goal of physics is to accurately describe 
nature. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. Your judgment will improve as you solve 
more and more physics problems, and it will become possible for you to make finer and finer judgments regarding whether nature is adequately 
described by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is 
reasonable, you have a deeper understanding of physics than just being able to mechanically solve a problem. 

When solving problems, we often perform these steps in different order, and we also tend to do several steps simultaneously. There is no rigid 
procedure that will work every time. Creativity and insight grow with experience, and the basics of problem solving become almost automatic. One 
way to get practice is to work out the text's examples for yourself as you read. Another is to work as many end-of-section problems as possible, 
starting with the easiest to build confidence and progressing to the more difficult. Once you become involved in physics, you will see it all around you, 
and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text. 

Unreasonable Results 

Physics must describe nature accurately. Some problems have results that are unreasonable because one premise is unreasonable or because 
certain premises are inconsistent with one another. The physical principle applied correctly then produces an unreasonable result. For example, if a 

person starting a foot race accelerates at 0.40 m/s for 100 s, his final speed will be 40 m/s (about 150 km/h) — clearly unreasonable because the 
time of 100 s is an unreasonable premise. The physics is correct in a sense, but there is more to describing nature than just manipulating equations 
correctly. Checking the result of a problem to see if it is reasonable does more than help uncover errors in problem solving — it also builds intuition in 
judging whether nature is being accurately described. 

Use the following strategies to determine whether an answer is reasonable and, if it is not, to determine what is the cause. 

Stepl 

Solve the problem using strategies as outlined and in the format followed in the worked examples in the text. In the example given in the preceding 
paragraph, you would identify the givens as the acceleration and time and use the equation below to find the unknown final velocity. That is, 

v = v + at = + (0.40 m/s 2 )(100 s) = 40 m/s. < 2 - 72 ) 

Step 2 

Check to see if the answer is reasonable. Is it too large or too small, or does it have the wrong sign, improper units, ...? In this case, you may need to 
convert meters per second into a more familiar unit, such as miles per hour. 

(2.73) 



f 40m Y 3.28ft V 1 mi Y 60 s Y 60 min ^i _ o Q mnh 
I s A m A5280ftAminA lh ) y p 



This velocity is about four times greater than a person can run — so it is too large. 

Step 3 

If the answer is unreasonable, look for what specifically could cause the identified difficulty. In the example of the runner, there are only two 
assumptions that are suspect. The acceleration could be too great or the time too long. First look at the acceleration and think about what the number 

means. If someone accelerates at 0.40 m/s , their velocity is increasing by 0.4 m/s each second. Does this seem reasonable? If so, the time must 
be too long. It is not possible for someone to accelerate at a constant rate of 0.40 m/s for 100 s (almost two minutes). 



62 CHAPTER 2 | KINEMATICS 

2.7 Falling Objects 

Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it 
and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations 
and learn much about gravity in the process. 

Gravity 

The most remarkable and unexpected fact about falling objects is that, if air resistance and friction are negligible, then in a given location all objects 
fall toward the center of Earth with the same constant acceleration, independent of their mass. This experimentally determined fact is unexpected, 
because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. 



i 



l 



h 




In air In a vacuum In a vacuum (the hard way) 

Figure 2.38 A hammer and a feather will fall with the same constant acceleration if air resistance is considered negligible. This is a general characteristic of gravity not unique 

to Earth, as astronaut David R. Scott demonstrated on the Moon in 1971, where the acceleration due to gravity is only 1.67 m/s . 

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball will reach the ground after 
a hard baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion 
of an object through the air, while friction between objects — such as between clothes and a laundry chute or between a stone and a pool into which it 
is dropped — also opposes motion between them. For the ideal situations of these first few chapters, an object falling without air resistance or friction 
is defined to be in free-fall. 

The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called the acceleration due 
to gravity. The acceleration due to gravity is constant, which means we can apply the kinematics equations to any falling object where air resistance 
and friction are negligible. This opens a broad class of interesting situations to us. The acceleration due to gravity is so important that its magnitude is 
given its own symbol, g . It is constant at any given location on Earth and has the average value 



g = 9.80 m/s 2 



(2.74) 



9 9 

Although g varies from 9.78 m/s to 9.83 m/s , depending on latitude, altitude, underlying geological formations, and local topography, the 

average value of 9.80 m/s will be used in this text unless otherwise specified. The direction of the acceleration due to gravity is downward 
(towards the center of Earth). In fact, its direction defines what we call vertical. Note that whether the acceleration a in the kinematic equations has 
the value +g or — g depends on how we define our coordinate system. If we define the upward direction as positive, then 

9 9 

a = — g = —9.80 m/s , and if we define the downward direction as positive, then a = g = 9.80 m/s . 

One-Dimensional Motion Involving Gravity 

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex 
ones. So we start by considering straight up and down motion with no air resistance or friction. These assumptions mean that the velocity (if there is 
any) is vertical. If the object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object 
is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g . We will also represent 

vertical displacement with the symbol y and use x for horizontal displacement. 



Kinematic Equations for Objects in Free-Fail where Acceleration = g 


v = v + gt 


(2.75) 


y = y + Vo t + lg t 2 


(2.76) 


y2 = v o + 2 s(y - yd) 


(2.77) 



Example 2.14 Calculating Position and Velocity of a Falling Object: A Rock Thrown Upward 



A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as 
it falls back to earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air 
resistance. 

Strategy 

Draw a sketch. 



CHAPTER 2 | KINEMATICS 63 



v = 13,0 m/s 



t 



= -9.8 m/s 2 



y 
k 



-*~x 



Figure 2.39 

We are asked to determine the position y at various times. It is reasonable to take the initial position y§ to be zero. This problem involves one- 
dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since 
up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so a is 
negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration 
due to gravity opposes the initial motion and will slow and eventually reverse it. 

Since we are asked for values of position and velocity at three times, we will refer to these as _y 1 and v 1 ; y 2 and v 2 ; and y$ and v 3 . 
Solution for Position y x 

1. Identify the knowns. We know that j = ; v = 13.0 m/s ; a = — g = -9.80 m/s ; and t = 1.00 s . 

1 2 

2. Identify the best equation to use. We will use y = y$ + vtf + ^at because it includes only one unknown, y (or j 1 , here), which is the 

value we want to find. 

3. Plug in the known values and solve for y^ . 

(2.78) 



y = + (13.0 m/s)(1.00 s) + ±(-9.80 m/s 2 )(1.00 s) 2 = 8.10 m 



Discussion 

The rock is 8.10 m above its starting point at t = 1.00 s, since j 1 > y$ . It could be moving up or down; the only way to tell is to calculate v 1 
and find out if it is positive or negative. 
Solution for Velocity v 1 

1. Identify the knowns. We know that y$ = 0; v = 13.0 m/s ; a = —g = —9.80 m/s ; and t = 1.00 s . We also know from the solution 
above that y± = 8.10 m . 

2. Identify the best equation to use. The most straightforward is v = v - gt (from v = v + at , where 
a = gravitational acceleration = —g). 

3. Plug in the knowns and solve. 

Vl = v - gt = 13.0 m/s - (9.80 m/s 2 )(1.00 s) = 3.20 m/s < 2 - 79 ) 

Discussion 

The positive value for v 1 means that the rock is still heading upward at t = 1.00 s . However, it has slowed from its original 13.0 m/s, as 

expected. 

Solution for Remaining Times 

The procedures for calculating the position and velocity at t = 2.00 s and 3.00 s are the same as those above. The results are summarized in 
Table 2.1 and illustrated in Figure 2.40. 



Table 2.1 Results 



Time, t Position, y Velocit 






1.00 s 


8.10 m 


3.20 m/s 


-9.80 m/s 2 


2.00 s 


6.40 m 


-6.60 m/s 


-9.80 m/s 2 


3.00 s 


-5.10m 


-16.4 m/s 


-9.80 m/s 2 



Graphing the data helps us understand it more clearly. 



64 CHAPTER 2 | KINEMATICS 



o 

1 



,6 



= 
o 

1 

1 



15 
10 

5 


-5 
-10 

-15 
-20 









-2 

-4 

-6 + 

-8 

-10 + 
-12 



Position vs. Time 




Time (s) 



Velocity vs. Time 



Time (s) 



Acceleration vs. Time 



-h 



-+- 



Time (s) 



Figure 2.40 Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly 
with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that 
the graph shows some horizontal motion — the shape of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The 
actual path of the rock in space is straight up, and straight down. 

Discussion 

The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since y± and Vj are both 
positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both y^ and v 3 
are negative, meaning the rock is below its starting point and continuing to move downward. Notice that when the rock is at its highest point (at 

9 9 

1.5 s), its velocity is zero, but its acceleration is still —9.80 m/s . Its acceleration is —9.80 m/s for the whole trip — while it is moving up and 
while it is moving down. Note that the values for y are the positions (or displacements) of the rock, not the total distances traveled. Finally, note 

that free-fall applies to upward motion as well as downward. Both have the same acceleration — the acceleration due to gravity, which remains 
constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we 
will discuss in more detail later. 



Making Connections: Take-Home Experiment — Reaction Time 



A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index finger, separated by 
about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedly, and try to catch it 
between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far 
would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brake was twice this reaction time? 



CHAPTER 2 | KINEMATICS 65 



Example 2.15 Calculating Velocity of a Falling Object: A Rock Thrown Down 



What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the 
rock when it is 5.10 m below the starting point, and has been thrown downward with an initial speed of 13.0 m/s. 

Strategy 

Draw a sketch. 



vq = -13.0 m/s 



H' 



= -9.8 m/s 2 



y 



->~x 



Figure 2.41 

Since up is positive, the final position of the rock will be negative because it finishes below the starting point at y$ = . Similarly, the initial 

velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will 
continue to move downward. 

Solution 

1. Identify the knowns. j = ; y± = — 5.10 m ; v = - 13.0 m/s ; a = —g = -9.80 m/s . 

9 9 

2. Choose the kinematic equation that makes it easiest to solve the problem. The equation v = v Q + 2a(y — y ) works well because the only 
unknown in it is v . (We will plug y± in for y .) 

3. Enter the known values 

v 2 = (-13.0 m/s) 2 + 2(-9.80 m/s 2 )(-5.10 m - m) = 268.96 m 2 /s 2 , ( 28 °) 

where we have retained extra significant figures because this is an intermediate result. 
Taking the square root, and noting that a square root can be positive or negative, gives 

v = ±16.4 m/s. (2.81) 

The negative root is chosen to indicate that the rock is still heading down. Thus, 

v = -16.4 m/s. (2.82) 

Discussion 

Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed. (See 
Example 2.14 and Figure 2.42(a).) This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, 
the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of 
the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 2.14) when the initial velocity is 13.0 m/s 
straight up, a result of ±3.20 m/s is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and 
heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction. 



66 CHAPTER 2 | KINEMATICS 



yi = 8.10 m 



i^ 



vq — 13 m/s 



yo = o 




• yi ; 



6.40 m 



■J3 = -5.10 m 



^3 — — 1 6.4 m/s 




^o = 



^o = — 1 3 m/s 



.7 = -5.10 m 



v — — 16.4 m/s 



(a) (b) 

Figure 2.42 (a) A person throws a rock straight up, as explored in Example 2.14. The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. (b) A person throws a rock 
straight down from a cliff with the same initial speed as before, as in Example 2.15. Note that at the same distance below the point of release, the rock has the same 
velocity in both cases. 

Another way to look at it is this: In Example 2.14, the rock is thrown up with an initial velocity of 13.0 m/s . It rises and then falls back down. 
When its position is y = on its way back down, its velocity is — 13.0 m/s . That is, it has the same speed on its way down as on its way up. 
We would then expect its velocity at a position of y = —5.10 m to be the same whether we have thrown it upwards at +13.0 m/s or thrown it 
downwards at — 13.0 m/s . The velocity of the rock on its way down from y = is the same whether we have thrown it up or down to start 
with, as long as the speed with which it was initially thrown is the same. 



Example 2.16 Find g from Data on a Falling Object 



The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a 
valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise 
acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object, usually a metal ball for 
which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure 2.43. Very 
precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time. 



CHAPTER 2 | KINEMATICS 67 







;N 




-0.049 



-0.196 



-0.441 



-0784 



-1.225 



' (m/s) 




-0.98 



-1.96 



-2.94 



-3.92 



-4.90 



*M 




0.1 



0.2 



0.3 



0.4 



0.5 



Position vs. Time for Falling Sphere 
Time t (s) 




Velocity vs. Time for Falling Sphere 
Time t (s) 



0.1 0.2 



0.4 0.5 0.6 




Acceleration vs. Time for Falling Sphere 




-2 
-4 
-6 



-10+ 



Time t (s) 

0.1 0.2 0.3 0.4 0.5 0.6 

— I 



~r 



~r 



~r 



n~ 



~i 



Figure 2.43 Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with time while displacement 
increases with time squared. Acceleration is a constant and is equal to gravitational acceleration. 

Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at 
this location? 

Strategy 

Draw a sketch. 



£>o = m/s 



• t 



y 



■ = ? 



-*~x 



Figure 2.44 

We need to solve for acceleration a . Note that in this case, displacement is downward and therefore negative, as is acceleration. 
Solution 



68 CHAPTER 2 | KINEMATICS 

1. Identify the knowns. y = ; y = -1.0000 m ; t = 0.45173 ; v = . 

2. Choose the equation that allows you to solve for a using the known values. 



3. Substitute for v and rearrange the equation to solve for a . Substituting for v yields 

Solving for a gives 

2(y-y ) (2.85) 



a 



t 2 



4. Substitute known values yields 



a = 2(- 1.0000 m-0) = _ 98010m/g 2 > (2-86) 

(0.45173 s) 2 



so, because a = —g with the directions we have chosen, 

g = 9.8010 m/s 2 . < 2 - 87 ) 

Discussion 

The negative value for a indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around 
the average value of 9.80 m/s , so 9.8010 m/s makes sense. Since the data going into the calculation are relatively precise, this value for 
g is more precise than the average value of 9.80 m/s ; it represents the local value for the acceleration due to gravity. 



Check Your Understanding 
















A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air resistance), 


how 


ong does 


it take to hit the water? 
















Solution 
















We know that initial position y§ = , 


final position y = —30.0 m , and 


a = -# = -9.80 m/s 2 


. We can 


then use 


the equation 






1 2 

y = yo + v o^ + o at t0 solve for t 


. Inserting a = —g , we obtain 














y 


= o+o-l^ 2 












(2.88) 


t 2 
t 


_ 2y 

-8 




;2.5s 










_ W2y _ ( J2(-30.0m)_ 
" 8 f -9.80 m/s 2 


= ±V6.12s 2 = 2.47 s* 


where we take the positive value as the physically relevant answer. Thus 


it takes about 2.5 seconds for the 


piece of 


ice to hit the water 





PhET Explorations: Equation Grapher 



Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (e.g. 
y = bx ) to see how they add to generate the polynomial curve. 



^ 



>■ 



<*r. 



.Jf 



PhET Interactive Simulation 



Figure 2.45 Equation Grapher (http://cnx.Org/content/m42102/l.3/equation-grapher_en.jar) 

2.8 Graphical Analysis of One-Dimensional Motion 

A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical 
quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate one-dimensional kinematics. 



CHAPTER 2 | KINEMATICS 69 

Slopes and General Relationships 

First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one 
another in such a graph, the horizontal axis is usually considered to be an independent variable and the vertical axis a dependent variable. If we 
call the horizontal axis the x -axis and the vertical axis the y -axis, as in Figure 2.46, a straight-line graph has the general form 



y = mx + b. 



(2.89) 



Here m is the slope, defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter b is used for the y-intercept, 
which is the point at which the line crosses the vertical axis. 



J 

Intercept 


, y = mx + b 

y- 

^r rise Ay 

y^ Slope = = A — m 
v^ r run m 

I b 


( 


> 



Figure 2.46 A straight-line graph. The equation for a straight line is y = mx + b . 

Graph of Displacement vs. Time (a = 0, so v is constant) 

Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, 
have x on the vertical axis and t on the horizontal axis. Figure 2.47 is just such a straight-line graph. It shows a graph of displacement versus time 
for a jet-powered car on a very flat dry lake bed in Nevada. 



i 

^ 2400 - 

^ 2000 - 

f 1600 - 
u 

I 1200 - 

"Si 800 - 
£ 525- 


i 


X = X() + vt. 








^n*-^ Mope — v — — — 

^ 00 0^ ^ At 


i 


X 

1 


**\~< 


\f *. 




iii ^ 


x = 400 


; i i ii 

i 1 2 3 4 5 6 t 7 

0.50 x . ,v 6.40 
Time t (s) 


I * 
8 



Figure 2.47 Graph of displacement versus time for a jet-powered car on the Bonneville Salt Flats. 

Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity v and the 
intercept is displacement at time zero — that is, Xq . Substituting these symbols into y = mx + b gives 



or 



x = vt + Xq 



x = x n + vt. 



(2.90) 



(2.91) 



Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical 
information about a specific situation. 



The Slope of x vs. t 

The slope of the graph of displacement x vs. time t is velocity v . 



Ar 

slope = -^ = v 



(2.92) 



70 CHAPTER 2 | KINEMATICS 

Notice that this equation is the same as that derived algebraically from other motion equations in Motion Equations for Constant Acceleration 
in One Dimension. 

From the figure we can see that the car has a displacement of 400 m at time 0.650 m at t = 1.0 s, and so on. Its displacement at times other than 
those listed in the table can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph. 



Example 2.17 Determining Average Velocity from a Graph of Displacement versus Time: Jet Car 



Find the average velocity of the car whose position is graphed in Figure 2.47. 

Strategy 

The slope of a graph of x vs. t is average velocity, since slope equals rise over run. In this case, rise = change in displacement and run = 

change in time, so that 

slope = Ai = v. < 2 - 93 ) 

Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two 
widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is 
larger.) 

Solution 

1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, 
that you could choose any two points.) 

2. Substitute the x and t values of the chosen points into the equation. Remember in calculating change (A) we always use final value minus 

initial value. 

- = M = 2000 m - 525 m (2.94) 

At 6.4 s -0.50 s ' 

yielding 

v = 250 m/s. (2.95) 

Discussion 

This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 
96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997. 

Graphs of Motion when a is constant but a ^ 

The graphs in Figure 2.48 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its 
acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m 
and 15 m/s, respectively. 



CHAPTER 2 | KINEMATICS 71 



3500 
~ 3000 
X 2500 
I 2000 
§ 1500 
"a 1000 



(a) 



(b) 



Jet Car Displacement 



Slope = v 




10 20 30 

Time, t (s) 



40 



180 i 
160 ■ 
^ 140 ■ 
1 120 

T ioo ■ 


Jet Car Velocity 




Slope = <i 






£ 80- 


VyT 






% 60 

5 40- 








20; 








oJ 

( 








) 10 20 30 


40 




Time, t (s) 









Jet Car Acceleration 








*1 


















N 

<3 


5< 
4 
















1 

u 


3- 

2 ■ 
1 ■ 
■ 


















( 


) 10 20 


30 


40 






Time, t (s) 







(c) 
Figure 2.48 Graphs of motion of a jet-powered car during the time span when its acceleration is constant, (a) The slope of an X vs. t graph is velocity. This is shown at two 
points, and the instantaneous velocities obtained are plotted in the next graph. Instantaneous velocity at any point is the slope of the tangent at that point, (b) The slope of the 

V vs. t graph is constant for this part of the motion, indicating constant acceleration, (c) Acceleration has the constant value of 5.0 m/s over the time interval plotted. 



72 CHAPTER 2 | KINEMATICS 




Figure 2.49 A U.S. Air Force jet car speeds down a track, (credit: Matt Trostle, Flickr) 

The graph of displacement versus time in Figure 2.48(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time 
progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous 
velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent 
lines are shown for two points in Figure 2.48(a). If this is done at every point on the curve and the values are plotted against time, then the graph of 
velocity versus time shown in Figure 2.48(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown 
in Figure 2.48(c). 



Example 2.18 Determining Instantaneous Velocity from the Slope at a Point: Jet Car 



Calculate the velocity of the jet car at a time of 25 s by finding the slope of the x vs. t graph in the graph below. 



1 


i 




& 3000 - 




2 2500 - 
1 2000 - 


q/ 

Slope = v / 


&Xq 


| 1500- 

J 1000- 

£ 500- 
Q 


p j( ~ Afe 




u n 


r f f i i i i 


I - 



Ms) 


x (m) 





200 


5 


338 


10 


600 


15 


988 


20 


1500 


25 


2138 


30 


2900 



5 10 15 20 25 30 
Time, t (s) 

Figure 2.50 The slope of an X vs. t graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point. 

Strategy 

The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 2.50, 
where Q is the point at t = 25 s . 

Solution 

1. Find the tangent line to the curve at t = 25 s . 

2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s. 

3. Plug these endpoints into the equation to solve for the slope, v . 



Thus, 



T A *Q (3120 m -1300 m) 

slope = v Q = _^= (32s _ 19s) 



= 1820 m = 14Q ^ 
^ 13 s 



(2.96) 



(2.97) 



Discussion 

This is the value given in this figure's table for v at t = 25 s . The value of 140 m/s for Vq is plotted in Figure 2.50. The entire graph of v vs. 
t can be obtained in this fashion. 



Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a v vs. t graph, 
rise = change in velocity Av and run = change in time A^ . 



CHAPTER 2 | KINEMATICS 73 



The Slope of vvs. t 



The slope of a graph of velocity v vs. time t is acceleration a . 

slope 



Av 
At 



(2.98) 



Since the velocity versus time graph in Figure 2.48(b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. 
Acceleration versus time is graphed in Figure 2.48(c). 

Additional general information can be obtained from Figure 2.50 and the expression for a straight line, y = mx + b . 

In this case, the vertical axis y is V , the intercept b is v , the slope m is a , and the horizontal axis x is t . Substituting these symbols yields 

v = v + at. (2.99) 

A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived 
algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension. 

It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to discover 
physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in 
anyway. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical 
relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships. 

Graphs of Motion Where Acceleration is Not Constant 

Now consider the motion of the jet car as it goes from 165 m/s to its top velocity of 250 m/s, graphed in Figure 2.51. Time again starts at zero, and 
the initial displacement and velocity are 2900 m and 165 m/s, respectively. (These were the final displacement and velocity of the car in the motion 

graphed in Figure 2.48.) Acceleration gradually decreases from 5.0 m/s to zero when the car hits 250 m/s. The slope of the x vs. t graph 
increases until t = 55 s , after which time the slope is constant. Similarly, velocity increases until 55 s and then becomes constant, since acceleration 
decreases to zero at 55 s and remains zero afterward. 



74 CHAPTER 2 | KINEMATICS 



25 



J 20 



(a) 



(b) 



Jet Car Displacement 




Jet Car Velocity 



Time, t (s) 



Jet Car Acceleration 




10 20 30 40 50 60 70 80 
Time, t (s) 




10 20 30 40 50 
Time, t (s) 



60 70 80 



(c) 

Figure 2.51 Graphs of motion of a jet-powered car as it reaches its top velocity. This motion begins where the motion in Figure 2.48 ends, (a) The slope of this graph is 
velocity; it is plotted in the next graph, (b) The velocity gradually approaches its top value. The slope of this graph is acceleration; it is plotted in the final graph, (c) Acceleration 
gradually declines to zero when velocity becomes constant. 



Example 2.19 Calculating Acceleration from a Graph of Velocity versus Time 



Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the v vs. t graph in Figure 2.51(b). 

Strategy 

The slope of the curve at t = 25 s is equal to the slope of the line tangent at that point, as illustrated in Figure 2.51(b). 

Solution 

Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope, a . 

, _ Av _ (260 m/s - 210 m/s) 
slope- Af - (51s _ 10s) 

a= 5Qjn/s = 10m/s 2 

50 s 



(2.100) 
(2.101) 



Discussion 



CHAPTER 2 | KINEMATICS 75 



Note that this value for a is consistent with the value plotted in Figure 2.51(c) at t = 25 s . 



A graph of displacement versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to 
generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a 
constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe 
both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical 
relationships is to graph them and look for underlying relationships. 



Check Your Understanding ^^^^H 








A graph of velocity vs. time of a ship coming into a harbor is shown below, (a) Describe the motion of the ship based on the graph. (b)What 
would a graph of the ship's acceleration look like? 




V 






t 


Figure 2.52 
Solution 

(a) The ship moves at constant velocity and then be 
maintains this lower deceleration rate until it stops it 

(b) A graph of acceleration vs. time would show zerc 
constant negative acceleration. 


gins to decelerate at a constant rate. At some point, its deceleration rate decreases. It 
oving. 

) acceleration in the first leg, large and constant negative acceleration in the second leg, and 




a 


i 








Figure 2.53 







Glossary 



acceleration due to gravity: acceleration of an object as a result of gravity 

acceleration: the rate of change in velocity; the change in velocity over time 

average acceleration: the change in velocity divided by the time over which it changes 

average speed: distance traveled divided by time during which motion occurs 

average velocity: displacement divided by time over which displacement occurs 

deceleration: acceleration in the direction opposite to velocity; acceleration that results in a decrease in velocity 

dependent variable: the variable that is being measured; usually plotted along the y -axis 

displacement: the change in position of an object 

distance traveled: the total length of the path traveled between two positions 

distance: the magnitude of displacement between two positions 

elapsed time: the difference between the ending time and beginning time 

free-fall: the state of movement that results from gravitational force only 

independent variable: the variable that the dependent variable is measured with respect to; usually plotted along the x -axis 

instantaneous acceleration: acceleration at a specific point in time 

instantaneous speed: magnitude of the instantaneous velocity 



76 CHAPTER 2 | KINEMATICS 

instantaneous velocity: velocity at a specific instant, or the average velocity over an infinitesimal time interval 

kinematics: the study of motion without considering its causes 

model: simplified description that contains only those elements necessary to describe the physics of a physical situation 

position: the location of an object at a particular time 

scalar: a quantity that is described by magnitude, but not direction 

slope: the difference in y -value (the rise) divided by the difference in x -value (the run) of two points on a straight line 

time: change, or the interval over which change occurs 

vector: a quantity that is described by both magnitude and direction 

y-intercept: the y- value when x = 0, or when the graph crosses the y -axis 



Section Summary 



2.1 Displacement 

• Kinematics is the study of motion without considering its causes. In this chapter, it is limited to motion along a straight line, called one- 
dimensional motion. 

• Displacement is the change in position of an object. 

• In symbols, displacement Ax is defined to be 

Ax — Xf — Xq, 

where Xq is the initial position and Xf is the final position. In this text, the Greek letter A (delta) always means "change in" whatever quantity 

follows it. The SI unit for displacement is the meter (m). Displacement has a direction as well as a magnitude. 

• When you start a problem, assign which direction will be positive. 

• Distance is the magnitude of displacement between two positions. 

• Distance traveled is the total length of the path traveled between two positions. 

2.2 Vectors, Scalars, and Coordinate Systems 

• A vector is any quantity that has magnitude and direction. 

• A scalar is any quantity that has magnitude but no direction. 

• Displacement and velocity are vectors, whereas distance and speed are scalars. 

• In one-dimensional motion, direction is specified by a plus or minus sign to signify left or right, up or down, and the like. 

2.3 Time, Velocity, and Speed 

• Time is measured in terms of change, and its SI unit is the second (s). Elapsed time for an event is 

At = t f - t , 

where tf is the final time and t^ is the initial time. The initial time is often taken to be zero, as if measured with a stopwatch; the elapsed time 
is then just t . 

• Average velocity v is defined as displacement divided by the travel time. In symbols, average velocity is 

- _ Ax _ x f ~ X Q 

At t f - t Q ' 

The SI unit for velocity is m/s. 

Velocity is a vector and thus has a direction. 

Instantaneous velocity v is the velocity at a specific instant or the average velocity for an infinitesimal interval. 

Instantaneous speed is the magnitude of the instantaneous velocity. 

Instantaneous speed is a scalar quantity, as it has no direction specified. 

Average speed is the total distance traveled divided by the elapsed time. (Average speed is not the magnitude of the average velocity.) Speed is 

a scalar quantity; it has no direction associated with it. 

2.4 Acceleration 

• Acceleration is the rate at which velocity changes. In symbols, average acceleration a is 

- = Av = v f~ v 

A^ t f - t ' 

• The SI unit for acceleration is m/s . 

• Acceleration is a vector, and thus has a both a magnitude and direction. 

• Acceleration can be caused by either a change in the magnitude or the direction of the velocity. 

• Instantaneous acceleration a is the acceleration at a specific instant in time. 

• Deceleration is an acceleration with a direction opposite to that of the velocity. 

2.5 Motion Equations for Constant Acceleration in One Dimension 

• To simplify calculations we take acceleration to be constant, so that a = a at all times. 

• We also take initial time to be zero. 



CHAPTER 2 | KINEMATICS 77 



• Initial position and velocity are given a subscript 0; final values have no subscript. Thus, 

A? = t 

Ax = x — Xq J 

Av = v — Vq 

• The following kinematic equations for motion with constant a are useful: 

X = Xq + vt 

- v + v 

V = Vq + at 

1 2 
X = Xq + Vq£ + ^rdt 

9 9 

V = Vq + 2a(x — Xq) 

• In vertical motion, y is substituted for x . 

2.6 Problem-Solving Basics for One-Dimensional Kinematics 

• The six basic problem solving steps for physics are: 

Step 1. Examine the situation to determine which physical principles are involved. 

Step 2. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). 

Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). 

Step 4. Find an equation or set of equations that can help you solve the problem. 

Step 5. Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units. 

Step 6. Check the answer to see if it is reasonable: Does it make sense? 

2.7 Falling Objects 

• An object in free-fall experiences constant acceleration if air resistance is negligible. 

• On Earth, all free-falling objects have an acceleration due to gravity g , which averages 

g = 9.80 m/s 2 . 

• Whether the acceleration a should be taken as +g or — g is determined by your choice of coordinate system. If you choose the upward 

9 9 

direction as positive, a = —g = -9.80 m/s is negative. In the opposite case, a = +g = 9.80 m/s is positive. Since acceleration is 
constant, the kinematic equations above can be applied with the appropriate +g or — g substituted for a . 

• For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration. 

2.8 Graphical Analysis of One-Dimensional Motion 

• Graphs of motion can be used to analyze motion. 

• Graphical solutions yield identical solutions to mathematical methods for deriving motion equations. 

• The slope of a graph of displacement x vs. time t is velocity v . 

• The slope of a graph of velocity v vs. time t graph is acceleration a . 

• Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs. 



Conceptual Questions 



2.1 Displacement 

1. Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Specifically identify 
each quantity in your example. 

2. Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which magnitude of displacement 
and displacement are exactly the same? 

3. Bacteria move back and forth by using their flagella (structures that look like little tails). Speeds of up to 50 |im/s [50x10 m/s) have been 
observed. The total distance traveled by a bacterium is large for its size, while its displacement is small. Why is this? 

2.2 Vectors, Scalars, and Coordinate Systems 

4. A student writes, "A bird that is diving for prey has a speed of — 10 ml s ." What is wrong with the student's statement? What has the student 
actually described? Explain. 

5. What is the speed of the bird in Exercise 2.4? 

6. Acceleration is the change in velocity over time. Given this information, is acceleration a vector or a scalar quantity? Explain. 

7. A weather forecast states that the temperature is predicted to be — 5°C the following day. Is this temperature a vector or a scalar quantity? 
Explain. 

2.3 Time, Velocity, and Speed 

8. Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time. 



78 CHAPTER 2 | KINEMATICS 

9. There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these 
two quantities. 

10. Does a car's odometer measure position or displacement? Does its speedometer measure speed or velocity? 

11. If you divide the total distance traveled on a car trip (as determined by the odometer) by the time for the trip, are you calculating the average 
speed or the magnitude of the average velocity? Under what circumstances are these two quantities the same? 

12. How are instantaneous velocity and instantaneous speed related to one another? How do they differ? 

2.4 Acceleration 

13. Is it possible for speed to be constant while acceleration is not zero? Give an example of such a situation. 

14. Is it possible for velocity to be constant while acceleration is not zero? Explain. 

15. Give an example in which velocity is zero yet acceleration is not. 

16. If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its acceleration? Is the 
acceleration positive or negative? 

17. Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that reduces the magnitude of 
a negative velocity? Of a positive velocity? 

2.6 Problem-Solving Basics for One-Dimensional Kinematics 

18. What information do you need in order to choose which equation or equations to use to solve a problem? Explain. 

19. What is the last thing you should do when solving a problem? Explain. 

2.7 Falling Objects 

20. What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down? 

21. An object that is thrown straight up falls back to Earth. This is one-dimensional motion, (a) When is its velocity zero? (b) Does its velocity change 
direction? (c) Does the acceleration due to gravity have the same sign on the way up as on the way down? 

22. Suppose you throw a rock nearly straight up at a coconut in a palm tree, and the rock misses on the way up but hits the coconut on the way 
down. Neglecting air resistance, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it 
had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain. 

23. If an object is thrown straight up and air resistance is negligible, then its speed when it returns to the starting point is the same as when it was 
released. If air resistance were not negligible, how would its speed upon return compare with its initial speed? How would the maximum height to 
which it rises be affected? 

24. The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration due to gravity being the same, how many 
times higher could a safe fall on the Moon be than on Earth (gravitational acceleration on the Moon is about 1/6 that of the Earth)? 

25. How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (gravitational 
acceleration on the Moon is about 1/6 of g on Earth)? 



2.8 Graphical Analysis of One-Dimensional Motion 

26. (a) Explain how you can use the graph of position versus time in Figure 2.54 to describe the change in velocity over time. Identify (b) the time ( t a 
, t h , t c , t d , or t Q ) at which the instantaneous velocity is greatest, (c) the time at which it is zero, and (d) the time at which it is negative. 



J 


L 














d 


e 


H 

C 

o 

(2 


a/ 












Time/ 







Figure 2.54 

27. (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 2.55. (b) Identify the time or 



times ( t a 



, etc.) at which the instantaneous velocity is greatest, (c) At which times is it zero? (d) At which times is it negative? 



CHAPTER 2 | KINEMATICS 79 




Figure 2.55 

28. (a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure 2.56. (b) Based on the 
graph, how does acceleration change over time? 



i 


i 






"l- 






/ 






~ "/Q 


£ 















l 
i 
i 


v \ -- 


1> 

i 




i 
l 
l 
i * 




i 

h 


Time t 


1 ^ 

h 



Figure 2.56 

29. (a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 2.57. (b) Identify the time or 



times ( t a , tu, t c , etc.) at which the acceleration is greatest, (c) At which times is it zero? (d) At which times is it negative? 



J 


v 




g, 


h 




& 












fr 




a 


iy* 




1 


'u 






e^tr 



















? 










\ j 
\k 


a 










\l 


Time t 



Figure 2.57 

30. Consider the velocity vs. time graph of a person in an elevator shown in Figure 2.58. Suppose the elevator is initially at rest. It then accelerates 
for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant 
so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, 
however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of (a) position vs. time and (b) acceleration vs. 
time for this trip. 



80 CHAPTER 2 | KINEMATICS 



Velocity vs. Time 




— I h- 

10 15 

Time t(&) 




25 



Figure 2.58 

31. A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the 
cylinder vs. time as it goes up and then down the plane. 



CHAPTER 2 | KINEMATICS 81 



Problems & Exercises 



2.1 Displacement 




Figure 2.59 

1. Find the following for path A in Figure 2.59: (a) The distance traveled, 
(b) The magnitude of the displacement from start to finish, (c) The 
displacement from start to finish. 

2. Find the following for path B in Figure 2.59: (a) The distance traveled, 
(b) The magnitude of the displacement from start to finish, (c) The 
displacement from start to finish. 

3. Find the following for path C in Figure 2.59: (a) The distance traveled, 
(b) The magnitude of the displacement from start to finish, (c) The 
displacement from start to finish. 

4. Find the following for path D in Figure 2.59: (a) The distance traveled, 
(b) The magnitude of the displacement from start to finish, (c) The 
displacement from start to finish. 

2.3 Time, Velocity, and Speed 

5. (a) Calculate Earth's average speed relative to the Sun. (b) What is its 
average velocity over a period of one year? 

6. A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 
5.00 m from the center of rotation, (a) Calculate the average speed of the 
blade tip in the helicopter's frame of reference, (b) What is its average 
velocity over one revolution? 

7. The North American and European continents are moving apart at a 
rate of about 3 cm/y. At this rate how long will it take them to drift 500 km 
farther apart than they are at present? 

8. Land west of the San Andreas fault in southern California is moving at 
an average velocity of about 6 cm/y northwest relative to land east of the 
fault. Los Angeles is west of the fault and may thus someday be at the 
same latitude as San Francisco, which is east of the fault. How far in the 
future will this occur if the displacement to be made is 590 km northwest, 
assuming the motion remains constant? 

9. On May 26, 1934, a streamlined, stainless steel diesel train called the 
Zephyr set the world's nonstop long-distance speed record for trains. Its 
run from Denver to Chicago took 13 hours, 4 minutes, 58 seconds, and 
was witnessed by more than a million people along the route. The total 
distance traveled was 1633.8 km. What was its average speed in km/h 
and m/s? 

10. Tidal friction is slowing the rotation of the Earth. As a result, the orbit 
of the Moon is increasing in radius at a rate of approximately 4 cm/year. 
Assuming this to be a constant rate, how many years will pass before the 

radius of the Moon's orbit increases by 3.84x10 m (1%)? 

11. A student drove to the university from her home and noted that the 
odometer reading of her car increased by 12.0 km. The trip took 18.0 
min. (a) What was her average speed? (b) If the straight-line distance 
from her home to the university is 10.3 km in a direction 25.0° south of 
east, what was her average velocity? (c) If she returned home by the 
same path 7 h 30 min after she left, what were her average speed and 
velocity for the entire trip? 



12. The speed of propagation of the action potential (an electrical signal) 
in a nerve cell depends (inversely) on the diameter of the axon (nerve 
fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m 
long, and the nerve impulse speed is 18 m/s, how long does it take for 
the nerve signal to travel this distance? 

13. Conversations with astronauts on the lunar surface were 
characterized by a kind of echo in which the earthbound person's voice 
was so loud in the astronaut's space helmet that it was picked up by the 
astronaut's microphone and transmitted back to Earth. It is reasonable to 
assume that the echo time equals the time necessary for the radio wave 
to travel from the Earth to the Moon and back (that is, neglecting any time 
delays in the electronic equipment). Calculate the distance from Earth to 
the Moon given that the echo time was 2.56 s and that radio waves travel 

at the speed of light (3.00X 10 8 m/s) . 

14. A football quarterback runs 15.0 m straight down the playing field in 
2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He 
breaks the tackle and runs straight forward another 21.0 m in 5.20 s. 
Calculate his average velocity (a) for each of the three intervals and (b) 
for the entire motion. 

15. The planetary model of the atom pictures electrons orbiting the 
atomic nucleus much as planets orbit the Sun. In this model you can view 
hydrogen, the simplest atom, as having a single electron in a circular 

orbit 1.06x10" m in diameter, (a) If the average speed of the 

electron in this orbit is known to be 2.20x 10 m/s , calculate the 
number of revolutions per second it makes about the nucleus, (b) What is 
the electron's average velocity? 

2.4 Acceleration 

16. A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. 
What is its acceleration? 

17. Professional Application 

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of 
extreme deceleration on the human body. On December 10, 1954, Stapp 
rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 
km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! 
Calculate his (a) acceleration and (b) deceleration. Express each in 

multiples of g (9.80 m/s ) by taking its ratio to the acceleration of 

gravity. 

18. A commuter backs her car out of her garage with an acceleration of 
1.40 m/s . (a) How long does it take her to reach a speed of 2.00 m/s? 

(b) If she then brakes to a stop in 0.800 s, what is her deceleration? 

19. Assume that an intercontinental ballistic missile goes from rest to a 
suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are 

classified). What is its average acceleration in m/s and in multiples of 
g (9.80 m/s 2 )? 

2.5 Motion Equations for Constant Acceleration in One 
Dimension 

20. An Olympic-class sprinter starts a race with an acceleration of 

4.50 m/s . (a) What is her speed 2.40 s later? (b) Sketch a graph of her 
position vs. time for this period. 

21. A well-thrown ball is caught in a well-padded mitt. If the deceleration 

of the ball is 2.10X10 4 m/s 2 , and 1.85 ms (1 ms = 10" 3 s) elapses 

from the time the ball first touches the mitt until it stops, what was the 
initial velocity of the ball? 

22. A bullet in a gun is accelerated from the firing chamber to the end of 
the barrel at an average rate of 6.20xl0 5 m/s 2 for 8.10xl0~ 4 s . 
What is its muzzle velocity (that is, its final velocity)? 

23. (a) A light-rail commuter train accelerates at a rate of 1.35 m/s . 
How long does it take to reach its top speed of 80.0 km/h, starting from 



82 CHAPTER 2 | KINEMATICS 



rest? (b) The same train ordinarily decelerates at a rate of 1.65 m/s . 
How long does it take to come to a stop from its top speed? (c) In 
emergencies the train can decelerate more rapidly, coming to rest from 

80.0 km/h in 8.30 s. What is its emergency deceleration in m/s ? 

24. While entering a freeway, a car accelerates from rest at a rate of 
2.40 m/s 2 for 12.0 s. (a) Draw a sketch of the situation, (b) List the 
knowns in this problem, (c) How far does the car travel in those 12.0 s? 
To solve this part, first identify the unknown, and then discuss how you 
chose the appropriate equation to solve for it. After choosing the 
equation, show your steps in solving for the unknown, check your units, 
and discuss whether the answer is reasonable, (d) What is the car's final 
velocity? Solve for this unknown in the same manner as in part (c), 
showing all steps explicitly. 

25. At the end of a race, a runner decelerates from a velocity of 9.00 m/s 
at a rate of 2.00 m/s . (a) How far does she travel in the next 5.00 s? 

(b) What is her final velocity? (c) Evaluate the result. Does it make 
sense? 

26. Professional Application: 

Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by 
the left ventricle of the heart, (a) Make a sketch of the situation, (b) List 
the knowns in this problem, (c) How long does the acceleration take? To 
solve this part, first identify the unknown, and then discuss how you 
chose the appropriate equation to solve for it. After choosing the 
equation, show your steps in solving for the unknown, checking your 
units, (d) Is the answer reasonable when compared with the time for a 
heartbeat? 

27. In a slap shot, a hockey player accelerates the puck from a velocity of 
8.00 m/s to 40.0 m/s in the same direction. If this shot takes 

3.33X 10 s , calculate the distance over which the puck accelerates. 

28. A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/ 
h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it 
travel in that time? 

29. Freight trains can produce only relatively small accelerations and 
decelerations, (a) What is the final velocity of a freight train that 

accelerates at a rate of 0.0500 m/s for 8.00 min, starting with an initial 

velocity of 4.00 m/s? (b) If the train can slow down at a rate of 

0.550 m/s , how long will it take to come to a stop from this velocity? 

(c) How far will it travel in each case? 

30. A fireworks shell is accelerated from rest to a velocity of 65.0 m/s 
over a distance of 0.250 m. (a) How long did the acceleration last? (b) 
Calculate the acceleration. 

31. A swan on a lake gets airborne by flapping its wings and running on 
top of the water, (a) If the swan must reach a velocity of 6.00 m/s to take 

off and it accelerates from rest at an average rate of 0.350 m/s , how 
far will it travel before becoming airborne? (b) How long does this take? 

32. Professional Application: 

A woodpecker's brain is specially protected from large decelerations by 
tendon-like attachments inside the skull. While pecking on a tree, the 
woodpecker's head comes to a stop from an initial velocity of 0.600 m/s 

in a distance of only 2.00 mm. (a) Find the acceleration in m/s and in 

multiples of g (g = 9.80 m/s J . (b) Calculate the stopping time, (c) The 

tendons cradling the brain stretch, making its stopping distance 4.50 mm 
(greater than the head and, hence, less deceleration of the brain). What 
is the brain's deceleration, expressed in multiples of g ? 

33. An unwary football player collides with a padded goalpost while 
running at a velocity of 7.50 m/s and comes to a full stop after 
compressing the padding and his body 0.350 m. (a) What is his 
deceleration? (b) How long does the collision last? 

34. In World War II, there were several reported cases of airmen who 
jumped from their flaming airplanes with no parachute to escape certain 
death. Some fell about 20,000 feet (6000 m), and some of them survived, 



with few life-threatening injuries. For these lucky pilots, the tree branches 
and snow drifts on the ground allowed their deceleration to be relatively 
small. If we assume that a pilot's speed upon impact was 123 mph (54 
m/s), then what was his deceleration? Assume that the trees and snow 
stopped him over a distance of 3.0 m. 

35. Consider a grey squirrel falling out of a tree to the ground, (a) If we 
ignore air resistance in this case (only for the sake of this problem), 
determine a squirrel's velocity just before hitting the ground, assuming it 
fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm 
through bending its limbs, compare its deceleration with that of the 
airman in the previous problem. 

36. An express train passes through a station. It enters with an initial 
velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s as it goes 
through. The station is 210 m long, (a) How long is the nose of the train in 
the station? (b) How fast is it going when the nose leaves the station? (c) 
If the train is 130 m long, when does the end of the train leave the 
station? (d) What is the velocity of the end of the train as it leaves? 

37. Dragsters can actually reach a top speed of 145 m/s in only 4.45 
s — considerably less time than given in Example 2.10 and Example 
2.11. (a) Calculate the average acceleration for such a dragster, (b) Find 
the final velocity of this dragster starting from rest and accelerating at the 
rate found in (a) for 402 m (a quarter mile) without using any information 
on time, (c) Why is the final velocity greater than that used to find the 
average acceleration? Hint. Consider whether the assumption of constant 
acceleration is valid for a dragster. If not, discuss whether the 
acceleration would be greater at the beginning or end of the run and what 
effect that would have on the final velocity. 

38. A bicycle racer sprints at the end of a race to clinch a victory. The 
racer has an initial velocity of 11.5 m/s and accelerates at the rate of 

0.500 m/s 2 for 7.00 s. (a) What is his final velocity? (b) The racer 
continues at this velocity to the finish line. If he was 300 m from the finish 
line when he started to accelerate, how much time did he save? (c) One 
other racer was 5.00 m ahead when the winner started to accelerate, but 
he was unable to accelerate, and traveled at 11.8 m/s until the finish line. 
How far ahead of him (in meters and in seconds) did the winner finish? 

39. In 1967, New Zealander Burt Munro set the world record for an Indian 
motorcycle, on the Bonneville Salt Flats in Utah, of 183.58 mi/h. The one- 
way course was 5.00 mi long. Acceleration rates are often described by 
the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and 
Burt accelerated at this rate until he reached his maximum speed, how 
long did it take Burt to complete the course? 

40. (a) A world record was set for the men's 100-m dash in the 2008 
Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt "coasted" 
across the finish line with a time of 9.69 s. If we assume that Bolt 
accelerated for 3.00 s to reach his maximum speed, and maintained that 
speed for the rest of the race, calculate his maximum speed and his 
acceleration, (b) During the same Olympics, Bolt also set the world 
record in the 200-m dash with a time of 19.30 s. Using the same 
assumptions as for the 100-m dash, what was his maximum speed for 
this race? 

2.7 Falling Objects 

41. Assume air resistance is negligible unless otherwise stated. 

42. Calculate the displacement and velocity at times of (a) 0.500, (b) 
1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial 
velocity of 15.0 m/s. Take the point of release to be ;y = . 

43. Calculate the displacement and velocity at times of (a) 0.500, (b) 
1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down 
with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in 
New York City. The roadway of this bridge is 70.0 m above the water. 

44. A basketball referee tosses the ball straight up for the starting tip-off. 
At what velocity must a basketball player leave the ground to rise 1.25 m 
above the floor in an attempt to get the ball? 

45. A rescue helicopter is hovering over a person whose boat has sunk. 
One of the rescuers throws a life preserver straight down to the victim 
with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to 



CHAPTER 2 | KINEMATICS 83 



reach the water, (a) List the knowns in this problem, (b) How high above 
the water was the preserver released? Note that the down draft of the 
helicopter reduces the effects of air resistance on the falling life 
preserver, so that an acceleration equal to that of gravity is reasonable. 

46. A dolphin in an aquatic show jumps straight up out of the water at a 
velocity of 13.0 m/s. (a) List the knowns in this problem, (b) How high 
does his body rise above the water? To solve this part, first note that the 
final velocity is now a known and identify its value. Then identify the 
unknown, and discuss how you chose the appropriate equation to solve 
for it. After choosing the equation, show your steps in solving for the 
unknown, checking units, and discuss whether the answer is reasonable, 
(c) How long is the dolphin in the air? Neglect any effects due to his size 
or orientation. 

47. A swimmer bounces straight up from a diving board and falls feet first 
into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 
1.80 m above the pool, (a) How long are her feet in the air? (b) What is 
her highest point above the board? (c) What is her velocity when her feet 
hit the water? 

48. (a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the 
ground when it is thrown straight up from the cliff with an initial velocity of 
8.00 m/s. (b) How long would it take to reach the ground if it is thrown 
straight down with the same speed? 

49. A very strong, but inept, shot putter puts the shot straight up vertically 
with an initial velocity of 11.0 m/s. How long does he have to get out of 
the way if the shot was released at a height of 2.20 m, and he is 1.80 m 
tall? 

50. You throw a ball straight up with an initial velocity of 15.0 m/s. It 
passes a tree branch on the way up at a height of 7.00 m. How much 
additional time will pass before the ball passes the tree branch on the 
way back down? 

51. A kangaroo can jump over an object 2.50 m high, (a) Calculate its 
vertical speed when it leaves the ground, (b) How long is it in the air? 

52. Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, 
Australia, a hiker hears a rock break loose from a height of 105 m. He 
can't see the rock right away but then does, 1.50 s later, (a) How far 
above the hiker is the rock when he can see it? (b) How much time does 
he have to move before the rock hits his head? 

53. An object is dropped from a height of 75.0 m above ground level, (a) 
Determine the distance traveled during the first second, (b) Determine 
the final velocity at which the object hits the ground, (c) Determine the 
distance traveled during the last second of motion before hitting the 
ground. 

54. There is a 250-m-high cliff at Half Dome in Yosemite National Park in 
California. Suppose a boulder breaks loose from the top of this cliff, (a) 
How fast will it be going when it strikes the ground? (b) Assuming a 
reaction time of 0.300 s, how long will a tourist at the bottom have to get 
out of the way after hearing the sound of the rock breaking loose 
(neglecting the height of the tourist, which would become negligible 
anyway if hit)? The speed of sound is 335 m/s on this day. 

55. A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m 
off the ground on its path up and takes 1.30 s to go past the window. 
What was the ball's initial velocity? 

56. Suppose you drop a rock into a dark well and, using precision 
equipment, you measure the time for the sound of a splash to return, (a) 
Neglecting the time required for sound to travel up the well, calculate the 
distance to the water if the sound returns in 2.0000 s. (b) Now calculate 
the distance taking into account the time for sound to travel up the well. 
The speed of sound is 332.00 m/s in this well. 

57. A steel ball is dropped onto a hard floor from a height of 1.50 m and 
rebounds to a height of 1.45 m. (a) Calculate its velocity just before it 
strikes the floor, (b) Calculate its velocity just after it leaves the floor on its 
way back up. (c) Calculate its acceleration during contact with the floor if 

that contact lasts 0.0800 ms (8.00xl0~ 5 s) . (d) How much did the ball 

compress during its collision with the floor, assuming the floor is 
absolutely rigid? 

58. A coin is dropped from a hot-air balloon that is 300 m above the 
ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum 



height reached, (b) its position and velocity 4.00 s after being released, 
and (c) the time before it hits the ground. 

59. A soft tennis ball is dropped onto a hard floor from a height of 1.50 m 
and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it 
strikes the floor, (b) Calculate its velocity just after it leaves the floor on its 
way back up. (c) Calculate its acceleration during contact with the floor if 

that contact lasts 3.50 ms (3.50xl0~ 3 s) . (d) How much did the ball 

compress during its collision with the floor, assuming the floor is 
absolutely rigid? 

2.8 Graphical Analysis of One-Dimensional Motion 

Note: There is always uncertainty in numbers taken from graphs. If your 
answers differ from expected values, examine them to see if they are 
within data extraction uncertainties estimated by you. 

60. (a) By taking the slope of the curve in Figure 2.60, verify that the 
velocity of the jet car is 115 m/s at t = 20 s . (b) By taking the slope of 
the curve at any point in Figure 2.61, verify that the jet car's acceleration 
is 5.0 m/s 2 . 



Position vs. Time 



3500 ■- 




15 20 
Time (a) 



35 



Figure 2.60 




Figure 2.61 

61. Take the slope of the curve in Figure 2.62 to verify that the velocity at 
t = 10 s is 207 m/s. 




Figure 2.62 

62. Take the slope of the curve in Figure 2.62 to verify that the velocity at 
t = 30.0 s is 238 m/s. 

63. By taking the slope of the curve in Figure 2.63, verify that the 
acceleration is 3.2 m/s at t = 10 s . 



84 CHAPTER 2 | KINEMATICS 




66. A graph of v(t) is shown for a world-class track sprinter in a 100-m 

race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? 
(b) What is his instantaneous velocity at t = 5 s ? (c) What is his 
average acceleration between and 4 s? (d) What is his time for the 
race? 



Figure 2.63 

64. Construct the displacement graph for the subway shuttle train as 
shown in Figure 2.48(a). You will need to use the information on 
acceleration and velocity given in the examples for this figure. 

65. (a) Take the slope of the curve in Figure 2.64 to find the jogger's 
velocity at t = 2.5 s . (b) Repeat at 7.5 s. These values must be 
consistent with the graph in Figure 2.65. 




Figure 2.64 



J 
^ 



Figure 2.65 



Position vs. Time 




10 15 

Time (s) 



Figure 2.67 

67. Figure 2.68 shows the displacement graph for a particle for 5 s. Draw 
the corresponding velocity and acceleration graphs. 

Position vs. Time 




Velocity vs. Time 




Time (s) 



Figure 2.68 




Figure 2.66 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 85 




Figure 3.1 Everyday motion that we experience is, thankfully, rarely as tortuous as a rollercoaster ride like this — the Dragon Khan in Spain's Universal Port Aventura 
Amusement Park. However, most motion is in curved, rather than straight-line, paths. Motion along a curved path is two- or three-dimensional motion, and can be described in 
a similar fashion to one-dimensional motion, (credit: Boris23/Wikimedia Commons) 



Learning Objectives 



3.1. Kinematics in Two Dimensions: An Introduction 

• Observe that motion in two dimensions consists of horizontal and vertical components. 

• Understand the independence of horizontal and vertical vectors in two-dimensional motion. 

3.2. Vector Addition and Subtraction: Graphical Methods 

• Understand the rules of vector addition, subtraction, and multiplication. 

• Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects. 

3.3. Vector Addition and Subtraction: Analytical Methods 

• Understand the rules of vector addition and subtraction using analytical methods. 

• Apply analytical methods to determine vertical and horizontal component vectors. 

• Apply analytical methods to determine the magnitude and direction of a resultant vector. 

3.4. Projectile Motion 

• Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory. 

• Determine the location and velocity of a projectile at different points in its trajectory. 

• Apply the principle of independence of motion to solve projectile motion problems. 

3.5. Addition of Velocities 

• Apply principles of vector addition to determine relative velocity. 

• Explain the significance of the observer in the measurement of velocity. 



86 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



Introduction to Two-Dimensional Kinematics 



The arc of a basketball, the orbit of a satellite, a bicycle rounding a curve, a swimmer diving into a pool, blood gushing out of a wound, and a puppy 
chasing its tail are but a few examples of motions along curved paths. In fact, most motions in nature follow curved paths rather than straight lines. 
Motion along a curved path on a flat surface or a plane (such as that of a ball on a pool table or a skater on an ice rink) is two-dimensional, and thus 
described by two-dimensional kinematics. Motion not confined to a plane, such as a car following a winding mountain road, is described by three- 
dimensional kinematics. Both two- and three-dimensional kinematics are simple extensions of the one-dimensional kinematics developed for straight- 
line motion in the previous chapter. This simple extension will allow us to apply physics to many more situations, and it will also yield unexpected 
insights about nature. 

3.1 Kinematics in Two Dimensions: An Introduction 




Figure 3.2 Walkers and drivers in a city like New York are rarely able to travel in straight lines to reach their destinations. Instead, they must follow roads and sidewalks, 
making two-dimensional, zigzagged paths, (credit: Margaret W. Carruthers) 

Two-Dimensional Motion: Walking in a City 

Suppose you want to walk from one point to another in a city with uniform square blocks, as pictured in Figure 3.3. 

Destination, 

y 

5 



Starting point 



ri I- 



h4 o 

c 

o 

2 :§ 



N 
n 



W 



t:: 



l i i i i i r 

123456789* 
9 blocks east 

Figure 3.3 A pedestrian walks a two-dimensional path between two points in a city. In this scene, all blocks are square and are the same size. 

The straight-line path that a helicopter might fly is blocked to you as a pedestrian, and so you are forced to take a two-dimensional path, such as the 
one shown. You walk 14 blocks in all, 9 east followed by 5 north. What is the straight-line distance? 

An old adage states that the shortest distance between two points is a straight line. The two legs of the trip and the straight-line path form a right 

9 9 9 

triangle, and so the Pythagorean theorem, a + b = c , can be used to find the straight-line distance. 



c = 




Figure 3.4 The Pythagorean theorem relates the length of the legs of a right triangle, labeled a and b , with the hypotenuse, labeled C . The relationship is given by: 



o o o / o o 

a + b = C . This can be rewritten, solving for C : C = \a + b . 



The hypotenuse of the triangle is the straight-line path, and so in this case its length in units of city blocks is 
V(9 blocks) + (5 blocks) = 10.3 blocks , considerably shorter than the 14 blocks you walked. (Note that we are using three significant figures in 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 87 

the answer. Although it appears that "9" and "5" have only one significant digit, they are discrete numbers. In this case "9 blocks" is the same as "9.0 
or 9.00 blocks." We have decided to use three significant figures in the answer in order to show the result more precisely.) 



012345678 9x 



5 
-4 
-3 

2 



N 
n 



W 



Starting point 

12 3 4 5 6 
9 blocks east 

Figure 3.5 The straight-line path followed by a helicopter between the two points is shorter than the 14 blocks walked by the pedestrian. All blocks are square and the same 
size. 

The fact that the straight-line distance (10.3 blocks) in Figure 3.5 is less than the total distance walked (14 blocks) is one example of a general 
characteristic of vectors. (Recall that vectors are quantities that have both magnitude and direction.) 

As for one-dimensional kinematics, we use arrows to represent vectors. The length of the arrow is proportional to the vector's magnitude. The arrow's 
length is indicated by hash marks in Figure 3.3 and Figure 3.5. The arrow points in the same direction as the vector. For two-dimensional motion, the 
path of an object can be represented with three vectors: one vector shows the straight-line path between the initial and final points of the motion, one 
vector shows the horizontal component of the motion, and one vector shows the vertical component of the motion. The horizontal and vertical 
components of the motion add together to give the straight-line path. For example, observe the three vectors in Figure 3.5. The first represents a 
9-block displacement east. The second represents a 5-block displacement north. These vectors are added to give the third vector, with a 10.3-block 
total displacement. The third vector is the straight-line path between the two points. Note that in this example, the vectors that we are adding are 
perpendicular to each other and thus form a right triangle. This means that we can use the Pythagorean theorem to calculate the magnitude of the 
total displacement. (Note that we cannot use the Pythagorean theorem to add vectors that are not perpendicular. We will develop techniques for 
adding vectors having any direction, not just those perpendicular to one another, in Vector Addition and Subtraction: Graphical Methods and 
Vector Addition and Subtraction: Analytical Methods.) 

The Independence of Perpendicular Motions 

The person taking the path shown in Figure 3.5 walks east and then north (two perpendicular directions). How far he or she walks east is only 
affected by his or her motion eastward. Similarly, how far he or she walks north is only affected by his or her motion northward. 

Independence of Motion 

The horizontal and vertical components of two-dimensional motion are independent of each other. Any motion in the horizontal direction does not 
affect motion in the vertical direction, and vice versa. 



This is true in a simple scenario like that of walking in one direction first, followed by another. It is also true of more complicated motion involving 
movement in two directions at once. For example, let's compare the motions of two baseballs. One baseball is dropped from rest. At the same 
instant, another is thrown horizontally from the same height and follows a curved path. A stroboscope has captured the positions of the balls at fixed 
time intervals as they fall. 

T'T- 

T f 

t r 

r r 

r 

Figure 3.6 This shows the motions of two identical balls — one falls from rest, the other has an initial horizontal velocity. Each subsequent position is an equal time interval. 
Arrows represent horizontal and vertical velocities at each position. The ball on the right has an initial horizontal velocity, while the ball on the left has no horizontal velocity. 
Despite the difference in horizontal velocities, the vertical velocities and positions are identical for both balls. This shows that the vertical and horizontal motions are 
independent. 

It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity implies that the vertical motion is 
independent of whether or not the ball is moving horizontally. (Assuming no air resistance, the vertical motion of a falling object is influenced by 
gravity only, and not by any horizontal forces.) Careful examination of the ball thrown horizontally shows that it travels the same horizontal distance 
between flashes. This is due to the fact that there are no additional forces on the ball in the horizontal direction after it is thrown. This result means 



88 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 

that the horizontal velocity is constant, and affected neither by vertical motion nor by gravity (which is vertical). Note that this case is true only for ideal 
conditions. In the real world, air resistance will affect the speed of the balls in both directions. 

The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional motions (horizontal and vertical). 
The key to analyzing such motion, called projectile motion, is to resolve (break) it into motions along perpendicular directions. Resolving two- 
dimensional motion into perpendicular components is possible because the components are independent. We shall see how to resolve vectors in 
Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods. We will find such techniques 
to be useful in many areas of physics. 

PhET Explorations: Ladybug Motion 2D 



Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the 
vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior. 




PhET Interactive Simulation 



Figure 3.7 Ladybug Motion 2D (http://cnx.Org/content/m42104/l.4/ladybug-motion-2d_en.jar) 



3.2 Vector Addition and Subtraction: Graphical Methods 




Figure 3.8 Displacement can be determined graphically using a scale map, such as this one of the Hawaiian Islands. A journey from Hawai'i to Moloka'i has a number of legs, 
or journey segments. These segments can be added graphically with a ruler to determine the total two-dimensional displacement of the journey, (credit: US Geological Survey) 

Vectors in Two Dimensions 

A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one- 
dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we 
specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector's 
magnitude and pointing in the direction of the vector. 

Figure 3.9 shows such a graphical representation of a vector, using as an example the total displacement for the person walking in a city considered 
in Kinematics in Two Dimensions: An Introduction. We shall use the notation that a boldface symbol, such as D , stands for a vector. Its 
magnitude is represented by the symbol in italics, D , and its direction by 6 . 



Vectors in this Text 

In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector F , which has 
both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as F , and the direction of the 
variable will be given by an angle 6 . 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 89 



Destination, 




i i i i i i r^n 

123456789* 



9 blocks east 

Figure 3.9 A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1° north of east. 




Figure 3.10 To describe the resultant vector for the person walking in a city considered in Figure 3.9 graphically, draw an arrow to represent the total displacement vector D . 
Using a protractor, draw a line at an angle 6 relative to the east-west axis. The length D of the arrow is proportional to the vector's magnitude and is measured along the 
line with a ruler. In this example, the magnitude D of the vector is 10.3 units, and the direction 6 is 29. 1° north of east. 



Vector Addition: Head-to-Tail Method 

The head -to-tail method is a graphical way to add vectors, described in Figure 3.11 below and in the steps following. The tail of the vector is the 
starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow. 

Y' 



0=0° 
I I I I 



r 



ft 



9 units 




(a) (b) (c) 

Figure 3.11 Head-to-Tail Method: The head-to-tail method of graphically adding vectors is illustrated for the two displacements of the person walking in a city considered in 
Figure 3.9. (a) Draw a vector representing the displacement to the east, (b) Draw a vector representing the displacement to the north. The tail of this vector should originate 
from the head of the first, east-pointing vector, (c) Draw a line from the tail of the east-pointing vector to the head of the north-pointing vector to form the sum or resultant 
vector D . The length of the arrow D is proportional to the vector's magnitude and is measured to be 10.3 units . Its direction, described as the angle with respect to the east 
(or horizontal axis) 6 is measured with a protractor to be 29.1° . 

Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor. 



90 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



y,i 



0=0° 



' x 

9 units 

(a) 

Figure 3.12 

Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector. 



(b) 



Figure 3.13 



Step 3. If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have only two vectors, so 
we have finished placing arrows tip to tail. 

Step 4. Draw an arrow from the tail of the first vector to the head of the last vector. This is the resultant, or the sum, of the other vectors. 




(c) 



Figure 3.14 



Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to 
determine this length.) 

Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, 
we will use trigonometric relationships to determine this angle.) 

The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring 
tools. It is valid for any number of vectors. 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 91 



Example 3.1 Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk 



Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on 
a flat field. First, she walks 25.0 m in a direction 49.0° north of east. Then, she walks 23.0 m heading 15.0° north of east. Finally, she turns 
and walks 32.0 m in a direction 68.0° south of east. 
Strategy 

Represent each displacement vector graphically with an arrow, labeling the first A , the second B , and the third C , making the lengths 
proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to 
determine the magnitude and direction of the resultant displacement, denoted R . 

Solution 

(1) Draw the three displacement vectors. 




Figure 3.15 

(2) Place the vectors head to tail retaining both their initial magnitude and direction. 

y T 




(a) 



Figure 3.16 

(3) Draw the resultant vector, R . 




Figure 3.17 

(4) Use a ruler to measure the magnitude of R , and a protractor to measure the direction of R . While the direction of the vector can be 
specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the 
resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and 
the vector. 



92 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



ya 




Figure 3.18 

In this case, the total displacement R is seen to have a magnitude of 50.0 m and to lie in a direction 7.0° south of east. By using its magnitude 
and direction, this vector can be expressed as R = 50.0 m and 6 = 7.0° south of east. 

Discussion 

The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent 
of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in Figure 3.19 and we will still get the 
same solution. 




Figure 3.19 

Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in every case and is an 
important characteristic of vectors. Vector addition is commutative. Vectors can be added in any order. 



A + B = B + A. 

(This is true for the addition of ordinary numbers as well — you get the same result whether you add 2 + 3 or 3 + 2 , for example). 



(3.1) 



Vector Subtraction 

Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract B from A , written A - B , we 
must first define what we mean by subtraction. The negative of a vector -B is defined to be B ; that is, graphically the negative of any vector has 
the same magnitude but the opposite direction, as shown in Figure 3.20. In other words, B has the same length as -B , but points in the opposite 
direction. Essentially, we just flip the vector so it points in the opposite direction. 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 93 




Figure 3.20 The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So B is the negative of — B ; it has the same length 
but opposite direction. 

The subtraction of vector B from vector A is then simply defined to be the addition of -B to A . Note that vector subtraction is the addition of a 
negative vector. The order of subtraction does not affect the results. 

A - B = A + (-B). (3.2) 

This is analogous to the subtraction of scalars (where, for example, 5-2 = 5 + (-2) ). Again, the result is independent of the order in which the 
subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates. 



Example 3.2 Subtracting Vectors Graphically: A Woman Sailing a Boat 



A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction 66.0° north of east from 
her current location, and then travel 30.0 m in a direction 1 12° north of east (or 22.0° west of north). If the woman makes a mistake and 
travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock. 







./ * 


V 




y i 


i 


/ A = 27.5 m 

X 66 ° 


V 


B = 30.0 m 
\112° 




X 


(a) 







Figure 3.21 
Strategy 

We can represent the first leg of the trip with a vector A , and the second leg of the trip with a vector B . The dock is located at a location 
A + B . If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance B (30.0 m) in the 
direction 180° - 112° = 68° south of east. We represent this as -B , as shown below. The vector -B has the same magnitude as B but is 
in the opposite direction. Thus, she will end up at a location A + (-B) , or A - B . 




Figure 3.22 

We will perform vector addition to compare the location of the dock, A + B , with the location at which the woman mistakenly arrives, 
A + (-B) . 



94 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 

Solution 

(1) To determine the location at which the woman arrives by accident, draw vectors A and -B . 

(2) Place the vectors head to tail. 

(3) Draw the resultant vector R . 

(4) Use a ruler and protractor to measure the magnitude and direction of R . 




Figure 3.23 

In this case, R = 23.0 m and 6 = 7.5° south of east. 

(5) To determine the location of the dock, we repeat this method to add vectors A and B . We obtain the resultant vector R' : 



y< 


A + B = R' 
R' 
R = 52.9 m 


\ B 

\^112° 

/a 




x 66° 
(c) 



Figure 3.24 

In this case R = 52.9 m and 6 = 90.1° north of east. 

We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip. 

Discussion 

Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works 
the same as for addition. 



Multiplication of Vectors and Scalars 

If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk 3 X 27.5 m , or 82.5 m, 
in a direction 66.0° north of east. This is an example of multiplying a vector by a positive scalar. Notice that the magnitude changes, but the 
direction stays the same. 

If the scalar is negative, then multiplying a vector by it changes the vector's magnitude and gives the new vector the opposite direction. For example, 
if you multiply by -2, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector A is 
multiplied by a scalar c , 

• the magnitude of the vector becomes the absolute value of c A , 

• if c is positive, the direction of the vector does not change, 

• if c is negative, the direction is reversed. 

In our case, c = 3 and A = 27.5 m . Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For 
example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by scalars are the same for division; simply 
treat the divisor as a scalar between and 1. 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 95 



Resolving a Vector into Components 

In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We 
will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular 
components of a single vector, for example the x- and y-components, or the north-south and east-west components. 

For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction 29.0° north of east and want to find 
out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east and 
north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. 
There are many applications in physics where this is a useful thing to do. We will see this soon in Projectile Motion, and much more when we cover 
forces in Dynamics: Newton's Laws of Motion. Most of these involve finding components along perpendicular axes (such as north and east), so 
that right triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal for finding 
vector components. 

PhET Explorations: Maze Game 



Learn about position, velocity, and acceleration in the "Arena of Pain". Use the green arrow to move the ball. Add more walls to the arena to 
make the game more difficult. Try to make a goal as fast as you can. 

<£ P PhET Interactive Simulation 

Figure 3.25 Maze Game (http://cnx.Org/content/m42127/l.6/maze-game_en.jar) 

3.3 Vector Addition and Subtraction: Analytical Methods 

Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical 
methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical 
methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. 
Analytical methods are limited only by the accuracy and precision with which physical quantities are known. 

Resolving a Vector into Perpendicular Components 

Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are 
independent. We very often need to separate a vector into perpendicular components. For example, given a vector like A in Figure 3.26, we may 
wish to find which two perpendicular vectors, A x and A y , add to produce it. 



y , 


Ay 


A x + 


A y = 
A y 




Ae 


r 





A x 

Figure 3.26 The vector A , with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, A x and A -y . These vectors form a right 
triangle. The analytical relationships among these vectors are summarized below. 

A x and A y are defined to be the components of A along the x- and y-axes. The three vectors A , A x , and A y form a right triangle: 



r\. y "T" r\. -y — r\.. 



(3.3) 



Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and 
direction). The relationship does not apply for the magnitudes alone. For example, if A x = 3 m east, A } , = 4m north, and A = 5 m north-east, 

then it is true that the vectors A x + A y = A . However, it is not true that the sum of the magnitudes of the vectors is also equal. That is, 



3m + 4m ^ 5m 



(3.4) 



Thus, 



96 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



A x + A y j,A (3.5) 

If the vector A is known, then its magnitude A (its length) and its angle 6 (its direction) are known. To find A x and A y , itsx- and y-components, 
we use the following relationships for a right triangle. 

A X = A cos (3.6) 

and 



A y = A sin 6. 



(3.7) 



flj/ I iH- y f^ 




A y = As\n$ 



A x = AcosO 

Figure 3.27 The magnitudes of the vector components A x and A-y can be related to the resultant vector A and the angle 6 with trigonometric identities. Here we see 
that A x = A COS 6 and A y = A sin 6 . 



Suppose, for example, that A is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two 
Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods. 

Destination > 



A y = A sin 8 

= (10.3 blocks)(sin 29.1°) 
= 5.0 blocks north 




Starting point 



A x = A cos $ 



(10.3 blocks) (cos 29.1°) 
= 9.0 blocks east 

Figure 3.28 We can use the relationships A x = A COS 6 and Ay = A sin 6 to determine the magnitude of the horizontal and vertical component vectors in this 
example. 

Then A = 10.3 blocks and = 29.1° , so that 



A x = A cos 6 = (10.3 blocks)(cos 29.1°) = 9.0 blocks 
A y = A sin = (10.3 blocks)(sin 29.1°) = 5.0 blocks. 



(3.8) 
(3.9) 



Calculating a Resultant Vector 

If the perpendicular components A x and A^ of a vector A are known, then A can also be found analytically. To find the magnitude A and 

direction 6 of a vector from its perpendicular components A x and A^ , we use the following relationships: 



= t2Ln-\A y /A x ). 



(3.10) 
(3.11) 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 97 




Figure 3.29 The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components A x and A y have been determined. 



Note that the equation A = \JA X + A y is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For 
example, if A x and A y are 9 and 5 blocks, respectively, then A = v9 +5—10.3 blocks, again consistent with the example of the person 
walking in a city. Finally, the direction is 6 = tan (5/9)=29.1° , as before. 

Determining Vectors and Vector Components with Analytical Methods 

Equations A x = A cos 6 and A y = A sin 6 are used to find the perpendicular components of a vector — that is, to go from A and 6 to A A 



and A y . Equations A = VA X + A y and 6 = tan (A y / A x ) are used to find a vector from its perpendicular components — that is, to go from 



A x and A y to A and 6 . Both processes are crucial to analytical methods of vector addition and subtraction. 



Adding Vectors Using Analytical Methods 

To see how to add vectors using perpendicular components, consider Figure 3.30, in which the vectors A and B are added to produce the 
resultant R . 

M 



w 



N 
4 




Figure 3.30 Vectors A and B are two legs of a walk, and R is the resultant or total displacement. You can use analytical methods to determine the magnitude and 
direction of R . 

If A and B represent two legs of a walk (two displacements), then R is the total displacement. The person taking the walk ends up at the tip of 
R. There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-direction. 
Those paths are the x- and y-components of the resultant, R x and R^ . If we know R x and R^ , we can find R and 6 using the equations 



A = \jA x + A y and 6 = tan (A y / A x ) . When you use the analytical method of vector addition, you can determine the components or the 

magnitude and direction of a vector. 

Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen 
perpendicular axes. Use the equations A x = A cos 6 and A y = A sin 6 to find the components. In Figure 3.31, these components are A x , A y , 

B x , and B y . The angles that vectors A and B make with the x-axis are A and # B , respectively. 



98 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



N 
i 



w 




s 

Figure 3.31 To add vectors A and B , first determine the horizontal and vertical components of each vector. These are the dotted vectors A x , Ay, B x and By shown 
in the image. 

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in 
Figure 3.32, 

(3.12) 



and 



Rx - ^X + ^JC 



Ry — Ay + By. 



(3.13) 



\l 
i 



W 




^ = At + &x 

Figure 3.32 The magnitude of the vectors A x and B x add to give the magnitude R x of the resultant vector in the horizontal direction. Similarly, the magnitudes of the 
vectors Ay and By add to give the magnitude Ry of the resultant vector in the vertical direction. 

Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The 
same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the 
second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes 
it easier to add them. Now that the components of R are known, its magnitude and direction can be found. 



Step 3. To get the magnitude R of the resultant, use the Pythagorean theorem: 



R = \lRi + R$. 



Step 4. To get the direction of the resultant: 



6 = tzm-\Ry/R x ). 
The following example illustrates this technique for adding vectors using perpendicular components. 



Example 3.3 Adding Vectors Using Analytical Methods 



(3.14) 



(3.15) 



Add the vector A to the vector B shown in Figure 3.33, using perpendicular components along the x- and y-axes. The x- and y-axes are along 
the east-west and north-south directions, respectively. Vector A represents the first leg of a walk in which a person walks 53.0 m in a 
direction 20.0° north of east. Vector B represents the second leg, a displacement of 34.0 m in a direction 63.0° north of east. 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 99 



N 
k 



W 




8 = 36.6* 



A = 53.0 rm 



A = 2(L0 U 



Figure 3.33 Vector A has magnitude 53.0 m and direction 20.0° north of the x-axis. Vector B has magnitude 34.0 m and direction 63.0° north of the x- 
axis. You can use analytical methods to determine the magnitude and direction of R . 

Strategy 

The components of A and B along the x- and y-axes represent walking due east and due north to get to the same ending point. Once found, 

they are combined to produce the resultant. 

Solution 

Following the method outlined above, we first find the components of A and B along the x- and y-axes. Note that A = 53.0 m , A = 20.0° , 

B = 34.0 m , and # B = 63.0° . We find the x-components by using A x = A cos 6 , which gives 



and 



A x = Acos6> A = (53.0m)(cos20.0°) 

= (53.0 m)(0.940) = 49.8 m 

B x = B cos B = (34.0 m)(cos 63.0°) 

= (34.0 m)(0.454) = 15.4 m. 



Similarly, the y-components are found using A y = A sin 6 A : 



A y = Asin6> A = (53.0m)(sin20.0°) 
= (53.0 m)(0.342) = 18.1m 



and 



By = £ sin 6> B = (34.0 m)(sin 63.0°) 
= (34.0 m)(0.891) = 30.3 m. 

The x- and y-components of the resultant are thus 

R X = A X + B X = 49.8 m + 15.4 m = 65.2 m 

and 

R y = Ay + B y = 18.1 m+30.3 m = 48.4 m. 

Now we can find the magnitude of the resultant by using the Pythagorean theorem: 



R = \IR 2 X + R 2 y = V(65.2) 2 + (48.4) 2 m 



*x ' "y 



so that 



Finally, we find the direction of the resultant: 



Thus, 



R = 81.2 m. 



6 = tan- 1 (7? y /^)=+tan _1 (48.4/65.2). 



-l 



<9 = tan" 1 (0.742) = 36.6 o . 



(3.16) 
(3.17) 

(3.18) 

(3.19) 

(3.20) 
(3.21) 
(3.22) 

(3.23) 
(3.24) 

(3.25) 



100 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



t 

i 




Wi _ 


y* 


ii : 

tfj E 

i il ; II 


i 


A + B = R j/f\ 


\ 






™ E 




-J x 


5 


A x = 49.2 m 0^=1154111 








fl, = ^ x + S # = 65,2 m 



Figure 3.34 Using analytical methods, we see that the magnitude of R is 81.2 m and its direction is 36.6° north of east. 

Discussion 

This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very 
similar — it is just the addition of a negative vector. 

Subtraction of vectors is accomplished by the addition of a negative vector. That is, A — B = A + (-B) . Thus, the method for the subtraction 

of vectors using perpendicular components is identical to that for addition. The components of -B are the negatives of the components of B . 

The x- and y-components of the resultant A — B = R are thus 

R x = A x + (-B x ) (3.26) 

and 

R y = A y + {-B y ) (3-27) 

and the rest of the method outlined above is identical to that for addition. (See Figure 3.35.) 

Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent 
of one another. The next module, Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and 
simplifies the physics. 



N 
1 




W 



Figure 3.35 The subtraction of the two vectors shown in Figure 3.30. The components of — B are the negatives of the components of B . The method of subtraction is the 
same as that for addition. 



PhET Explorations: Vector Addition 

Learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and 
components of each vector can be displayed in several formats. 



M 



<-<T 



Vf 



PhET Interactive Simulation 



Figure 3.36 Vector Addition (http://cnx.Org/content/m42128/l.10/vector-addition_en.jar) 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 101 

3.4 Projectile Motion 

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a 
projectile, and its path is called its trajectory. The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional 
Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two- 
dimensional projectile motion, such as that of a football or other object for which air resistance is negligible. 

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact 
was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to 
analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice 
of axes is the most sensible, because acceleration due to gravity is vertical — thus, there will be no acceleration along the horizontal axis when air 
resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis they-axis. Figure 3.37 illustrates the notation for 
displacement, where s is defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. 

The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation A to represent a vector with components A x 

and Ay . If we continued this format, we would call displacement s with components s x and s y . However, to simplify the notation, we will simply 

represent the component vectors as x and y .) 

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- 
and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of 

acceleration are then very simple: a y = - g = - 9.80 m/s . (Note that this definition assumes that the upwards direction is defined as the 

positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a 
positive value.) Because gravity is vertical, a x = . Both accelerations are constant, so the kinematic equations can be used. 



Review of Kinematic Equations (constant a ] 



x = x n + vt 





v -t- v 
V= 2 




v = Vq + at 


X = 


1 2 

= Xq + vtf + ^at 


v 2 -- 


<-> 
= Vq + 2a(x — x ) 



(3.28) 
(3.29) 

(3.30) 
(3.31) 

(3.32) 




Figure 3.37 The total displacement S of a soccer ball at a point along its path. The vector S has components X and y along the horizontal and vertical axes. Its magnitude 
is S , and it makes an angle 6 with the horizontal. 

Given these assumptions, the following steps are then used to analyze projectile motion: 

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so 
A x = A cos 6 and A y = A sin 6 are used. The magnitude of the components of displacement s along these axes are x and y. The 

magnitudes of the components of the velocity v are v x = v cos 6 and v y = v sin 0, where v is the magnitude of the velocity and 6 is its 

direction, as shown in Figure 3.38. Initial values are denoted with a subscript 0, as usual. 

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal 
and vertical motion take the following forms: 

Horizontal Motion^ = 0) (3.33) 

x = x + v x t (3.34) 

v x = v 0x = v x = velocity is a constant. (3.35) 



Vertical Motion (assuming positive is up a 



y " 



-g = -9.80m/s z ) 



(3.36) 



102 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



y = yo + \(yo y + v y )t 

Vj = v 0y - gt 
y = y + v 0y t - ±gt 2 

v y = v l y " 2 s(y - yo)- 



(3.37) 

(3.38) 
(3.39) 

(3.40) 



Step 3. Solve for the unknowns in the two separate motions — one horizontal and one vertical. Note that the only common variable between the 

motions is time t . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples 

below. 

Step 4. Recombine the two motions to find the total displacement s and velocity v . Because the x - and y -motions are perpendicular, we determine 



these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing A = VA X + A y and 
6 = tan - (A y I A x ) in the following form, where 6 is the direction of the displacement s and 6 V is the direction of the velocity v : 



Total displacement and velocity 



it 



S = \] X 2 + y 2 

= tan- l (y/x) 

v = iv 2 x + v 2 y 
6 V = tan" (vy/v x ). 

V u 



Tty 






v . 



fib 



*t 



(a) 






(3.41) 
(3.42) 
(3.43) 

(3.44) 



v x v x v x 



(b] 



V* V* v x 



4 



IdJ 



Figure 3.38 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes, (b) The 
horizontal motion is simple, because CL X = and V x is thus constant, (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the 

vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical 
velocity, (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory. 



Example 3.4 A Fireworks Projectile Explodes High and Away 



During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in 
Figure 3.39. The fuse is timed to ignite the shell just as it reaches its highest point above the ground, (a) Calculate the height at which the shell 
explodes, (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell 
when it explodes? 

Strategy 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 103 



Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into 
horizontal and vertical motions in which a x = and a y = - g . We can then define Xq and y$ to be zero and solve for the desired 

quantities. 
Solution for (a) 

By "height" we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached 
when Vy = . Since we know the initial and final velocities as well as the initial position, we use the following equation to find y : 



Vy = 1 


vl y - 2g(y - Jo)- 






y 

h = 233 m 


' 

^ 






\V 






t° 








N^ = 75° 






x= 125 m 


X 



(3.45) 



Figure 3.39 The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m 
away horizontally. 



Because y Q and v^ are both zero, the equation simplifies to 



= vl y -2 8 y. (3-46) 



Solving for y gives 



Vly (3 ' 47) 

Now we must find v , the component of the initial velocity in the y-direction. It is given by v = v sin 6 , where v is the initial velocity of 
70.0 m/s, and 6q = 75.0° is the initial angle. Thus, 

v 0y = v sin <9 = (70.0 m/s)(sin 75°) = 67.6 m/s. (3.48) 

and y is 



so that 



(67.6 m/s) 2 
y 2(9.80 m/s 2 )' 



y = 233m. 



(3.49) 



(3.50) 



Discussion for (a) 

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note 
also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical 
component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large 
fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the 
initial velocity would have to be somewhat larger than that given to reach the same height. 

Solution for (b) 

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use 
y = yo + 9"( v 0y + v y)t ■ Because Jo ' s zero > tnis equation reduces to simply 



y = ^( v 0y + v y) L 



(3.51) 



104 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



Note that the final vertical velocity, 


v y , at the highest point is zero. Thus, 






t 2y 2(233 m) 


(3.52) 


(v 0y + v y ) (67.6 m/s) 




= 6.90 s. 




Discussion for (b) 






This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass 


before 


the shell explodes. (Another way of 


1 2 

finding the time is by using y = y^ 4- Vq t — -^-gt , and solving the quadratic equation for t .) 




Solution for (c) 






Because air resistance is negligible 


, a x = and the horizontal velocity is constant, as discussed above. The horizontal displacement 


is 


horizontal velocity multiplied by time as given by x = Xq + v x t , where Xq is equal to zero: 






x = v x t, 


(3.53) 


where v x is the x-component of the velocity, which is given by v x = Vq cos 6q . Now, 






v x = v cos <9 = (70.0 m/s)(cos 75.0°) = 18.1 m/s. 


(3.54) 


The time t for both motions is the same, and so x is 






x = (18.1 m/s)(6.90 s) = 125 m. 


(3.55) 


Discussion for (c) 






The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping 
the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land 
directly below. 



In solving part (a) of the preceding example, the expression we found for y is valid for any projectile motion where air resistance is negligible. Call 
the maximum height y = h; then, 

v 2 (3.56) 

2g- 

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity. 

Defining a Coordinate System 

It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin 
for the x and y positions. Often, it is convenient to choose the initial position of the object as the origin such that Xq = and Jq = . It is 

also important to define the positive and negative directions in the x and y directions. Typically, we define the positive vertical direction as 

upwards, and the positive horizontal direction is usually the direction of the object's motion. When this is the case, the vertical acceleration, g , 

takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. 
For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction 
downwards since the motion of the ball is solely in the downwards direction. If this is the case, g takes a positive value. 



Example 3.5 Calculating Projectile Motion: Hot Rock Projectile 



Kilauea in Hawaii is the world's most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather 
than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0° above the horizontal, as 
shown in Figure 3.40. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point, (a) Calculate the time it takes the 
rock to follow this path, (b) What are the magnitude and direction of the rock's velocity at impact? 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 105 




Figure 3.40 The trajectory of a rock ejected from the Kilauea volcano. 

Strategy 

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The 
time a projectile is in the air is governed by its vertical motion alone. We will solve for t first. While the rock is rising and falling vertically, the 
horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be 
recombined to obtain v and 6 V at the final time t determined in the first part of the example. 

Solution for (a) 

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using 



y = yo + v o/ - \& 2 - 



(3.57) 



If we take the initial position y§ to be zero, then the final position is y = —20.0 m. Now the initial vertical velocity is the vertical component of 
the initial velocity, found from v = v sin 6 = ( 25.0 m/s )( sin 35.0° ) = 14.3 m/s . Substituting known values yields 

-20.0 m = (14.3 m/s)t - (4.90 m/s 2 ) 2 . ( 3 - 58 ) 

Rearranging terms gives a quadratic equation in t : 

(4.90 m/s 2 ) 2 - (14.3 m/s); - (20.0 m) = 0. < 3 - 59 ) 

This expression is a quadratic equation of the form at + bt + c = , where the constants are a = 4.90 , b = - 14.3 , and c = - 20.0. Its 
solutions are given by the quadratic formula: 

(3.60) 



t = 



-b ± ib 1 - 4ac 



2a 



This equation yields two solutions: t = 3.96 and t = - 1.03 . (It is left as an exercise for the reader to verify these solutions.) The time is 
t = 3.96 s or - 1.03 s . The negative value of time implies an event before the start of motion, and so we discard it. Thus, 

t = 3.96 s. (3.61) 

Discussion for (a) 

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and 
lands 20.0 m below its starting altitude will spend 3.96 s in the air. 

Solution for (b) 

From the information now in hand, we can find the final horizontal and vertical velocities v x and v y and combine them to find the total velocity 
v and the angle 6q it makes with the horizontal. Of course, v x is constant so we can solve for it at any horizontal location. In this case, we 
chose the starting point since we know both the initial velocity and initial angle. Therefore: 

v x = v cos <9 = (25.0 m/s)(cos 35°) = 20.5 m/s. (3.62) 

The final vertical velocity is given by the following equation: 

v y = v 0y ~ St, (3.63) 

where v 0y was found in part (a) to be 14.3 m/s . Thus, 



v y = 14.3 m/s - (9.80 m/s 2 )(3.96 s) 



so that 



vy 



-24.5 m/s. 



(3.64) 



(3.65) 



To find the magnitude of the final velocity v we combine its perpendicular components, using the following equation: 



106 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 





v = \jv 2 x + v 2 = V(20.5 m/s) 2 + ( - 24.5 m/s) 2 , (3-bb) 


which gives 


v = 31.9 m/s. (3.67) 


The direction # v is found from the equation: 


v = tzn-\vy/v x ) (3 - 68) 


so that 


6 V = tan _1 ( - 24.5/20.5) = tan _1 ( - 1.19). < 3 - 69 ) 


Thus, 


6> v = -50.1°. (3.70) 


Discussion for (b) 


The negative angle means that the velocity is 50.1° below the horizontal. This result is consistent with the fact that the final vertical velocity is 


negative and hence downward — as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.40.) 



One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the 
first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the 
horizontal distance R traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military 
purposes — such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits 
of satellites around the Earth. Let us consider projectile range further. 




-fl=255m- 



M 

Figure 3.41 Trajectories of projectiles on level ground, (a) The greater the initial speed Vq , the greater the range for a given initial angle, (b) The effect of initial angle 6q on 

the range of a projectile with a given initial speed. Note that the range is the same for 15° and 75° , although the maximum heights of those paths are different. 

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed Vq , the greater the range, as shown in Figure 

3.41(a). The initial angle 0q also has a dramatic effect on the range, as illustrated in Figure 3.41(b). For a fixed initial speed, such as might be 

produced by a cannon, the maximum range is obtained with 6q = 45° . This is true only for conditions neglecting air resistance. If air resistance is 

considered, the maximum angle is approximately 38° . Interestingly, for every initial angle except 45° , there are two angles that give the same 
range — the sum of those angles is 90° . The range also depends on the value of the acceleration of gravity g . The lunar astronaut Alan Shepherd 

was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R of a projectile on level ground for which air 
resistance is negligible is given by 

vl sin 20 o < 3 - 71 ) 

R = 1 , 

where v is the initial speed and 6q is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem 
(hints are given), but it does fit the major features of projectile range as described. 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 107 

When we speak of the range of a projectile on level ground, we assume that R is very small compared with the circumference of the Earth. If, 
however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is 
larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.42.) If 
the initial speed is great enough, the projectile goes into orbit. This is called escape velocity. This possibility was recognized centuries before it could 
be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls 
continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in 
greater depth later in this text. 

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of 
Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights 
beyond the immediate topic. 




Figure 3.42 Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range 
increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved. 

PhET Explorations: Projectile Motion 

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. 
Make a game out of this simulation by trying to hit a target. 

ct t^PhET Interactive Simulation 

Figure 3.43 Projectile Motion (http://cnx.Org/content/m42042/l.8/projectile-motion_en.jar) 

3.5 Addition of Velocities 



Relative Velocity 

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the 
shore, as in Figure 3.44. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat 
downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in 
which it is pointed, as illustrated in Figure 3.45. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the 
ground carries it sideways. 



108 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 




Figure 3.44 A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the 
sum of its velocity relative to the river plus the velocity of the river relative to the shore. 



y (north) i 

X 






,.'i 

X. 

\K 


[ 


- \ 

(eastl I \ 







Figure 3.45 An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it 
points; rather, it moves in the direction of its total velocity (solid arrow). 

In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on 
solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 3.44 and Figure 3.45. 
These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then 
consider certain aspects of what relative velocity means. 

How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Vector Addition and 
Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for 
any other vectors. In one-dimensional motion, the addition of velocities is simple — they add like ordinary numbers. For example, if a field hockey 
player is moving at 5 m/s straight toward the goal and drives the ball in the same direction with a velocity of 30 m/s relative to her body, then the 
velocity of the ball is 35 m/s relative to the stationary, profusely sweating goalkeeper standing in front of the goal. 

In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The 
following equations give the relationships between the magnitude and direction of velocity ( v and 6 ) and its components ( v x and v y ) along the x- 

and y-axes of an appropriately chosen coordinate system: 

v x = v cos 6 

v y = v sin 6 



v = 



^ + Vy 



(3.72) 
(3.73) 
(3.74) 



6 = tan \VylVx). 



(3.75) 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 109 




Figure 3.46 The velocity, V , of an object traveling at an angle 6 to the horizontal axis is the sum of component vectors Y x and V-y . 



These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a 
velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are 
known. 

Take-Home Experiment: Relative Velocity of a Boat 

Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to drain. Try pushing the 
boat from one side of the tub to the other and perpendicular to the flow of water. Which way do you need to push the boat so that it ends up 
immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual velocity of the boat. 



Example 3.6 Adding Velocities: A Boat on a River 



= 1 2 m/s 



v t>oal 

viau = 0.75 itVs 




Figure 3.47 A boat attempts to travel straight across a river at a speed 0.75 m/s. The current in the river, however, flows at a speed of 1.20 m/s to the right. What is the 
total displacement of the boat relative to the shore? 

Refer to Figure 3.47, which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction of the boat's velocity 
relative to an observer on the shore, v tot . The velocity of the boat, v boat , is 0.75 m/s in the y -direction relative to the river and the velocity of 

the river, v river , is 1.20 m/s to the right. 

Strategy 

We start by choosing a coordinate system with its x -axis parallel to the velocity of the river, as shown in Figure 3.47. Because the boat is 
directed straight toward the other shore, its velocity relative to the water is parallel to the y -axis and perpendicular to the velocity of the river. 

Thus, we can add the two velocities by using the equations v tot = Vv^ + v y and 6 = tan" (v y /v x ) directly. 



Solution 

The magnitude of the total velocity is 

where 



v tot = \l v l + v r 



v * = v river = 1.20 m/s 



(3.76) 



(3.77) 



and 



110 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 





v y = v boat = 0.750 m/s. 








(3.78) 


Thus, 










(3.79) 


v tot = V(1.20 m/s) 2 + (0.750 m/s) 2 


yielding 


v tot = 1.42 m/s. 








(3.80) 


The direction of the total velocity 6 is 


given by: 

6 = tan-\v y /v x ) = tan" ! (0.750/ 1.20). 








(3.81) 


This equation gives 


= 32.0°. 








(3.82) 


Discussion 












Both the magnitude v and the direction 6 of the total velocity are consistent with Figure 3.47. Note that because the velocity 
large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced by the small angle (only 
velocity has relative to the riverbank. 


i of the 
32.0° 


; river is 
) the total 



Example 3.7 Calculating Velocity: Wind Velocity Causes an Airplane to Drift 



Calculate the wind velocity for the situation shown in Figure 3.48. The plane is known to be moving at 45.0 m/s due north relative to the air 
mass, while its velocity relative to the ground (its total velocity) is 38.0 m/s in a direction 20.0° west of north. 



TP% 




:v n 



ifa = 38,0 m/s \ v, 

y (north) 



vj, = 45.0 m/s 



20.0° 



x (east) 



70° 




110° 



Figure 3.48 An airplane is known to be heading north at 45.0 m/s, though its velocity relative to the ground is 38.0 m/s at an angle west of north. What is the speed and 
direction of the wind? 

Strategy 

In this problem, somewhat different from the previous example, we know the total velocity v tot and that it is the sum of two other velocities, v w 

(the wind) and v p (the plane relative to the air mass). The quantity v p is known, and we are asked to find v w . None of the velocities are 

perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we can find the components of v w , then 

we can combine them to solve for its magnitude and direction. As shown in Figure 3.48, we choose a coordinate system with its x-axis due east 
and its y-axis due north (parallel to v p ). (You may wish to look back at the discussion of the addition of vectors using perpendicular components 

in Vector Addition and Subtraction: Analytical Methods.) 
Solution 

Because v tot is the vector sum of the v w and v p , its x- and y-components are the sums of the x- and y-components of the wind and plane 
velocities. Note that the plane only has vertical component of velocity so v pjc = and v p>7 = v p . That is, 

v tot;c = v wx (3-83) 

and 



v toty = v wjc + v p . 



(3.84) 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 111 



We can use the first of these two equations to find v WJC : 






v W jc = v totJC = v tot cos 110°. 


(3.85) 


Because v tot = 38.0 m/s and cos 110° = - 0.342 we have 






v wx = (38.0 m/s)(-0.342)=- 


-13.0 m/s. 


(3.86) 


The minus sign indicates motion west which is consistent with the diagram. 






Now, to find v wy we note that 






v toty = v wjc + v p 




(3.87) 


Here v toty = v tot sin 110° ; thus, 






Vwy = (38.0 m/s)(0.940) - 45.0 m/s = -9.29 m/s. 


(3.88) 


This minus sign indicates motion south which is consistent with the diagram. 






Now that the perpendicular components of the wind velocity v wx and v w ^ are 


known, we can find the magnitude and direction of v w . 


First, 


the magnitude is 




(3.89) 


^w = / V V w^ + v wj 


= V( - 13.0 m/s) 2 + ( - 


9.29 m/s) 2 


so that 






v w = 16.0 m/s. 




(3.90) 


The direction is: 






6 = tan -1 (v w>7 /v WJC ) = tan _1 ( - 


•9.29/ -13.0) 


(3.91) 


giving 






= 35.6°. 




(3.92) 


Discussion 






The wind's speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in Figure 3.48. 
Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity 
relative to the air mass as well as heading in a different direction. 



Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to 
one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile 
motion we always use a coordinate system with one axis parallel to gravity. 

Relative Velocities and Classical Relativity 

When adding velocities, we have been careful to specify that the velocity is relative to some reference frame. These velocities are called relative 
velocities. For example, the velocity of an airplane relative to an air mass is different from its velocity relative to the ground. Both are quite different 
from the velocity of an airplane relative to its passengers (which should be close to zero). Relative velocities are one aspect of relativity, which is 
defined to be the study of how different observers moving relative to each other measure the same phenomenon. 

Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879-1955), the greatest physicist of the 20th century. 
Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later chapters. The relative velocities in this 
section are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations 
where speeds are less than about 1% of the speed of light — that is, less than 3,000 km/s . Most things we encounter in daily life move slower than 
this speed. 

Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a 
moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving 
forward? The answer is that if air resistance is negligible, the binoculars will hit at the base of the mast at a point directly below its point of release. 
Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars 
have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See Figure 3.49.) To the observer 
on shore, the binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This 
observer sees the curved path shown in Figure 3.49. Although the paths look different to the different observers, each sees the same result — the 
binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities relative to the 
observer. 



112 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 




Figure 3.49 Classical relativity. The same motion as viewed by two different observers. An observer on the moving ship sees the binoculars dropped from the top of its mast 
fall straight down. An observer on shore sees the binoculars take the curved path, moving forward with the ship. Both observers see the binoculars strike the deck at the base 
of the mast. The initial horizontal velocity is different relative to the two observers. (The ship is shown moving rather fast to emphasize the effect.) 



Example 3.8 Calculating Relative Velocity: An Airline Passenger Drops a Coin 



An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the floor 1.50 m below its 
point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth? 




t 



(b) 



Figure 3.50 The motion of a coin dropped inside an airplane as viewed by two different observers, (a) An observer in the plane sees the coin fall straight down, (b) An 
observer on the ground sees the coin move almost horizontally. 

Strategy 

Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin is zero relative to the 
plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/s horizontal relative to the Earth and 
gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a coordinate system with vertical and horizontal axes. 

Solution for (a) 

Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final velocity can be found 
using the equation: 

(3.93) 



v v " = v (U r-2g(y-y ). 



Vy 



v Qy 



Substituting known values into the equation, we get 



2/„2 



= Z - 2(9.80 m/s z )( - 1.50 m - m) = 29.4 m z /s 



(3.94) 



yielding 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 113 



i^ = -5.42 m/s. (3.95) 



We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the velocity is directed 
downwards, and we have defined the positive direction to be upwards. There is no initial horizontal velocity relative to the plane and no horizontal 
acceleration, and so the motion is straight down relative to the plane. 

Solution for (b) 

Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the final vertical velocity 
for the coin relative to the ground is v y = — 5.42 m/s , the same as found in part (a). In contrast to part (a), there now is a horizontal 

component of the velocity. However, since there is no horizontal acceleration, the initial and final horizontal velocities are the same and 
v x = 260 m/s . The x- and y-components of velocity can be combined to find the magnitude of the final velocity: 



2__„ 2 (3.96) 

v — y v 

Thus, 



V x +V y 



v = V(260 m/s) 2 + ( - 5.42 m/s) 2 (3-97) 

yielding 

v = 260.06 m/s. (3.98) 

The direction is given by: 

6 = taxr\v y /Vjd = tan _1 ( - 5.42/260) ( 3 - 99 ) 

so that 

= tan _1 ( - 0.0208) = -1.19°. ( 310 °) 

Discussion 

In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth and fell 1.50 m. This 
result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as when it is at rest on the ground. This result 
is also true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The plane is moving so fast 
horizontally to begin with that its final velocity is barely greater than the initial velocity. Once again, we see that in two dimensions, vectors do not 
add like ordinary numbers — the final velocity v in part (b) is not (260 - 5.42) m/s ; rather, it is 260.06 m/s . The velocity's magnitude had to be 

calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on 
the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity 
of the plane is much larger, so that the two observers see very different paths. (See Figure 3.50.) In addition, both observers see the coin fall 
1.50 m vertically, but the one on the ground also sees it move forward 144 m (this calculation is left for the reader). Thus, one observer sees a 
vertical path, the other a nearly horizontal path. 

Making Connections: Relativity and Einstein 

Because Einstein was able to clearly define how measurements are made (some involve light) and because the speed of light is the same 
for all observers, the outcomes are spectacularly unexpected. Time varies with observer, energy is stored as increased mass, and more 
surprises await. 



PhET Exploratio ns: Motion in 2D 

Try the new "Ladybug Motion 2D" simulation for the latest updated version. Learn about position, velocity, and acceleration vectors. Move the 
ball with the mouse or let the simulation move the ball in four types of motion (2 types of linear, simple harmonic, circle). 



«* 



& 



4f 



PhET Interactive Simulation 



Figure 3.51 Motion in 2D (http://cnx.Org/content/m42045/l.7/motion-2d_en.jar) 



Glossary 



air resistance: a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance 
is assumed to be zero 

analytical method: the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric 
identities 



114 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 

classical relativity: the study of relative velocities in situations where speeds are less than about 1% of the speed of light — that is, less than 3000 
km/s 

commutative: refers to the interchangeability of order in a function; vector addition is commutative because the order in which vectors are added 
together does not affect the final sum 

component (of a 2-d vector): a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as 
a sum of two vertical and horizontal vector components 

direction (of a vector): the orientation of a vector in space 

head (of a vector): the end point of a vector; the location of the tip of the vector's arrowhead; also referred to as the "tip" 

head-to-tail method: a method of adding vectors in which the tail of each vector is placed at the head of the previous vector 

kinematics: the study of motion without regard to mass or force 

magnitude (of a vector): the length or size of a vector; magnitude is a scalar quantity 

motion: displacement of an object as a function of time 

projectile motion: the motion of an object that is subject only to the acceleration of gravity 

projectile: an object that travels through the air and experiences only acceleration due to gravity 

range: the maximum horizontal distance that a projectile travels 

relative velocity: the velocity of an object as observed from a particular reference frame 

relativity: the study of how different observers moving relative to each other measure the same phenomenon 

resultant vector: the vector sum of two or more vectors 

resultant: the sum of two or more vectors 

scalar: a quantity with magnitude but no direction 

tail: the start point of a vector; opposite to the head or tip of the arrow 

trajectory: the path of a projectile through the air 

vector addition: the rules that apply to adding vectors together 

vector: a quantity that has both magnitude and direction; an arrow used to represent quantities with both magnitude and direction 

velocity: speed in a given direction 



Section Summary 



3.1 Kinematics in Two Dimensions: An Introduction 

• The shortest path between any two points is a straight line. In two dimensions, this path can be represented by a vector with horizontal and 
vertical components. 

• The horizontal and vertical components of a vector are independent of one another. Motion in the horizontal direction does not affect motion in 
the vertical direction, and vice versa. 

3.2 Vector Addition and Subtraction: Graphical Methods 

• The graphical method of adding vectors A and B involves drawing vectors on a graph and adding them using the head-to-tail method. The 
resultant vector R is defined such that A + B = R . The magnitude and direction of R are then determined with a ruler and protractor, 
respectively. 

• The graphical method of subtracting vector B from A involves adding the opposite of vector B , which is defined as — B . In this case, 
A - B = A + (-B) = R . Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector R . 

• Addition of vectors is commutative such that A + B = B + A . 

• The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at 
the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector. 

• If a vector A is multiplied by a scalar quantity c , the magnitude of the product is given by cA . If c is positive, the direction of the product 
points in the same direction as A ; if c is negative, the direction of the product points in the opposite direction as A . 

3.3 Vector Addition and Subtraction: Analytical Methods 

• The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the 
magnitude and direction of a resultant vector. 

• The steps to add vectors A and B using the analytical method are as follows: 

Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the 
equations 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 115 



A x = A cos 6 

and 



B x = BcosO 



A y = A sin 6 
By = BsinO. 
Step 2: Add the horizontal and vertical components of each vector to determine the components R x and R y of the resultant vector, R : 

Rx = A x + B x 
and 

Ry = Ay + By 

Step 3: Use the Pythagorean theorem to determine the magnitude, R , of the resultant vector R : 

Step 4: Use a trigonometric identity to determine the direction, 6 , of R : 

e = tm~\R y /R x ). 

3.4 Projectile Motion 

• Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity. 

• To solve projectile motion problems, perform the following steps: 

1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The 
components of position s are given by the quantities x and y , and the components of the velocity v are given by v x = v cos 6 and 

Vy = v sin 6 , where v is the magnitude of the velocity and 6 is its direction. 

2. Analyze the motion of the projectile in the horizontal direction using the following equations: 

Horizontal motion^ = 0) 

x = x + v x t 
v x = v 0x = v x = velocity is a constant. 

3. Analyze the motion of the projectile in the vertical direction using the following equations: 

Vertical motion (Assuming positive direction is up; a y = —g = -9.80 m/s ) 

y = yo + \(yo y + v y )t 

v y = v 0y - St 

y = yo + v / - ^gt 2 
v y = v l y - 2 s(y - yo)- 

4. Recombine the horizontal and vertical components of location and/or velocity using the following equations: 

s = \lx 2 + y 2 
6 = tzm-\y/x) 



v 2 x + v 2 



6 y = tan l (v y /v x ). 
The maximum height h of a projectile launched with initial vertical velocity v 0>7 is given by 

The maximum horizontal distance traveled by a projectile is called the range. The range R of a projectile on level ground launched at an angle 
#0 above the horizontal with initial speed v is given by 

2 



r ^ v sin 2(9 Q 



3.5 Addition of Velocities 

• Velocities in two dimensions are added using the same analytical vector techniques, which are rewritten as 

v x = v cos 6 

v y = v sin 6 

v 2 x + v 2 y 



116 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



6 = tan (v y /v x ). 



Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame. 
Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one 
another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s). 



Conceptual Questions 



3.2 Vector Addition and Subtraction: Graphical Methods 

1. Which of the following is a vector: a person's height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this 
book, the Earth's population, the acceleration of gravity? 

2. Give a specific example of a vector, stating its magnitude, units, and direction. 

3. What do vectors and scalars have in common? How do they differ? 

4. Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total 
distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each camper? 

Path 2 




Figure 3.52 

5. If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up anywhere on the 
circle shown in Figure 3.53. What other information would he need to get to Sacramento? 

jFoinl Arena \ 




Figure 3.53 

6. Suppose you take two steps A and B (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? 
More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting 
point A + B the sum of the lengths of the two steps? 

7. Explain why it is not possible to add a scalar to a vector. 

8. If you take two steps of different sizes, can you end up at your starting point? More generally, can two vectors with different magnitudes ever add to 
zero? Can three or more? 

3.3 Vector Addition and Subtraction: Analytical Methods 

9. Suppose you add two vectors A and B . What relative direction between them produces the resultant with the greatest magnitude? What is the 
maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude? 

10. Give an example of a nonzero vector that has a component of zero. 

11. Explain why a vector cannot have a component greater than its own magnitude. 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 117 

12. If the vectors A and B are perpendicular, what is the component of A along the direction of B ? What is the component of B along the 
direction of A ? 

3.4 Projectile Motion 

13. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0° nor 90° 
): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time 
other than at t = ? (d) Can the speed ever be the same as the initial speed at a time other than at t = ? 

14. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0° nor 90° 
): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite 
in direction to a component of velocity? 

15. For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that 
give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer 
to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the 
higher trajectory? 

16. During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, 
simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor 
at the same time. 

3.5 Addition of Velocities 

17. What frame or frames of reference do you instinctively use when driving a car? When flying in a commercial jet airplane? 

18. A basketball player dribbling down the court usually keeps his eyes fixed on the players around him. He is moving fast. Why doesn't he need to 
keep his eyes on the ball? 

19. If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall straight down as viewed 
by a person standing at the side of the road? Under what condition would this occur? How would the motion of the ball appear to the person who 
threw it? 

20. The hat of a jogger running at constant velocity falls off the back of his head. Draw a sketch showing the path of the hat in the jogger's frame of 
reference. Draw its path as viewed by a stationary observer. 

21. A clod of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the direction of its velocity 
relative to the truck just before it hits? Is this the same as the direction of its velocity relative to ground just before it hits? Explain your answers. 



118 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



Problems & Exercises 



3.2 Vector Addition and Subtraction: Graphical Methods 

Use graphical methods to solve these problems. You may assume 
data taken from graphs is accurate to three digits. 

1. Find the following for path A in Figure 3.54: (a) the total distance 
traveled, and (b) the magnitude and direction of the displacement from 
start to finish. 



W 



Start- 



A r— t 

XLz 






m m 

— — — .* 



Figure 3.54 The various lines represent paths taken by different people walking in a 
city. All blocks are 120 m on a side. 

2. Find the following for path B in Figure 3.54: (a) the total distance 
traveled, and (b) the magnitude and direction of the displacement from 
start to finish. 

3. Find the north and east components of the displacement for the hikers 
shown in Figure 3.52. 

4. Suppose you walk 18.0 m straight west and then 25.0 m straight north. 
How far are you from your starting point, and what is the compass 
direction of a line connecting your starting point to your final position? (If 
you represent the two legs of the walk as vector displacements A and 
B , as in Figure 3.55, then this problem asks you to find their sum 

R = A + B ) 



B 




A+ B = R 



W 



N 
4 



Figure 3.55 The two displacements A and B add to give a total displacement R 
having magnitude R and direction 6 . 

5. Suppose you first walk 12.0 m in a direction 20° west of north and 
then 20.0 m in a direction 40.0° south of west. How far are you from 
your starting point, and what is the compass direction of a line connecting 
your starting point to your final position? (If you represent the two legs of 
the walk as vector displacements A and B , as in Figure 3.56, then this 
problem finds their sum R = A + B .) 



= 20 m 




Figure 3.56 

6. Repeat the problem above, but reverse the order of the two legs of the 
walk; show that you get the same final result. That is, you first walk leg 

B , which is 20.0 m in a direction exactly 40° south of west, and then 
leg A , which is 12.0 m in a direction exactly 20° west of north. (This 
problem shows that A + B = B + A.) 

7. (a) Repeat the problem two problems prior, but for the second leg you 
walk 20.0 m in a direction 40.0° north of east (which is equivalent to 
subtracting B from A — that is, to finding R' = A — B ). (b) Repeat 
the problem two problems prior, but now you first walk 20.0 m in a 
direction 40.0° south of west and then 12.0 m in a direction 20.0° east 
of south (which is equivalent to subtracting A from B — that is, to 
finding R" = B - A = - R ). Show that this is the case. 

8. Show that the order of addition of three vectors does not affect their 
sum. Show this property by choosing any three vectors A , B , and C , 
all having different lengths and directions. Find the sum A + B + C 
then find their sum when added in a different order and show the result is 
the same. (There are five other orders in which A , B , and C can be 
added; choose only one.) 

9. Show that the sum of the vectors discussed in Example 3.2 gives the 
result shown in Figure 3.24. 

10. Find the magnitudes of velocities v A and v B in Figure 3.57 

V,ot=V A +V B 




Figure 3.57 The two velocities V^ and Vg add to give a total V tot . 

11. Find the components of v tot along the x- and y-axes in Figure 3.57. 

12. Find the components of v tot along a set of perpendicular axes 
rotated 30° counterclockwise relative to those in Figure 3.57. 

3.3 Vector Addition and Subtraction: Analytical Methods 

13. Find the following for path C in Figure 3.58: (a) the total distance 
traveled and (b) the magnitude and direction of the displacement from 
start to finish. In this part of the problem, explicitly show how you follow 
the steps of the analytical method of vector addition. 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 119 



N 



W 



-»—*■ 



s ijT a 
Start =_ •>» I — 






■ ■ 



Figure 3.58 The various lines represent paths taken by different people walking in a 
city. All blocks are 120 m on a side. 

14. Find the following for path D in Figure 3.58: (a) the total distance 
traveled and (b) the magnitude and direction of the displacement from 
start to finish. In this part of the problem, explicitly show how you follow 
the steps of the analytical method of vector addition. 

15. Find the north and east components of the displacement from San 
Francisco to Sacramento shown in Figure 3.59. 




Figure 3.59 

16. Solve the following problem using analytical techniques: Suppose you 
walk 18.0 m straight west and then 25.0 m straight north. How far are you 
from your starting point, and what is the compass direction of a line 
connecting your starting point to your final position? (If you represent the 
two legs of the walk as vector displacements A and B , as in Figure 
3.60, then this problem asks you to find their sum R = A + B .) 



B 




A+ B = R 



W 



N 
i 



Figure 3.60 The two displacements A and B add to give a total displacement R 
having magnitude R and direction 6 . 

Note that you can also solve this graphically. Discuss why the analytical 
technique for solving this problem is potentially more accurate than the 
graphical technique. 

17. Repeat Exercise 3.16 using analytical techniques, but reverse the 
order of the two legs of the walk and show that you get the same final 
result. (This problem shows that adding them in reverse order gives the 



same result — that is, B + A = A + B .) Discuss how taking another 

path to reach the same point might help to overcome an obstacle 
blocking you other path. 

18. You drive 7.50 km in a straight line in a direction 15° east of north, 
(a) Find the distances you would have to drive straight east and then 
straight north to arrive at the same point. (This determination is 
equivalent to find the components of the displacement along the east and 
north directions.) (b) Show that you still arrive at the same point if the 
east and north legs are reversed in order. 

19. Do Exercise 3.16 again using analytical techniques and change the 
second leg of the walk to 25.0 m straight south. (This is equivalent to 
subtracting B from A — that is, finding R' = A - B ) (b) Repeat 
again, but now you first walk 25.0 m north and then 18.0 m east. (This 
is equivalent to subtract A from B — that is, to find A = B + C . Is 
that consistent with your result?) 

20. A new landowner has a triangular piece of flat land she wishes to 
fence. Starting at the west corner, she measures the first side to be 80.0 
m long and the next to be 105 m. These sides are represented as 
displacement vectors A from B in Figure 3.61. She then correctly 
calculates the length and orientation of the third side C . What is her 
result? 

A + B + C = 



105 m 




N 
i 



W 



80 m 



Figure 3.61 

21. You fly 32.0 km in a straight line in still air in the direction 35.0° 
south of west, (a) Find the distances you would have to fly straight south 
and then straight west to arrive at the same point. (This determination is 
equivalent to finding the components of the displacement along the south 
and west directions.) (b) Find the distances you would have to fly first in a 
direction 45.0° south of west and then in a direction 45.0° west of 
north. These are the components of the displacement along a different 
set of axes — one rotated 45° . 

22. A farmer wants to fence off his four-sided plot of flat land. He 
measures the first three sides, shown as A, B, and C in Figure 3.62, 
and then correctly calculates the length and orientation of the fourth side 
D . What is his result? 



A = 470 km 



A + B + C + D 
C = 3.02 km 



N 
ii 




W 



B = 2.48 km 



Figure 3.62 

23. In an attempt to escape his island, Gilligan builds a raft and sets to 
sea. The wind shifts a great deal during the day, and he is blown along 



120 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



the following straight lines: 2.50 km 45.0° north of west; then 
4.70 km 60.0° south of east; then 1.30 km 25.0° south of west; 
then 5.10 km straight east; then 1.70 km 5.00° east of north; then 
7.20 km 55.0° south of west; and finally 2.80 km 10.0° north of 
east. What is his final position relative to the island? 

24. Suppose a pilot flies 40.0 km in a direction 60° north of east and 
then flies 30.0 km in a direction 15° north of east as shown in Figure 
3.63. Find her total distance R from the starting point and the direction 
6 of the straight-line path to the final position. Discuss qualitatively how 
this flight would be altered by a wind from the north and how the effect of 
the wind would depend on both wind speed and the speed of the plane 



relative to the air mass. 



B = 15° 



A + B = R 




Figure 3.63 

3.4 Projectile Motion 

25. A projectile is launched at ground level with an initial speed of 50.0 
m/s at an angle of 30.0° above the horizontal. It strikes a target above 
the ground 3.00 seconds later. What are the x and y distances from 
where the projectile was launched to where it lands? 

26. A ball is kicked with an initial velocity of 16 m/s in the horizontal 
direction and 12 m/s in the vertical direction, (a) At what speed does the 
ball hit the ground? (b) For how long does the ball remain in the air? 
(c)What maximum height is attained by the ball? 

27. A ball is thrown horizontally from the top of a 60.0-m building and 
lands 100.0 m from the base of the building. Ignore air resistance, (a) 
How long is the ball in the air? (b) What must have been the initial 
horizontal component of the velocity? (c) What is the vertical component 
of the velocity just before the ball hits the ground? (d) What is the velocity 
(including both the horizontal and vertical components) of the ball just 
before it hits the ground? 

28. (a) A daredevil is attempting to jump his motorcycle over a line of 
buses parked end to end by driving up a 32° ramp at a speed of 
40.0 m/s (144 km/h) . How many buses can he clear if the top of the 

takeoff ramp is at the same height as the bus tops and the buses are 
20.0 m long? (b) Discuss what your answer implies about the margin of 
error in this act — that is, consider how much greater the range is than the 
horizontal distance he must travel to miss the end of the last bus. 
(Neglect air resistance.) 

29. An archer shoots an arrow at a 75.0 m distant target; the bull's-eye of 
the target is at same height as the release height of the arrow, (a) At what 
angle must the arrow be released to hit the bull's-eye if its initial speed is 
35.0 m/s? In this part of the problem, explicitly show how you follow the 
steps involved in solving projectile motion problems, (b) There is a large 
tree halfway between the archer and the target with an overhanging 
horizontal branch 3.50 m above the release height of the arrow. Will the 
arrow go over or under the branch? 

30. A rugby player passes the ball 7.00 m across the field, where it is 
caught at the same height as it left his hand, (a) At what angle was the 
ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of 
the two possible angles was used? (b) What other angle gives the same 
range, and why would it not be used? (c) How long did this pass take? 



31. Verify the ranges for the projectiles in Figure 3.41(a) for 6 = 45° 
and the given initial velocities. 

32. Verify the ranges shown for the projectiles in Figure 3.41(b) for an 
initial velocity of 50 m/s at the given initial angles. 

33. The cannon on a battleship can fire a shell a maximum distance of 
32.0 km. (a) Calculate the initial velocity of the shell, (b) What maximum 
height does it reach? (At its highest, the shell is above 60% of the 
atmosphere — but air resistance is not really negligible as assumed to 
make this problem easier.) (c) The ocean is not flat, because the Earth is 

curved. Assume that the radius of the Earth is 6.37X 10 km . How 
many meters lower will its surface be 32.0 km from the ship along a 
horizontal line parallel to the surface at the ship? Does your answer imply 
that error introduced by the assumption of a flat Earth in projectile motion 
is significant here? 

34. An arrow is shot from a height of 1.5 m toward a cliff of height H . It 
is shot with a velocity of 30 m/s at an angle of 60° above the horizontal. 
It lands on the top edge of the cliff 4.0 s later, (a) What is the height of the 
cliff? (b) What is the maximum height reached by the arrow along its 
trajectory? (c) What is the arrow's impact speed just before hitting the 
cliff? 

35. In the standing broad jump, one squats and then pushes off with the 
legs to see how far one can jump. Suppose the extension of the legs 
from the crouch position is 0.600 m and the acceleration achieved from 
this position is 1.25 times the acceleration due to gravity, g . How far can 

they jump? State your assumptions. (Increased range can be achieved 
by swinging the arms in the direction of the jump.) 

36. The world long jump record is 8.95 m (Mike Powell, USA, 1991). 
Treated as a projectile, what is the maximum range obtainable by a 
person if he has a take-off speed of 9.5 m/s? State your assumptions. 

37. Serving at a speed of 170 km/h, a tennis player hits the ball at a 
height of 2.5 m and an angle 6 below the horizontal. The service line is 
11.9 m from the net, which is 0.91 m high. What is the angle 6 such that 
the ball just crosses the net? Will the ball land in the service box, whose 
out line is 6.40 m from the net? 

38. A football quarterback is moving straight backward at a speed of 200 
m/s when he throws a pass to a player 18.0 m straight downfield. (a) If 
the ball is thrown at an angle of 25° relative to the ground and is caught 
at the same height as it is released, what is its initial speed relative to the 
ground? (b) How long does it take to get to the receiver? (c) What is its 
maximum height above its point of release? 

39. Gun sights are adjusted to aim high to compensate for the effect of 
gravity, effectively making the gun accurate only for a specific range, (a) If 
a gun is sighted to hit targets that are at the same height as the gun and 
100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 
m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss 
qualitatively how a larger muzzle velocity would affect this problem and 
what would be the effect of air resistance. 

40. An eagle is flying horizontally at a speed of 3.00 m/s when the fish in 
her talons wiggles loose and falls into the lake 5.00 m below. Calculate 
the velocity of the fish relative to the water when it hits the water. 

41. An owl is carrying a mouse to the chicks in its nest. Its position at that 
time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter 
nest. The owl is flying east at 3.50 m/s at an angle 30.0° below the 
horizontal when it accidentally drops the mouse. Is the owl lucky enough 
to have the mouse hit the nest? To answer this question, calculate the 
horizontal position of the mouse when it has fallen 12.0 m. 

42. Suppose a soccer player kicks the ball from a distance 30 m toward 
the goal. Find the initial speed of the ball if it just passes over the goal, 
2.4 m above the ground, given the initial direction to be 40° above the 
horizontal. 

43. Can a goalkeeper at her/ his goal kick a soccer ball into the 
opponent's goal without the ball touching the ground? The distance will 
be about 95 m. A goalkeeper can give the ball a speed of 30 m/s. 



CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 121 



44. The free throw line in basketball is 4.57 m (15 ft) from the basket, 
which is 3.05 m (10 ft) above the floor. A player standing on the free 
throw line throws the ball with an initial speed of 7.15 m/s, releasing it at 
a height of 2.44 m (8 ft) above the floor. At what angle above the 
horizontal must the ball be thrown to exactly hit the basket? Note that 
most players will use a large initial angle rather than a flat shot because it 
allows for a larger margin of error. Explicitly show how you follow the 
steps involved in solving projectile motion problems. 

45. In 2007, Michael Carter (U.S.) set a world record in the shot put with 
a throw of 24.77 m. What was the initial speed of the shot if he released it 
at a height of 2.10 m and threw it at an angle of 38.0° above the 
horizontal? (Although the maximum distance for a projectile on level 
ground is achieved at 45° when air resistance is neglected, the actual 
angle to achieve maximum range is smaller; thus, 38° will give a longer 
range than 45° in the shot put.) 

46. A basketball player is running at 5.00 m/s directly toward the basket 
when he jumps into the air to dunk the ball. He maintains his horizontal 
velocity, (a) What vertical velocity does he need to rise 0.750 m above 
the floor? (b) How far from the basket (measured in the horizontal 
direction) must he start his jump to reach his maximum height at the 
same time as he reaches the basket? 

47. A football player punts the ball at a 45.0° angle. Without an effect 
from the wind, the ball would travel 60.0 m horizontally, (a) What is the 
initial speed of the ball? (b) When the ball is near its maximum height it 
experiences a brief gust of wind that reduces its horizontal velocity by 
1.50 m/s. What distance does the ball travel horizontally? 

48. Prove that the trajectory of a projectile is parabolic, having the form 
y = ax + bx .To obtain this expression, solve the equation x = v$ x t 

for t and substitute it into the expression for y = v t - (1 I2)gt 
(These equations describe the x and y positions of a projectile that 
starts at the origin.) You should obtain an equation of the form 
y = ax + bx where a and b are constants. 

v n sin 20 

49. Derive R = — - — ■= for the range of a projectile on level ground 

by finding the time t at which y becomes zero and substituting this 
value of t into the expression for x — Xq , noting that R = x — Xq 

50. Unreasonable Results (a) Find the maximum range of a super 
cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable 
about the range you found? (c) Is the premise unreasonable or is the 
available equation inapplicable? Explain your answer, (d) If such a 
muzzle velocity could be obtained, discuss the effects of air resistance, 
thinning air with altitude, and the curvature of the Earth on the range of 
the super cannon. 

51. Construct Your Own Problem Consider a ball tossed over a fence. 
Construct a problem in which you calculate the ball's needed initial 
velocity to just clear the fence. Among the things to determine are; the 
height of the fence, the distance to the fence from the point of release of 
the ball, and the height at which the ball is released. You should also 
consider whether it is possible to choose the initial speed for the ball and 
just calculate the angle at which it is thrown. Also examine the possibility 
of multiple solutions given the distances and heights you have chosen. 

3.5 Addition of Velocities 

52. Bryan Allen pedaled a human-powered aircraft across the English 
Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) 
He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° 
south of east. What was his total displacement? (b) Allen encountered a 
headwind averaging 2.00 m/s almost precisely in the opposite direction of 
his motion relative to the Earth. What was his average velocity relative to 
the air? (c) What was his total displacement relative to the air mass? 

53. A seagull flies at a velocity of 9.00 m/s straight into the wind, (a) If it 
takes the bird 20.0 min to travel 6.00 km relative to the Earth, what is the 



velocity of the wind? (b) If the bird turns around and flies with the wind, 
how long will he take to return 6.00 km? (c) Discuss how the wind affects 
the total round-trip time compared to what it would be with no wind. 

54. Near the end of a marathon race, the first two runners are separated 
by a distance of 45.0 m. The front runner has a velocity of 3.50 m/s, and 
the second a velocity of 4.20 m/s. (a) What is the velocity of the second 
runner relative to the first? (b) If the front runner is 250 m from the finish 
line, who will win the race, assuming they run at constant velocity? (c) 
What distance ahead will the winner be when she crosses the finish line? 

55. Verify that the coin dropped by the airline passenger in the Example 
3.8 travels 144 m horizontally while falling 1.50 m in the frame of 
reference of the Earth. 

56. A football quarterback is moving straight backward at a speed of 2.00 
m/s when he throws a pass to a player 18.0 m straight downfield. The 
ball is thrown at an angle of 25.0° relative to the ground and is caught at 
the same height as it is released. What is the initial velocity of the ball 
relative to the quarterback ? 

57. A ship sets sail from Rotterdam, The Netherlands, heading due north 
at 7.00 m/s relative to the water. The local ocean current is 1.50 m/s in a 
direction 40.0° north of east. What is the velocity of the ship relative to 
the Earth? 

58. A jet airplane flying from Darwin, Australia, has an air speed of 260 
m/s in a direction 5.0° south of west. It is in the jet stream, which is 
blowing at 35.0 m/s in a direction 15° south of east. What is the velocity 
of the airplane relative to the Earth? (b) Discuss whether your answers 
are consistent with your expectations for the effect of the wind on the 
plane's path. 

59. (a) In what direction would the ship in Exercise 3.57 have to travel in 
order to have a velocity straight north relative to the Earth, assuming its 
speed relative to the water remains 7.00 m/s ? (b) What would its speed 
be relative to the Earth? 

60. (a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s 
in a direction 20° south of east (as in Exercise 3.58). Its direction of 
motion relative to the Earth is 45.0° south of west, while its direction of 
travel relative to the air is 5.00° south of west. What is the airplane's 
speed relative to the air mass? (b) What is the airplane's speed relative 
to the Earth? 

61. A sandal is dropped from the top of a 15.0-m-high mast on a ship 
moving at 1.75 m/s due south. Calculate the velocity of the sandal when 
it hits the deck of the ship: (a) relative to the ship and (b) relative to a 
stationary observer on shore, (c) Discuss how the answers give a 
consistent result for the position at which the sandal hits the deck. 

62. The velocity of the wind relative to the water is crucial to sailboats. 
Suppose a sailboat is in an ocean current that has a velocity of 2.20 m/s 
in a direction 30.0° east of north relative to the Earth. It encounters a 
wind that has a velocity of 4.50 m/s in a direction of 50.0° south of west 
relative to the Earth. What is the velocity of the wind relative to the water? 

63. The great astronomer Edwin Hubble discovered that all distant 
galaxies are receding from our Milky Way Galaxy with velocities 
proportional to their distances. It appears to an observer on the Earth that 
we are at the center of an expanding universe. Figure 3.64 illustrates this 
for five galaxies lying along a straight line, with the Milky Way Galaxy at 
the center. Using the data from the figure, calculate the velocities: (a) 
relative to galaxy 2 and (b) relative to galaxy 5. The results mean that 
observers on all galaxies will see themselves at the center of the 
expanding universe, and they would likely be aware of relative velocities, 
concluding that it is not possible to locate the center of expansion with 
the given information. 



Galaxy 1 

300 Mly 



Galaxy 2 Galaxy 3 Galaxy 4 

150 Mly MW 190 Mly 



Galaxy 5 
450 Mly 



-4500 km/s * = -2200 km/s 



v = 2S30 km/s v. = 6700 km/s 



122 CHAPTER 3 | TWO-DIMENSIONAL KINEMATICS 



Figure 3.64 Five galaxies on a straight line, showing their distances and velocities 
relative to the Milky Way (MW) Galaxy. The distances are in millions of light years 
(Mly), where a light year is the distance light travels in one year. The velocities are 
nearly proportional to the distances. The sizes of the galaxies are greatly 
exaggerated; an average galaxy is about 0.1 Mly across. 

64. (a) Use the distance and velocity data in Figure 3.64 to find the rate 
of expansion as a function of distance. 

(b) If you extrapolate back in time, how long ago would all of the galaxies 
have been at approximately the same position? The two parts of this 
problem give you some idea of how the Hubble constant for universal 
expansion and the time back to the Big Bang are determined, 
respectively. 

65. An athlete crosses a 25-m-wide river by swimming perpendicular to 
the water current at a speed of 0.5 m/s relative to the water. He reaches 
the opposite side at a distance 40 m downstream from his starting point. 
How fast is the water in the river flowing with respect to the ground? 
What is the speed of the swimmer with respect to a friend at rest on the 
ground? 

66. A ship sailing in the Gulf Stream is heading 25.0° west of north at a 
speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is 
4.80 m/s 5.00° west of north. What is the velocity of the Gulf Stream? 
(The velocity obtained is typical for the Gulf Stream a few hundred 
kilometers off the east coast of the United States.) 

67. An ice hockey player is moving at 8.00 m/s when he hits the puck 
toward the goal. The speed of the puck relative to the player is 29.0 m/s. 
The line between the center of the goal and the player makes a 90.0° 
angle relative to his path as shown in Figure 3.65. What angle must the 
puck's velocity make relative to the player (in his frame of reference) to 
hit the center of the goal? 



29.0 m/s 




Goal 



Figure 3.65 An ice hockey player moving across the rink must shoot backward to give 
the puck a velocity toward the goal. 

68. Unreasonable Results Suppose you wish to shoot supplies straight 
up to astronauts in an orbit 36,000 km above the surface of the Earth, (a) 
At what velocity must the supplies be launched? (b) What is 
unreasonable about this velocity? (c) Is there a problem with the relative 
velocity between the supplies and the astronauts when the supplies 
reach their maximum height? (d) Is the premise unreasonable or is the 
available equation inapplicable? Explain your answer. 

69. Unreasonable Results A commercial airplane has an air speed of 
280 m/s due east and flies with a strong tailwind. It travels 3000 km in a 
direction 5° south of east in 1.50 h. (a) What was the velocity of the 
plane relative to the ground? (b) Calculate the magnitude and direction of 
the tailwind's velocity, (c) What is unreasonable about both of these 
velocities? (d) Which premise is unreasonable? 

70. Construct Your Own Problem Consider an airplane headed for a 
runway in a cross wind. Construct a problem in which you calculate the 
angle the airplane must fly relative to the air mass in order to have a 
velocity parallel to the runway. Among the things to consider are the 
direction of the runway, the wind speed and direction (its velocity) and the 
speed of the plane relative to the air mass. Also calculate the speed of 
the airplane relative to the ground. Discuss any last minute maneuvers 
the pilot might have to perform in order for the plane to land with its 
wheels pointing straight down the runway. 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 123 



DYNAMICS: FORCE AND NEWTON'S LAWS OF 
MOTION 




Figure 4.1 Newton's laws of motion describe the motion of the dolphin's path, (credit: Jin Jang) 



Learning Objectives 



4.1. Development of Force Concept 

• Understand the definition of force. 

4.2. Newton's First Law of Motion: Inertia 

• Define mass and inertia. 

• Understand Newton's first law of motion. 

4.3. Newton's Second Law of Motion: Concept of a System 

• Define net force, external force, and system. 

• Understand Newton's second law of motion. 

• Apply Newton's second law to determine the weight of an object. 

4.4. Newton's Third Law of Motion: Symmetry in Forces 

• Understand Newton's third law of motion. 

• Apply Newton's third law to define systems and solve problems of motion. 

4.5. Normal, Tension, and Other Examples of Forces 

• Define normal and tension forces. 

• Apply Newton's laws of motion to solve problems involving a variety of forces. 

• Use trigonometric identities to resolve weight into components. 

4.6. Problem-Solving Strategies 

• Understand and apply a problem-solving procedure to solve problems using Newton's laws of motion. 

4.7. Further Applications of Newton's Laws of Motion 

• Apply problem-solving techniques to solve for quantities in more complex systems of forces. 

• Integrate concepts from kinematics to solve problems using Newton's laws of motion. 

4.8. Extended Topic: The Four Basic Forces — An Introduction 

• Understand the four basic forces that underlie the processes in nature. 



124 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 



Introduction to Dynamics: Newton's Laws of Motion 



Motion draws our attention. Motion itself can be beautiful, causing us to marvel at the forces needed to achieve spectacular motion, such as that of a 
dolphin jumping out of the water, or a pole vaulter, or the flight of a bird, or the orbit of a satellite. The study of motion is kinematics, but kinematics 
only describes the way objects move — their velocity and their acceleration. Dynamics considers the forces that affect the motion of moving objects 
and systems. Newton's laws of motion are the foundation of dynamics. These laws provide an example of the breadth and simplicity of principles 
under which nature functions. They are also universal laws in that they apply to similar situations on Earth as well as in space. 

Issac Newton's (1642-1727) laws of motion were just one part of the monumental work that has made him legendary. The development of Newton's 
laws marks the transition from the Renaissance into the modern era. This transition was characterized by a revolutionary change in the way people 
thought about the physical universe. For many centuries natural philosophers had debated the nature of the universe based largely on certain rules of 
logic with great weight given to the thoughts of earlier classical philosophers such as Aristotle (384-322 BC). Among the many great thinkers who 
contributed to this change were Newton and Galileo. 











PHILOSOPHIC 

NATUR.ALIS 

PRINCIPI A 

MATHEMATICA- 




IMPRIMATUR. 

s. PEPTS, ^.StPR^HS. 


L y H D I n I, 

Jnflu SKHiMit kw* K Tjpii '&<& #*** P™^" fP* 







Figure 4.2 Issac Newton's monumental work, Philosophiae Naturalis Principia Mathematica, was published in 1687. It proposed scientific laws that are still used today to 
describe the motion of objects, (credit: Service commun de la documentation de I'Universite de Strasbourg) 

Galileo was instrumental in establishing observation as the absolute determinant of truth, rather than "logical" argument. Galileo's use of the 
telescope was his most notable achievement in demonstrating the importance of observation. He discovered moons orbiting Jupiter and made other 
observations that were inconsistent with certain ancient ideas and religious dogma. For this reason, and because of the manner in which he dealt 
with those in authority, Galileo was tried by the Inquisition and punished. He spent the final years of his life under a form of house arrest. Because 
others before Galileo had also made discoveries by observing the nature of the universe, and because repeated observations verified those of 
Galileo, his work could not be suppressed or denied. After his death, his work was verified by others, and his ideas were eventually accepted by the 
church and scientific communities. 

Galileo also contributed to the formation of what is now called Newton's first law of motion. Newton made use of the work of his predecessors, which 
enabled him to develop laws of motion, discover the law of gravity, invent calculus, and make great contributions to the theories of light and color. It is 
amazing that many of these developments were made with Newton working alone, without the benefit of the usual interactions that take place among 
scientists today. 

It was not until the advent of modern physics early in the 20th century that it was discovered that Newton's laws of motion produce a good 
approximation to motion only when the objects are moving at speeds much, much less than the speed of light and when those objects are larger than 

the size of most molecules (about 10~ m in diameter). These constraints define the realm of classical mechanics, as discussed in Introduction to 

the Nature of Science and Physics. At the beginning of the 20 th century, Albert Einstein (1879-1955) developed the theory of relativity and, along 
with many other scientists, developed quantum theory. This theory does not have the constraints present in classical physics. All of the situations we 
consider in this chapter, and all those preceding the introduction of relativity in Special Relativity, are in the realm of classical physics. 

Making Connections: Past and Present Philosophy 



The importance of observation and the concept of cause and effect were not always so entrenched in human thinking. This realization was a part 
of the evolution of modern physics from natural philosophy. The achievements of Galileo, Newton, Einstein, and others were key milestones in 
the history of scientific thought. Most of the scientific theories that are described in this book descended from the work of these scientists. 



4.1 Development of Force Concept 



Dynamics is the study of the forces that cause objects and systems to move. To understand this, we need a working definition of force. Our intuitive 
definition of force — that is, a push or a pull — is a good place to start. We know that a push or pull has both magnitude and direction (therefore, it is a 
vector quantity) and can vary considerably in each regard. For example, a cannon exerts a strong force on a cannonball that is launched into the air. 
In contrast, Earth exerts only a tiny downward pull on a flea. Our everyday experiences also give us a good idea of how multiple forces add. If two 
people push in different directions on a third person, as illustrated in Figure 4.3, we might expect the total force to be in the direction shown. Since 
force is a vector, it adds just like other vectors, as illustrated in Figure 4.3(a) for two ice skaters. Forces, like other vectors, are represented by arrows 
and can be added using the familiar head-to-tail method or by trigonometric methods. These ideas were developed in Two-Dimensional 
Kinematics. 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 125 




Free-body diagram 



Fi 

(a) (b) 

Figure 4.3 Part (a) shows an overhead view of two ice skaters pushing on a third. Forces are vectors and add like other vectors, so the total force on the third skater is in the 
direction shown. In part (b), we see a free-body diagram representing the forces acting on the third skater. 

Figure 4.3(b) is our first example of a free-body diagram, which is a technique used to illustrate all the external forces acting on a body. The body 
is represented by a single isolated point (or free body), and only those forces acting on the body from the outside (external forces) are shown. (These 
forces are the only ones shown, because only external forces acting on the body affect its motion. We can ignore any internal forces within the body.) 
Free-body diagrams are very useful in analyzing forces acting on a system and are employed extensively in the study and application of Newton's 
laws of motion. 

A more quantitative definition of force can be based on some standard force, just as distance is measured in units relative to a standard distance. 
One possibility is to stretch a spring a certain fixed distance, as illustrated in Figure 4.4, and use the force it exerts to pull itself back to its relaxed 
shape — called a restoring force — as a standard. The magnitude of all other forces can be stated as multiples of this standard unit of force. Many other 
possibilities exist for standard forces. (One that we will encounter in Magnetism is the magnetic force between two wires carrying electric current.) 
Some alternative definitions of force will be given later in this chapter. 



(a) 



» H — Ax — H 
(b) 



H 



o[ 



|-* — Ax — H 
(c) 

Figure 4.4 The force exerted by a stretched spring can be used as a standard unit of force, (a) This spring has a length X when undistorted. (b) When stretched a distance 

Ax , the spring exerts a restoring force, F restore , which is reproducible, (c) A spring scale is one device that uses a spring to measure force. The force F restore is 
exerted on whatever is attached to the hook. Here F restore has a magnitude of 6 units in the force standard being employed. 

Take-Hom e Experi ment: Force Standards 

To investigate force standards and cause and effect, get two identical rubber bands. Hang one rubber band vertically on a hook. Find a small 
household item that could be attached to the rubber band using a paper clip, and use this item as a weight to investigate the stretch of the rubber 
band. Measure the amount of stretch produced in the rubber band with one, two, and four of these (identical) items suspended from the rubber 
band. What is the relationship between the number of items and the amount of stretch? How large a stretch would you expect for the same 
number of items suspended from two rubber bands? What happens to the amount of stretch of the rubber band (with the weights attached) if the 
weights are also pushed to the side with a pencil? 

4.2 Newton's First Law of Motion: Inertia 

Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and stop unless some effort 
is made to keep it moving. What Newton's first law of motion states, however, is the following: 

Newton's First Law of Motion 



A body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. 



Note the repeated use of the verb "remains." We can think of this law as preserving the status quo of motion. 

Rather than contradicting our experience, Newton's first law of motion states that there must be a cause (which is a net external force) for there to 
be any change in velocity (either a change in magnitude or direction). We will define net external force in the next section. An object sliding across a 
table or floor slows down due to the net force of friction acting on the object. If friction disappeared, would the object still slow down? 

The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, consider what happens to an object 
sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray the surface with talcum powder to make the surface smoother, 



126 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 

the object slides farther. If we make the surface even smoother by rubbing lubricating oil on it, the object slides farther yet. Extrapolating to a 
frictionless surface, we can imagine the object sliding in a straight line indefinitely. Friction is thus the cause of the slowing (consistent with Newton's 
first law). The object would not slow down at all if friction were completely eliminated. Consider an air hockey table. When the air is turned off, the 
puck slides only a short distance before friction slows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and the 
puck glides long distances without slowing down. Additionally, if we know enough about the friction, we can accurately predict how quickly the object 
will slow down. Friction is an external force. 

Newton's first law is completely general and can be applied to anything from an object sliding on a table to a satellite in orbit to blood pumped from 
the heart. Experiments have thoroughly verified that any change in velocity (speed or direction) must be caused by an external force. The idea of 
generally applicable or universal laws is important not only here — it is a basic feature of all laws of physics. Identifying these laws is like recognizing 
patterns in nature from which further patterns can be discovered. The genius of Galileo, who first developed the idea for the first law, and Newton, 
who clarified it, was to ask the fundamental question, "What is the cause?" Thinking in terms of cause and effect is a worldview fundamentally 
different from the typical ancient Greek approach when questions such as "Why does a tiger have stripes?" would have been answered in Aristotelian 
fashion, "That is the nature of the beast." True perhaps, but not a useful insight. 

Mass 

The property of a body to remain at rest or to remain in motion with constant velocity is called inertia. Newton's first law is often called the law of 
inertia. As we know from experience, some objects have more inertia than others. It is obviously more difficult to change the motion of a large 
boulder than that of a basketball, for example. The inertia of an object is measured by its mass. Roughly speaking, mass is a measure of the amount 
of "stuff" (or matter) in something. The quantity or amount of matter in an object is determined by the numbers of atoms and molecules of various 
types it contains. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon. 
In practice, it is very difficult to count and identify all of the atoms and molecules in an object, so masses are not often determined in this manner. 
Operationally, the masses of objects are determined by comparison with the standard kilogram. 



Check Your Understanding 



Which has more mass: a kilogram of cotton balls or a kilogram of gold? 

Solution 

They are equal. A kilogram of one substance is equal in mass to a kilogram of another substance. The quantities that might differ between them 
are volume and density. 

4.3 Newton's Second Law of Motion: Concept of a System 

Newton's second law of motion is closely related to Newton's first law of motion. It mathematically states the cause and effect relationship between 
force and changes in motion. Newton's second law of motion is more quantitative and is used extensively to calculate what happens in situations 
involving a force. Before we can write down Newton's second law as a simple equation giving the exact relationship of force, mass, and acceleration, 
we need to sharpen some ideas that have already been mentioned. 

First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity 
means, by definition, that there is an acceleration. Newton's first law says that a net external force causes a change in motion; thus, we see that a 
net external force causes acceleration. 

Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct — an external force acts from 
outside the system of interest. For example, in Figure 4.5(a) the system of interest is the wagon plus the child in it. The two forces exerted by the 
other children are external forces. An internal force acts between elements of the system. Again looking at Figure 4.5(a), the force the child in the 
wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion of a 
system, according to Newton's first law. (The internal forces actually cancel, as we shall see in the next section.) You must define the boundaries of 
the system before you can determine which forces are external. Sometimes the system is obvious, whereas other times identifying the boundaries of 
a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton's laws. This 
concept will be revisited many times on our journey through physics. 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 127 




Free-body diagram 



_^ 



F 



Each force acting F n et 
on the system 

adds to produce 
a net force, F^. 



<b) 




Figure 4.5 Different forces exerted on the same mass produce different accelerations, (a) Two children push a wagon with a child in it. Arrows representing all external forces 
are shown. The system of interest is the wagon and its rider. The weight W of the system and the support of the ground N are also shown for completeness and are 
assumed to cancel. The vector f represents the friction acting on the wagon, and it acts to the left, opposing the motion of the wagon, (b) All of the external forces acting on 
the system add together to produce a net force, F net . The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass 
of the system. Each force vector extends from this dot. Because there are two forces acting to the right, we draw the vectors collinearly. (c) A larger net external force produces 
a larger acceleration (a > a ) when an adult pushes the child. 

Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external force acting on a 
system. This assumption has been verified experimentally and is illustrated in Figure 4.5. In part (a), a smaller force causes a smaller acceleration 
than the larger force illustrated in part (c). For completeness, the vertical forces are also shown; they are assumed to cancel since there is no 
acceleration in the vertical direction. The vertical forces are the weight w and the support of the ground N , and the horizontal force f represents 
the force of friction. These will be discussed in more detail in later sections. For now, we will define friction as a force that opposes the motion past 
each other of objects that are touching. Figure 4.5(b) shows how vectors representing the external forces add together to produce a net force, F net . 



To obtain an equation for Newton's second law, we first write the relationship of acceleration and net external force as the proportionality 

a oc F npt , 



(4.1) 



where the symbol ex means "proportional to," and F net is the net external force. (The net external force is the vector sum of all external forces 

and can be determined graphically, using the head-to-tail method, or analytically, using components. The techniques are the same as for the addition 
of other vectors, and are covered in Two-Dimensional Kinematics.) This proportionality states what we have said in words — acceleration is directly 
proportional to the net external force. Once the system of interest is chosen, it is important to identify the external forces and ignore the internal ones. 
It is a tremendous simplification not to have to consider the numerous internal forces acting between objects within the system, such as muscular 
forces within the child's body, let alone the myriad of forces between atoms in the objects, but by doing so, we can easily solve some very complex 
problems with only minimal error due to our simplification 

Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the 
inertia), the smaller the acceleration produced by a given force. And indeed, as illustrated in Figure 4.6, the same net external force applied to a car 
produces a much smaller acceleration than when applied to a basketball. The proportionality is written as 



a oc 



1 



(4.2) 



where m is the mass of the system. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is exactly linearly 
proportional to the net external force. 



128 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 





(a) (b) 

The free-body diagrams for both objects are the same. 



(c) 

Figure 4.6 The same force exerted on systems of different masses produces different accelerations, (a) A basketball player pushes on a basketball to make a pass. (The effect 
of gravity on the ball is ignored.) (b) The same player exerts an identical force on a stalled SUV and produces a far smaller acceleration (even if friction is negligible), (c) The 
free-body diagrams are identical, permitting direct comparison of the two situations. A series of patterns for the free-body diagram will emerge as you do more problems. 

It has been found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two 
proportionalities just given yields Newton's second law of motion. 



Newton's Second Law of Motion 



The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely 
proportional to its mass. 

In equation form, Newton's second law of motion is 



This is often written in the more familiar form 



a m ' 



F net = ma. 



When only the magnitude of force and acceleration are considered, this equation is simply 



F nQt = ma. 



(4.3) 

(4.4) 
(4.5) 



Although these last two equations are really the same, the first gives more insight into what Newton's second law means. The law is a cause and 
effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is completely based on 
experimental verification. 

Units of Force 

F net = ma is used to define the units of force in terms of the three basic units for mass, length, and time. The SI unit of force is called the newton 



(abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of lm/s . That is, since F net = ma , 

1 N = 1 ke • m/s 2 . 



(4.6) 



While almost the entire world uses the newton for the unit of force, in the United States the most familiar unit of force is the pound (lb), where 1 N = 
0.225 lb. 

Weight and the Gravitational Force 

When an object is dropped, it accelerates toward the center of Earth. Newton's second law states that a net force on an object is responsible for its 
acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight w . Weight can be 
denoted as a vector w because it has a direction; down is, by definition, the direction of gravity, and hence weight is a downward force. The 
magnitude of weight is denoted as w . Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same 
acceleration g . Using Galileo's result and Newton's second law, we can derive an equation for weight. 

Consider an object with mass m falling downward toward Earth. It experiences only the downward force of gravity, which has magnitude w . 
Newton's second law states that the magnitude of the net external force on an object is F net = ma . 

Since the object experiences only the downward force of gravity, F nQt = w . We know that the acceleration of an object due to gravity is g , or 
a = g . Substituting these into Newton's second law gives 



Weight 



This is the equation for weight — the gravitational force on a mass m : 

w = mg. 



(4.7) 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 129 



Since g = 9.80 m/s on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see: 

w = mg = (1.0 kg)(9.80 m/s 2 ) = 9.8 N. < 4 - 8 ) 

Recall that g can take a positive or negative value, depending on the positive direction in the coordinate system. Be sure to take this into 
consideration when solving problems with weight. 

When the net external force on an object is its weight, we say that it is in free-fall. That is, the only force acting on the object is the force of gravity. In 
the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is always some upward force from the air 
acting on the object. 

The acceleration due to gravity g varies slightly over the surface of Earth, so that the weight of an object depends on location and is not an intrinsic 

property of the object. Weight varies dramatically if one leaves Earth's surface. On the Moon, for example, the acceleration due to gravity is only 

1.67 m/s . A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon. 

The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body, such as Earth, 
the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition 
of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of "weightlessness" and "microgravity," 
they are really referring to the phenomenon we call "free-fall" in physics. We shall use the above definition of weight, and we will make careful 
distinctions between free-fall and actual weightlessness. 

It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is the quantity of matter 
(how much "stuff") and does not vary in classical physics, whereas weight is the gravitational force and does vary depending on gravity. It is tempting 
to equate the two, since most of our examples take place on Earth, where the weight of an object only varies a little with the location of the object. 
Furthermore, the terms mass and weight are used interchangeably in everyday language; for example, our medical records often show our "weight" 
in kilograms, but never in the correct units of newtons. 

Common Misconceptions: Mass vs. Weight 



Mass and weight are often used interchangeably in everyday language. However, in science, these terms are distinctly different from one 
another. Mass is a measure of how much matter is in an object. The typical measure of mass is the kilogram (or the "slug" in English units). 
Weight, on the other hand, is a measure of the force of gravity acting on an object. Weight is equal to the mass of an object ( m ) multiplied by 
the acceleration due to gravity ( g ). Like any other force, weight is measured in terms of newtons (or pounds in English units). 

Assuming the mass of an object is kept intact, it will remain the same, regardless of its location. However, because weight depends on the 
acceleration due to gravity, the weight of an object can change when the object enters into a region with stronger or weaker gravity. For example, 

9 9 

the acceleration due to gravity on the Moon is 1.67 m/s (which is much less than the acceleration due to gravity on Earth, 9.80 m/s ). If you 
measured your weight on Earth and then measured your weight on the Moon, you would find that you "weigh" much less, even though you do 
not look any skinnier. This is because the force of gravity is weaker on the Moon. In fact, when people say that they are "losing weight," they 
really mean that they are losing "mass" (which in turn causes them to weigh less). 

Take-Home Experiment: Mass and Weight 

What do bathroom scales measure? When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains 
springs that compress in proportion to your weight — similar to rubber bands expanding when pulled. The springs provide a measure of your 
weight (for an object which is not accelerating). This is a force in newtons (or pounds). In most countries, the measurement is divided by 9.80 to 
give a reading in mass units of kilograms. The scale measures weight but is calibrated to provide information about mass. While standing on a 
bathroom scale, push down on a table next to you. What happens to the reading? Why? Would your scale measure the same "mass" on Earth as 
on the Moon? 



Example 4.1 What Acceleration Can a Person Produce when Pushing a Lawn Mower'* 



Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb) parallel to the ground. The mass of the 
mower is 24 kg. What is its acceleration? 



130 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 




Figure 4.7 The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right? 

Strategy 

Since F net and m are given, the acceleration can be calculated directly from Newton's second law as stated in F net = ma . 

Solution 

The magnitude of the acceleration a is a = — ^p . Entering known values gives 

24 kg 



(4.9) 



Substituting the units kg • m/s for N yields 



51 kg • m/s 
24 kg 



2.1m/s z . 



(4.10) 



Discussion 

The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this 
example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the 
force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moves forward), 
and the vertical forces must cancel if there is to be no acceleration in the vertical direction (the mower is moving only horizontally). The 
acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long because the person's 
top speed would soon be reached. 



Example 4.2 What Rocket Thrust Accelerates This Sled? 



Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high 
speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force 
exerted by each rocket, called its thrust T , for the four-rocket propulsion system shown in Figure 4.8. The sled's initial acceleration is 

49 m/s , the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N. 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 131 



a 








T » tiy — - 




T 




V , 






W 












T 1 .. 

^-IVL 


S^'^™ t 




f w 

T T 


N 

Free-body diagram 
N 


T *^ 


1 

W 



Figure 4.8 A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust T . As in other situations where there is only 
horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in magnitude and opposite in direction to its weight, 
W . The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction ( f ) is drawn larger 
than scale. 

Strategy 

Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This 
leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken 
as the positive direction. See the free-body diagram in the figure. 

Solution 

Since acceleration, mass, and the force of friction are given, we start with Newton's second law and look for ways to find the thrust of the 
engines. Since we have defined the direction of the force and acceleration as acting "to the right," we need to consider only the magnitudes of 
these quantities in the calculations. Hence we begin with 

^net = ma, (4.11) 

where F net is the net force along the horizontal direction. We can see from Figure 4.8 that the engine thrusts add, while friction opposes the 
thrust. In equation form, the net external force is 

Fnel = *T-f. (4.12) 

Substituting this into Newton's second law gives 

F net = ma = 4T-f. (4.13) 

Using a little algebra, we solve for the total thrust 47: 

AT = ma + /. (4.14) 

Substituting known values yields 



So the total thrust is 



and the individual thrusts are 



AT = ma + / = (2100 kg)(49 m/s 2 ) + 650 N. 



AT= l.OxKTN, 



r= LQxlQiN X10 4 N. 

4 



(4.15) 



(4.16) 



(4.17) 



Discussion 



The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of 
human endurance and the setup designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, 

with accelerations of 45 g 's. (Recall that g , the acceleration due to gravity, is 9.80 m/s . When we say that an acceleration is 45 g 's, it is 

9 9 

45x9.80 m/s , which is approximately 440 m/s .) While living subjects are not used any more, land speeds of 10,000 km/h have been 
obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that 
choosing the system of interest is crucial — and the choice is not always obvious. 



132 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 

Newton's second law of motion is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make 
predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about 
nature. The next section introduces the third and final law of motion. 

4.4 Newton's Third Law of Motion: Symmetry in Forces 

There is a passage in the musical Man of la Mancha that relates to Newton's third law of motion. Sancho, in describing a fight with his wife to Don 
Quixote, says, "Of course I hit her back, Your Grace, but she's a lot harder than me and you know what they say, 'Whether the stone hits the pitcher 
or the pitcher hits the stone, it's going to be bad for the pitcher.'" This is exactly what happens whenever one body exerts a force on another — the first 
also experiences a force (equal in magnitude and opposite in direction). Numerous common experiences, such as stubbing a toe or throwing a ball, 
confirm this. It is precisely stated in Newton's third law of motion. 

Newton's Third Law of Motion 



Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to 
the force that it exerts. 



This law represents a certain symmetry in nature: Forces always occur in pairs, and one body cannot exert a force on another without experiencing a 
force itself. We sometimes refer to this law loosely as "action-reaction," where the force exerted is the action and the force experienced as a 
consequence is the reaction. Newton's third law has practical uses in analyzing the origin of forces and understanding which forces are external to a 
system. 

We can readily see Newton's third law at work by taking a look at how people move about. Consider a swimmer pushing off from the side of a pool, 
as illustrated in Figure 4.9. She pushes against the pool wall with her feet and accelerates in the direction opposite to that of her push. The wall has 
exerted an equal and opposite force back on the swimmer. You might think that two equal and opposite forces would cancel, but they do not because 
they act on different systems. In this case, there are two systems that we could investigate: the swimmer or the wall. If we select the swimmer to be 
the system of interest, as in the figure, then F wall on feet is an external force on this system and affects its motion. The swimmer moves in the 

direction of F wall on feet . In contrast, the force F feet on wall acts on the wall and not on our system of interest. Thus F feet on wall does not directly 

affect the motion of the system and does not cancel F wall on feet . Note that the swimmer pushes in the direction opposite to that in which she wishes 

to move. The reaction to her push is thus in the desired direction. 

SV^?rri erf interest ,^^^^« ^_ ^. 

Free-twiy Diagram 




Figure 4.9 When the swimmer exerts a force Ff eet on wa jj on the wall, she accelerates in the direction opposite to that of her push. This means the net external force on her 
is in the direction opposite to Ff eet on wa jj . This opposition occurs because, in accordance with Newton's third law of motion, the wall exerts a force F wa jj on f eet on her, 
equal in magnitude but in the direction opposite to the one she exerts on it. The line around the swimmer indicates the system of interest. Note that Ff eet on wa jj does not 
act on this system (the swimmer) and, thus, does not cancel F wa ^ on f eet . Thus the free-body diagram shows only F wa jj on f eet , W , the gravitational force, and BF , 
the buoyant force of the water supporting the swimmer's weight. The vertical forces W and BF cancel since there is no vertical motion. 

Other examples of Newton's third law are easy to find. As a professor paces in front of a whiteboard, she exerts a force backward on the floor. The 
floor exerts a reaction force forward on the professor that causes her to accelerate forward. Similarly, a car accelerates because the ground pushes 
forward on the drive wheels in reaction to the drive wheels pushing backward on the ground. You can see evidence of the wheels pushing backward 
when tires spin on a gravel road and throw rocks backward. In another example, rockets move forward by expelling gas backward at high velocity. 
This means the rocket exerts a large backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction 
force forward on the rocket. This reaction force is called thrust. It is a common misconception that rockets propel themselves by pushing on the 
ground or on the air behind them. They actually work better in a vacuum, where they can more readily expel the exhaust gases. Helicopters similarly 
create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly by exerting force on air in a direction 
opposite to that of whatever force they need. For example, the wings of a bird force air downward and backward in order to get lift and move forward. 
An octopus propels itself in the water by ejecting water through a funnel from its body, similar to a jet ski. In a situation similar to Sancho's, 
professional cage fighters experience reaction forces when they punch, sometimes breaking their hand by hitting an opponent's body. 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 133 



Example 4.3 Getting Up To Speed: Choosing the Correct System 



A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure 4.10. Her mass is 65.0 kg, the cart's is 12.0 
kg, and the equipment's is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All 
forces opposing the motion, such as friction on the cart's wheels and air resistance, total 24.0 N. 



F*w botfy diagrams 



w 

Systam I 

IN' 




System 2 

Figure 4.10 A professor pushes a cart of demonstration equipment. The lengths of the arrows are proportional to the magnitudes of the forces (except for f , since it is 
too small to draw to scale). Different questions are asked in each example; thus, the system of interest must be defined differently for each. System 1 is appropriate for 
Example 4.4, since it asks for the acceleration of the entire group of objects. Only F^ oor and f are external forces acting on System 1 along the line of motion. All 

other forces either cancel or act on the outside world. System 2 is chosen for this example so that F pro f will be an external force and enter into Newton's second law. 

Note that the free-body diagrams, which allow us to apply Newton's second law, vary with the system chosen. 

Strategy 

Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in Figure 4.10. The professor 
pushes backward with a force F foot of 150 N. According to Newton's third law, the floor exerts a forward reaction force F floor of 150 N on 

System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction. The problem is therefore one- 
dimensional along the horizontal direction. As noted, f opposes the motion and is thus in the opposite direction of F floor . Note that we do not 

include the forces F f or F cart because these are internal forces, and we do not include F foot because it acts on the floor, not on the 

system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use 
Newton's second law to find the acceleration as requested. See the free-body diagram in the figure. 

Solution 

Newton's second law is given by 

n _ ^net ( 4 -!8) 

u m ' 

The net external force on System 1 is deduced from Figure 4.10 and the discussion above to be 

^net = ffloor - / = 150 N - 24.0 N = 126 N. (4.19) 

The mass of System 1 is 

m = (65.0 + 12.0 + 7.0) kg = 84 kg. (4.20) 

These values of F nQt and m produce an acceleration of 

, _ ^net ( 4 - 21 ) 

84 kg 

Discussion 

None of the forces between components of System 1, such as between the professor's hands and the cart, contribute to the net external force 
because they are internal to System 1. Another way to look at this is to note that forces between components of a system cancel because they 
are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite 
force back on her. In this case both forces act on the same system and, therefore, cancel. Thus internal forces (between components of a 
system) cancel. Choosing System 1 was crucial to solving this problem. 



134 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 



Example 4.4 Force on the Cart — Choosing a New System 




Calculate the force the professor exerts on the cart in Figure 4.10 using data from the previous example if needed. 


Strategy 




If we now define the system of interest to be the cart plus equipment (System 2 in Figure 4.10), then the net external force on System 2 is the 
force the professor exerts on the cart minus friction. The force she exerts on the cart, F f , is an external force acting on System 2. F f 


was internal to System 1, but it is external to System 2 and will enter Newton's second law for System 2. 




Solution 




Newton's second law can be used to find F f . Starting with 




a = FnQt 
u m 


(4.22) 


and noting that the magnitude of the net external force on System 2 is 




r net = r prof — / » 


(4.23) 


we solve for F prof , the desired quantity: 




** prof = ** net "■" / • 


(4.24) 


The value of / is given, so we must calculate net F net . That can be done since both the acceleration and mass of System 2 are known. Using 


Newton's second law we see that 




F nQt = ma, 


(4.25) 


where the mass of System 2 is 19.0 kg ( m - 12.0 kg + 7.0 kg) and its acceleration was found to be a = 
Thus, 


1.5 m/s in the previous example. 


F nQt = ma, 


(4.26) 


F nQt = (19.0 kg)(1.5 m/s 2 ) = 29 N. 


(4.27) 


Now we can find the desired force: 




** prof = r net * J » 


(4.28) 


F prof = 29 N+24.0 N = 53 N. 


(4.29) 


Discussion 




It is interesting that this force is significantly less than the 150-N force the professor exerted backward on 
transmitted to the cart; some of it accelerates the professor. 


the floor. Not all of that 150-N force is 


The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation 
(which is not necessarily the same thing). 



PhET Explorations: Gravity Force Lab 

Visualize the gravitational force that two objects exert on each other. Change properties of the objects in order to see how it changes the gravity 
force. 



M 



<<; 'PhET Interactive Simulation 

Figure 4.11 Gravity Force Lab (http://cnx.Org/content/m42074/l.3/gravity-force-lab_en.jar) 

4.5 Normal, Tension, and Other Examples of Forces 

Forces are given many names, such as push, pull, thrust, lift, weight, friction, and tension. Traditionally, forces have been grouped into several 
categories and given names relating to their source, how they are transmitted, or their effects. The most important of these categories are discussed 
in this section, together with some interesting applications. Further examples of forces are discussed later in this text. 



Normal Force 

Weight (also called force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You definitely 
notice that you must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustrated in Figure 4.12(a). But how do 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 135 



inanimate objects like a table support the weight of a mass placed on them, such as shown in Figure 4.12(b)? When the bag of dog food is placed on 
the table, the table actually sags slightly under the load. This would be noticeable if the load were placed on a card table, but even rigid objects 
deform when a force is applied to them. Unless the object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or 
trampoline or diving board). The greater the deformation, the greater the restoring force. So when the load is placed on the table, the table sags until 
the restoring force becomes as large as the weight of the load. At this point the net external force on the load is zero. That is the situation when the 
load is stationary on the table. The table sags quickly, and the sag is slight so we do not notice it. But it is similar to the sagging of a trampoline when 
you climb onto it. 







<b> 



hand 



W 



W 



FnHhbody diagrams 

Figure 4.12 (a) The person holding the bag of dog food must supply an upward force Fj ian( j equal in magnitude and opposite in direction to the weight of the food W . (b) 

The card table sags when the dog food is placed on it, much like a stiff trampoline. Elastic restoring forces in the table grow as it sags until they supply a force N equal in 
magnitude and opposite in direction to the weight of the load. 

We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed 
in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this force 
is defined to be a normal force and here is given the symbol N . (This is not the unit for force N.) The word normal means perpendicular to a 
surface. The normal force can be less than the object's weight if the object is on an incline, as you will see in the next example. 

Common M isconception: Normal Force (N) vs. Newton (N) 

In this section we have introduced the quantity normal force, which is represented by the variable N . This should not be confused with the 
symbol for the newton, which is also represented by the letter N. These symbols are particularly important to distinguish because the units of a 
normal force ( N ) happen to be newtons (N). For example, the normal force N that the floor exerts on a chair might be N = 100 N . One 
important difference is that normal force is a vector, while the newton is simply a unit. Be careful not to confuse these letters in your calculations! 
You will encounter more similarities among variables and units as you proceed in physics. Another example of this is the quantity work ( W) and 
the unit watts (W). 



Example 4.5 Weight on an Incline, a Two-Dimensional Problem 



Consider the skier on a slope shown in Figure 4.13. Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is 
negligible? (b) What is her acceleration if friction is known to be 45.0 N? 



136 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 




Figure 4.13 Since motion and friction are parallel to the slope, it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope 
and the other is perpendicular (axes shown to left of skier). N is perpendicular to the slope and f is parallel to the slope, but W has components along both axes, 
namely Wj_ and W ■■ . N is equal in magnitude to Wj_ , so that there is no motion perpendicular to the slope, but f is less than W ■■ , so that there is a 
downslope acceleration (along the parallel axis). 

Strategy 

This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two- 
dimensional kinematics also works very well here. Choose a convenient coordinate system and project the vectors onto its axes, creating two 
connected o/ie-dimensional problems to solve. The most convenient coordinate system for motion on an incline is one that has one coordinate 
parallel to the slope and one perpendicular to the slope. (Remember that motions along mutually perpendicular axes are independent.) We use 
the symbols _L and || to represent perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because 

there is no motion perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external forces 
acting on the system are the skier's weight, friction, and the support of the slope, respectively labeled w , f , and N in Figure 4.13. N is 
always perpendicular to the slope, and f is parallel to it. But w is not in the direction of either axis, and so the first step we take is to project it 
into components along the chosen axes, defining w n to be the component of weight parallel to the slope and w_|_ the component of weight 

perpendicular to the slope. Once this is done, we can consider the two separate problems of forces parallel to the slope and forces perpendicular 
to the slope. 

Solution 

The magnitude of the component of the weight parallel to the slope is w u = w sin (25°) = mg sin (25°) , and the magnitude of the 
component of the weight perpendicular to the slope is w_|_ = w cos (25°) = mg cos (25°) . 

(a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to 
the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier's weight parallel 
to the slope w m and friction / . Using Newton's second law, with subscripts to denote quantities parallel to the slope, 

F nM ,1 (4.30) 



where F n 



= mg sin (25°) , assuming no friction for this part, so that 



a I, =■ 



_ mg sm \z,j ) _ 

~ m ~~ 



g sin (25°) 



(4.31) 



(9.80 m/s 2 )(0.4226) = 4.14 m/s 2 



is the acceleration. 

(b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes motion between 
surfaces in contact. So the net external force is now 



F mt II - w II " /» 



(4.32) 



and substituting this into Newton's second law, a n = — -^- , gives 

„ _ F net| | _W|| -/_mgsin(25°)-/ 



(4.33) 



m 



m 



m 



We substitute known values to obtain 



(60.0 kg)(9.80 m/s 2 )(0.4226) - 45.0 N 
60.0 kg 



(4.34) 



which yields 



a I, = 3.39 m/s 2 



(4.35) 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 137 

which is the acceleration parallel to the incline when there is 45.0 N of opposing friction. 

Discussion 

Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a 
general result that if friction on an incline is negligible, then the acceleration down the incline is a = g sin# , regardless of mass. This is related 

to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of 
mass, slide down a frictionless incline with the same acceleration (if the angle is the same). 



Resolving Weight into Components 




Wii = wsin(#) = mg s\n(8) 
ivj. = wcos(&) = mgco$(&) 



Figure 4.14 An object rests on an incline that makes an angle with the horizontal. 

When an object rests on an incline that makes an angle 6 with the horizontal, the force of gravity acting on the object is divided into two 
components: a force acting perpendicular to the plane, w_|_ , and a force acting parallel to the plane, w m . The perpendicular force of weight, 

Wj_ , is typically equal in magnitude and opposite in direction to the normal force, N . The force acting parallel to the plane, w m , causes the 

object to accelerate down the incline. The force of friction, f , opposes the motion of the object, so it acts upward along the plane. 

It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle 6 to the horizontal, 
then the magnitudes of the weight components are 

w II = w sin (6) = mg sin (0) (4.36) 



and 



Wj_ = w cos (0) = mg cos (0). 



(4.37) 



Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the 
three weight vectors. Notice that the angle 6 of the incline is the same as the angle formed between w and w_|_ . Knowing this property, you 

can use trigonometry to determine the magnitude of the weight components: 

(4.38) 



cos(0) = %- 



Wj_ 



w cos (6) = mg cos (6) 



W II 

sin (6) = -J- 

w || = w sin (6) = mg sin (6) 



(4.39) 



Take-Home Experiment: Force Parallel 



To investigate how a force parallel to an inclined plane changes, find a rubber band, some objects to hang from the end of the rubber band, and 
a board you can position at different angles. How much does the rubber band stretch when you hang the object from the end of the board? Now 
place the board at an angle so that the object slides off when placed on the board. How much does the rubber band extend if it is lined up 
parallel to the board and used to hold the object stationary on the board? Try two more angles. What does this show? 



Tension 

A tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. The word "tension" comes 
from a Latin word meaning "to stretch." Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons. 
Any flexible connector, such as a string, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a force carried by a flexible 
connector is a tension with direction parallel to the connector. It is important to understand that tension is a pull in a connector. In contrast, consider 
the phrase: "You can't push a rope." The tension force pulls outward along the two ends of a rope. 

Consider a person holding a mass on a rope as shown in Figure 4.15. 



138 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 




Free-body diagram 



It 
I" 



Figure 4.15 When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force T , that force must be parallel to the length of the rope, 
as shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force but in opposite directions on the hand and the supported mass 
(neglecting the weight of the rope). This is an example of Newton's third law. The rope is the medium that carries the equal and opposite forces between the two objects. The 
tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations 
along the rope. 

Tension in the rope must equal the weight of the supported mass, as we can prove using Newton's second law. If the 5.00-kg mass in the figure is 
stationary, then its acceleration is zero, and thus F net = . The only external forces acting on the mass are its weight w and the tension T 

supplied by the rope. Thus, 

^net = T - W = 0, (4.40) 

where T and w are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here. Thus, just as you would 
expect, the tension equals the weight of the supported mass: 

T = w = mg. (4.41) 

For a 5.00-kg mass, then (neglecting the mass of the rope) we see that 

T = mg = (5.00 kg)(9.80 m/s 2 ) = 49.0 N. < 4 - 42 ) 

If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and 
measure of the tension force in the rope. 

Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If 
there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector. This is 
illustrated in Figure 4.16 (a) and (b). 

Extensor muscle _ 

Tendon 




(a) Flexor muscle Tendon 




Figure 4.16 (a) Tendons in the finger carry force T from the muscles to other parts of the finger, usually changing the force's direction, but not its magnitude (the tendons are 
relatively friction free), (b) The brake cable on a bicycle carries the tension T from the handlebars to the brake mechanism. Again, the direction but not the magnitude of T 
is changed. 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 139 



Example 4.6 What Is the Tension in a Tightrope? 



Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 4.17. 




Figure 4.17 The weight of a tightrope walker causes a wire to sag by 5.0 degrees. The system of interest here is the point in the wire at which the tightrope walker is 
standing. 

Strategy 

As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person's weight. Thus, the tension on either 
side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having 
the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces 
acting on him are his weight w and the two tensions T L (left tension) and T R (right tension), as illustrated. It is reasonable to neglect the 

weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. 
One conclusion is possible at the outset — we can see from part (b) of the figure that the magnitudes of the tensions T L and T R must be equal. 

This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are T L and T R . Thus, the 

magnitude of those forces must be equal so that they cancel each other out. 

Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient 
coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. 
We call the horizontal the x -axis and the vertical the y -axis. 

Solution 

First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all 
of the horizontal and vertical components of each force acting on the system. 



j »Crr. 


T L 




^50 a 


Free-body diagram 
fT L , 


5.0*^ 


T^ 


T R 


D Tr ' 


\ 

^ i 

net F x =0; 


net F y = 






Tr* 









Figure 4.18 When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the tightrope walker is 
stationary. The small angle results in T being much greater than W . 

Consider the horizontal components of the forces (denoted with a subscript x ): 

P netx = ■*■ Lx ~ * Rx' 
The net external horizontal force F nQtx = , since the person is stationary. Thus, 

^netjc = = T Lx — T^x 



'Lx 



'Rx- 



Now, observe Figure 4.18. You can use trigonometry to determine the magnitude of T L and T R . Notice that: 

cos (5.0°) 



_ T Lx 



(4.43) 



(4.44) 



(4.45) 



[ Lx 



= T L cos (5.0°) 



cos (5.0°) = -=P* 



Rjc 



= T R cos (5.0°). 



Equating T Lx and T^ : 



140 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 



T L cos (5.0°) = T R cos (5.0°). 


(4.46) 


Thus, 




T L = T R = T, 


(4.47) 


as predicted. Now, considering the vertical components (denoted by a subscript y ), we can solve for T . Again, 


since the person is stationary, 


Newton's second law implies that net F y = . Thus, as illustrated in the free-body diagram in Figure 4.18, 




F nety = T Ly + T Ry -W = 0. 


(4.48) 


Observing Figure 4.18, we can use trigonometry to determine the relationship between T L , T R , and T . As 


we determined from the 


analysis in the horizontal direction, T L = T R = T : 




T 

sin (5.0°) = -^ 

2 L 


(4.49) 


T Ly = T L sin (5.0°) = T sin (5.0°) 




T 

sin (5.0°) = -^ 

2 R 




T Ry = T R sin (5.0°) = T sin (5.0°). 




Now, we can substitute the values for T L and T R , into the net force equation in the vertical direction: 




F nety = T L y + T Ry ~ w = ° 


(4.50) 


F nQty = T sin (5.0°) + T sin (5.0°) - w = 




2 T sin (5.0°) - w = 




2 T sin (5.0°) = w 




and 




T _ W Mg 


(4.51) 


2 sin (5.0°) 2 sin (5.0°)' 


so that 




(70.0 kg)(9.80 m/s 2 ) 
2(0.0872) 


(4.52) 


and the tension is 




T = 3900 N. 


(4.53) 


Discussion 




Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times 
the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the 
tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to 
support the weight of the tightrope walker. 



If we wish to create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in Figure 4.19. As we 
saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to 
the weight of the tightrope walker in the following way: 

T = W. (4.54) 

2 sin (0)' 

We can extend this expression to describe the tension T created when a perpendicular force ( Fj^ ) is exerted at the middle of a flexible connector: 

T = 



F\ (4.55) 



2 sin (<9) ' 

Note that 6 is the angle between the horizontal and the bent connector. In this case, T becomes very large as 6 approaches zero. Even the 
relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., 6 = and 
sin 6 = ). (See Figure 4.19.) 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 141 




£ 







Figure 4.19 We can create a very large tension in the chain by pushing on it perpendicular to its length, as shown. Suppose we wish to pull a car out of the mud when no tow 
truck is available. Each time the car moves forward, the chain is tightened to keep it as nearly straight as possible. The tension in the chain is given by T = -= — : — j~- ; 

since 6 is small, T is very large. This situation is analogous to the tightrope walker shown in Figure 4.17, except that the tensions shown here are those transmitted to the 
car and the tree rather than those acting at the point where Fj_ is applied. 




Figure 4.20 Unless an infinite tension is exerted, any flexible connector — such as the chain at the bottom of the picture — will sag under its own weight, giving a characteristic 
curve when the weight is evenly distributed along the length. Suspension bridges — such as the Golden Gate Bridge shown in this image — are essentially very heavy flexible 
connectors. The weight of the bridge is evenly distributed along the length of flexible connectors, usually cables, which take on the characteristic shape, (credit: Leaflet, 
Wikimedia Commons) 

Extended Topic: Real Forces and Inertial Frames 

There is another distinction among forces in addition to the types already mentioned. Some forces are real, whereas others are not. Real forces are 
those that have some physical origin, such as the gravitational pull. Contrastingly, fictitious forces are those that arise simply because an observer is 
in an accelerating frame of reference, such as one that rotates (like a merry-go-round) or undergoes linear acceleration (like a car slowing down). For 
example, if a satellite is heading due north above Earth's northern hemisphere, then to an observer on Earth it will appear to experience a force to the 
west that has no physical origin. Of course, what is happening here is that Earth is rotating toward the east and moves east under the satellite. In 
Earth's frame this looks like a westward force on the satellite, or it can be interpreted as a violation of Newton's first law (the law of inertia). An 
inertial frame of reference is one in which all forces are real and, equivalently, one in which Newton's laws have the simple forms given in this 
chapter. 

Earth's rotation is slow enough that Earth is nearly an inertial frame. You ordinarily must perform precise experiments to observe fictitious forces and 
the slight departures from Newton's laws, such as the effect just described. On the large scale, such as for the rotation of weather systems and ocean 
currents, the effects can be easily observed. 

The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known inertial frame. Unless 
stated otherwise, all phenomena discussed in this text are considered in inertial frames. 

All the forces discussed in this section are real forces, but there are a number of other real forces, such as lift and thrust, that are not discussed in this 
section. They are more specialized, and it is not necessary to discuss every type of force. It is natural, however, to ask where the basic simplicity we 
seek to find in physics is in the long list of forces. Are some more basic than others? Are some different manifestations of the same underlying force? 
The answer to both questions is yes, as will be seen in the next (extended) section and in the treatment of modern physics later in the text. 



PhET Explorations: Forces in 1 Dimension 



Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting 
on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including 
gravitational and normal forces). 



M 



4m 



j 



PhET Interactive Simulation 



Figure 4.21 Forces in 1 Dimension (http://cnx.Org/content/m42075/l.4/forces-ld_en.jar) 



142 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 

4.6 Problem-Solving Strategies 

Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing 
exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton's laws of 
motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies 
are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop. 

Problem-Solving Strategy for Newton's Laws of Motion 

Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton's laws of motion are involved (if the 
problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 4.22(a). Then, as in 
Figure 4.22(b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent 
(whenever sufficient information exists). 



Th*& force is ncKl a force on 
Ihe system oF internal sines it 
S erarl&d on th& outside woUd. 
It mu&1 be omitled from tha 
free-body diagram 

k 






gj 



I l> 



System of intsres-l 



Free-totfy 

diagram 



J 



Sketch Idemify forces 



(c) 



Oefme syslem of Nerest 



Ttiase forces must be equal 
and opposite since the 
flel external fttte is zert. 
Thus T ■ -w 

W 

Add forces 



Figure 4.22 (a) A sketch of Tarzan hanging from a vine, (b) Arrows are used to represent all forces. T is the tension in the vine above Tarzan, F-p is the force he exerts on 
the vine, and W is his weight. All other forces, such as the nudge of a breeze, are assumed negligible, (c) Suppose we are given the ape man's mass and asked to find the 
tension in the vine. We then define the system of interest as shown and draw a free-body diagram. F-p is no longer shown, because it is not a force acting on the system of 

interest; rather, F-p acts on the outside world, (d) Showing only the arrows, the head-to-tail method of addition is used. It is apparent that T = - W , if Tarzan is stationary. 

Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and 
unknowns. Then carefully determine the system of interest. This decision is a crucial step, since Newton's second law involves only external forces. 
Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step 
to employ Newton's second law. (See Figure 4.22(c).) Newton's third law may be used to identify whether forces are exerted between components of 
a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on 
what question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in 
clearly defining systems will be beneficial in later chapters as well. 

A diagram showing the system of interest and all of the external forces is called a free-body diagram. Only forces are shown on free-body diagrams, 
not acceleration or velocity. We have drawn several of these in worked examples. Figure 4.22(c) shows a free-body diagram for the system of 
interest. Note that no internal forces are shown in a free-body diagram. 

Step 3. Once a free-body diagram is drawn, Newton's second law can be applied to solve the problem. This is done in Figure 4.22(d) for a particular 
situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation 
form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional — that is, if all forces are parallel — then they add 
like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the 
force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, 
when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always 
convenient to make one axis parallel to the direction of motion, if this is known. 



Applying Newton's Second Law 



Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If the acceleration is 
zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is nonzero in a particular direction, then the 
net force is described by the equation: F net = ma . 

For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then you will have the 
following conclusions: 



F nQtx = ma, 



(4.56) 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 143 

(4.57) 
You will need this information in order to determine unknown forces acting in a system. 



^net y = 0. 



Step 4. As always, check the solution to see whether it is reasonable. In some cases, this is obvious. For example, it is reasonable to find that friction 
causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, 
and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the 
units. If you are solving for force and end up with units of m/s, then you have made a mistake. 

4.7 Further Applications of Newton's Laws of Motion 

There are many interesting applications of Newton's laws of motion, a few more of which are presented in this section. These serve also to illustrate 
some further subtleties of physics and to help build problem-solving skills. 



Example 4.7 Drag Force on a Barge 



Suppose two tugboats push on a barge at different angles, as shown in Figure 4.23. The first tugboat exerts a force of 2.7X 10 N in the x- 
direction, and the second tugboat exerts a force of 3.6x 10 N in the y-direction. 




F r = 2.7* KPN 



/f„ 



F. 



F v = 3.6 * 1(T H 




f, y 



i_ 



(a) 



M 



Figure 4.23 (a) A view from above of two tugboats pushing on a barge, (b) The free-body diagram for the ship contains only forces acting in the plane of the water. It 
omits the two vertical forces — the weight of the barge and the buoyant force of the water supporting it cancel and are not shown. Since the applied forces are 
perpendicular, thex- and y-axes are in the same direction as F x and F-y . The problem quickly becomes a one-dimensional problem along the direction of F a pp , 

since friction is in the direction opposite to F a pp . 

If the mass of the barge is 5.0x 10 kg and its acceleration is observed to be 7.5x 10" m/s in the direction shown, what is the drag force 

of the water on the barge resisting the motion? (Note: drag force is a frictional force exerted by fluids, such as air or water. The drag force 
opposes the motion of the object.) 

Strategy 

The directions and magnitudes of acceleration and the applied forces are given in Figure 4.23(a). We will define the total force of the tugboats 
on the barge as F app so that: 



app 



=F X + F 



y 



(4.58) 



Since the barge is flat bottomed, the drag of the water F D will be in the direction opposite to F app , as shown in the free-body diagram in 

Figure 4.23(b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Our strategy is to find the 
magnitude and direction of the net applied force F app , and then apply Newton's second law to solve for the drag force F D . 

Solution 

Since ¥ x and F-y are perpendicular, the magnitude and direction of F app are easily found. First, the resultant magnitude is given by the 
Pythagorean theorem: 



^app — y^x + Fy 



(4.59) 



app 



= V(2.7xl0 :) N) 2 + (3.6xl0 :) N) 2 = 4.5xl(rN. 



The angle is given by 



144 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 

(4.60) 



= tan 
6 = tan 



■if 3.6X10^ = S3 o 



which we know, because of Newton's first law, is the same direction as the acceleration. F D is in the opposite direction of F app , since it acts to 

slow down the acceleration. Therefore, the net external force is in the same direction as F app , but its magnitude is slightly less than F app . The 
problem is now one-dimensional. From Figure 4.23(b), we can see that 

F ne t = F app -FD- < 4 - 61 ) 

But Newton's second law states that 

F nQt = ma. (4.62) 

Thus, 

This can be solved for the magnitude of the drag force of the water F D in terms of known quantities: 

^D = ^app - ma - ( 4 - 64 ) 

Substituting known values gives 

F D = (4.5X 10 5 N) - (5.0X 10 6 kg)(7.5x 1(T 2 m/s 2 ) = 7.5x 10 4 N. (465) 

The direction of F D has already been determined to be in the direction opposite to F app , or at an angle of 53° south of west. 

Discussion 

The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with 
tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low 
speeds, consistent with the answer to this example, where F D is less than l/600th of the weight of the ship. 

In the earlier example of a tightrope walker we noted that the tensions in wires supporting a mass were equal only because the angles on either side 
were equal. Consider the following example, where the angles are not equal; slightly more trigonometry is involved. 



Example 4.8 Different Tensions at Different Angles 



Consider the traffic light (mass 15.0 kg) suspended from two wires as shown in Figure 4.24. Find the tension in each wire, neglecting the 
masses of the wires. 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 145 



Just some or lh* fences are 
shown here. T 

1 i J 




Oniy tc-rces on the system 
ere shown. 



m 







T- 



i 



T. 



w 



(0) 



Free-body diagram 
4 T 



T„-»- 



i- 



I 



The ntf vartloflJ 

force is zero, so 

T, r -T„ = 



w 



*.TV 



w 



The net horizontal 
tor-Do is zero, so 

Tia ■ "Tfc, 



ioj 



Figure 4.24 A traffic light is suspended from two wires, (b) Some of the forces involved, (c) Only forces acting on the system are shown here. The free-body diagram for 
the traffic light is also shown, (d) The forces projected onto vertical (y) and horizontal (x) axes. The horizontal components of the tensions must cancel, and the sum of the 
vertical components of the tensions must equal the weight of the traffic light, (e) The free-body diagram shows the vertical and horizontal forces acting on the traffic light. 

Strategy 

The system of interest is the traffic light, and its free-body diagram is shown in Figure 4.24(c). The three forces involved are not parallel, and so 
they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the 
vector projections on it are shown in part (d) of the figure. There are two unknowns in this problem ( T 1 and T 2 ), so two equations are needed 

to find them. These two equations come from applying Newton's second law along the vertical and horizontal axes, noting that the net external 
force is zero along each axis because acceleration is zero. 

Solution 



First consider the horizontal or x-axis: 



Thus, as you might expect, 



[ 2x' 



lx 



0. 



T lx - T 2x- 



(4.66) 



(4.67) 



This gives us the following relationship between T 1 and T 2 : 

T x cos (30°) = T 2 cos (45°). (4.68) 

Thus, 

T 2 = (1.225)^. (4.69) 

Note that Ti and T 2 are not equal in this case, because the angles on either side are not equal. It is reasonable that T 2 ends up being 
greater than T± , because it is exerted more vertically than T± . 

Now consider the force components along the vertical or y-axis: 

F mty = T ly + T 2y -w = 0. (4.70) 



146 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 



This implies 


T ly + T 2y = w - 








(4.71) 


Substituting the expressions for the vertical components gives 












T x sin (30°) + T 2 sin (45°) 


= w. 






(4.72) 


There are two unknowns in this 


equation, but substituting the expression for T 2 in terms of T x reduces this to 


one equation with 


one 


unknown: 




7^(0.500) + (1.2257^X0.707) = 


w = mg, 






(4.73) 


which yields 














(1.366)7^ = (15.0 kg)(9.80 m/s 2 ). 






(4.74) 


Solving this last equation gives 


the magnitude of T l to be 

T x = 108 N. 








(4.75) 


Finally, the magnitude of T 2 is 


determined using the relationship between them, 

T 2 = 132 N. 


T 2 = 1.225 T x , found above 


Thus we obtain 




(4.76) 


Discussion 












Both tensions would be larger if both wires were more horizontal, and they will be 
they were in the earlier example of a tightrope walker). 


equal if and only if the angles on either side are 


the 


same (as 



The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward 
to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? 
Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed: will the 
scale still read more than your weight at rest? Consider the following example. 



Example 4.9 What Does the Bathroom Scale Read in an Elevator'* 



Figure 4.25 shows a 75.0-kg man (weight of about 165 lb) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the 
elevator accelerates upward at a rate of 1.20 m/s , and (b) if the elevator moves upward at a constant speed of 1 m/s. 



1 



fi2 



Fjf 






k" 



F- 




System of Interest 

Free-body diagram 



k 



y 



(a) (b) 

Figure 4.25 (a) The various forces acting when a person stands on a bathroom scale in an elevator. The arrows are approximately correct for when the elevator is 
accelerating upward — broken arrows represent forces too large to be drawn to scale. T is the tension in the supporting cable, W is the weight of the person, W s is 

the weight of the scale, W e is the weight of the elevator, F s is the force of the scale on the person, Fp is the force of the person on the scale, F t is the force of the 

scale on the floor of the elevator, and N is the force of the floor upward on the scale, (b) The free-body diagram shows only the external forces acting on the designated 
system of interest — the person. 



Strategy 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 147 

If the scale is accurate, its reading will equal Fp , the magnitude of the force the person exerts downward on it. Figure 4.25(a) shows the 

numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person 
is chosen to be the system of interest and a free-body diagram is drawn as in Figure 4.25(b). Analysis of the free-body diagram using Newton's 
laws can produce answers to both parts (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on 
the person are his weight w and the upward force of the scale F s . According to Newton's third law F p and F s are equal in magnitude and 

opposite in direction, so that we need to find F s in order to find what the scale reads. We can do this, as usual, by applying Newton's second 
law, 

i^net = rna. (4.77) 

From the free-body diagram we see that F net = F s — w , so that 

F s - w = ma. (4.78) 



Solving for F s gives an equation with only one unknown: 
or, because w = mg , simply 



F s = ma + w, (4.79) 



F s = ma + mg. (4.80) 



No assumptions were made about the acceleration, and so this solution should be valid for a variety of accelerations in addition to the ones in 
this exercise. 

Solution for (a) 

In this part of the problem, a = 1.20 m/s , so that 

F s = (75.0 kg)(1.20 m/s 2 ) + (75.0 kg)(9.80 m/s 2 ), ( 4 - 81 ) 

yielding 

F s = 825 N. (4.82) 

Discussion for (a) 

This is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be 
equal to his weight: 

F nQt = ma = = F s -w (4.83) 

F s = w = mg 

F s = (75.0 kg)(9.80 m/s 2 ) 

F s = 735 N. 

So, the scale reading in the elevator is greater than his 735-N (165 lb) weight. This means that the scale is pushing up on the person with a force 
greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale 
reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. 

Solution for (b) 

Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant 
velocity — up, down, or stationary — acceleration is zero because a = -^ , and Av = . 

Thus, 

F s = ma + mg = + mg. (4.84) 

Now 

F s = (75.0 kg)(9.80 m/s 2 ), ( 4 - 85 ) 

which gives 

F s = 735 N. (4.86) 

Discussion for (b) 

The scale reading is 735 N, which equals the person's weight. This will be the case whenever the elevator has a constant velocity — moving up, 
moving down, or stationary. 

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a 
is negative, and the scale reading is less than the weight of the person, until a constant downward velocity is reached, at which time the scale reading 



148 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 

again becomes equal to the person's weight. If the elevator is in free-fall and accelerating downward at g , then the scale reading will be zero and the 
person will appear to be weightless. 

Integrating Concepts: Newton's Laws of Motion and Kinematics 

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton's 
laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, 
forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters. When approaching problems that involve various 
types of forces, acceleration, velocity, and/or position, use the following steps to approach the problem: 

Problem-Solving Strategy 

Step 1. Identify which physical principles are involved. Listing the givens and the quantities to be calculated will allow you to identify the principles 

involved. 

Step 2. Solve the problem using strategies outlined in the text. If these are available for the specific topic, you should refer to them. You should also 

refer to the sections of the text that deal with a particular topic. The following worked example illustrates how these strategies are applied to an 

integrated concept problem. 



Example 4.10 What Force Must a Soccer Player Exert to Reach Top Speech 



A soccer player starts from rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What was his average acceleration? (b) 
What average force did he exert backward on the ground to achieve this acceleration? The player's mass is 70.0 kg, and air resistance is 
negligible. 

Strategy 

1. To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters in which they are 
found. Part (a) of this example considers acceleration along a straight line. This is a topic of kinematics. Part (b) deals with force, a topic of 
dynamics found in this chapter. 

2. The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve 
identifying knowns and unknowns, checking to see if the answer is reasonable, and so forth. 

Solution for (a) 

We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Av = 8.00 m/s . We are given the 
elapsed time, and so At = 2.50 s . The unknown is acceleration, which can be found from its definition: 



n _ Av (4.87) 

At' 



Substituting the known values yields 



n _ 8.00 m/s (4.88) 

2.50 s 

= 3.20 m/s 2 . 
Discussion for (a) 

This is an attainable acceleration for an athlete in good condition. 

Solution for (b) 

Here we are asked to find the average force the player exerts backward to achieve this forward acceleration. Neglecting air resistance, this 
would be equal in magnitude to the net external force on the player, since this force causes his acceleration. Since we now know the player's 
acceleration and are given his mass, we can use Newton's second law to find the force exerted. That is, 



Substituting the known values of m and a gives 



^net — 



(70.0kg)(3.20m/s z ) 

= 224 N. 



(4.89) 



(4.90) 



Discussion for (b) 

This is about 50 pounds, a reasonable average force. 

This worked example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is 
to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. 
These strategies are found throughout the text, and many worked examples show how to use them for single topics. You will find these 
techniques for integrated concept problems useful in applications of physics outside of a physics course, such as in your profession, in other 
science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles. 



4.8 Extended Topic: The Four Basic Forces — An Introduction 

One of the most remarkable simplifications in physics is that only four distinct forces account for all known phenomena. In fact, nearly all of the forces 
we experience directly are due to only one basic force, called the electromagnetic force. (The gravitational force is the only force we experience 
directly that is not electromagnetic.) This is a tremendous simplification of the myriad of apparently different forces we can list, only a few of which 
were discussed in the previous section. As we will see, the basic forces are all thought to act through the exchange of microscopic carrier particles, 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 149 

and the characteristics of the basic forces are determined by the types of particles exchanged. Action at a distance, such as the gravitational force of 
Earth on the Moon, is explained by the existence of a force field rather than by "physical contact." 

The four basic forces are the gravitational force, the electromagnetic force, the weak nuclear force, and the strong nuclear force. Their properties are 
summarized in Table 4.1. Since the weak and strong nuclear forces act over an extremely short range, the size of a nucleus or less, we do not 
experience them directly, although they are crucial to the very structure of matter. These forces determine which nuclei are stable and which decay, 
and they are the basis of the release of energy in certain nuclear reactions. Nuclear forces determine not only the stability of nuclei, but also the 
relative abundance of elements in nature. The properties of the nucleus of an atom determine the number of electrons it has and, thus, indirectly 
determine the chemistry of the atom. More will be said of all of these topics in later chapters. 

Concept Connections: Th e Four B asic Forces 

The four basic forces will be encountered in more detail as you progress through the text. The gravitational force is defined in Uniform Circular 
Motion and Gravitation, electric force in Electric Charge and Electric Field, magnetic force in Magnetism, and nuclear forces in 
Radioactivity and Nuclear Physics. On a macroscopic scale, electromagnetism and gravity are the basis for all forces. The nuclear forces are 
vital to the substructure of matter, but they are not directly experienced on the macroscopic scale. 



Table 4.1 Properties of the Four Basic Forces 1 



[i] 







Approximate Relative Strengths 


Range Attraction/Repulsion 


Carrier Particle 


Gravitational 


1Q -38 


CO 


attractive only 


Graviton 


Electromagnetic 


io- 2 


CO 


attractive and repulsive 


Photon 


Weak nuclear 


10 " 13 


< 10" 18 m 


attractive and repulsive 


W + , W" , z° 


Strong nuclear 


1 


< l(T 15 m 


attractive and repulsive 


gluons 



The gravitational force is surprisingly weak — it is only because gravity is always attractive that we notice it at all. Our weight is the gravitational force 
due to the entire Earth acting on us. On the very large scale, as in astronomical systems, the gravitational force is the dominant force determining the 
motions of moons, planets, stars, and galaxies. The gravitational force also affects the nature of space and time. As we shall see later in the study of 
general relativity, space is curved in the vicinity of very massive bodies, such as the Sun, and time actually slows down near massive bodies. 

Electromagnetic forces can be either attractive or repulsive. They are long-range forces, which act over extremely large distances, and they nearly 
cancel for macroscopic objects. (Remember that it is the net external force that is important.) If they did not cancel, electromagnetic forces would 
completely overwhelm the gravitational force. The electromagnetic force is a combination of electrical forces (such as those that cause static 
electricity) and magnetic forces (such as those that affect a compass needle). These two forces were thought to be quite distinct until early in the 19th 
century, when scientists began to discover that they are different manifestations of the same force. This discovery is a classical case of the unification 
of forces. Similarly, friction, tension, and all of the other classes of forces we experience directly (except gravity, of course) are due to electromagnetic 
interactions of atoms and molecules. It is still convenient to consider these forces separately in specific applications, however, because of the ways 
they manifest themselves. 



Concept Connections: Unifying Forces 

Attempts to unify the four basic forces are discussed in relation to elementary particles later in this text. By "unify" we mean finding connections 
between the forces that show that they are different manifestations of a single force. Even if such unification is achieved, the forces will retain 
their separate characteristics on the macroscopic scale and may be identical only under extreme conditions such as those existing in the early 
universe. 

Physicists are now exploring whether the four basic forces are in some way related. Attempts to unify all forces into one come under the rubric of 
Grand Unified Theories (GUTs), with which there has been some success in recent years. It is now known that under conditions of extremely high 
density and temperature, such as existed in the early universe, the electromagnetic and weak nuclear forces are indistinguishable. They can now be 
considered to be different manifestations of one force, called the electroweak force. So the list of four has been reduced in a sense to only three. 
Further progress in unifying all forces is proving difficult — especially the inclusion of the gravitational force, which has the special characteristics of 
affecting the space and time in which the other forces exist. 

While the unification of forces will not affect how we discuss forces in this text, it is fascinating that such underlying simplicity exists in the face of the 
overt complexity of the universe. There is no reason that nature must be simple — it simply is. 



Action at a Distance: Concept of a Field 

All forces act at a distance. This is obvious for the gravitational force. Earth and the Moon, for example, interact without coming into contact. It is also 
true for all other forces. Friction, for example, is an electromagnetic force between atoms that may not actually touch. What is it that carries forces 
between objects? One way to answer this question is to imagine that a force field surrounds whatever object creates the force. A second object 
(often called a test object) placed in this field will experience a force that is a function of location and other variables. The field itself is the "thing" that 
carries the force from one object to another. The field is defined so as to be a characteristic of the object creating it; the field does not depend on the 
test object placed in it. Earth's gravitational field, for example, is a function of the mass of Earth and the distance from its center, independent of the 



1. The graviton is a proposed particle, though it has not yet been observed by scientists. See the discussion of gravitational waves later in this section. 
The particles W + , W~ , and Z are called vector bosons; these were predicted by theory and first observed in 1983. There are eight types of 
gluons proposed by scientists, and their existence is indicated by meson exchange in the nuclei of atoms. 



150 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 

presence of other masses. The concept of a field is useful because equations can be written for force fields surrounding objects (for gravity, this 
yields w = mg at Earth's surface), and motions can be calculated from these equations. (See Figure 4.26.) 




Figure 4.26 The electric force field between a positively charged particle and a negatively charged particle. When a positive test charge is placed in the field, the charge will 
experience a force in the direction of the force field lines. 

Concept Connections: Force Fields 



The concept of a force field is also used in connection with electric charge and is presented in Electric Charge and Electric Field. It is also a 
useful idea for all the basic forces, as will be seen in Particle Physics. Fields help us to visualize forces and how they are transmitted, as well as 
to describe them with precision and to link forces with subatomic carrier particles. 



The field concept has been applied very successfully; we can calculate motions and describe nature to high precision using field equations. As useful 
as the field concept is, however, it leaves unanswered the question of what carries the force. It has been proposed in recent decades, starting in 1935 
with Hideki Yukawa's (1907-1981) work on the strong nuclear force, that all forces are transmitted by the exchange of elementary particles. We can 
visualize particle exchange as analogous to macroscopic phenomena such as two people passing a basketball back and forth, thereby exerting a 
repulsive force without touching one another. (See Figure 4.27.] 




trim 



i*\ 






((&- 



Direction of 

■■::>:■■ 



meson 



M 

Figure 4.27 The exchange of masses resulting in repulsive forces, (a) The person throwing the basketball exerts a force F„^ on it toward the other person and feels a 

reaction force Fg away from the second person, (b) The person catching the basketball exerts a force F p 2 on it to stop the ball and feels a reaction force F g away from 

the first person, (c) The analogous exchange of a meson between a proton and a neutron carries the strong nuclear forces Fg^ and F exc j 1 between them. An attractive 

force can also be exerted by the exchange of a mass — if person 2 pulled the basketball away from the first person as he tried to retain it, then the force between them would 
be attractive. 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 151 



This idea of particle exchange deepens rather than contradicts field concepts. It is more satisfying philosophically to think of something physical 
actually moving between objects acting at a distance. Table 4.1 lists the exchange or carrier particles, both observed and proposed, that carry the 
four forces. But the real fruit of the particle-exchange proposal is that searches for Yukawa's proposed particle found it and a number of others that 
were completely unexpected, stimulating yet more research. All of this research eventually led to the proposal of quarks as the underlying 
substructure of matter, which is a basic tenet of GUTs. If successful, these theories will explain not only forces, but also the structure of matter itself. 
Yet physics is an experimental science, so the test of these theories must lie in the domain of the real world. As of this writing, scientists at the CERN 
laboratory in Switzerland are starting to test these theories using the world's largest particle accelerator: the Large Hadron Collider. This accelerator 
(27 km in circumference) allows two high-energy proton beams, traveling in opposite directions, to collide. An energy of 14 million electron volts will 
be available. It is anticipated that some new particles, possibly force carrier particles, will be found. (See Figure 4.28.) One of the force carriers of 
high interest that researchers hope to detect is the Higgs boson. The observation of its properties might tell us why different particles have different 
masses. 



jHHM|i*n;: 


-s; •mm l n , m^^' t *v- 


* "sjjSBI 









Figure 4.28 The world's largest particle accelerator spans the border between Switzerland and France. Two beams, traveling in opposite directions close to the speed of light, 
collide in a tube similar to the central tube shown here. External magnets determine the beam's path. Special detectors will analyze particles created in these collisions. 
Questions as broad as what is the origin of mass and what was matter like the first few seconds of our universe will be explored. This accelerator began preliminary operation 
in 2008. (credit: Frank Hommes) 

Tiny particles also have wave-like behavior, something we will explore more in a later chapter. To better understand force-carrier particles from 
another perspective, let us consider gravity. The search for gravitational waves has been going on for a number of years. Almost 100 years ago, 
Einstein predicted the existence of these waves as part of his general theory of relativity. Gravitational waves are created during the collision of 
massive stars, in black holes, or in supernova explosions — like shock waves. These gravitational waves will travel through space from such sites 
much like a pebble dropped into a pond sends out ripples — except these waves move at the speed of light. A detector apparatus has been built in the 
U.S., consisting of two large installations nearly 3000 km apart — one in Washington state and one in Louisiana! The facility is called the Laser 
Interferometer Gravitational-Wave Observatory (LIGO). Each installation is designed to use optical lasers to examine any slight shift in the relative 
positions of two masses due to the effect of gravity waves. The two sites allow simultaneous measurements of these small effects to be separated 
from other natural phenomena, such as earthquakes. Initial operation of the detectors began in 2002, and work is proceeding on increasing their 
sensitivity. Similar installations have been built in Italy (VIRGO), Germany (GEO600), and Japan (TAMA300) to provide a worldwide network of 
gravitational wave detectors. 

International collaboration in this area is moving into space with the joint EU/US project LISA (Laser Interferometer Space Antenna). Earthquakes and 
other Earthly noises will be no problem for these monitoring spacecraft. LISA will complement LIGO by looking at much more massive black holes 
through the observation of gravitational-wave sources emitting much larger wavelengths. Three satellites will be placed in space above Earth in an 
equilateral triangle (with 5,000,000-km sides) (Figure 4.29). The system will measure the relative positions of each satellite to detect passing 
gravitational waves. Accuracy to within 10% of the size of an atom will be needed to detect any waves. The launch of this project might be as early as 
2018. 

"I'm sure LIGO will tell us something about the universe that we didn't know before. The history of science tells us that any time you go where you 
haven't been before, you usually find something that really shakes the scientific paradigms of the day. Whether gravitational wave astrophysics will do 
that, only time will tell." — David Reitze, LIGO Input Optics Manager, University of Florida 







'-—--«» ^ 4 "^*ii 










^ 


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aa 


„arr* 


^ESb^B 




■■ 


K^£? - 


S|2 


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Bk~ 






■■ 


mmSm 





Figure 4.29 Space-based future experiments for the measurement of gravitational waves. Shown here is a drawing of LISA'S orbit. Each satellite of LISA will consist of a laser 
source and a mass. The lasers will transmit a signal to measure the distance between each satellite's test mass. The relative motion of these masses will provide information 
about passing gravitational waves, (credit: NASA) 

The ideas presented in this section are but a glimpse into topics of modern physics that will be covered in much greater depth in later chapters. 



152 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 



Glossary 



acceleration: the rate at which an object's velocity changes over a period of time 

carrier particle: a fundamental particle of nature that is surrounded by a characteristic force field; photons are carrier particles of the 
electromagnetic force 

dynamics: the study of how forces affect the motion of objects and systems 

external force: a force acting on an object or system that originates outside of the object or system 

force field: a region in which a test particle will experience a force 

force: a push or pull on an object with a specific magnitude and direction; can be represented by vectors; can be expressed as a multiple of a 
standard force 

free-body diagram: a sketch showing all of the external forces acting on an object or system; the system is represented by a dot, and the forces 
are represented by vectors extending outward from the dot 

free-fall: a situation in which the only force acting on an object is the force due to gravity 

friction: a force past each other of objects that are touching; examples include rough surfaces and air resistance 

inertia: the tendency of an object to remain at rest or remain in motion 

inertial frame of reference: a coordinate system that is not accelerating; all forces acting in an inertial frame of reference are real forces, as 
opposed to fictitious forces that are observed due to an accelerating frame of reference 

law of inertia: see Newton's first law of motion 

mass: the quantity of matter in a substance; measured in kilograms 

Newton's first law of motion: a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net 
external force; also known as the law of inertia 

Newton's second law of motion: the net external force F net on an object with mass m is proportional to and in the same direction as the 

F 

acceleration of the object, a , and inversely proportional to the mass; defined mathematically as a = — ^ 

Newton's third law of motion: whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude 
and opposite in direction to the force that the first body exerts 

net external force: the vector sum of all external forces acting on an object or system; causes a mass to accelerate 

normal force: the force that a surface applies to an object to support the weight of the object; acts perpendicular to the surface on which the 
object rests 

system: defined by the boundaries of an object or collection of objects being observed; all forces originating from outside of the system are 
considered external forces 

tension: the pulling force that acts along a medium, especially a stretched flexible connector, such as a rope or cable; when a rope supports the 
weight of an object, the force on the object due to the rope is called a tension force 

thrust: a reaction force that pushes a body forward in response to a backward force; rockets, airplanes, and cars are pushed forward by a thrust 
reaction force 

weight: the force w due to gravity acting on an object of mass m ; defined mathematically as: w = mg , where g is the magnitude and direction 
of the acceleration due to gravity 



Section Summary 



4.1 Development of Force Concept 

• Dynamics is the study of how forces affect the motion of objects. 

• Force is a push or pull that can be defined in terms of various standards, and it is a vector having both magnitude and direction. 

• External forces are any outside forces that act on a body. A free-body diagram is a drawing of all external forces acting on a body. 

4.2 Newton's First Law of Motion: Inertia 

• Newton's first law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on 
by a net external force. This is also known as the law of inertia. 

• Inertia is the tendency of an object to remain at rest or remain in motion. Inertia is related to an object's mass. 

• Mass is the quantity of matter in a substance. 

4.3 Newton's Second Law of Motion: Concept of a System 

• Acceleration, a , is defined as a change in velocity, meaning a change in its magnitude or direction, or both. 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 153 

• An external force is one acting on a system from outside the system, as opposed to internal forces, which act between components within the 
system. 

• Newton's second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external 
force acting on the system, and inversely proportional to its mass. 

F 

• In equation form, Newton's second law of motion is a = — ^ . 

• This is often written in the more familiar form: F net = ma . 

• The weight w of an object is defined as the force of gravity acting on an object of mass m . The object experiences an acceleration due to 
gravity g : 

w = mg. 

• If the only force acting on an object is due to gravity, the object is in free fall. 

• Friction is a force that opposes the motion past each other of objects that are touching. 

4.4 Newton's Third Law of Motion: Symmetry in Forces 

• Newton's third law of motion represents a basic symmetry in nature. It states: Whenever one body exerts a force on a second body, the first 
body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts. 

• A thrust is a reaction force that pushes a body forward in response to a backward force. Rockets, airplanes, and cars are pushed forward by a 
thrust reaction force. 

4.5 Normal, Tension, and Other Examples of Forces 

• When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting force acts 
perpendicular to and away from the surface. It is called a normal force, N . 

• When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object: 

N = mg. 

• When objects rest on an inclined plane that makes an angle 6 with the horizontal surface, the weight of the object can be resolved into 
components that act perpendicular ( w_|_ ) and parallel ( w n ) to the surface of the plane. These components can be calculated using: 

w II = w sin (6) = mg sin (6) 

w_ L = w cos (6) = mg cos (6). 

• The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, T . When a rope supports the weight 
of an object that is at rest, the tension in the rope is equal to the weight of the object: 

T = mg. 

• In any inertial frame of reference (one that is not accelerated or rotated), Newton's laws have the simple forms given in this chapter and all 
forces are real forces having a physical origin. 

4.6 Problem-Solving Strategies 

• To solve problems involving Newton's laws of motion, follow the procedure described: 

1. Draw a sketch of the problem. 

2. Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the 
forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions 
from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and 
draw them on the free-body diagram. 

3. Write Newton's second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not 
accelerate in a particular direction (for example, the x -direction) then F mtx = . If the object does accelerate in that direction, 

F nQtx = ma. 

4. Check your answer. Is the answer reasonable? Are the units correct? 

4.7 Further Applications of Newton's Laws of Motion 

• Newton's laws of motion can be applied in numerous situations to solve problems of motion. 

• Some problems will contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors 
into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you 
can determine whether F nQt = ma or F nQt = . 

• The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating, the normal force will be 
less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force will always be less than the full 
weight of the object. 

• Some problems will contain various physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from 
kinematics and dynamics in order to solve these problems of motion. 

4.8 Extended Topic: The Four Basic Forces — An Introduction 

• The various types of forces that are categorized for use in many applications are all manifestations of the four basic forces in nature. 

• The properties of these forces are summarized in Table 4.1. 

• Everything we experience directly without sensitive instruments is due to either electromagnetic forces or gravitational forces. The nuclear 
forces are responsible for the submicroscopic structure of matter, but they are not directly sensed because of their short ranges. Attempts are 
being made to show all four forces are different manifestations of a single unified force. 

• A force field surrounds an object creating a force and is the carrier of that force. 



154 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 



Conceptual Questions 



4.1 Development of Force Concept 

1. Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the 
same force repeatedly. 

2. What properties do forces have that allow us to classify them as vectors? 

4.2 Newton's First Law of Motion: Inertia 

3. How are inertia and mass related? 

4. What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body? 

4.3 Newton's Second Law of Motion: Concept of a System 

5. Which statement is correct? (a) Net force causes motion, (b) Net force causes change in motion. Explain your answer and give an example. 

6. Why can we neglect forces such as those holding a body together when we apply Newton's second law of motion? 

7. Explain how the choice of the "system of interest" affects which forces must be considered when applying Newton's second law of motion. 

8. Describe a situation in which the net external force on a system is not zero, yet its speed remains constant. 

9. A system can have a nonzero velocity while the net external force on it is zero. Describe such a situation. 

10. A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory? 

11. (a) Give an example of different net external forces acting on the same system to produce different accelerations, (b) Give an example of the 
same net external force acting on systems of different masses, producing different accelerations, (c) What law accurately describes both effects? 
State it in words and as an equation. 

12. If the acceleration of a system is zero, are no external forces acting on it? What about internal forces? Explain your answers. 

13. If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the object? 

14. The gravitational force on the basketball in Figure 4.6 is ignored. When gravity is taken into account, what is the direction of the net external force 
on the basketball — above horizontal, below horizontal, or still horizontal? 

4.4 Newton's Third Law of Motion: Symmetry in Forces 

15. When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat — is there 
really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.) 

16. A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the "ballistocardiograph." What physics principle(s) 
are involved here to measure the force of cardiac contraction? How might we construct such a device? 

17. Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal in magnitude and 
opposite in direction. Which of Newton's laws of motion apply? 

18. Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe how Newton's third law 
applies when one is fired. Can you safely stand close behind one when it is fired? 

19. An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will 
experience an equal and opposite force from the other player. Use Newton's laws and draw a free-body diagram of an appropriate system to explain 
how he can still out-push the opposition if he is strong enough. 

20. Newton's third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the "system of 
interest" affects whether one such pair of forces cancels. 

4.5 Normal, Tension, and Other Examples of Forces 

21. If a leg is suspended by a traction setup as shown in Figure 4.30, what is the tension in the rope? 

@ ® 







CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 155 

Figure 4.30 A leg is suspended by a traction system in which wires are used to transmit forces. Frictionless pulleys change the direction of the force T without changing its 
magnitude. 

22. In a traction setup for a broken bone, with pulleys and rope available, how might we be able to increase the force along the femur using the same 
weight? (See Figure 4.30.) (Note that the femur is the shin bone shown in this image.) 

4.7 Further Applications of Newton's Laws of Motion 

23. To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at g . 

Why will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference 
between their apparent weightlessness in orbit and in the aircraft? 

24. A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward ride. Can this really 
happen without the person being tied to the floor of the elevator? Explain your answer. 

4.8 Extended Topic: The Four Basic Forces — An Introduction 

25. Explain, in terms of the properties of the four basic forces, why people notice the gravitational force acting on their bodies if it is such a 
comparatively weak force. 

26. What is the dominant force between astronomical objects? Why are the other three basic forces less significant over these very large distances? 

27. Give a detailed example of how the exchange of a particle can result in an attractive force. (For example, consider one child pulling a toy out of 
the hands of another.) 



156 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 



Problems & Exercises 



4.3 Newton's Second Law of Motion: Concept of a 
System 

You may assume data taken from illustrations is accurate to three 
digits. 

1. A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s . 
What is the net external force on him? 

2. If the sprinter from the previous problem accelerates at that rate for 20 
m, and then maintains that velocity for the remainder of the 100-m dash, 
what will be his time for the race? 

3. A cleaner pushes a 4.50-kg laundry cart in such a way that the net 
external force on it is 60.0 N. Calculate its acceleration. 

4. Since astronauts in orbit are apparently weightless, a clever method of 
measuring their masses is needed to monitor their mass gains or losses 
to adjust diets. One way to do this is to exert a known force on an 
astronaut and measure the acceleration produced. Suppose a net 
external force of 50.0 N is exerted and the astronaut's acceleration is 

measured to be 0.893 m/s . (a) Calculate her mass, (b) By exerting a 
force on the astronaut, the vehicle in which they orbit experiences an 
equal and opposite force. Discuss how this would affect the 
measurement of the astronaut's acceleration. Propose a method in which 
recoil of the vehicle is avoided. 

5. In Figure 4.7, the net external force on the 24-kg mower is stated to be 
51 N. If the force of friction opposing the motion is 24 N, what force F (in 
newtons) is the person exerting on the mower? Suppose the mower is 
moving at 1.5 m/s when the force F is removed. How far will the mower 
go before stopping? 

6. The same rocket sled drawn in Figure 4.31 is decelerated at a rate of 
196 m/s . What force is necessary to produce this deceleration? 

Assume that the rockets are off. The mass of the system is 2100 kg. 




Figure 4.31 

7. (a) If the rocket sled shown in Figure 4.32 starts with only one rocket 
burning, what is its acceleration? Assume that the mass of the system is 
2100 kg, and the force of friction opposing the motion is known to be 650 
N. (b) Why is the acceleration not one-fourth of what it is with all rockets 
burning? 




Figure 4.32 

8. What is the deceleration of the rocket sled if it comes to rest in 1.1 s 
from a speed of 1000 km/h? (Such deceleration caused one test subject 
to black out and have temporary blindness.) 

9. Suppose two children push horizontally, but in exactly opposite 
directions, on a third child in a wagon. The first child exerts a force of 
75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of 
the third child plus wagon is 23.0 kg. (a) What is the system of interest if 
the acceleration of the child in the wagon is to be calculated? (b) Draw a 



free-body diagram, including all forces acting on the system, (c) Calculate 
the acceleration, (d) What would the acceleration be if friction were 15.0 
N? 

10. A powerful motorcycle can produce an acceleration of 3.50 m/s 
while traveling at 90.0 km/h. At that speed the forces resisting motion, 
including friction and air resistance, total 400 N. (Air resistance is 
analogous to air friction. It always opposes the motion of an object.) What 
force does the motorcycle exert backward on the ground to produce its 
acceleration if the mass of the motorcycle with rider is 245 kg? 

11. The rocket sled shown in Figure 4.33 accelerates at a rate of 
49.0 m/s . Its passenger has a mass of 75.0 kg. (a) Calculate the 
horizontal component of the force the seat exerts against his body. 
Compare this with his weight by using a ratio, (b) Calculate the direction 
and magnitude of the total force the seat exerts against his body. 




Figure 4.33 

12. Repeat the previous problem for the situation in which the rocket sled 
decelerates at a rate of 201 m/s . In this problem, the forces are 
exerted by the seat and restraining belts. 

13. The weight of an astronaut plus his space suit on the Moon is only 
250 N. How much do they weigh on Earth? What is the mass on the 
Moon? On Earth? 

14. Suppose the mass of a fully loaded module in which astronauts take 
off from the Moon is 10,000 kg. The thrust of its engines is 30,000 N. (a) 
Calculate its acceleration in a vertical takeoff from the Moon, (b) Could it 
lift off from Earth? If not, why not? If it could, calculate its acceleration. 

4.4 Newton's Third Law of Motion: Symmetry in Forces 

15. What net external force is exerted on a 1100-kg artillery shell fired 
from a battleship if the shell is accelerated at 2.40xl0 4 m/s 2 ? What 
force is exerted on the ship by the artillery shell? 

16. A brave but inadequate rugby player is being pushed backward by an 
opposing player who is exerting a force of 800 N on him. The mass of the 
losing player plus equipment is 90.0 kg, and he is accelerating at 

1.20 m/s backward, (a) What is the force of friction between the losing 
player's feet and the grass? (b) What force does the winning player exert 
on the ground to move forward if his mass plus equipment is 110 kg? (c) 
Draw a sketch of the situation showing the system of interest used to 
solve each part. For this situation, draw a free-body diagram and write 
the net force equation. 

4.5 Normal, Tension, and Other Examples of Forces 

17. Two teams of nine members each engage in a tug of war. Each of the 
first team's members has an average mass of 68 kg and exerts an 
average force of 1350 N horizontally. Each of the second team's 
members has an average mass of 73 kg and exerts an average force of 
1365 N horizontally, (a) What is the acceleration of the two teams? (b) 
What is the tension in the section of rope between the teams? 

18. What force does a trampoline have to apply to a 45.0-kg gymnast to 
accelerate her straight up at 7.50 m/s ? Note that the answer is 
independent of the velocity of the gymnast — she can be moving either up 
or down, or be stationary. 

19. (a) Calculate the tension in a vertical strand of spider web if a spider 

of mass 8.00x10 kg hangs motionless on it. (b) Calculate the 

tension in a horizontal strand of spider web if the same spider sits 
motionless in the middle of it much like the tightrope walker in Figure 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 157 



4.17. The strand sags at an angle of 12° below the horizontal. Compare 
this with the tension in the vertical strand (find their ratio). 

20. Suppose a 60.0-kg gymnast climbs a rope, (a) What is the tension in 
the rope if he climbs at a constant speed? (b) What is the tension in the 

rope if he accelerates upward at a rate of 1.50 m/s ? 

21. Show that, as stated in the text, a force Fj_ exerted on a flexible 

medium at its center and perpendicular to its length (such as on the 
tightrope wire in Figure 4.17) gives rise to a tension of magnitude 

T = 



2 sin (0) 



22. Consider the baby being weighed in Figure 4.34. (a) What is the 
mass of the child and basket if a scale reading of 55 N is observed? (b) 
What is the tension T 1 in the cord attaching the baby to the scale? (c) 

What is the tension T 2 in the cord attaching the scale to the ceiling, if 

the scale has a mass of 0.500 kg? (d) Draw a sketch of the situation 
indicating the system of interest used to solve each part. The masses of 
the cords are negligible. 




Figure 4.34 A baby is weighed using a spring scale. 



4.6 Problem-Solving Strategies 

23. A 5.00x10 -kg rocket is accelerating straight up. Its engines 

produce 1.250X10 7 N of thrust, and air resistance is 4.50xl0 6 N. 
What is the rocket's acceleration? Explicitly show how you follow the 
steps in the Problem-Solving Strategy for Newton's laws of motion. 

24. The wheels of a midsize car exert a force of 2100 N backward on the 
road to accelerate the car in the forward direction. If the force of friction 
including air resistance is 250 N and the acceleration of the car is 

1.80 m/s , what is the mass of the car plus its occupants? Explicitly 
show how you follow the steps in the Problem-Solving Strategy for 
Newton's laws of motion. For this situation, draw a free-body diagram 
and write the net force equation. 



25. Calculate the force a 70.0-kg high jumper must exert on the ground to 
produce an upward acceleration 4.00 times the acceleration due to 
gravity. Explicitly show how you follow the steps in the Problem-Solving 
Strategy for Newton's laws of motion. 

26. When landing after a spectacular somersault, a 40.0-kg gymnast 
decelerates by pushing straight down on the mat. Calculate the force she 
must exert if her deceleration is 7.00 times the acceleration due to 
gravity. Explicitly show how you follow the steps in the Problem-Solving 
Strategy for Newton's laws of motion. 

27. A freight train consists of two 8.00X 10 -kg engines and 45 cars 

with average masses of 5.50x10 kg . (a) What force must each 
engine exert backward on the track to accelerate the train at a rate of 
5.00X 10" 2 m/s 2 if the force of friction is 7.50x 10 5 N , assuming the 
engines exert identical forces? This is not a large frictional force for such 
a massive system. Rolling friction for trains is small, and consequently 
trains are very energy-efficient transportation systems, (b) What is the 
force in the coupling between the 37th and 38th cars (this is the force 
each exerts on the other), assuming all cars have the same mass and 
that friction is evenly distributed among all of the cars and engines? 

28. Commercial airplanes are sometimes pushed out of the passenger 
loading area by a tractor, (a) An 1800-kg tractor exerts a force of 

1.75x10 N backward on the pavement, and the system experiences 
forces resisting motion that total 2400 N. If the acceleration is 

0.150 m/s , what is the mass of the airplane? (b) Calculate the force 
exerted by the tractor on the airplane, assuming 2200 N of the friction is 
experienced by the airplane, (c) Draw two sketches showing the systems 
of interest used to solve each part, including the free-body diagrams for 
each. 

29. A 1100-kg car pulls a boat on a trailer, (a) What total force resists the 
motion of the car, boat, and trailer, if the car exerts a 1900-N force on the 

road and produces an acceleration of 0.550 m/s ? The mass of the 
boat plus trailer is 700 kg. (b) What is the force in the hitch between the 
car and the trailer if 80% of the resisting forces are experienced by the 
boat and trailer? 

30. (a) Find the magnitudes of the forces F 1 and F 2 that add to give 

the total force F tot shown in Figure 4.35. This may be done either 

graphically or by using trigonometry, (b) Show graphically that the same 
total force is obtained independent of the order of addition of F 1 and 

F 2 . (c) Find the direction and magnitude of some other pair of vectors 

that add to give F tot . Draw these to scale on the same drawing used in 

part (b) or a similar picture. 



F M = 20 N 




Free-body diagram 



F, 



Figure 4.35 



31. Two children pull a third child on a snow saucer sled exerting forces 
F 1 and F 2 as shown from above in Figure 4.36. Find the acceleration 

of the 49.00-kg sled and child system. Note that the direction of the 
frictional force is unspecified; it will be in the opposite direction of the sum 

of Fj and F 2 . 



158 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 

F, = 10 N 




Figure 4.36 An overhead view of the horizontal forces acting on a child's snow saucer 
sled. 

32. Suppose your car was mired deeply in the mud and you wanted to 
use the method illustrated in Figure 4.37 to pull it out. (a) What force 
would you have to exert perpendicular to the center of the rope to 
produce a force of 12,000 N on the car if the angle is 2.00°? In this part, 
explicitly show how you follow the steps in the Problem-Solving Strategy 
for Newton's laws of motion, (b) Real ropes stretch under such forces. 
What force would be exerted on the car if the angle increases to 7.00° 
and you still apply the force found in part (a) to its center? 




Figure 4.37 

33. What force is exerted on the tooth in Figure 4.38 if the tension in the 
wire is 25.0 N? Note that the force applied to the tooth is smaller than the 
tension in the wire, but this is necessitated by practical considerations of 
how force can be applied in the mouth. Explicitly show how you follow 
steps in the Problem-Solving Strategy for Newton's laws of motion. 




Figure 4.38 Braces are used to apply forces to teeth to realign them. Shown in this 
figure are the tensions applied by the wire to the protruding tooth. The total force 
applied to the tooth by the wire, F a pp , points straight toward the back of the mouth. 

34. Figure 4.39 shows Superhero and Trusty Sidekick hanging 
motionless from a rope. Superhero's mass is 90.0 kg, while Trusty 
Sidekick's is 55.0 kg, and the mass of the rope is negligible, (a) Draw a 
free-body diagram of the situation showing all forces acting on 
Superhero, Trusty Sidekick, and the rope, (b) Find the tension in the rope 
above Superhero, (c) Find the tension in the rope between Superhero 
and Trusty Sidekick. Indicate on your free-body diagram the system of 
interest used to solve each part. 




I* 



Figure 4.39 Superhero and Trusty Sidekick hang motionless on a rope as they try to 
figure out what to do next. Will the tension be the same everywhere in the rope? 

35. A nurse pushes a cart by exerting a force on the handle at a 
downward angle 35.0° below the horizontal. The loaded cart has a 
mass of 28.0 kg, and the force of friction is 60.0 N. (a) Draw a free-body 
diagram for the system of interest, (b) What force must the nurse exert to 
move at a constant velocity? 

36. Construct Your Own Problem Consider the tension in an elevator 
cable during the time the elevator starts from rest and accelerates its load 
upward to some cruising velocity. Taking the elevator and its load to be 
the system of interest, draw a free-body diagram. Then calculate the 
tension in the cable. Among the things to consider are the mass of the 
elevator and its load, the final velocity, and the time taken to reach that 
velocity. 

37. Construct Your Own Problem Consider two people pushing a 
toboggan with four children on it up a snow-covered slope. Construct a 
problem in which you calculate the acceleration of the toboggan and its 
load. Include a free-body diagram of the appropriate system of interest as 
the basis for your analysis. Show vector forces and their components and 
explain the choice of coordinates. Among the things to be considered are 
the forces exerted by those pushing, the angle of the slope, and the 
masses of the toboggan and children. 

38. Unreasonable Results (a) Repeat Exercise 4.29, but assume an 
acceleration of 1.20 m/s is produced, (b) What is unreasonable about 
the result? (c) Which premise is unreasonable, and why is it 
unreasonable? 

39. Unreasonable Results (a) What is the initial acceleration of a rocket 
that has a mass of 1.50x10 kg at takeoff, the engines of which 

produce a thrust of 2.00X 10 6 N ? Do not neglect gravity, (b) What is 
unreasonable about the result? (This result has been unintentionally 
achieved by several real rockets.) (c) Which premise is unreasonable, or 
which premises are inconsistent? (You may find it useful to compare this 
problem to the rocket problem earlier in this section.) 



CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 159 



4.7 Further Applications of Newton's Laws of Motion 

40. A flea jumps by exerting a force of 1.20x10" N straight down on 
the ground. A breeze blowing on the flea parallel to the ground exerts a 
force of 0.500x10" N on the flea. Find the direction and magnitude 
of the acceleration of the flea if its mass is 6.00X 10 kg . Do not 
neglect the gravitational force. 

41. Two muscles in the back of the leg pull upward on the Achilles 
tendon, as shown in Figure 4.40. (These muscles are called the medial 
and lateral heads of the gastrocnemius muscle.) Find the magnitude and 
direction of the total force on the Achilles tendon. What type of movement 
could be caused by this force? 



F 2 (200 N) ^(200 N) 







Figure 4.40 Achilles tendon 



42. A 76.0-kg person is being pulled away from a burning building as 
shown in Figure 4.41. Calculate the tension in the two ropes if the person 
is momentarily motionless. Include a free-body diagram in your solution. 




Figure 4.41 The force T2 needed to hold steady the person being rescued from the 
fire is less than her weight and less than the force T ^ in the other rope, since the 
more vertical rope supports a greater part of her weight (a vertical force). 

43. Integrated Concepts A 35.0-kg dolphin decelerates from 12.0 to 
7.50 m/s in 2.30 s to join another dolphin in play. What average force was 
exerted to slow him if he was moving horizontally? (The gravitational 
force is balanced by the buoyant force of the water.) 

44. Integrated Concepts When starting a foot race, a 70.0-kg sprinter 
exerts an average force of 650 N backward on the ground for 0.800 s. (a) 
What is his final speed? (b) How far does he travel? 

45. Integrated Concepts A large rocket has a mass of 2.00X 10 kg at 

n 

takeoff, and its engines produce a thrust of 3.50X 10 N . (a) Find its 
initial acceleration if it takes off vertically, (b) How long does it take to 
reach a velocity of 120 km/h straight up, assuming constant mass and 
thrust? (c) In reality, the mass of a rocket decreases significantly as its 
fuel is consumed. Describe qualitatively how this affects the acceleration 
and time for this motion. 

46. Integrated Concepts A basketball player jumps straight up for a ball. 
To do this, he lowers his body 0.300 m and then accelerates through this 
distance by forcefully straightening his legs. This player leaves the floor 
with a vertical velocity sufficient to carry him 0.900 m above the floor, (a) 
Calculate his velocity when he leaves the floor, (b) Calculate his 
acceleration while he is straightening his legs. He goes from zero to the 
velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force 
he exerts on the floor to do this, given that his mass is 110 kg. 

47. Integrated Concepts A 2.50-kg fireworks shell is fired straight up 
from a mortar and reaches a height of 110 m. (a) Neglecting air 
resistance (a poor assumption, but we will make it for this example), 
calculate the shell's velocity when it leaves the mortar, (b) The mortar 
itself is a tube 0.450 m long. Calculate the average acceleration of the 
shell in the tube as it goes from zero to the velocity found in (a), (c) What 
is the average force on the shell in the mortar? Express your answer in 
newtons and as a ratio to the weight of the shell. 

48. Integrated Concepts Repeat Exercise 4.47 for a shell fired at an 
angle 10.0° from the vertical. 

49. Integrated Concepts An elevator filled with passengers has a mass 
of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 

1.20 m/s for 1.50 s. Calculate the tension in the cable supporting the 
elevator, (b) The elevator continues upward at constant velocity for 8.50 
s. What is the tension in the cable during this time? (c) The elevator 

decelerates at a rate of 0.600 m/s for 3.00 s. What is the tension in 



160 CHAPTER 4 | DYNAMICS: FORCE AND NEWTON'S LAWS OF MOTION 

the cable during deceleration? (d) How high has the elevator moved 
above its original starting point, and what is its final velocity? 

50. Unreasonable Results (a) What is the final velocity of a car originally 
traveling at 50.0 km/h that decelerates at a rate of 0.400 m/s 2 for 50.0 
s? (b) What is unreasonable about the result? (c) Which premise is 
unreasonable, or which premises are inconsistent? 

51. Unreasonable Results A 75.0-kg man stands on a bathroom scale 
in an elevator that accelerates from rest to 30.0 m/s in 2.00 s. (a) 
Calculate the scale reading in newtons and compare it with his weight. 
(The scale exerts an upward force on him equal to its reading.) (b) What 
is unreasonable about the result? (c) Which premise is unreasonable, or 
which premises are inconsistent? 

4.8 Extended Topic: The Four Basic Forces — An 
Introduction 

52. (a) What is the strength of the weak nuclear force relative to the 
strong nuclear force? (b) What is the strength of the weak nuclear force 
relative to the electromagnetic force? Since the weak nuclear force acts 
at only very short distances, such as inside nuclei, where the strong and 
electromagnetic forces also act, it might seem surprising that we have 
any knowledge of it at all. We have such knowledge because the weak 
nuclear force is responsible for beta decay, a type of nuclear decay not 
explained by other forces. 

53. (a) What is the ratio of the strength of the gravitational force to that of 
the strong nuclear force? (b) What is the ratio of the strength of the 
gravitational force to that of the weak nuclear force? (c) What is the ratio 
of the strength of the gravitational force to that of the electromagnetic 
force? What do your answers imply about the influence of the 
gravitational force on atomic nuclei? 

54. What is the ratio of the strength of the strong nuclear force to that of 
the electromagnetic force? Based on this ratio, you might expect that the 
strong force dominates the nucleus, which is true for small nuclei. Large 
nuclei, however, have sizes greater than the range of the strong nuclear 
force. At these sizes, the electromagnetic force begins to affect nuclear 
stability. These facts will be used to explain nuclear fusion and fission 
later in this text. 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 161 



FURTHER APPLICATIONS OF NEWTON'S LAWS: 
FRICTION, DRAG, AND ELASTICITY 




Figure 5.1 Total hip replacement surgery has become a common procedure. The head (or ball) of the patient's femur fits into a cup that has a hard plastic-like inner lining, 
(credit: National Institutes of Health, via Wikimedia Commons) 



Learning Objectives 



5.1. Friction 

• Discuss the general characteristics of friction. 

• Describe the various types of friction. 

• Calculate the magnitude of static and kinetic friction. 

5.2. Drag Forces 

• Express mathematically the drag force. 

• Discuss the applications of drag force. 

• Define terminal velocity. 

• Determine the terminal velocity given mass. 

5.3. Elasticity: Stress and Strain 

• State Hooke's law. 

• Explain Hooke's law using graphical representation between deformation and applied force. 

• Discuss the three types of deformations such as changes in length, sideways shear and changes in volume. 

• Describe with examples the young's modulus, shear modulus and bulk modulus. 

• Determine the change in length given mass, length and radius. 



162 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 

Introduction: Further Applications of Newton's Laws 



Describe the forces on the hip joint. What means are taken to ensure that this will be a good movable joint? From the photograph (for an adult) in 
Figure 5.1, estimate the dimensions of the artificial device. 

It is difficult to categorize forces into various types (aside from the four basic forces discussed in previous chapter). We know that a net force affects 
the motion, position, and shape of an object. It is useful at this point to look at some particularly interesting and common forces that will provide 
further applications of Newton's laws of motion. We have in mind the forces of friction, air or liquid drag, and deformation. 

5.1 Friction 

Friction is a force that is around us all the time that opposes relative motion between systems in contact but also allows us to move (which you have 
discovered if you have ever tried to walk on ice). While a common force, the behavior of friction is actually very complicated and is still not completely 
understood. We have to rely heavily on observations for whatever understandings we can gain. However, we can still deal with its more elementary 
general characteristics and understand the circumstances in which it behaves. 

Friction 



Friction is a force that opposes relative motion between systems in contact. 

One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion 
or attempted motion of the systems relative to each other. If two systems are in contact and moving relative to one another, then the friction between 
them is called kinetic friction. For example, friction slows a hockey puck sliding on ice. But when objects are stationary, static friction can act 
between them; the static friction is usually greater than the kinetic friction between the objects. 

Kinetic Friction 

If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction. 

Imagine, for example, trying to slide a heavy crate across a concrete floor — you may push harder and harder on the crate and not move it at all. This 
means that the static friction responds to what you do — it increases to be equal to and in the opposite direction of your push. But if you finally push 
hard enough, the crate seems to slip suddenly and starts to move. Once in motion it is easier to keep it in motion than it was to get it started, 
indicating that the kinetic friction force is less than the static friction force. If you add mass to the crate, say by placing a box on top of it, you need to 
push even harder to get it started and also to keep it moving. Furthermore, if you oiled the concrete you would find it to be easier to get the crate 
started and keep it going (as you might expect). 

Figure 5.2 is a crude pictorial representation of how friction occurs at the interface between two objects. Close-up inspection of these surfaces shows 
them to be rough. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips 
of the surface hitting, break off the points, or do both. A considerable force can be resisted by friction with no apparent motion. The harder the 
surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. Part of the friction is due to 
adhesive forces between the surface molecules of the two objects, which explain the dependence of friction on the nature of the substances. 
Adhesion varies with substances in contact and is a complicated aspect of surface physics. Once an object is moving, there are fewer points of 
contact (fewer molecules adhering), so less force is required to keep the object moving. At small but nonzero speeds, friction is nearly independent of 
speed. 

Direction of motion 
or attempted motion 




Figure 5.2 Frictional forces, such as / , always oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the 

surfaces in contact, as seen in the expanded view. In order for the object to move, it must rise to where the peaks can skip along the bottom surface. Thus a force is required 
just to set the object in motion. Some of the peaks will be broken off, also requiring a force to maintain motion. Much of the friction is actually due to attractive forces between 
molecules making up the two objects, so that even perfectly smooth surfaces are not friction-free. Such adhesive forces also depend on the substances the surfaces are made 
of, explaining, for example, why rubber-soled shoes slip less than those with leather soles. 

The magnitude of the frictional force has two forms: one for static situations (static friction), the other for when there is motion (kinetic friction). 
When there is no motion between the objects, the magnitude of static friction f s is 

/s < PsN, (5.1) 

where ju s is the coefficient of static friction and N is the magnitude of the normal force (the force perpendicular to the surface). 

Magnitude of Static Friction 

Magnitude of static friction / s is 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 163 



/s < VsN, 

where jh s is the coefficient of static friction and N is the magnitude of the normal force. 



(5.2) 



The symbol < means less than or equal to, implying that static friction can have a minimum and a maximum value of ju s N . Static friction is a 
responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds 
/s(max) > tne 0D J ect wi " move. Thus 

/s(max) = A/s^V. ( 5 - 3 ) 

Once an object is moving, the magnitude of kinetic friction f k is given by 

/k = Pk^ ( 5 - 4 ) 

where /i k is the coefficient of kinetic friction. A system in which / k = ju k N is described as a system in which friction behaves simply. 

Mag nitude of Kinetic Friction 



The magnitude of kinetic friction / k is given by 



where ju k is the coefficient of kinetic friction. 



fk = V*N> 



(5.5) 



As seen in Table 5.1, the coefficients of kinetic friction are less than their static counterparts. That values of ju in Table 5.1 are stated to only one or, 
at most, two digits is an indication of the approximate description of friction given by the above two equations. 

Table 5.1 Coefficients of Static and Kinetic Friction 



System 




Ms 


Mk 


Rubber on dry concrete 


1.0 


0.7 


Rubber on wet concrete 


0.7 


0.5 


Wood on wood 


0.5 


0.3 


Waxed wood on wet snow 


0.14 


0.1 


Metal on wood 


0.5 


0.3 


Steel on steel (dry) 


0.6 


0.3 


Steel on steel (oiled) 


0.05 


0.03 


Teflon on steel 


0.04 


0.04 


Bone lubricated by synovial fluid 


0.016 


0.015 


Shoes on wood 


0.9 


0.7 


Shoes on ice 


0.1 


0.05 


Ice on ice 


0.1 


0.03 


Steel on ice 


0.4 


0.02 



The equations given earlier include the dependence of friction on materials and the normal force. The direction of friction is always opposite that of 
motion, parallel to the surface between objects, and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel 

to the floor) has a mass of 100 kg, then the normal force would be equal to its weight, W = mg = (100 kg)(9.80 m/s ) = 980 N , perpendicular to 

the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than 

/s(max) = A^s^ — (0.45)(980 N) = 440 N to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 

0.30, so that a force of only 290 N ( f k = /u k N = (0.30)(980 N) = 290 N ) would keep it moving at a constant speed. If the floor is lubricated, both 

coefficients are considerably less than they would be without lubrication. Coefficient of friction is a unit less quantity with a magnitude usually between 
and 1.0. The coefficient of the friction depends on the two surfaces that are in contact. 

Take-Home Experiment 

Find a small plastic object (such as a food container) and slide it on a kitchen table by giving it a gentle tap. Now spray water on the table, 
simulating a light shower of rain. What happens now when you give the object the same-sized tap? Now add a few drops of (vegetable or olive) 
oil on the surface of the water and give the same tap. What happens now? This latter situation is particularly important for drivers to note, 
especially after a light rain shower. Why? 



164 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 

Many people have experienced the slipperiness of walking on ice. However, many parts of the body, especially the joints, have much smaller 
coefficients of friction — often three or four times less than ice. A joint is formed by the ends of two bones, which are connected by thick tissues. The 
knee joint is formed by the lower leg bone (the tibia) and the thighbone (the femur). The hip is a ball (at the end of the femur) and socket (part of the 
pelvis) joint. The ends of the bones in the joint are covered by cartilage, which provides a smooth, almost glassy surface. The joints also produce a 
fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic joint can be replaced by an artificial joint (Figure 5.3). These replacements 
can be made of metals (stainless steel or titanium) or plastic (polyethylene), also with very small coefficients of friction. 




Figure 5.3 Artificial knee replacement is a procedure that has been performed for more than 20 years. In this figure, we see the post-op x rays of the right knee joint 
replacement, (credit: Mike Baird, Flickr) 

Other natural lubricants include saliva produced in our mouths to aid in the swallowing process, and the slippery mucus found between organs in the 
body, allowing them to move freely past each other during heartbeats, during breathing, and when a person moves. Artificial lubricants are also 
common in hospitals and doctor's clinics. For example, when ultrasonic imaging is carried out, a gel is used to lubricate the surface between the 
transducer and the skin — thereby reducing the coefficient of friction between the two surfaces. This allows the transducer to mover freely over the 
skin. 



Example 5.1 Skiing Exercise 



A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N. 

Strategy 

The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force N as / k = jiiyJV ; thus, the coefficient 

of kinetic friction can be found if we can find the normal force of the skier on a slope. The normal force is always perpendicular to the surface, 
and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier's weight perpendicular to the 
slope. (See the skier and free-body diagram in Figure 5.4.) 




Figure 5.4 The motion of the skier and friction are parallel to the slope and so it is most convenient to project all forces onto a coordinate system where one axis is 
parallel to the slope and the other is perpendicular (axes shown to left of skier). N (the normal force) is perpendicular to the slope, and f (the friction) is parallel to the 
slope, but W (the skier's weight) has components along both axes, namely Wj_ and W// . N is equal in magnitude to Wj_ , so there is no motion perpendicular to 

the slope. However, f is less than W// in magnitude, so there is acceleration down the slope (along the x-axis). 



That is, 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 165 

N = w ± = wcos25° = mgcos25°. (5.6) 

Substituting this into our expression for kinetic friction, we get 

/ k = A* k mgcos25°, (5.7) 

which can now be solved for the coefficient of kinetic friction ju^ . 

Solution 

Solving for ju^ gives 

„ _/ k _ / k _ A < 5 - 8 ) 

w cos 25° mg cos 25°. 

Substituting known values on the right-hand side of the equation, 

A/k = 410N^ = o.082. < 5 - 9 > 

(62 kg)(9.80 m/s 2 )(0.906) 

Discussion 

This result is a little smaller than the coefficient listed in Table 5.1 for waxed wood on snow, but it is still reasonable since values of the 
coefficients of friction can vary greatly. In situations like this, where an object of mass m slides down a slope that makes an angle 6 with the 
horizontal, friction is given by / k = jnyng cos 6 . All objects will slide down a slope with constant acceleration under these circumstances. Proof 
of this is left for this chapter's Problems and Exercises. 

Take-Home Experiment 

An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the 
coefficient of kinetic friction between two objects. As shown in Example 5.1, the kinetic friction on a slope / k = jUyjng cos 6 . The component of 

the weight down the slope is equal to mg sin# (see the free-body diagram in Figure 5.4). These forces act in opposite directions, so when they 

have equal magnitude, the acceleration is zero. Writing these out: 

f k = Fg x (5.10) 

(JL^mg cos 6 = mg sin 6. (5.11) 



Solving for ju^ , we find that 



mg sin6> , n (5.12) 

ll^ = — 77 = tan 6. 

K mg cos 6 



Put a coin on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to 
move. Measure the angle of tilt relative to the horizontal and find /i k . Note that the coin will not start to slide at all until an angle greater than 6 

is attained, since the coefficient of static friction is larger than the coefficient of kinetic friction. Discuss how this may affect the value for /i k and 

its uncertainty. 

We have discussed that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. 
Furthermore, simple friction is always proportional to the normal force. 

Making Connection s: Submicrosco pic Explanations of Friction 

The simpler aspects of friction dealt with so far are its macroscopic (large-scale) characteristics. Great strides have been made in the atomic- 
scale explanation of friction during the past several decades. Researchers are finding that the atomic nature of friction seems to have several 
fundamental characteristics. These characteristics not only explain some of the simpler aspects of friction — they also hold the potential for the 
development of nearly friction-free environments that could save hundreds of billions of dollars in energy which is currently being converted 
(unnecessarily) to heat. 

Figure 5.5 illustrates one macroscopic characteristic of friction that is explained by microscopic (small-scale) research. We have noted that friction is 
proportional to the normal force, but not to the area in contact, a somewhat counterintuitive notion. When two rough surfaces are in contact, the 
actual contact area is a tiny fraction of the total area since only high spots touch. When a greater normal force is exerted, the actual contact area 
increases, and it is found that the friction is proportional to this area. 



166 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 




Small normal 
force 



Large normal 
force 



Figure 5.5 Two rough surfaces in contact have a much smaller area of actual contact than their total area. When there is a greater normal force as a result of a greater applied 
force, the area of actual contact increases as does friction. 

But the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is now being 
determined. In other words, why do surfaces get warmer when rubbed? Essentially, atoms are linked with one another to form lattices. When 
surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate — essentially creating sound waves that penetrate the material. The 
sound waves diminish with distance and their energy is converted into heat. Chemical reactions that are related to frictional wear can also occur 
between atoms and molecules on the surfaces. Figure 5.6 shows how the tip of a probe drawn across another material is deformed by atomic-scale 
friction. The force needed to drag the tip can be measured and is found to be related to shear stress, which will be discussed later in this chapter. The 

variation in shear stress is remarkable (more than a factor of 10 ) and difficult to predict theoretically, but shear stress is yielding a fundamental 
understanding of a large-scale phenomenon known since ancient times — friction. 

-■ Motion of probe 



Friction 




Figure 5.6 The tip of a probe is deformed sideways by frictional force as the probe is dragged across a surface. Measurements of how the force varies for different materials 
are yielding fundamental insights into the atomic nature of friction. 

PhET Explorations: Forces and Motion 



Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting 
on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. Draw a free-body diagram of all the forces (including 
gravitational and normal forces). 



n 



% 



^PhET Interactive Simulation 



Figure 5.7 Forces and Motion (http://cnx.Org/content/m42139/l.3/forces-and-motion_en.jar) 



5.2 Drag Forces 

Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force 
when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the 
harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air — you have decreased the area of your 
hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is 
proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its 
size, its velocity, and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the drag 

force F D is found to be proportional to the square of the speed of the object. We can write this relationship mathematically as F D oc v . When 

taking into account other factors, this relationship becomes 



■ = ICpAv 2 , 



(5.13) 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 167 

where C is the drag coefficient, A is the area of the object facing the fluid, and p is the density of the fluid. (Recall that density is mass per unit 

volume.) This equation can also be written in a more generalized fashion as F D = bAv , where b is a constant equivalent to 0.5CpA . We have 

set the exponent n for these equations as 2 because, when an object is moving at high velocity through air, the magnitude of the drag force is 
proportional to the square of the speed. As we shall see in a few pages on fluid dynamics, for small particles moving at low speeds in a fluid, the 
exponent n is equal to 1. 



Drag Force 


Drag force F D is found to be proportional to the square of the speed of the object. Mathematically 




F D ocv 2 


(5.14) 


F D = \CpAv 2 , 


(5.15) 


where C is the drag coefficient, A is the area of the object facing the fluid, and p is the density of the fluid. 





Athletes as well as car designers seek to reduce the drag force to lower their race times. (See Figure 5.8). "Aerodynamic" shaping of an automobile 
can reduce the drag force and so increase a car's gas mileage. 




Figure 5.8 From racing cars to bobsled racers, aerodynamic shaping is crucial to achieving top speeds. Bobsleds are designed for speed. They are shaped like a bullet with 
tapered fins, (credit: U.S. Army, via Wikimedia Commons) 

The value of the drag coefficient, C , is determined empirically, usually with the use of a wind tunnel. (See Figure 5.9). 




Figure 5.9 NASA researchers test a model plane in a wind tunnel, (credit: NASA/Ames) 

The drag coefficient can depend upon velocity, but we will assume that it is a constant here. Table 5.2 lists some typical drag coefficients for a variety 
of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over 50% of the power of a car is used to overcome air 
drag. The most fuel-efficient cruising speed is about 70-80 km/h (about 45-50 mi/h). For this reason, during the 1970s oil crisis in the United States, 
maximum speeds on highways were set at about 90 km/h (55 mi/h). 



168 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 

Table 5.2 Drag Coefficient 
Values Typical values of 
drag coefficient C . 







Airfoil 


0.05 


Toyota Camry 


0.28 


Ford Focus 


0.32 


Honda Civic 


0.36 


Ferrari Testarossa 


0.37 


Dodge Ram pickup 


0.43 


Sphere 


0.45 


Hummer H2 SUV 


0.64 


Skydiver (feet first) 


0.70 


Bicycle 


0.90 


Skydiver (horizontal) 


1.0 


Circular flat plate 


1.12 



Substantial research is under way in the sporting world to minimize drag. The dimples on golf balls are being redesigned as are the clothes that 
athletes wear. Bicycle racers and some swimmers and runners wear full bodysuits. Australian Cathy Freeman wore a full body suit in the 2000 
Sydney Olympics, and won the gold medal for the 400 m race. Many swimmers in the 2008 Beijing Olympics wore (Speedo) body suits; it might have 
made a difference in breaking many world records (See Figure 5.10). Most elite swimmers (and cyclists) shave their body hair. Such innovations can 
have the effect of slicing away milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is that 
careful and precise guidelines must be continuously developed to maintain the integrity of the sport. 




Figure 5.10 Body suits, such as this LZR Racer Suit, have been credited with many world records after their release in 2008. Smoother "skin" and more compression forces on 
a swimmer's body provide at least 10% less drag, (credit: NASA/Kathy Barnstorff) 

Some interesting situations connected to Newton's second law occur when considering the effects of drag forces upon a moving object. For instance, 
consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring 
the buoyant force). The downward force of gravity remains constant regardless of the velocity at which the person is moving. However, as the 
person's velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus 
producing a net force of zero. A zero net force means that there is no acceleration, as given by Newton's second law. At this point, the person's 
velocity remains constant and we say that the person has reached his terminal velocity ( v t ). Since F D is proportional to the speed, a heavier 

skydiver must go faster for F D to equal his weight. Let's see how this works out more quantitatively. 



At the terminal velocity, 



Thus, 



Using the equation for drag force, we have 



mg - F L 



mg = F D . 



1 2 

mg = -jpCAv . 



(5.16) 
(5.17) 
(5.18) 



Solving for the velocity, we obtain 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 169 

_ K \2mg (5-19) 



V ' \pCA' 

Assume the density of air is p = 1.21 kg/m . A 75-kg skydiver descending head first will have an area approximately A = 0.18 m and a drag 
coefficient of approximately C = 0.70 . We find that 

2(75 kg)(9.80 m/s 2 ) (5 ' 20) 



(1.21 kg/m 3 )(0.70)(0. 18 m 2 ) 



= 98 m/s 
= 350km/h. 

This means a skydiver with a mass of 75 kg achieves a maximum terminal velocity of about 350 km/h while traveling in a pike (head first) position, 
minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about 200 km/h as the area increases. This 
terminal velocity becomes much smaller after the parachute opens. 

Take-Home Experiment 

This interesting activity examines the effect of weight upon terminal velocity. Gather together some nested coffee filters. Leaving them in their 
original shape, measure the time it takes for one, two, three, four, and five nested filters to fall to the floor from the same height (roughly 2 m). 
(Note that, due to the way the filters are nested, drag is constant and only mass varies.) They obtain terminal velocity quite quickly, so find this 

2 

velocity as a function of mass. Plot the terminal velocity v versus mass. Also plot v versus mass. Which of these relationships is more linear? 
What can you conclude from these graphs? 



Example 5.2 A Terminal Velocity 








Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position. 






Strategy 








At terminal velocity, F nQt = . Thus the drag force on the skydiver must equal the force of gravity (the person's weight). Using the equation of 


1 2 

drag force, we find mg = -^pCAv . 








Thus the terminal velocity v t can be written 


as 








v - \l 2mg 




(5.21) 


Solution 








All quantities are known except the person's 


projected area. This is an adult (82 kg) falling spread eagle. 


We can estimate the frontal 


area as 




A = (2 m)(0.35 m) = 0.70 m 2 . 




(5.22) 


Using our equation for v t , we find that 






(5.23) 


J 2(85 kg)(9.80 m/s 2 ) 


Vt f (1.21 kg/m 3 )(1.0)(0.70 m 2 ) 
= 44 m/s. 


Discussion 








This result is consistent with the value for v t 


mentioned earlier. The 75-kg skydiver going feet first had a v = 98 m/s . He weighed 


less but 


had a smaller frontal area and so a smaller d 


rag due to the air. 







The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m high branch of a tree, you will 
likely get hurt — possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You don't reach a terminal velocity in 
such a short distance, but the squirrel does. 

The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J.B.S. Haldane, titled "On Being the 
Right Size." 

To the mouse and any smaller animal, [gravity] presents practically no clangers. You can drop a mouse down a thousand-yard mine shaft; and, on 
arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse 
splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal's length, breadth, 
and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small 
animal is relatively ten times greater than the driving force. 

The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than 
air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes' law, which states that 



170 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 



F s = 6nrt]v, 
where r is the radius of the object, r\ is the viscosity of the fluid, and v is the object's velocity. 



(5.24) 



Stokes' Law 



F s = 6jrrrjv, 
where r is the radius of the object, r\ is the viscosity of the fluid, and v is the object's velocity. 



(5.25) 



Good examples of this law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so small, we find that many 
of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria (size about 1 urn ) can be about 2 |im/s . To 

move at a greater speed, many bacteria swim using flagella (organelles shaped like little tails) that are powered by little motors embedded in the cell. 
Sediment in a lake can move at a greater terminal velocity (about 5 |im/s ), so it can take days to reach the bottom of the lake after being deposited 

on the surface. 

If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fishes, dolphins, and even massive whales 
are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large distances often have particular features 
such as long necks. Flocks of birds fly in the shape of a spear head as the flock forms a streamlined pattern (see Figure 5.11). In humans, one 
important example of streamlining is the shape of sperm, which need to be efficient in their use of energy. 




Figure 5.11 Geese fly in a V formation during their long migratory travels. This shape reduces drag and energy consumption for individual birds, and also allows them a better 
way to communicate, (credit: Julo, Wikimedia Commons) 

Galileo's Experiment 

Galileo is said to have dropped two objects of different masses from the Tower of Pisa. He measured how long it took each to reach the ground. 
Since stopwatches weren't readily available, how do you think he measured their fall time? If the objects were the same size, but with different 
masses, what do you think he should have observed? Would this result be different if done on the Moon? 

PhET Explorations: Masses & Springs 

A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. 
Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energy for each spring. 






PhET Interactive Simulation 



Figure 5.12 Masses & Springs (http://cnx.Org/content/m42080/l.4/mass-spring-lab_en.jar) 



5.3 Elasticity: Stress and Strain 

We now move from consideration of forces that affect the motion of an object (such as friction and drag) to those that affect an object's shape. If a 
bulldozer pushes a car into a wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a 
deformation. Even very small forces are known to cause some deformation. For small deformations, two important characteristics are observed. 
First, the object returns to its original shape when the force is removed — that is, the deformation is elastic for small deformations. Second, the size of 
the deformation is proportional to the force — that is, for small deformations, Hooke's law is obeyed. In equation form, Hooke's law is given by 

F = £AL, (5.26) 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 171 

where AL is the amount of deformation (the change in length, for example) produced by the force F , and k is a proportionality constant that 
depends on the shape and composition of the object and the direction of the force. Note that this force is a function of the deformation AL — it is not 
constant as a kinetic friction force is. Rearranging this to 

k 

makes it clear that the deformation is proportional to the applied force. Figure 5.13 shows the Hooke's law relationship between the extension AL of 
a spring or of a human bone. For metals or springs, the straight line region in which Hooke's law pertains is much larger. Bones are brittle and the 
elastic region is small and the fracture abrupt. Eventually a large enough stress to the material will cause it to break or fracture. 



Hooke's Law 



F = kAL, 



(5.28) 



where AL is the amount of deformation (the change in length, for example) produced by the force F , and k is a proportionality constant that 
depends on the shape and composition of the object and the direction of the force. 

(5.29) 




*- Hooke's law -*-] 
obeyed ' 
Elastic region 

Figure 5.13 A graph of deformation AL versus applied force F . The straight segment is the linear region where Hooke's law is obeyed. The slope of the straight region is 
y . For larger forces, the graph is curved but the deformation is still elastic — AL will return to zero if the force is removed. Still greater forces permanently deform the object 

until it finally fractures. The shape of the curve near fracture depends on several factors, including how the force F is applied. Note that in this graph the slope increases just 
before fracture, indicating that a small increase in F is producing a large increase in L near the fracture. 

The proportionality constant k depends upon a number of factors for the material. For example, a guitar string made of nylon stretches when it is 
tightened, and the elongation AL is proportional to the force applied (at least for small deformations). Thicker nylon strings and ones made of steel 
stretch less for the same applied force, implying they have a larger k (see Figure 5.14). Finally, all three strings return to their normal lengths when 
the force is removed, provided the deformation is small. Most materials will behave in this manner if the deformation is less that about 0.1% or about 
1 part in 10 3 . 




Figure 5.14 The same force, in this case a weight ( W ), applied to three different guitar strings of identical length produces the three different deformations shown as shaded 
segments. The string on the left is thin nylon, the one in the middle is thicker nylon, and the one on the right is steel. 



172 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 



Stretch Yourself a Little 



How would you go about measuring the proportionality constant k of a rubber band? If a rubber band stretched 3 cm when a 100-g mass was 
attached to it, then how much would it stretch if two similar rubber bands were attached to the same mass — even if put together in parallel or 
alternatively if tied together in series? 



We now consider three specific types of deformations: changes in length (tension and compression), sideways shear (stress), and changes in 
volume. All deformations are assumed to be small unless otherwise stated. 

Changes in Length — Tension and Compression: Elastic Modulus 

A change in length AL is produced when a force is applied to a wire or rod parallel to its length Lq , either stretching it (a tension) or compressing it. 

(See Figure 5.15.) 




(a) (b) 

Figure 5.15 (a) Tension. The rod is stretched a length AL when a force is applied parallel to its length, (b) Compression. The same rod is compressed by forces with the 
same magnitude in the opposite direction. For very small deformations and uniform materials, AL is approximately the same for the same magnitude of tension or 
compression. For larger deformations, the cross-sectional area changes as the rod is compressed or stretched. 

Experiments have shown that the change in length ( AL ) depends on only a few variables. As already noted, AL is proportional to the force F and 
depends on the substance from which the object is made. Additionally, the change in length is proportional to the original length Lq and inversely 

proportional to the cross-sectional area of the wire or rod. For example, a long guitar string will stretch more than a short one, and a thick string will 
stretch less than a thin one. We can combine all these factors into one equation for AL : 



al = 1£l 0> 



(5.30) 



where AL is the change in length, F the applied force, Y is a factor, called the elastic modulus or Young's modulus, that depends on the 
substance, A is the cross-sectional area, and Lq is the original length. Table 5.3 lists values of Y for several materials — those with a large Y are 
said to have a large tensile strength because they deform less for a given tension or compression. 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 173 



Table 5.3 Elastic Moduli [1] 









(10 9 N/m 2 ) 


(10 9 N/m 2 ) 


(10 9 N/m 2 ) 




Material 










Aluminum 


70 


25 


75 


Bone -tension 


16 


80 


8 


Bone- 
compression 


9 






Brass 


90 


35 


75 


Brick 


15 






Concrete 


20 






Glass 


70 


20 


30 


Granite 


45 


20 


45 


Hair (human) 


10 






Hardwood 


15 


10 




Iron, cast 


100 


40 


90 


Lead 


16 


5 


50 


Marble 


60 


20 


70 


Nylon 


5 






Polystyrene 


3 






Silk 


6 






Spider thread 


3 






Steel 


210 


80 


130 


Tendon 


1 






Acetone 






0.7 


Ethanol 






0.9 


Glycerin 






4.5 


Mercury 






25 


Water 






2.2 



Young's moduli are not listed for liquids and gases in Table 5.3 because they cannot be stretched or compressed in only one direction. Note that 
there is an assumption that the object does not accelerate, so that there are actually two applied forces of magnitude F acting in opposite directions. 
For example, the strings in Figure 5.15 are being pulled down by a force of magnitude w and held up by the ceiling, which also exerts a force of 
magnitude w . 



Example 5.3 The Stretch of a Long Cable 



Suspension cables are used to carry gondolas at ski resorts. (See Figure 5.16) Consider a suspension cable that includes an unsupported span 
of 3 km. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of 5.6 cm and the maximum tension it can 

withstand is 3.0xl0 6 N. 



L 




Figure 5.16 Gondolas travel along suspension cables at the Gala Yuzawa ski resort in Japan, (credit: Rudy Herman, Flickr) 
Strategy 



1. Approximate and average values. Young's moduli Y for tension and compression sometimes differ but are averaged here. Bone has significantly 
different Young's moduli for tension and compression. 



174 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 



The force is equal to the maximum tension, or F = 3. Ox 10 N . The cross-sectional area is nr = 2.46x 10 m . The equation 



AL = 



IF 



^rL can be used to find the change in length. 

Y J\ 



Solution 

All quantities are known. Thus, 



AL 



-h 



3.0x10° N 
k210xl0 9 N/m 2 A2.46xl0" 3 m 2 
18 m. 



(3020 m) 



(5.31) 



Discussion 

This is quite a stretch, but only about 0.6% of the unsupported length. Effects of temperature upon length might be important in these 
environments. 

Bones, on the whole, do not fracture due to tension or compression. Rather they generally fracture due to sideways impact or bending, resulting in 
the bone shearing or snapping. The behavior of bones under tension and compression is important because it determines the load the bones can 
carry. Bones are classified as weight-bearing structures such as columns in buildings and trees. Weight-bearing structures have special features; 
columns in building have steel-reinforcing rods while trees and bones are fibrous. The bones in different parts of the body serve different structural 
functions and are prone to different stresses. Thus the bone in the top of the femur is arranged in thin sheets separated by marrow while in other 
places the bones can be cylindrical and filled with marrow or just solid. Overweight people have a tendency toward bone damage due to sustained 
compressions in bone joints and tendons. 

Another biological example of Hooke's law occurs in tendons. Functionally, the tendon (the tissue connecting muscle to bone) must stretch easily at 
first when a force is applied, but offer a much greater restoring force for a greater strain. Figure 5.17 shows a stress-strain relationship for a human 
tendon. Some tendons have a high collagen content so there is relatively little strain, or length change; others, like support tendons (as in the leg) can 
change length up to 10%. Note that this stress-strain curve is nonlinear, since the slope of the line changes in different regions. In the first part of the 
stretch called the toe region, the fibers in the tendon begin to align in the direction of the stress — this is called uncrimping. In the linear region, the 
fibrils will be stretched, and in the failure region individual fibers begin to break. A simple model of this relationship can be illustrated by springs in 
parallel: different springs are activated at different lengths of stretch. Examples of this are given in the problems at end of this chapter. Ligaments 
(tissue connecting bone to bone) behave in a similar way. 

M 




Toe unear Failure f 

region region region 

Figure 5.17 Typical stress-strain curve for mammalian tendon. Three regions are shown: (1) toe region (2) linear region, and (3) failure region. 

Unlike bones and tendons, which need to be strong as well as elastic, the arteries and lungs need to be very stretchable. The elastic properties of the 
arteries are essential for blood flow. The pressure in the arteries increases and arterial walls stretch when the blood is pumped out of the heart. When 
the aortic valve shuts, the pressure in the arteries drops and the arterial walls relax to maintain the blood flow. When you feel your pulse, you are 
feeling exactly this — the elastic behavior of the arteries as the blood gushes through with each pump of the heart. If the arteries were rigid, you would 
not feel a pulse. The heart is also an organ with special elastic properties. The lungs expand with muscular effort when we breathe in but relax freely 
and elastically when we breathe out. Our skins are particularly elastic, especially for the young. A young person can go from 100 kg to 60 kg with no 
visible sag in their skins. The elasticity of all organs reduces with age. Gradual physiological aging through reduction in elasticity starts in the early 
20s. 



Example 5.4 Calculating Deformation: 


How Much Does Your Leg Shorten When You Stand on It? | 




Calculate the change in length of the upper leg bone (the femur) when a 70.0 kg man supports 62.0 kg of his mass on it, 
be equivalent to a uniform rod that is 40.0 cm long and 2.00 cm in radius. 


assuming 


the bone to 


Strategy 








The force is equal to the weight supported, or 








F = 


mg = (62.0 kg)(9.80 m/s 2 ) = 607.6 N, 




(5.32) 


and the cross-sectional area is nr = 1.257x10 


3 2 1 /7 

m . The equation AL = ttjLq can ,3e usecl t0 fincl tne cnan 9 e 

Y A 


n length. 




Solution 









CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 175 

All quantities except AL are known. Note that the compression value for Young's modulus for bone must be used here. Thus, 

(5.33) 



AL = ( J -Y 607-6 N \o.4QOm) 

V9xl0 9 N/m 2 Al.257xl0- 3 m 2 / 

= 2xl0" 5 m. 



Discussion 



This small change in length seems reasonable, consistent with our experience that bones are rigid. In fact, even the rather large forces 
encountered during strenuous physical activity do not compress or bend bones by large amounts. Although bone is rigid compared with fat or 
muscle, several of the substances listed in Table 5.3 have larger values of Young's modulus Y . In other words, they are more rigid and have 
greater tensile strength. 

The equation for change in length is traditionally rearranged and written in the following form: 

E = Y AL (5-34) 

A *L ' 



The ratio of force to area, ^- , is defined as stress (measured in N/m ), and the ratio of the change in length to length, -y^ , is defined as strain (a 
unitless quantity). In other words, 



A L 



stress = Yx strain. (5.35) 

In this form, the equation is analogous to Hooke's law, with stress analogous to force and strain analogous to deformation. If we again rearrange this 
equation to the form 

F=YA AL, (5-36) 

L 

we see that it is the same as Hooke's law with a proportionality constant 

k= YA (5-37) 

W 

This general idea — that force and the deformation it causes are proportional for small deformations — applies to changes in length, sideways bending, 
and changes in volume. 

Stress 

F 7 

The ratio of force to area, 4r > is defined as stress measured in N/m . 
A 

Strarn 

The ratio of the change in length to length, -y^ , is defined as strain (a unitless quantity). In other words, 

L 

stress = Fx strain. (5.38) 

Sideways Stress: Shear Modulus 

Figure 5.18 illustrates what is meant by a sideways stress or a shearing force. Here the deformation is called Ax and it is perpendicular to Lq , 

rather than parallel as with tension and compression. Shear deformation behaves similarly to tension and compression and can be described with 
similar equations. The expression for shear deformation is 



= IHt (5-39) 

SA 1 " 



Ax = ^L , 



where S is the shear modulus (see Table 5.3) and F is the force applied perpendicular to Lq and parallel to the cross-sectional area A . Again, to 

keep the object from accelerating, there are actually two equal and opposite forces F applied across opposite faces, as illustrated in Figure 5.18. 
The equation is logical — for example, it is easier to bend a long thin pencil (small A ) than a short thick one, and both are more easily bent than 
similar steel rods (large S). 

Shear Deformation 

A, _ lF_r (5.40) 

where S is the shear modulus and F is the force applied perpendicular to Lq and parallel to the cross-sectional area A . 



176 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 




Figure 5.18 Shearing forces are applied perpendicular to the length Lq and parallel to the area A , producing a deformation Ax . Vertical forces are not shown, but it 

should be kept in mind that in addition to the two shearing forces, F , there must be supporting forces to keep the object from rotating. The distorting effects of these 
supporting forces are ignored in this treatment. The weight of the object also is not shown, since it is usually negligible compared with forces large enough to cause significant 
deformations. 

Examination of the shear moduli in Table 5.3 reveals some telling patterns. For example, shear moduli are less than Young's moduli for most 
materials. Bone is a remarkable exception. Its shear modulus is not only greater than its Young's modulus, but it is as large as that of steel. This is 
one reason that bones can be long and relatively thin. Bones can support loads comparable to that of concrete and steel. Most bone fractures are not 
caused by compression but by excessive twisting and bending. 

The spinal column (consisting of 26 vertebral segments separated by discs) provides the main support for the head and upper part of the body. The 
spinal column has normal curvature for stability, but this curvature can be increased, leading to increased shearing forces on the lower vertebrae. 
Discs are better at withstanding compressional forces than shear forces. Because the spine is not vertical, the weight of the upper body exerts some 
of both. Pregnant women and people that are overweight (with large abdomens) need to move their shoulders back to maintain balance, thereby 
increasing the curvature in their spine and so increasing the shear component of the stress. An increased angle due to more curvature increases the 
shear forces along the plane. These higher shear forces increase the risk of back injury through ruptured discs. The lumbosacral disc (the wedge 
shaped disc below the last vertebrae) is particularly at risk because of its location. 

The shear moduli for concrete and brick are very small; they are too highly variable to be listed. Concrete used in buildings can withstand 
compression, as in pillars and arches, but is very poor against shear, as might be encountered in heavily loaded floors or during earthquakes. Modern 
structures were made possible by the use of steel and steel-reinforced concrete. Almost by definition, liquids and gases have shear moduli near zero, 
because they flow in response to shearing forces. 



Example 5.5 Calculating Force Required to Deform: That Nail Does Not Bend Much Under a Load 



Find the mass of the picture hanging from a steel nail as shown in Figure 5.19, given that the nail bends only 1.80 u,m . (Assume the shear 
modulus is known to two significant figures.) 



1 ,50 mm 




L D = 5,00 mm 



Figure 5.19 Side view of a nail with a picture hung from it. The nail flexes very slightly (shown much larger than actual) because of the shearing effect of the supported 
weight. Also shown is the upward force of the wall on the nail, illustrating that there are equal and opposite forces applied across opposite cross sections of the nail. See 
Example 5.5 for a calculation of the mass of the picture. 

Strategy 

The force F on the nail (neglecting the nail's own weight) is the weight of the picture w . If we can find w , then the mass of the picture is just 

Y . The equation Ax = ttj^o can be solved for F ■ 

Solution 

-1Z/ 



Solving the equation Ax = -tttLo for ^ , we see that all other quantities can be found: 

SA 

U 



F = #Ax. 



(5.41) 



S is found in Table 5.3 and is S = 80x10 N/m . The radius r is 0.750 mm (as seen in the figure), so the cross-sectional area is 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 177 



A = jtr 2 = 1.77xl0" 6 m 2 . 


(5.42) 


The value for Lq is also shown in the figure. Thus, 




p = (80X10 9 N/m 2 )(1.77xl0- 6 m 2 ) (1 80xlQ _ 6 = 5J R 
(5.00X10 -3 m) 


(5.43) 


This 51 N force is the weight w of the picture, so the picture's mass is 




m = | = £ = 5.2kg. 


(5.44) 


Discussion 




This is a fairly massive picture, and it is impressive that the nail flexes only 1.80 \im — an amount undetectable to the unaided eye. 





Changes in Volume: Bulk Modulus 

An object will be compressed in all directions if inward forces are applied evenly on all its surfaces as in Figure 5.20. It is relatively easy to compress 
gases and extremely difficult to compress liquids and solids. For example, air in a wine bottle is compressed when it is corked. But if you try corking a 
brim-full bottle, you cannot compress the wine — some must be removed if the cork is to be inserted. The reason for these different compressibilities is 
that atoms and molecules are separated by large empty spaces in gases but packed close together in liquids and solids. To compress a gas, you 
must force its atoms and molecules closer together. To compress liquids and solids, you must actually compress their atoms and molecules, and very 
strong electromagnetic forces in them oppose this compression. 



Volume V 




Figure 5.20 An inward force on all surfaces compresses this cube. Its change in volume is proportional to the force per unit area and its original volume, and is related to the 
compressibility of the substance. 

We can describe the compression or volume deformation of an object with an equation. First, we note that a force "applied evenly" is defined to have 
the same stress, or ratio of force to area ^- on all surfaces. The deformation produced is a change in volume A V , which is found to behave very 

similarly to the shear, tension, and compression previously discussed. (This is not surprising, since a compression of the entire object is equivalent to 
compressing each of its three dimensions.) The relationship of the change in volume to other physical quantities is given by 



AV =ij v °> 



(5.45) 



where B is the bulk modulus (see Table 5.3), Vq is the original volume, and ^- is the force per unit area applied uniformly inward on all surfaces. 

Note that no bulk moduli are given for gases. 

What are some examples of bulk compression of solids and liquids? One practical example is the manufacture of industrial-grade diamonds by 
compressing carbon with an extremely large force per unit area. The carbon atoms rearrange their crystalline structure into the more tightly packed 
pattern of diamonds. In nature, a similar process occurs deep underground, where extremely large forces result from the weight of overlying material. 
Another natural source of large compressive forces is the pressure created by the weight of water, especially in deep parts of the oceans. Water 
exerts an inward force on all surfaces of a submerged object, and even on the water itself. At great depths, water is measurably compressed, as the 
following example illustrates. 



Example 5.6 Calculating Change in Volume with Deformation: How Much Is Water Compressed at Great Ocean 
Depths? 



A V 7 2 

Calculate the fractional decrease in volume (-^fr-) for seawater at 5.00 km depth, where the force per unit area is 5.00x10 N/m 
Strategy 



178 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 



Equation AV - 


1 F 
= -jrrVo is the correct physical relationship. 


All quantities in the equation except 


AV 
^0 


are 


known. 




Solution 
















Solving for the i 


unknown ^z Qives 
v 


AV _ IF 
V BA' 










(5.46) 


Substituting known values with the value for the bulk modulus B from Table 5.3, 














AV 


_ 5.00xl0 7 N/m 2 










(5.47) 




Vo 


2.2xl0 9 N/m 2 
= 0.023 = 2.3%. 












Discussion 
















Although measurable, this is not a significant decrease in volume considering that the force 
pounds per square foot). Liquids and solids are extraordinarily difficult to compress. 


per unit area is about 500 atmospheres (1 million 



Conversely, very large forces are created by liquids and solids when they try to expand but are constrained from doing so — which is equivalent to 
compressing them to less than their normal volume. This often occurs when a contained material warms up, since most materials expand when their 
temperature increases. If the materials are tightly constrained, they deform or break their container. Another very common example occurs when 
water freezes. Water, unlike most materials, expands when it freezes, and it can easily fracture a boulder, rupture a biological cell, or crack an engine 
block that gets in its way. 

Other types of deformations, such as torsion or twisting, behave analogously to the tension, shear, and bulk deformations considered here. 



Glossary 



deformation: change in shape due to the application of force 

drag force: F D , found to be proportional to the square of the speed of the object; mathematically 

F D (xv 2 
F D = ±CpAv 2 , 
where C is the drag coefficient, A is the area of the object facing the fluid, and p is the density of the fluid 

friction: a force that opposes relative motion or attempts at motion between systems in contact 

Hooke's law: proportional relationship between the force F on a material and the deformation AL it causes, F = kAL 
kinetic friction: a force that opposes the motion of two systems that are in contact and moving relative to one another 
magnitude of kinetic friction: f k = jj k N , where /i k is the coefficient of kinetic friction 

magnitude of static friction: f s < jli s N , where ju s is the coefficient of static friction and N is the magnitude of the normal force 

Stokes' law: F s = 6nrrjv , where r is the radius of the object, r\ is the viscosity of the fluid, and v is the object's velocity 

shear deformation: deformation perpendicular to the original length of an object 

static friction: a force that opposes the motion of two systems that are in contact and are not moving relative to one another 

strain: ratio of change in length to original length 

stress: ratio of force to area 

tensile strength: measure of deformation for a given tension or compression 



Section Summary 



5.1 Friction 

• Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the 
normal force N pushing the systems together. (A normal force is always perpendicular to the contact surface between systems.) Friction 
depends on both of the materials involved. The magnitude of static friction / s between systems stationary relative to one another is given by 

/s < ^ 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 179 

where /u s is the coefficient of static friction, which depends on both of the materials. 

• The kinetic friction force / k between systems moving relative to one another is given by 

/k = A*k# • 

where /i k is the coefficient of kinetic friction, which also depends on both materials. 

5.2 Drag Forces 

• Drag forces acting on an object moving in a fluid oppose the motion. For larger objects (such as a baseball) moving at a velocity v in air, the 
drag force is given by 

F D = \CpAv 1 , 

where C is the drag coefficient (typical values are given in Table 5.2), A is the area of the object facing the fluid, and p is the fluid density. 

• For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes' law, 

F s = 6jir]rv, 

where r is the radius of the object, rj is the fluid viscosity, and v is the object's velocity. 

5.3 Elasticity: Stress and Strain 

• Hooke's law is given by 

F = kAL, 

where AL is the amount of deformation (the change in length), F is the applied force, and k is a proportionality constant that depends on the 
shape and composition of the object and the direction of the force. The relationship between the deformation and the applied force can also be 
written as 

where Y is Young's modulus, which depends on the substance, A is the cross-sectional area, and Lq is the original length. 

F ? 

• The ratio of force to area, ^- , is defined as stress, measured in N/m . 

A 

• The ratio of the change in length to length, -y^ , is defined as strain (a unitless quantity). In other words, 

stress = Yx strain. 

• The expression for shear deformation is 

SA 1 
where S is the shear modulus and F is the force applied perpendicular to Lq and parallel to the cross-sectional area A . 
The relationship of the change in volume to other physical quantities is given by 

F 

where B is the bulk modulus, Vq is the original volume, and ^- is the force per unit area applied uniformly inward on all surfaces. 



Ax = ±±L , 



Conceptual Questions 



5.1 Friction 

1. Define normal force. What is its relationship to friction when friction behaves simply? 

2. The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to 
vertical walls and even to ceilings. 

3. When you learn to drive, you discover that you need to let up slightly on the brake pedal as you come to a stop or the car will stop with a jerk. 
Explain this in terms of the relationship between static and kinetic friction. 

4. When you push a piece of chalk across a chalkboard, it sometimes screeches because it rapidly alternates between slipping and sticking to the 
board. Describe this process in more detail, in particular explaining how it is related to the fact that kinetic friction is less than static friction. (The same 
slip-grab process occurs when tires screech on pavement.) 

5.2 Drag Forces 

5. Athletes such as swimmers and bicyclists wear body suits in competition. Formulate a list of pros and cons of such suits. 

6. Two expressions were used for the drag force experienced by a moving object in a liquid. One depended upon the speed, while the other was 
proportional to the square of the speed. In which types of motion would each of these expressions be more applicable than the other one? 

7. As cars travel, oil and gasoline leaks onto the road surface. If a light rain falls, what does this do to the control of the car? Does a heavy rain make 
any difference? 

8. Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall? 

5.3 Elasticity: Stress and Strain 



180 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 

9. The elastic properties of the arteries are essential for blood flow. Explain the importance of this in terms of the characteristics of the flow of blood 
(pulsating or continuous). 

10. What are you feeling when you feel your pulse? Measure your pulse rate for 10 s and for 1 min. Is there a factor of 6 difference? 

11. Examine different types of shoes, including sports shoes and thongs. In terms of physics, why are the bottom surfaces designed as they are? 
What differences will dry and wet conditions make for these surfaces? 

12. Would you expect your height to be different depending upon the time of day? Why or why not? 

13. Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall? 

14. Explain why pregnant women often suffer from back strain late in their pregnancy. 

15. An old carpenter's trick to keep nails from bending when they are pounded into hard materials is to grip the center of the nail firmly with pliers. 
Why does this help? 

16. When a glass bottle full of vinegar warms up, both the vinegar and the glass expand, but vinegar expands significantly more with temperature 
than glass. The bottle will break if it was filled to its tightly capped lid. Explain why, and also explain how a pocket of air above the vinegar would 
prevent the break. (This is the function of the air above liquids in glass containers.) 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 181 



Problems & Exercises 



5.1 Friction 

1. A physics major is cooking breakfast when he notices that the frictional 
force between his steel spatula and his Teflon frying pan is only 0.200 N. 
Knowing the coefficient of kinetic friction between the two materials, he 
quickly calculates the normal force. What is it? 

2. (a) When rebuilding her car's engine, a physics major must exert 300 
N of force to insert a dry steel piston into a steel cylinder. What is the 
normal force between the piston and cylinder? (b) What force would she 
have to exert if the steel parts were oiled? 

3. (a) What is the maximum frictional force in the knee joint of a person 
who supports 66.0 kg of her mass on that knee? (b) During strenuous 
exercise it is possible to exert forces to the joints that are easily ten times 
greater than the weight being supported. What is the maximum force of 
friction under such conditions? The frictional forces in joints are relatively 
small in all circumstances except when the joints deteriorate, such as 
from injury or arthritis. Increased frictional forces can cause further 
damage and pain. 

4. Suppose you have a 120-kg wooden crate resting on a wood floor, (a) 
What maximum force can you exert horizontally on the crate without 
moving it? (b) If you continue to exert this force once the crate starts to 
slip, what will its acceleration then be? 

5. (a) If half of the weight of a small 1.00xl0 3 kg utility truck is 

supported by its two drive wheels, what is the maximum acceleration it 
can achieve on dry concrete? (b) Will a metal cabinet lying on the 
wooden bed of the truck slip if it accelerates at this rate? (c) Solve both 
problems assuming the truck has four-wheel drive. 

6. A team of eight dogs pulls a sled with waxed wood runners on wet 
snow (mush!). The dogs have average masses of 19.0 kg, and the 
loaded sled with its rider has a mass of 210 kg. (a) Calculate the 
acceleration starting from rest if each dog exerts an average force of 185 
N backward on the snow, (b) What is the acceleration once the sled 
starts to move? (c) For both situations, calculate the force in the coupling 
between the dogs and the sled. 

7. Consider the 65.0-kg ice skater being pushed by two others shown in 
Figure 5.21. (a) Find the direction and magnitude of F tot , the total force 

exerted on her by the others, given that the magnitudes F± and F 2 are 

26.4 N and 18.6 N, respectively, (b) What is her initial acceleration if she 
is initially stationary and wearing steel-bladed skates that point in the 
direction of F tot ? (c) What is her acceleration assuming she is already 

moving in the direction of F tot ? (Remember that friction always acts in 

the direction opposite that of motion or attempted motion between 
surfaces in contact.] 




Free-body diagram 






tb> 



to) 

Figure 5.21 

8. Show that the acceleration of any object down a frictionless incline that 
makes an angle 6 with the horizontal is a = g sin 6 . (Note that this 

acceleration is independent of mass.) 

9. Show that the acceleration of any object down an incline where friction 
behaves simply (that is, where / k = jUyJV ) is 

a = g( sin 6 — ^cos 6). Note that the acceleration is independent of 



mass and reduces to the expression found in the previous problem when 
friction becomes negligibly small (^ k = 0). 

10. Calculate the deceleration of a snow boarder going up a 5.0° , slope 
assuming the coefficient of friction for waxed wood on wet snow. The 
result of Exercise 5.1 may be useful, but be careful to consider the fact 
that the snow boarder is going uphill. Explicitly show how you follow the 
steps in Problem-Solving Strategies. 

11. (a) Calculate the acceleration of a skier heading down a 10.0° slope, 
assuming the coefficient of friction for waxed wood on wet snow, (b) Find 
the angle of the slope down which this skier could coast at a constant 
velocity. You can neglect air resistance in both parts, and you will find the 
result of Exercise 5.1 to be useful. Explicitly show how you follow the 
steps in the Problem-Solving Strategies. 

12. If an object is to rest on an incline without slipping, then friction must 
equal the component of the weight of the object parallel to the incline. 
This requires greater and greater friction for steeper slopes. Show that 
the maximum angle of an incline above the horizontal for which an object 

will not slide down is 6 = tan" |i s . You may use the result of the 

previous problem. Assume that a = and that static friction has 
reached its maximum value. 

13. Calculate the maximum deceleration of a car that is heading down a 
6° slope (one that makes an angle of 6° with the horizontal) under the 
following road conditions. You may assume that the weight of the car is 
evenly distributed on all four tires and that the coefficient of static friction 
is involved — that is, the tires are not allowed to slip during the 
deceleration. (Ignore rolling.) Calculate for a car: (a) On dry concrete, (b) 
On wet concrete, (c) On ice, assuming that ju s = 0.100 , the same as 

for shoes on ice. 

14. Calculate the maximum acceleration of a car that is heading up a 4° 
slope (one that makes an angle of 4° with the horizontal) under the 
following road conditions. Assume that only half the weight of the car is 
supported by the two drive wheels and that the coefficient of static friction 
is involved — that is, the tires are not allowed to slip during the 
acceleration. (Ignore rolling.) (a) On dry concrete, (b) On wet concrete, 
(c) On ice, assuming that |i s = 0.100 , the same as for shoes on ice. 

15. Repeat Exercise 5.2 for a car with four-wheel drive. 

16. A freight train consists of two 8.00X 10 -kg engines and 45 cars 

with average masses of 5.50X 10 kg . (a) What force must each 
engine exert backward on the track to accelerate the train at a rate of 
5.00xl0" 2 m/s 2 if the force of friction is 7.50X10 5 N , assuming 
the engines exert identical forces? This is not a large frictional force for 
such a massive system. Rolling friction for trains is small, and 
consequently trains are very energy-efficient transportation systems, (b) 
What is the force in the coupling between the 37th and 38th cars (this is 
the force each exerts on the other), assuming all cars have the same 
mass and that friction is evenly distributed among all of the cars and 
engines? 

17. Consider the 52.0-kg mountain climber in Figure 5.22. (a) Find the 
tension in the rope and the force that the mountain climber must exert 
with her feet on the vertical rock face to remain stationary. Assume that 
the force is exerted parallel to her legs. Also, assume negligible force 
exerted by her arms, (b) What is the minimum coefficient of friction 
between her shoes and the cliff? 



182 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 




Figure 5.22 Part of the climber's weight is supported by her rope and part by friction 
between her feet and the rock face. 

18. A contestant in a winter sporting event pushes a 45.0-kg block of ice 
across a frozen lake as shown in Figure 5.23(a). (a) Calculate the 
minimum force F he must exert to get the block moving, (b) What is its 
acceleration once it starts to move, if that force is maintained? 

19. Repeat Exercise 5.3 with the contestant pulling the block of ice with 
a rope over his shoulder at the same angle above the horizontal as 
shown in Figure 5.23(b). 




(a) 




(b) 

Figure 5.23 Which method of sliding a block of ice requires less force — (a) pushing or 
(b) pulling at the same angle above the horizontal? 

5.2 Drag Forces 

20. The terminal velocity of a person falling in air depends upon the 
weight and the area of the person facing the fluid. Find the terminal 
velocity (in meters per second and kilometers per hour) of an 80.0-kg 
skydiver falling in a pike (headfirst) position with a surface area of 

0.140 m 2 . 

21. A 60-kg and a 90-kg skydiver jump from an airplane at an altitude of 
6000 m, both falling in the pike position. Make some assumption on their 
frontal areas and calculate their terminal velocities. How long will it take 
for each skydiver to reach the ground (assuming the time to reach 



terminal velocity is small)? Assume all values are accurate to three 
significant digits. 

22. A 560-g squirrel with a surface area of 930 cm falls from a 5.0-m 
tree to the ground. Estimate its terminal velocity. (Use a drag coefficient 
for a horizontal skydiver.) What will be the velocity of a 56-kg person 
hitting the ground, assuming no drag contribution in such a short 
distance? 

23. To maintain a constant speed, the force provided by a car's engine 
must equal the drag force plus the force of friction of the road (the rolling 
resistance), (a) What are the drag forces at 70 km/h and 100 km/h for a 

Toyota Camry? (Drag area is 0.70 m ) (b) What is the drag force at 70 

km/h and 100 km/h for a Hummer H2? (Drag area is 2.44 m ) Assume 
all values are accurate to three significant digits. 

24. By what factor does the drag force on a car increase as it goes from 
65 to 110 km/h? 

25. Calculate the velocity a spherical rain drop would achieve falling from 
5.00 km (a) in the absence of air drag (b) with air drag. Take the size 

across of the drop to be 4 mm, the density to be 1.00x10 kg/m , and 

9 

the surface area to be nr . 

26. Using Stokes' law, verify that the units for viscosity are kilograms per 
meter per second. 

27. Find the terminal velocity of a spherical bacterium (diameter 

2.00 |im ) falling in water. You will first need to note that the drag force is 

equal to the weight at terminal velocity. Take the density of the bacterium 
to be LlOxlO 3 kg/m 3 . 

28. Stokes' law describes sedimentation of particles in liquids and can be 
used to measure viscosity. Particles in liquids achieve terminal velocity 
quickly. One can measure the time it takes for a particle to fall a certain 
distance and then use Stokes' law to calculate the viscosity of the liquid. 

Suppose a steel ball bearing (density 7.8x10 kg/m , diameter 

3.0 mm ) is dropped in a container of motor oil. It takes 12 s to fall a 
distance of 0.60 m. Calculate the viscosity of the oil. 

5.3 Elasticity: Stress and Strain 

29. During a circus act, one performer swings upside down hanging from 
a trapeze holding another, also upside-down, performer by the legs. If the 
upward force on the lower performer is three times her weight, how much 
do the bones (the femurs) in her upper legs stretch? You may assume 
each is equivalent to a uniform rod 35.0 cm long and 1.80 cm in radius. 
Her mass is 60.0 kg. 

30. During a wrestling match, a 150 kg wrestler briefly stands on one 
hand during a maneuver designed to perplex his already moribund 
adversary. By how much does the upper arm bone shorten in length? 
The bone can be represented by a uniform rod 38.0 cm in length and 
2.10 cm in radius. 



31. (a) The "lead" in pencils is a graphite composition with a Young's 
modulus of about 1x10 N/m . Calculate the change in length of the 
lead in an automatic pencil if you tap it straight into the pencil with a force 
of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long, (b) Is the 
answer reasonable? That is, does it seem to be consistent with what you 
have observed when using pencils? 

32. TV broadcast antennas are the tallest artificial structures on Earth. In 
1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the 
top of one 610-m high antenna to perform gravity experiments. By how 
much was the antenna compressed, if we consider it to be equivalent to a 
steel cylinder 0.150 m in radius? 

33. (a) By how much does a 65.0-kg mountain climber stretch her 
0.800-cm diameter nylon rope when she hangs 35.0 m below a rock 
outcropping? (b) Does the answer seem to be consistent with what you 
have observed for nylon ropes? Would it make sense if the rope were 
actually a bungee cord? 



CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 183 



34. A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a 
solid cylinder 4.00 cm in diameter. A strong wind bends the pole much as 
a horizontal force of 900 N exerted at the top would. How far to the side 
does the top of the pole flex? 

35. As an oil well is drilled, each new section of drill pipe supports its own 
weight and that of the pipe and drill bit beneath it. Calculate the stretch in 
a new 6.00 m length of steel pipe that supports 3.00 km of pipe having a 
mass of 20.0 kg/m and a 100-kg drill bit. The pipe is equivalent in 
strength to a solid cylinder 5.00 cm in diameter. 

36. Calculate the force a piano tuner applies to stretch a steel piano wire 
8.00 mm, if the wire is originally 0.850 mm in diameter and 1.35 m long. 

37. A vertebra is subjected to a shearing force of 500 N. Find the shear 
deformation, taking the vertebra to be a cylinder 3.00 cm high and 4.00 
cm in diameter. 

38. A disk between vertebrae in the spine is subjected to a shearing force 
of 600 N. Find its shear deformation, taking it to have the shear modulus 

of 1x10 N/m . The disk is equivalent to a solid cylinder 0.700 cm 
high and 4.00 cm in diameter. 

39. When using a pencil eraser, you exert a vertical force of 6.00 N at a 
distance of 2.00 cm from the hardwood-eraser joint. The pencil is 6.00 
mm in diameter and is held at an angle of 20.0° to the horizontal, (a) By 
how much does the wood flex perpendicular to its length? (b) How much 
is it compressed lengthwise? 

40. To consider the effect of wires hung on poles, we take data from 
Example 4.8, in which tensions in wires supporting a traffic light were 
calculated. The left wire made an angle 30.0° below the horizontal with 
the top of its pole and carried a tension of 108 N. The 12.0 m tall hollow 
aluminum pole is equivalent in strength to a 4.50 cm diameter solid 
cylinder, (a) How far is it bent to the side? (b) By how much is it 
compressed? 

41. A farmer making grape juice fills a glass bottle to the brim and caps it 
tightly. The juice expands more than the glass when it warms up, in such 

away that the volume increases by 0.2% (that is, AV/V = 2xl0~ 3 ) 
relative to the space available. Calculate the force exerted by the juice 
per square centimeter if its bulk modulus is 1.8x10 N/m .assuming 
the bottle does not break. In view of your answer, do you think the bottle 
will survive? 

42. (a) When water freezes, its volume increases by 9.05% (that is, 

A V / Vq = 9.05x 10 ). What force per unit area is water capable of 

exerting on a container when it freezes? (It is acceptable to use the bulk 
modulus of water in this problem.) (b) Is it surprising that such forces can 
fracture engine blocks, boulders, and the like? 

43. This problem returns to the tightrope walker studied in Example 4.6, 
who created a tension of 3.94x10 N in a wire making an angle 5.0° 
below the horizontal with each supporting pole. Calculate how much this 
tension stretches the steel wire if it was originally 15 m long and 0.50 cm 
in diameter. 

44. The pole in Figure 5.24 is at a 90.0° bend in a power line and is 
therefore subjected to more shear force than poles in straight parts of the 
line. The tension in each line is 4.00X 10 N , at the angles shown. The 
pole is 15.0 m tall, has an 18.0 cm diameter, and can be considered to 
have half the strength of hardwood, (a) Calculate the compression of the 
pole, (b) Find how much it bends and in what direction, (c) Find the 
tension in a guy wire used to keep the pole straight if it is attached to the 
top of the pole at an angle of 30.0° with the vertical. (Clearly, the guy 
wire must be in the opposite direction of the bend.) 




Figure 5.24 This telephone pole is at a 90° bend in a power line. A guy wire is 
attached to the top of the pole at an angle of 30° with the vertical. 



184 CHAPTER 5 | FURTHER APPLICATIONS OF NEWTON'S LAWS: FRICTION, DRAG, AND ELASTICITY 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 185 




Figure 6.1 This Australian Grand Prix Formula 1 race car moves in a circular path as it makes the turn. Its wheels also spin rapidly — the latter completing many revolutions, the 
former only part of one (a circular arc). The same physical principles are involved in each, (credit: Richard Munckton) 



Learning Objectives 



6.1. Rotation Angle and Angular Velocity 

Define arc length, rotation angle, radius of curvature and angular velocity. 
Calculate the angular velocity of a car wheel spin. 

6.2. Centripetal Acceleration 
Establish the expression for centripetal acceleration. 
Explain the centrifuge. 

6.3. Centripetal Force 
Calculate coefficient of friction on a car tire. 
Calculate ideal speed and angle of a car on a turn. 

Fictitious Forces and Non-inertial Frames: The Coriolis Force 

Discuss the inertial frame of reference. 

Discuss the non-inertial frame of reference. 

Describe the effects of the Coriolis force. 
Newton's Universal Law of Gravitation 

Explain Earth's gravitational force. 

Describe the gravitational effect of the Moon on Earth. 

Discuss weightlessness in space. 

Examine the Cavendish experiment 
6.6. Satellites and Kepler's Laws: An Argument for Simplicity 

State Kepler's laws of planetary motion. 

Derive the third Kepler's law for circular orbits. 

Discuss the Ptolemaic model of the universe. 



6.4 



6.5 



186 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 



Introduction to Uniform Circular Motion and Gravitation 



Many motions, such as the arc of a bird's flight or Earth's path around the Sun, are curved. Recall that Newton's first law tells us that motion is along 
a straight line at constant speed unless there is a net external force. We will therefore study not only motion along curves, but also the forces that 
cause it, including gravitational forces. In some ways, this chapter is a continuation of Dynamics: Newton's Laws of Motion as we study more 
applications of Newton's laws of motion. 

This chapter deals with the simplest form of curved motion, uniform circular motion, motion in a circular path at constant speed. Studying this topic 
illustrates most concepts associated with rotational motion and leads to the study of many new topics we group under the name rotation. Pure 
rotational motion occurs when points in an object move in circular paths centered on one point. Pure translational motion is motion with no rotation. 
Some motion combines both types, such as a rotating hockey puck moving along ice. 

6.1 Rotation Angle and Angular Velocity 

In Kinematics, we studied motion along a straight line and introduced such concepts as displacement, velocity, and acceleration. Two-Dimensional 
Kinematics dealt with motion in two dimensions. Projectile motion is a special case of two-dimensional kinematics in which the object is projected 
into the air, while being subject to the gravitational force, and lands a distance away. In this chapter, we consider situations where the object does not 
land but moves in a curve. We begin the study of uniform circular motion by defining two angular quantities needed to describe rotational motion. 

Rotation Angle 

When objects rotate about some axis — for example, when the CD (compact disc) in Figure 6.2 rotates about its center — each point in the object 
follows a circular arc. Consider a line from the center of the CD to its edge. Each pit used to record sound along this line moves through the same 
angle in the same amount of time. The rotation angle is the amount of rotation and is analogous to linear distance. We define the rotation angle A6 
to be the ratio of the arc length to the radius of curvature: 



A0 = 4^. 



(6.1) 




Figure 6.2 All points on a CD travel in circular arcs. The pits along a line from the center to the edge all move through the same angle A# in a time A^ . 




A0 = 



As 



Figure 6.3 The radius of a circle is rotated through an angle A# . The arc length As is described on the circumference. 

The arc length As is the distance traveled along a circular path as shown in Figure 6.3 Note that r is the radius of curvature of the circular path. 

We know that for one complete revolution, the arc length is the circumference of a circle of radius r . The circumference of a circle is 2%r . Thus for 
one complete revolution the rotation angle is 



&0 = 2ZL = 2n . 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 187 

(6.2) 



This result is the basis for defining the units used to measure rotation angles, A6 to be radians (rad), defined so that 

27C rad = 1 revolution. 
A comparison of some useful angles expressed in both degrees and radians is shown in Table 6.1. 



Table 6.1 Comparison of Angular Units 



(6.3) 



Degree Measures Radian Measure 


30° 


71 

6 


60° 


n 
3 


90° 


n 

2 


120° 


2tc 
3 


135° 


3tc 
4 


180° 


it 



A8- 



A0 = 




AS2 



Figure 6.4 Points 1 and 2 rotate through the same angle ( A6 ), but point 2 moves through a greater arc length (As) because it is at a greater distance from the center of 
rotation (r) . 

If A6 = 2n rad, then the CD has made one complete revolution, and every point on the CD is back at its original position. Because there are 360° 
in a circle or one revolution, the relationship between radians and degrees is thus 

2n rad = 360° (6.4) 

so that 

lrad= 3|01 = 57.30 (6.5) 

2% 

Angular Velocity 

How fast is an object rotating? We define angular velocity co as the rate of change of an angle. In symbols, this is 

a) = M (6-6) 

At 

where an angular rotation A6 takes place in a time At . The greater the rotation angle in a given amount of time, the greater the angular velocity. 
The units for angular velocity are radians per second (rad/s). 

Angular velocity w is analogous to linear velocity v . To get the precise relationship between angular and linear velocity, we again consider a pit on 
the rotating CD. This pit moves an arc length As in a time A^ , and so it has a linear velocity 

V = M (6-7) 

At' 



188 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 

From A6 = ^jr- we see that As = rA6 . Substituting this into the expression for v gives 

rAO (6.8) 

At 

We write this relationship in two different ways and gain two different insights: 

v = roo or oo = j. ( 6 - 9 ) 

The first relationship in v = roo or oo = j states that the linear velocity v is proportional to the distance from the center of rotation, thus, it is largest 
for a point on the rim (largest r ), as you might expect. We can also call this linear speed v of a point on the rim the tangential speed. The second 
relationship in v = roo or oo = y can be illustrated by considering the tire of a moving car. Note that the speed of a point on the rim of the tire is the 

same as the speed v of the car. See Figure 6.5. So the faster the car moves, the faster the tire spins — large v means a large oo , because 
v = roo . Similarly, a larger-radius tire rotating at the same angular velocity ( oo ) will produce a greater linear speed ( v ) for the car. 




Figure 6.5 A car moving at a velocity V to the right has a tire rotating with an angular velocity 00 The speed of the tread of the tire relative to the axle is V , the same as if 
the car were jacked up. Thus the car moves forward at linear velocity V = roo , where r is the tire radius. A larger angular velocity for the tire means a greater velocity for 
the car. 



Example 6.1 How Fast Does a Car Tire Spin? 



Calculate the angular velocity of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h ). See Figure 6.5. 

Strategy 

Because the linear speed of the tire rim is the same as the speed of the car, we have v = 15.0 m/s. The radius of the tire is given to be 

r = 0.300 m. Knowing v and r , we can use the second relationship in v = roo, oo = j to calculate the angular velocity. 

Solution 

To calculate the angular velocity, we will use the following relationship: 



oo — r . 



L (6.10) 

Substituting the knowns, 

oo = ^°F /S = 50.0 rad/s. (611) 

0.300 m 

Discussion 

When we cancel units in the above calculation, we get 50.0/s. But the angular velocity must have units of rad/s. Because radians are actually 
unitless (radians are defined as a ratio of distance), we can simply insert them into the answer for the angular velocity. Also note that if an earth 
mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would 
have an angular velocity 

oo = (15.0 m/s) /(1.20 m) = 12.5 rad/s. (6.12) 

Both oo and v have directions (hence they are angular and linear velocities, respectively). Angular velocity has only two directions with respect to 
the axis of rotation — it is either clockwise or counterclockwise. Linear velocity is tangent to the path, as illustrated in Figure 6.6. 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 189 



Take-Home Experiment 



Tie an object to the end of a string and swing it around in a horizontal circle above your head (swing at your wrist). Maintain uniform speed as the 
object swings and measure the angular velocity of the motion. What is the approximate speed of the object? Identify a point close to your hand 
and take appropriate measurements to calculate the linear speed at this point. Identify other circular motions and measure their angular 
velocities. 




Figure 6.6 As an object moves in a circle, here a fly on the edge of an old-fashioned vinyl record, its instantaneous velocity is always tangent to the circle. The direction of the 
angular velocity is clockwise in this case. 

PhET Explorations: Ladybug Revolution 

PI 

tZ "^PhET Interactive Simulation 

Figure 6.7 Ladybug Revolution (http://cnx.Org/content/m42083/l.4/rotation_en.jar) 

Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or 
angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs. 



6.2 Centripetal Acceleration 

We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the 
direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be 
constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at 
constant speed, you are in uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The 
sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we examine the direction and 
magnitude of that acceleration. 

Figure 6.8 shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the 
path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This 
pointing is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion (resulting from a net 
external force) the centripetal acceleration a c ); centripetal means "toward the center" or "center seeking." 



190 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 



AV = V 2 - V 




Figure 6.8 The directions of the velocity of an object at two different points are shown, and the change in velocity Av is seen to point directly toward the center of curvature. 
(See small inset.) Because a c = Av/ A^ , the acceleration is also toward the center; a c is called centripetal acceleration. (Because A6 is very small, the arc length 
As is equal to the chord length Ar for small time differences.) 

The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity 
vectors and the one formed by the radii r and As are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The two 
equal sides of the velocity vector triangle are the speeds Vj = v 2 = v . Using the properties of two similar triangles, we obtain 



Av 
v 

Acceleration is Av / At , and so we first solve this expression for Av : 



r ' 



Then we divide this by At , yielding 



Av = %As. 



Av v v As 

At r At' 



(6.13) 



(6.14) 



(6.15) 



Finally, noting that Av I At = a c and that As/ At = v , the linear or tangential speed, we see that the magnitude of the centripetal acceleration is 



a c = y F , 



(6.16) 



which is the acceleration of an object in a circle of radius r at a speed v . So, centripetal acceleration is greater at high speeds and in sharp curves 
(smaller radius), as you have noticed when driving a car. But it is a bit surprising that a c is proportional to speed squared, implying, for example, that 
it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that a c is greater for tighter turns, as you 
have probably noticed. 

9 9 

It is also useful to express a c in terms of angular velocity. Substituting v = rw into the above expression, we find a c = (rco) I r = rco . We can 
express the magnitude of centripetal acceleration using either of two equations: 

V 2 2 

a c = -y-; a c = rco . 

Recall that the direction of a c is toward the center. You may use whichever expression is more convenient, as illustrated in examples below. 



(6.17) 



A centrifuge (see Figure 6.9b) is a rotating device used to separate specimens of different densities. High centripetal acceleration significantly 
decreases the time it takes for separation to occur, and makes separation possible with small samples. Centrifuges are used in a variety of 
applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid 
medium and the separation of macromolecules, such as DNA and protein, from a solution. Centrifuges are often rated in terms of their centripetal 
acceleration relative to acceleration due to gravity (g) ; maximum centripetal acceleration of several hundred thousand g is possible in a vacuum. 

Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of 
Earth's gravity. 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 191 



Example 6.2 How Does the Centripetal Acceleration of a Car Around a Curve Compare with That Due to 
Gravity? 



What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? 
Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 6.9(a). 

Strategy 

2 
Because v and r are given, the first expression in a c = ^-; a c = rco is the most convenient to use. 

Solution 

Entering the given values of v = 25.0 m/s and r = 500 m into the first expression for a c gives 

- -^- (25-0 m/s) 2 _ L25m/s2 _ 



(6.18) 



c r 500 m 

Discussion 

To compare this with the acceleration due to gravity (g = 9.80 m/s ) , we take the ratio of a c / g = (1 .25 m/s 1/(9.80 m/s 1 = 0.128 . Thus, 
a c = 0.128 g and is noticeable especially if you were not wearing a seat belt. 




(a) Car around corner 



m 







(b) Centrifuge 

Figure 6.9 (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this centripetal acceleration is found 
in Example 6.2. (b) A particle of mass in a centrifuge is rotating at constant angular velocity . It must be accelerated perpendicular to its velocity or it would continue in a 
straight line. The magnitude of the necessary acceleration is found in Example 6.3. 



Example 6.3 How Big Is the Centripetal Acceleration in an Ultracentrifuge . 



Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at 7.5 X 10 rev/min. Determine the 

ratio of this acceleration to that due to gravity. See Figure 6.9(b). 

Strategy 



192 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 

The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity co . Because r is 

v 2 2 

given, we can use the second expression in the equation a c = -y-; a c = rco to calculate the centripetal acceleration. 

Solution 

To convert 7.50x10 rev/min to radians per second, we use the facts that one revolution is 27crad and one minute is 60.0 s. Thus, 

co = 7.50X 10 4 J^x^p^X-L^ = 7854 rad/s. (619) 

mm 1 rev 60.0 s 

2 
Now the centripetal acceleration is given by the second expression in a c = ^-; a c = rco as 



a c = rco 1 . < 6 - 2 °) 



Converting 7.50 cm to meters and substituting known values gives 



2 = 4^v10 6 m/c 2 ( 6 - 21 ) 



a c = (0.0750 m)(7854 rad/s) z = 4.63x10° m/s z . 
Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of a c to g yields 

a c _ 4.63X10 6 _ 47?y1 5 ( 6 - 22 ) 

8 ~ 9.80 " 4 - /ZXiU ' 

Discussion 

This last result means that the centripetal acceleration is 472,000 times as strong as g . It is no wonder that such high co centrifuges are called 

ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other 
materials. 

Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is 
needed to cause a centripetal acceleration. In Centripetal Force, we will consider the forces involved in circular motion. 

PhET Explorations: Ladybug Motion 2D 



Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the 
vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior. 



«T 



«# 



vf 



PhET Interactive Simulation 



Figure 6.10 Ladybug Motion 2D (http://cnx.Org/content/m42084/l.5/ladybug-motion-2d_en.jar) 

6.3 Centripetal Force 

Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the 
force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a 
spinning centrifuge. 

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the 
same as the direction of centripetal acceleration. According to Newton's second law of motion, net force is mass times acceleration: net F = ma . 
For uniform circular motion, the acceleration is the centripetal acceleration — a = a c . Thus, the magnitude of centripetal force F c is 

F c = ma c . (6.23) 

2 
By using the expressions for centripetal acceleration a c from a c = ^-; a c = rco , we get two expressions for the centripetal force F c in terms of 

mass, velocity, angular velocity, and radius of curvature: 

F c = ra-y-; F c = mrco . 

You may use whichever expression for centripetal force is more convenient. Centripetal force F c is always perpendicular to the path and pointing to 
the center of curvature, because a c is perpendicular to the velocity and pointing to the center of curvature. 

Note that if you solve the first expression for r , you get 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 193 

(6.25) 



This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature — that is, a tight curve. 

Path 




F c ts parallel to a since F c = ma c 




Figure 6.11 Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the F c , the smaller the radius of curvature r and the sharper the 
curve. The second curve has the same V , but a larger F c produces a smaller r' . 



Example 6.4 What Coefficient of Friction Do Car Tires Need on a Flat Curve? 



(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s. 

(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason 
that keeps the car from slipping (see Figure 6.12). 

Strategy and Solution for (a) 



We know that F c = ^^- . Thus, 



_ mv 2 _ (900 kg)(25.0 m/s) 2 _ 
Fc ~ — - (500^ -H25N. 



(6.26) 



Strategy for (b) 

Figure 6.12 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and 
because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction 
(at which the tires roll but do not slip) is ju s N , where ju s is the static coefficient of friction and N is the normal force. The normal force equals 

the car's weight on level ground, so that TV = mg . Thus the centripetal force in this situation is 



F c = f = A<sN = jLi s mg. 
Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for F c from the equation 



F c = m^ 



) 



m^j- = ju s mg. 



(6.27) 



(6.28) 



(6.29) 



We solve this for /u s , noting that mass cancels, and obtain 



194 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 



Solution for (b) 

Substituting the knowns, 



rs ~ rg- 



(25.0 m/s)" 



= 0.13. 



(6.30) 



(6.31) 



(500 m)(9.80 m/s z ) 

(Because coefficients of friction are approximate, the answer is given to only two digits.] 
Discussion 



We could also solve part (a) using the first expression in ^ c 



V 

— m—^ 



F c = mrco 



\ 



because m, v, and r are given. The coefficient of friction found in 



part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, 
because static friction is a responsive force, being able to assume a value less than but no more than jli s N . A higher coefficient would also 

allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that 
mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction 
is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would 
be less as will be discussed below. 



Free-body 
diagram 




Fc = /i s N 



Figure 6.12 This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires 
and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway. 

Let us now consider banked curves, where the slope of the road helps you negotiate the curve. See Figure 6.13. The greater the angle 6 , the 
faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an "ideally banked curve," the 
angle 6 is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an 
expression for 6 for an ideally banked curve and consider an example related to it. 

For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in 
the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, 
it is most convenient to consider components along perpendicular axes — in this case, the vertical and horizontal directions. 

Figure 6.13 shows a free body diagram for a car on a frictionless banked curve. If the angle 6 is ideal for the speed and radius, then the net external 
force will equal the necessary centripetal force. The only two external forces acting on the car are its weight w and the normal force of the road N . 
(A frictionless surface can only exert a force perpendicular to the surface — that is, a normal force.) These two forces must add to give a net external 
force that is horizontal toward the center of curvature and has magnitude mv /r . Because this is the crucial force and it is horizontal, we use a 
coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal 
force— that is, 

yvsin^ = ^. (6 ' 32) 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 195 



Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external 
forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is N cos 6 , 
and the only other vertical force is the car's weight. These must be equal in magnitude; thus, 

N cos 6 = mg. (6.33) 

Now we can combine the last two equations to eliminate N and get an expression for 6 , as desired. Solving the second equation for 
JSf = mg/ (cos 6) , and substituting this into the first yields 



mg- 



sin (9 
cos 6 



mv 
r 



mg tan(0) = m ^r- 



(6.34) 
(6.35) 



Taking the inverse tangent gives 



6 = tan 



■») 



tan<9 



(ideally banked curve, no friction). 



(6.36) 



This expression can be understood by considering how 6 depends on v and r . A large 6 will be obtained for a large v and a small r . That is, 
roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed 
than if the curve is frictionless. Note that 6 does not depend on the mass of the vehicle. 




NsrnO = F c 
= ^net 



Figure 6.13 The car on this banked curve is moving away and turning to the left. 



Example 6.5 What Is the Ideal Speed to Take a Steeply Banked Tight Curve? ^^^^^^^^^^^^^^^^^H 


Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, 
with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at 
which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless. 


Strategy 


We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so 
that speed appears on the left-hand side and then substitute known quantities. 


Solution 


Starting with 


v 2 ( 6 - 37 ) 
tan 6 = -^ 


we get 


v = (rgtan<9) 1/2 . (6 - 38) 


Noting that tan 65.0° = 2.14, we obtain 


r 1/2 (6.39) 
v = [(100m)(9.80m/s 2 )(2.14)] 


= 45.8 m/s. 


Discussion 



196 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 

This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at 
significantly higher speeds. 

Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is 
involved — a number of these are presented in this chapter's Problems and Exercises. 

Take-Home Experiment 

Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the 
end of the club or racquet. You may choose to do this in slow motion. 

PhET Explorations: Gravity and Orbits 

Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances 
between different heavenly bodies, and turn off gravity to see what would happen without it! 



V* s *) 



<£ ~> PhET Interactive Simulation 



Figure 6.14 Gravity and Orbits (http://cnx.Org/content/m42086/l.5/gravity-and-orbits_en.jar) 



6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force 

What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of a tropical cyclone have in common? 
Each exhibits fictitious forces — unreal forces that arise from motion and may seem real, because the observer's frame of reference is accelerating or 
rotating. 

When taking off in a jet, most people would agree it feels as if you are being pushed back into the seat as the airplane accelerates down the runway. 
Yet a physicist would say that you tend to remain stationary while the seat pushes forward on you, and there is no real force backward on you. An 
even more common experience occurs when you make a tight curve in your car — say, to the right. You feel as if you are thrown (that is, forced) 
toward the left relative to the car. Again, a physicist would say that you are going in a straight line but the car moves to the right, and there is no real 
force on you to the left. Recall Newton's first law. 




(a) 



(b) 



Figure 6.15 (a) The car driver feels herself forced to the left relative to the car when she makes a right turn. This is a fictitious force arising from the use of the car as a frame of 
reference, (b) In the Earth's frame of reference, the driver moves in a straight line, obeying Newton's first law, and the car moves to the right. There is no real force to the left 
on the driver relative to Earth. There is a real force to the right on the car to make it turn. 

We can reconcile these points of view by examining the frames of reference used. Let us concentrate on people in a car. Passengers instinctively use 
the car as a frame of reference, while a physicist uses Earth. The physicist chooses Earth because it is very nearly an inertial frame of 
reference — one in which all forces are real (that is, in which all forces have an identifiable physical origin). In such a frame of reference, Newton's 
laws of motion take the form given in Dynamics: Newton's Laws of Motion The car is a non-inertial frame of reference because it is accelerated 
to the side. The force to the left sensed by car passengers is a fictitious force having no physical origin. There is nothing real pushing them left — the 
car, as well as the driver, is actually accelerating to the right. 

Let us now take a mental ride on a merry-go-round — specifically, a rapidly rotating playground merry-go-round. You take the merry-go-round to be 
your frame of reference because you rotate together. In that non-inertial frame, you feel a fictitious force, named centrifugal force (not to be 
confused with centripetal force), trying to throw you off. You must hang on tightly to counteract the centrifugal force. In Earth's frame of reference, 
there is no force trying to throw you off. Rather you must hang on to make yourself go in a circle because otherwise you would go in a straight line, 
right off the merry-go-round. 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 197 





■"net ""centripetal 



F - F 

1 f ict — ■ centrifugal 







Merry-go-round's rotating frame of reference 
(a) 



Inertial frame of reference 
(b) 



Figure 6.16 (a) A rider on a merry-go-round feels as if he is being thrown off. This fictitious force is called the centrifugal force — it explains the rider's motion in the rotating 
frame of reference, (b) In an inertial frame of reference and according to Newton's laws, it is his inertia that carries him off and not a real force (the unshaded rider has 
^net = ar| d heads in a straight line). A real force, ^"centripetal • ' s nee ded to cause a circular path. 



This inertial effect, carrying you away from the center of rotation if there is no centripetal force to cause circular motion, is put to good use in 
centrifuges (see Figure 6.17). A centrifuge spins a sample very rapidly, as mentioned earlier in this chapter. Viewed from the rotating frame of 
reference, the fictitious centrifugal force throws particles outward, hastening their sedimentation. The greater the angular velocity, the greater the 
centrifugal force. But what really happens is that the inertia of the particles carries them along a line tangent to the circle while the test tube is forced 
in a circular path by a centripetal force. 



centrifugal 
force 



y 




inertial 
force 



Particle continues to left 
as test tube moves up, 
Therefore particle moves 
down in tube by virtue of 
its inertia. 

Figure 6.17 Centrifuges use inertia to perform their task. Particles in the fluid sediment come out because their inertia carries them away from the center of rotation. The large 
angular velocity of the centrifuge quickens the sedimentation. Ultimately, the particles will come into contact with the test tube walls, which will then supply the centripetal force 
needed to make them move in a circle of constant radius. 

Let us now consider what happens if something moves in a frame of reference that rotates. For example, what if you slide a ball directly away from 
the center of the merry-go-round, as shown in Figure 6.18? The ball follows a straight path relative to Earth (assuming negligible friction) and a path 
curved to the right on the merry-go-round's surface. A person standing next to the merry-go-round sees the ball moving straight and the merry-go- 
round rotating underneath it. In the merry-go-round's frame of reference, we explain the apparent curve to the right by using a fictitious force, called 
the Coriolis force, that causes the ball to curve to the right. The fictitious Coriolis force can be used by anyone in that frame of reference to explain 
why objects follow curved paths and allows us to apply Newton's Laws in non-inertial frames of reference. 



198 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 




Path relative to 
the earth 



(a) 




Path relative to 
the merry-go-round 



(b) 
Figure 6.18 Looking down on the counterclockwise rotation of a merry-go-round, we see that a ball slid straight toward the edge follows a path curved to the right. The person 
slides the ball toward point B, starting at point A. Both points rotate to the shaded positions (A' and B') shown in the time that the ball follows the curved path in the rotating 
frame and a straight path in Earth's frame. 

Up until now, we have considered Earth to be an inertial frame of reference with little or no worry about effects due to its rotation. Yet such effects do 
exist — in the rotation of weather systems, for example. Most consequences of Earth's rotation can be qualitatively understood by analogy with the 
merry-go-round. Viewed from above the North Pole, Earth rotates counterclockwise, as does the merry-go-round in Figure 6.18. As on the merry-go- 
round, any motion in Earth's northern hemisphere experiences a Coriolis force to the right. Just the opposite occurs in the southern hemisphere; 
there, the force is to the left. Because Earth's angular velocity is small, the Coriolis force is usually negligible, but for large-scale motions, such as 
wind patterns, it has substantial effects. 

The Coriolis force causes hurricanes in the northern hemisphere to rotate in the counterclockwise direction, while the tropical cyclones (what 
hurricanes are called below the equator) in the southern hemisphere rotate in the clockwise direction. The terms hurricane, typhoon, and tropical 
storm are regionally-specific names for tropical cyclones, storm systems characterized by low pressure centers, strong winds, and heavy rains. 
Figure 6.19 helps show how these rotations take place. Air flows toward any region of low pressure, and tropical cyclones contain particularly low 
pressures. Thus winds flow toward the center of a tropical cyclone or a low-pressure weather system at the surface. In the northern hemisphere, 
these inward winds are deflected to the right, as shown in the figure, producing a counterclockwise circulation at the surface for low-pressure zones 
of any type. Low pressure at the surface is associated with rising air, which also produces cooling and cloud formation, making low-pressure patterns 
quite visible from space. Conversely, wind circulation around high-pressure zones is clockwise in the northern hemisphere but is less visible because 
high pressure is associated with sinking air, producing clear skies. 

The rotation of tropical cyclones and the path of a ball on a merry-go-round can just as well be explained by inertia and the rotation of the system 
underneath. When non-inertial frames are used, fictitious forces, such as the Coriolis force, must be invented to explain the curved path. There is no 
identifiable physical source for these fictitious forces. In an inertial frame, inertia explains the path, and no force is found to be without an identifiable 
source. Either view allows us to describe nature, but a view in an inertial frame is the simplest and truest, in the sense that all forces have real origins 
and explanations. 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 199 




Figure 6.19 (a) The counterclockwise rotation of this northern hemisphere hurricane is a major consequence of the Coriolis force, (credit: NASA) (b) Without the Coriolis force, 
air would flow straight into a low-pressure zone, such as that found in tropical cyclones, (c) The Coriolis force deflects the winds to the right, producing a counterclockwise 
rotation, (d) Wind flowing away from a high-pressure zone is also deflected to the right, producing a clockwise rotation, (e) The opposite direction of rotation is produced by the 
Coriolis force in the southern hemisphere, leading to tropical cyclones, (credit: NASA) 



6.5 Newton's Universal Law of Gravitation 

What do aching feet, a falling apple, and the orbit of the Moon have in common? Each is caused by the gravitational force. Our feet are strained by 
supporting our weight — the force of Earth's gravity on us. An apple falls from a tree because of the same force acting a few meters above Earth's 
surface. And the Moon orbits Earth because gravity is able to supply the necessary centripetal force at a distance of hundreds of millions of meters. 
In fact, the same force causes planets to orbit the Sun, stars to orbit the center of the galaxy, and galaxies to cluster together. Gravity is another 
example of underlying simplicity in nature. It is the weakest of the four basic forces found in nature, and in some ways the least understood. It is a 
force that acts at a distance, without physical contact, and is expressed by a formula that is valid everywhere in the universe, for masses and 
distances that vary from the tiny to the immense. 

Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and 
astronomical motions. See Figure 6.20. But Newton was not the first to suspect that the same force caused both our weight and the motion of 
planets. His forerunner Galileo Galilei had contended that falling bodies and planetary motions had the same cause. Some of Newton's 
contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But 
Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic 
sections — circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph — it had been known for some time that moons, 
planets, and comets follow such paths, but no one had been able to propose a mechanism that caused them to follow these paths and not others. 



200 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 




Figure 6.20 According to early accounts, Newton was inspired to make the connection between falling bodies and astronomical motions when he saw an apple fall from a tree 
and realized that if the gravitational force could extend above the ground to a tree, it might also reach the Sun. The inspiration of Newton's apple is a part of worldwide folklore 
and may even be based in fact. Great importance is attached to it because Newton's universal law of gravitation and his laws of motion answered very old questions about 
nature and gave tremendous support to the notion of underlying simplicity and unity in nature. Scientists still expect underlying simplicity to emerge from their ongoing inquiries 
into nature. 

The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Stated 
in modern language, Newton's universal law of gravitation states that every particle in the universe attracts every other particle with a force along 
a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between 
them. 





CM 

Figure 6.21 Gravitational attraction is along a line joining the centers of mass of these two bodies. The magnitude of the force is the same on each, consistent with Newton's 
third law. 



Misconception Alert 

The magnitude of the force on each object (one has larger mass than the other) is the same, consistent with Newton's third law. 



The bodies we are dealing with tend to be large. To simplify the situation we assume that the body acts as if its entire mass is concentrated at one 
specific point called the center of mass (CM), which will be further explored in Linear Momentum and Collisions. For two bodies having masses 
m and M with a distance r between their centers of mass, the equation for Newton's universal law of gravitation is 



F = G 



mM 



(6.40) 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 201 

where F is the magnitude of the gravitational force and G is a proportionality factor called the gravitational constant. G is a universal 
gravitational constant — that is, it is thought to be the same everywhere in the universe. It has been measured experimentally to be 

-llN-m 2 < 6 - 41 ) 



G = 6.673x10" 



kg^ 



in SI units. Note that the units of G are such that a force in newtons is obtained from F = G^BM. ) w hen considering masses in kilograms and 

distance in meters. For example, two 1.000 kg masses separated by 1.000 m will experience a gravitational attraction of 6.673x10" N . This is 
an extraordinarily small force. The small magnitude of the gravitational force is consistent with everyday experience. We are unaware that even large 
objects like mountains exert gravitational forces on us. In fact, our body weight is the force of attraction of the entire Earth on us with a mass of 

6xl0 24 kg. 

Recall that the acceleration due to gravity g is about 9.80 m/s on Earth. We can now determine why this is so. The weight of an object mg is the 
gravitational force between it and Earth. Substituting mg for F in Newton's universal law of gravitation gives 

mg = G^f, < 6 - 42 > 

r 

where m is the mass of the object, M is the mass of Earth, and r is the distance to the center of Earth (the distance between the centers of mass 
of the object and Earth). See Figure 6.22. The mass m of the object cancels, leaving an equation for g : 



o-rM 
g - G—^. 

r 
Substituting known values for Earth's mass and radius (to three significant figures), 



g=6.67xl0~ 



24! 



HN-m 2 L5.98xlO Z4 kg 



6 W %2' 



kg z ) (6.38xl0°m) 



and we obtain a value for the acceleration of a falling body: 



g = 9.80 m/s 2 . 



(6.43) 



(6.44) 



(6.45) 




Figure 6.22 The distance between the centers of mass of Earth and an object on its surface is very nearly the same as the radius of Earth, because Earth is so much larger 
than the object. 

This is the expected value and is independent of the body's mass. Newton's law of gravitation takes Galileo's observation that all masses fall with the 
same acceleration a step further, explaining the observation in terms of a force that causes objects to fall — in fact, in terms of a universally existing 
force of attraction between masses. 

Take-Home Experiment 

Take a marble, a ball, and a spoon and drop them from the same height. Do they hit the floor at the same time? If you drop a piece of paper as 
well, does it behave like the other objects? Explain your observations. 

Making Connections 



Attempts are still being made to understand the gravitational force. As we shall see in Particle Physics, modern physics is exploring the 
connections of gravity to other forces, space, and time. General relativity alters our view of gravitation, leading us to think of gravitation as 
bending space and time. 



In the following example, we make a comparison similar to one made by Newton himself. He noted that if the gravitational force caused the Moon to 
orbit Earth, then the acceleration due to gravity should equal the centripetal acceleration of the Moon in its orbit. Newton found that the two 
accelerations agreed "pretty nearly." 



202 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 



Example 6.6 Earth's Gravitational Force Is the Centripetal Force Making the Moon Move in a Curved Path 



(a) Find the acceleration due to Earth's gravity at the distance of the Moon. 

(b) Calculate the centripetal acceleration needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth), and compare it with 
the value of the acceleration due to Earth's gravity that you have just found. 

Strategy for (a) 

This calculation is the same as the one finding the acceleration due to gravity at Earth's surface, except that r is the distance from the center of 

o 

Earth to the center of the Moon. The radius of the Moon's nearly circular orbit is 3.84x 10 m . 

Solution for (a) 

Substituting known values into the expression for g found above, remembering that M is the mass of Earth not the Moon, yields 

= G4=f6.67xlO-"N^A 5.98X10* kg <M9 



r 2 {' kg 2 ) (3.84x10 s m) 2 

2.70xl0- 3 m/s. 2 



Strategy for (b) 

Centripetal acceleration can be calculated using either form of 



a =£ 

u c — y 

a r = rco 



i 



(6.47) 



We choose to use the second form: 



rw\ < 6 - 48 ) 



where w is the angular velocity of the Moon about Earth. 

Solution for (b) 

Given that the period (the time it takes to make one complete rotation) of the Moon's orbit is 27.3 days, (d) and using 

1 dx24^x60m%<60^ = 86,400 s (649) 

d hr mm 



we see that 



The centripetal acceleration is 



rn - A(9 _ 2jtrad _ 9 ^ v i n -6rad (6.50) 

W ~ At ~ (27.3 d)(86,400 s/d) " z ' DDXiU s ■ 



2c = rw 2 = (3.84x10 s m)(2.66xl0" 6 rad/s) 2 (651) 



= 2.72xl0" 3 m/s. 2 

The direction of the acceleration is toward the center of the Earth. 

Discussion 

The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth's gravity found in (a). This 
agreement is approximate because the Moon's orbit is slightly elliptical, and Earth is not stationary (rather the Earth-Moon system rotates about 
its center of mass, which is located some 1700 km below Earth's surface). The clear implication is that Earth's gravitational force causes the 
Moon to orbit Earth. 

Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton's third law, if Earth exerts a force on the 
Moon, then the Moon should exert an equal and opposite force on Earth (see Figure 6.23). We do not sense the Moon's effect on Earth's motion, 
because the Moon's gravity moves our bodies right along with Earth but there are other signs on Earth that clearly show the effect of the Moon's 
gravitational force as discussed in Satellites and Kepler's Laws: An Argument for Simplicity. 



[nolto scale) 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 203 



Earth's path has 
"wiggles" (exaggerated) 




Earth-moon center of mass 
follows elliptical path 



(a) (b) 

Figure 6.23 (a) Earth and the Moon rotate approximately once a month around their common center of mass, (b) Their center of mass orbits the Sun in an elliptical orbit, but 
Earth's path around the Sun has "wiggles" in it. Similar wiggles in the paths of stars have been observed and are considered direct evidence of planets orbiting those stars. 
This is important because the planets' reflected light is often too dim to be observed. 



Tides 

Ocean tides are one very observable result of the Moon's gravity acting on Earth. Figure 6.24 is a simplified drawing of the Moon's position relative to 
the tides. Because water easily flows on Earth's surface, a high tide is created on the side of Earth nearest to the Moon, where the Moon's 
gravitational pull is strongest. Why is there also a high tide on the opposite side of Earth? The answer is that Earth is pulled toward the Moon more 
than the water on the far side, because Earth is closer to the Moon. So the water on the side of Earth closest to the Moon is pulled away from Earth, 
and Earth is pulled away from water on the far side. As Earth rotates, the tidal bulge (an effect of the tidal forces between an orbiting natural satellite 
and the primary planet that it orbits) keeps its orientation with the Moon. Thus there are two tides per day (the actual tidal period is about 12 hours 
and 25.2 minutes), because the Moon moves in its orbit each day as well). 

Low tide 





Moon 



Low tide 



(not to scale) 



Figure 6.24 The Moon causes ocean tides by attracting the water on the near side more than Earth, and by attracting Earth more than the water on the far side. The distances 
and sizes are not to scale. For this simplified representation of the Earth-Moon system, there are two high and two low tides per day at any location, because Earth rotates 
under the tidal bulge. 

The Sun also affects tides, although it has about half the effect of the Moon. However, the largest tides, called spring tides, occur when Earth, the 
Moon, and the Sun are aligned. The smallest tides, called neap tides, occur when the Sun is at a 90° angle to the Earth-Moon alignment. 




A 



Moon 



Sun 



spring tide 



(a) 




Moon spring tide 



(b) 



Sun 



r^ 




neap tide 



(c) 



Sun 



Moon 

Figure 6.25 (a, b) Spring tides: The highest tides occur when Earth, the Moon, and the Sun are aligned. (c)Neap tide: The lowest tides occur when the Sun lies at 90° to the 
Earth-Moon alignment. Note that this figure is not drawn to scale. 



204 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 

Tides are not unique to Earth but occur in many astronomical systems. The most extreme tides occur where the gravitational force is the strongest 
and varies most rapidly, such as near black holes (see Figure 6.26). A few likely candidates for black holes have been observed in our galaxy. These 
have masses greater than the Sun but have diameters only a few kilometers across. The tidal forces near them are so great that they can actually 
tear matter from a companion star. 




Figure 6.26 A black hole is an object with such strong gravity that not even light can escape it. This black hole was created by the supernova of one star in a two-star system. 
The tidal forces created by the black hole are so great that it tears matter from the companion star. This matter is compressed and heated as it is sucked into the black hole, 
creating light and X-rays observable from Earth. 



"Weightlessness" and Microgravity 

In contrast to the tremendous gravitational force near black holes is the apparent gravitational field experienced by astronauts orbiting Earth. What is 
the effect of "weightlessness" upon an astronaut who is in orbit for months? Or what about the effect of weightlessness upon plant growth? 
Weightlessness doesn't mean that an astronaut is not being acted upon by the gravitational force. There is no "zero gravity" in an astronaut's orbit. 
The term just means that the astronaut is in free-fall, accelerating with the acceleration due to gravity. If an elevator cable breaks, the passengers 
inside will be in free fall and will experience weightlessness. You can experience short periods of weightlessness in some rides in amusement parks. 




Figure 6.27 Astronauts experiencing weightlessness on board the International Space Station, (credit: NASA) 

Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface. 
Many interesting biology and physics topics have been studied over the past three decades in the presence of microgravity. Of immediate concern is 
the effect on astronauts of extended times in outer space, such as at the International Space Station. Researchers have observed that muscles will 
atrophy (waste away) in this environment. There is also a corresponding loss of bone mass. Study continues on cardiovascular adaptation to space 
flight. On Earth, blood pressure is usually higher in the feet than in the head, because the higher column of blood exerts a downward force on it, due 
to gravity. When standing, 70% of your blood is below the level of the heart, while in a horizontal position, just the opposite occurs. What difference 
does the absence of this pressure differential have upon the heart? 

Some findings in human physiology in space can be clinically important to the management of diseases back on Earth. On a somewhat negative 
note, spaceflight is known to affect the human immune system, possibly making the crew members more vulnerable to infectious diseases. 
Experiments flown in space also have shown that some bacteria grow faster in microgravity than they do on Earth. However, on a positive note, 
studies indicate that microbial antibiotic production can increase by a factor of two in space-grown cultures. One hopes to be able to understand 
these mechanisms so that similar successes can be achieved on the ground. In another area of physics space research, inorganic crystals and 
protein crystals have been grown in outer space that have much higher quality than any grown on Earth, so crystallography studies on their structure 
can yield much better results. 

Plants have evolved with the stimulus of gravity and with gravity sensors. Roots grow downward and shoots grow upward. Plants might be able to 
provide a life support system for long duration space missions by regenerating the atmosphere, purifying water, and producing food. Some studies 
have indicated that plant growth and development are not affected by gravity, but there is still uncertainty about structural changes in plants grown in 
a microgravity environment. 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 205 

The Cavendish Experiment: Then and Now 

As previously noted, the universal gravitational constant G is determined experimentally. This definition was first done accurately by Henry 
Cavendish (1731-1810), an English scientist, in 1798, more than 100 years after Newton published his universal law of gravitation. The 
measurement of G is very basic and important because it determines the strength of one of the four forces in nature. Cavendish's experiment was 
very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most), using apparatus 
like that in Figure 6.28. Remarkably, his value for G differs by less than 1% from the best modern value. 

One important consequence of knowing G was that an accurate value for Earth's mass could finally be obtained. This was done by measuring the 
acceleration due to gravity as accurately as possible and then calculating the mass of Earth M from the relationship Newton's universal law of 
gravitation gives 

r 

where m is the mass of the object, M is the mass of Earth, and r is the distance to the center of Earth (the distance between the centers of mass 
of the object and Earth). See Figure 6.21. The mass m of the object cancels, leaving an equation for g : 



^ _ jM (6.53) 

Rearranging to solve for M yields 



r 



„,2 (6.54) 

M =V- 

So M can be calculated because all quantities on the right, including the radius of Earth r , are known from direct measurements. We shall see in 
Satellites and Kepler's Laws: An Argument for Simplicity that knowing G also allows for the determination of astronomical masses. Interestingly, 
of all the fundamental constants in physics, G is by far the least well determined. 

The Cavendish experiment is also used to explore other aspects of gravity. One of the most interesting questions is whether the gravitational force 
depends on substance as well as mass — for example, whether one kilogram of lead exerts the same gravitational pull as one kilogram of water. A 
Hungarian scientist named Roland von Eotvos pioneered this inquiry early in the 20th century. He found, with an accuracy of five parts per billion, that 
the gravitational force does not depend on the substance. Such experiments continue today, and have improved upon Eotvos' measurements. 
Cavendish-type experiments such as those of Eric Adelberger and others at the University of Washington, have also put severe limits on the 
possibility of a fifth force and have verified a major prediction of general relativity — that gravitational energy contributes to rest mass. Ongoing 
measurements there use a torsion balance and a parallel plate (not spheres, as Cavendish used) to examine how Newton's law of gravitation works 
over sub-millimeter distances. On this small-scale, do gravitational effects depart from the inverse square law? So far, no deviation has been 
observed. 




Figure 6.28 Cavendish used an apparatus like this to measure the gravitational attraction between the two suspended spheres ( Hi ) and the two on the stand ( M ) by 
observing the amount of torsion (twisting) created in the fiber. Distance between the masses can be varied to check the dependence of the force on distance. Modern 
experiments of this type continue to explore gravity. 

6.6 Satellites and Kepler's Laws: An Argument for Simplicity 

Examples of gravitational orbits abound. Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris. The Moon's orbit 
about Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets about the Sun are no less interesting. 
If we look further, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through 
gravity. 

All these motions are governed by gravitational force, and it is possible to describe them to various degrees of precision. Precise descriptions of 
complex systems must be made with large computers. However, we can describe an important class of orbits without the use of computers, and we 
shall find it instructive to study them. These orbits have the following characteristics: 



206 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 



1. A small mass m orbits a much larger mass M . This allows us to view the motion as if M were stationary — in fact, as if from an inertial frame 
of reference placed on M — without significant error. Mass m is the satellite of M , if the orbit is gravitational ly bound. 

2. The system is isolated from other masses. This allows us to neglect any small effects due to outside masses. 

The conditions are satisfied, to good approximation, by Earth's satellites (including the Moon), by objects orbiting the Sun, and by the satellites of 
other planets. Historically, planets were studied first, and there is a classical set of three laws, called Kepler's laws of planetary motion, that describe 
the orbits of all bodies satisfying the two previous conditions (not just planets in our solar system). These descriptive laws are named for the German 
astronomer Johannes Kepler (1571-1630), who devised them after careful study (over some 20 years) of a large amount of meticulously recorded 
observations of planetary motion done by Tycho Brahe (1546-1601). Such careful collection and detailed recording of methods and data are 
hallmarks of good science. Data constitute the evidence from which new interpretations and meanings can be constructed. 

Kepler's Laws of Planetary Motion 
Kepler's First Law 

The orbit of each planet about the Sun is an ellipse with the Sun at one focus. 



y \ 


1 

\ 


m 




i 1[ is^ 








\ f, 


/ 


2 


f X 



Planet _ ft V 




Figure 6.29 (a) An ellipse is a closed curve such that the sum of the distances from a point on the curve to the two foci ( f± and f^ ) is a constant. You can draw an ellipse 

as shown by putting a pin at each focus, and then placing a string around a pencil and the pins and tracing a line on paper. A circle is a special case of an ellipse in which the 
two foci coincide (thus any point on the circle is the same distance from the center), (b) For any closed gravitational orbit, m follows an elliptical path with M at one focus. 
Kepler's first law states this fact for planets orbiting the Sun. 

Kepler's Second Law 

Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times (see Figure 6.30). 

Kepler's Third Law 

The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun. In 
equation form, this is 

Tl = rJ_ < 6 - 55 ) 

T 2"J' 



where T is the period (time for one orbit) and r is the average radius. This equation is valid only for comparing two small masses orbiting the same 
large one. Most importantly, this is a descriptive equation only, giving no information as to the cause of the equality. 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 207 



Planet 




Figure 6.30 The shaded regions have equal areas. It takes equal times for m to go from A to B, from C to D, and from E to F. The mass m moves fastest when it is closest 
to M . Kepler's second law was originally devised for planets orbiting the Sun, but it has broader validity. 

Note again that while, for historical reasons, Kepler's laws are stated for planets orbiting the Sun, they are actually valid for all bodies satisfying the 
two previously stated conditions. 



Example 6.7 Find the Time for One Orbit of an Earth Satellite 



Given that the Moon orbits Earth each 27.3 d and that it is an average distance of 3.84x10 m from the center of Earth, calculate the period of 

an artificial satellite orbiting at an average altitude of 1500 km above Earth's surface. 

Strategy 

T 2 r 3 
The period, or time for one orbit, is related to the radius of the orbit by Kepler's third law, given in mathematical form in —^r = -\ . Let us use 

1 2 r 2 

the subscript 1 for the Moon and the subscript 2 for the satellite. We are asked to find T 2 . The given information tells us that the orbital radius of 

o 

the Moon is r 1 = 3.84x 10 m , and that the period of the Moon is T 1 = 27.3 d . The height of the artificial satellite above Earth's surface is 
given, and so we must add the radius of Earth (6380 km) to get r 2 = (1500 + 6380) km = 7880 km . Now all quantities are known, and so 
T 2 can be found. 

Solution 

Kepler's third law is 

T 2 r 3 ' 

1 2 r 2 

To solve for T 2 , we cross-multiply and take the square root, yielding 



ir \ 3/2 



Substituting known values yields 



T 2 = 27.3dx^2Jlx| 

d V3.84xl0 5 kmy 

= 1.93 h. 



3/2 
J 7880 km \ 

V3.84xl0 5 km/ 



(6.57) 
(6.58) 

(6.59) 



Discussion This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will orbit in the same 
amount of time. This fact is related to the condition that the satellite's mass is small compared with that of Earth. 



People immediately search for deeper meaning when broadly applicable laws, like Kepler's, are discovered. It was Newton who took the next giant 
step when he proposed the law of universal gravitation. While Kepler was able to discover what was happening, Newton discovered that gravitational 
force was the cause. 



208 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 

Derivation of Kepler's Third Law for Circular Orbits 

We shall derive Kepler's third law, starting with Newton's laws of motion and his universal law of gravitation. The point is to demonstrate that the force 
of gravity is the cause for Kepler's laws (although we will only derive the third one). 

Let us consider a circular orbit of a small mass m around a large mass M , satisfying the two conditions stated at the beginning of this section. 
Gravity supplies the centripetal force to mass m . Starting with Newton's second law applied to circular motion, 

v 2 ( 6 - 6 °) 

r net = ma c = m-y-. 

The net external force on mass m is gravity, and so we substitute the force of gravity for F 



:tnM _ w v_ 

r 2 - m r 



net ■ 
„ „2 (6^61) 

r 
The mass m cancels, yielding 

Gf = v 2 . < 6 - 62 > 

The fact that m cancels out is another aspect of the oft-noted fact that at a given location all masses fall with the same acceleration. Here we see 
that at a given orbital radius r , all masses orbit at the same speed. (This was implied by the result of the preceding worked example.) Now, to get at 
Kepler's third law, we must get the period T into the equation. By definition, period T is the time for one complete orbit. Now the average speed v 
is the circumference divided by the period — that is, 

(6.63) 



Substituting this into the previous equation gives 



9 

Solving for T yields 



V = 


. 2%r 
T ' 


GM. = 


_4;c 2 r 2 
T 2 ' 


T 2 = 


4tt 2 3 
GM ' 


aking the ratio o 







Using subscripts 1 and 2 to denote two different satellites, and taking the ratio of the last equation for satellite 1 to satellite 2 yields 



(6.64) 



(6.65) 



(6.66) 



This is Kepler's third law. Note that Kepler's third law is valid only for comparing satellites of the same parent body, because only then does the mass 
of the parent body M cancel. 

1 A.TT 3 ^9 

Now consider what we get if we solve T = y^—r for the ratio r IT . We obtain a relationship that can be used to determine the mass M of a 
parent body from the orbits of its satellites: 

A = ^M (6 " 67) 

T 2 4jc 2 

If r and T are known for a satellite, then the mass M of the parent can be calculated. This principle has been used extensively to find the masses 
of heavenly bodies that have satellites. Furthermore, the ratio r IT should be a constant for all satellites of the same parent body (because 
r 3 IT 2 = GM I 4jc 2 ). (See Table 6.2). 

It is clear from Table 6.2 that the ratio of r I T is constant, at least to the third digit, for all listed satellites of the Sun, and for those of Jupiter. Small 
variations in that ratio have two causes — uncertainties in the r and T data, and perturbations of the orbits due to other bodies. Interestingly, those 
perturbations can be — and have been — used to predict the location of new planets and moons. This is another verification of Newton's universal law 
of gravitation. 

Making Connections 

Newton's universal law of gravitation is modified by Einstein's general theory of relativity, as we shall see in Particle Physics. Newton's gravity is 
not seriously in error — it was and still is an extremely good approximation for most situations. Einstein's modification is most noticeable in 
extremely large gravitational fields, such as near black holes. However, general relativity also explains such phenomena as small but long-known 
deviations of the orbit of the planet Mercury from classical predictions. 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 209 

The Case for Simplicity 

The development of the universal law of gravitation by Newton played a pivotal role in the history of ideas. While it is beyond the scope of this text to 
cover that history in any detail, we note some important points. The definition of planet set in 2006 by the International Astronomical Union (IAU) 
states that in the solar system, a planet is a celestial body that: 

1. is in orbit around the Sun, 

2. has sufficient mass to assume hydrostatic equilibrium and 

3. has cleared the neighborhood around its orbit. 

A non-satellite body fulfilling only the first two of the above criteria is classified as "dwarf planet." 

In 2006, Pluto was demoted to a 'dwarf planet' after scientists revised their definition of what constitutes a "true" planet. 



Table 6.2 Orbital Data and Kepler's Third Law 



Parent Satellite Average orbital radius r(km) Period T(y) j 3 / j 2 (km 3 / y 2 ) 












Earth 


Moon 


3.84X10 5 


0.07481 


l.OlxlO 18 


Sun 


Mercury 


5.79X10 7 


0.2409 


3.34X10 24 




Venus 


1.082x10 s 


0.6150 


3.35X10 24 


Earth 


1.496x10 s 


1.000 


3.35X10 24 


Mars 


2.279x10 s 


1.881 


3.35X10 24 


Jupiter 


7.783x10 s 


11.86 


3.35X10 24 


Saturn 


1.427X10 9 


29.46 


3.35X10 24 


Neptune 


4.497X10 9 


164.8 


3.35X10 24 


Pluto 


5.90X10 9 


248.3 


3.33X10 24 


Jupiter 


lo 


4.22x10 s 


0.00485 (1.77 d) 


3.19X10 21 




Europa 


6.71X10 5 


0.00972 (3.55 d) 


3.20X10 21 


Ganymede 


1.07X10 6 


0.0196 (7.16 d) 


3.19X10 21 


Callisto 


1.88X10 6 


0.0457 (16.19 d) 


3.20X10 21 



The universal law of gravitation is a good example of a physical principle that is very broadly applicable. That single equation for the gravitational 
force describes all situations in which gravity acts. It gives a cause for a vast number of effects, such as the orbits of the planets and moons in the 
solar system. It epitomizes the underlying unity and simplicity of physics. 

Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth as shown in Figure 
6.31(a). This is called the Ptolemaic view, for the Greek philosopher who lived in the second century AD. This model is characterized by a list of facts 
for the motions of planets with no cause and effect explanation. There tended to be a different rule for each heavenly body and a general lack of 
simplicity. 

Figure 6.31(b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all 
motions in the solar system, but all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. As our 
knowledge of nature has grown, the basic simplicity of its laws has become ever more evident. 



210 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 



?<\mary rnoti 0n 

„,-~-Path-^ 
of Mars 



•- Stars 



YJVIars \ 

f\V Epicycle 
/ / ,Moqn.venus< \ \ m\ 

Mercury ; ' • 
y / : /^Saturn 

Jupiter 



Ptolemaic view Copernican view 

(a) (b) 

Figure 6.31 (a) The Ptolemaic model of the universe has Earth at the center with the Moon, the planets, the Sun, and the stars revolving about it in complex superpositions of 
circular paths. This geocentric model, which can be made progressively more accurate by adding more circles, is purely descriptive, containing no hints as to what are the 
causes of these motions, (b) The Copernican model has the Sun at the center of the solar system. It is fully explained by a small number of laws of physics, including Newton's 
universal law of gravitation. 




Glossary 



angular velocity: w , the rate of change of the angle with which an object moves on a circular path 

arc length: As , the distance traveled by an object along a circular path 

banked curve: the curve in a road that is sloping in a manner that helps a vehicle negotiate the curve 

Coriolis force: the fictitious force causing the apparent deflection of moving objects when viewed in a rotating frame of reference 

center of mass: the point where the entire mass of an object can be thought to be concentrated 

centrifugal force: a fictitious force that tends to throw an object off when the object is rotating in a non-inertial frame of reference 

centripetal acceleration: the acceleration of an object moving in a circle, directed toward the center 

centripetal force: any net force causing uniform circular motion 

fictitious force: a force having no physical origin 

gravitational constant, G: a proportionality factor used in the equation for Newton's universal law of gravitation; it is a universal constant — that is, 
it is thought to be the same everywhere in the universe 

ideal angle: the angle at which a car can turn safely on a steep curve, which is in proportion to the ideal speed 

ideal banking: the sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at a certain speed without 
the aid of friction between the tires and the road; the net external force on the vehicle equals the horizontal centripetal force in the absence of 
friction 

ideal speed: the maximum safe speed at which a vehicle can turn on a curve without the aid of friction between the tire and the road 

microgravity: an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface 

Newton's universal law of gravitation: every particle in the universe attracts every other particle with a force along a line joining them; the force 
is directly proportional to the product of their masses and inversely proportional to the square of the distance between them 

non-inertial frame of reference: an accelerated frame of reference 

pit: a tiny indentation on the spiral track moulded into the top of the polycarbonate layer of CD 

radians: a unit of angle measurement 

radius of curvature: radius of a circular path 

rotation angle: the ratio of the arc length to the radius of curvature on a circular path: 



A0 = 



As 



ultracentrifuge: a centrifuge optimized for spinning a rotor at very high speeds 
uniform circular motion: the motion of an object in a circular path at constant speed 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 211 



Section Summary 



6.1 Rotation Angle and Angular Velocity 

• Uniform circular motion is motion in a circle at constant speed. The rotation angle A6 is defined as the ratio of the arc length to the radius of 
curvature: 

where arc length As is distance traveled along a circular path and r is the radius of curvature of the circular path. The quantity A6 is 
measured in units of radians (rad), for which 

27C rad = 360°= 1 revolution. 

• The conversion between radians and degrees is 1 rad = 57.3° . 

• Angular velocity w is the rate of change of an angle, 

where a rotation A6 takes place in a time At . The units of angular velocity are radians per second (rad/s). Linear velocity v and angular 
velocity oo are related by 

v = rw or w = j. 

6.2 Centripetal Acceleration 

• Centripetal acceleration a c is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is 

perpendicular to the linear velocity v and has the magnitude 

V 2 2 

a c = -y-; a c = rco . 

• The unit of centripetal acceleration is m/s . 

6.3 Centripetal Force 

• Centripetal force F c is any force causing uniform circular motion. It is a "center-seeking" force that always points toward the center of rotation. 

It is perpendicular to linear velocity v and has magnitude 

F c = ma c , 



which can also be expressed as 

F c = m\- 

or 
F c = mrco 

6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force 

• Rotating and accelerated frames of reference are non-inertial. 

• Fictitious forces, such as the Coriolis force, are needed to explain motion in such frames. 

6.5 Newton's Universal Law of Gravitation 

• Newton's universal law of gravitation: Every particle in the universe attracts every other particle with a force along a line joining them. The force 
is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In equation form, 
this is 



r 

—11 9 9 

where F is the magnitude of the gravitational force. G is the gravitational constant, given by G = 6.673x10 N • m /kg . 

• Newton's law of gravitation applies universally. 

6.6 Satellites and Kepler's Laws: An Argument for Simplicity 

• Kepler's laws are stated for a small mass m orbiting a larger mass M in near-isolation. Kepler's laws of planetary motion are then as follows: 
Kepler's first law 

The orbit of each planet about the Sun is an ellipse with the Sun at one focus. 

Kepler's second law 

Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times. 

Kepler's third law 

The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the 

Sun: 

T 2 r 3 

T 2 r V 

1 2 r 2 



where T is the period (time for one orbit) and r is the average radius of the orbit. 
The period and radius of a satellite's orbit about a larger body M are related by 



212 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 



or 



4;r 2 r 3 
GM 



T 2 4% 2 



M. 



Conceptual Questions 



6.1 Rotation Angle and Angular Velocity 

1. There is an analogy between rotational and linear physical quantities. What rotational quantities are analogous to distance and velocity? 

6.2 Centripetal Acceleration 

2. Can centripetal acceleration change the speed of circular motion? Explain. 

6.3 Centripetal Force 

3. If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain. 

4. Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any 
combination of forces be a centripetal force? 

5. If centripetal force is directed toward the center, why do you feel that you are 'thrown' away from the center as a car goes around a curve? Explain. 

6. Race car drivers routinely cut corners as shown in Figure 6.32. Explain how this allows the curve to be taken at the greatest speed. 




Path 2 
Pathl \ 

Figure 6.32 Two paths around a race track curve are shown. Race car drivers will take the inside path (called cutting the corner) whenever possible because it allows them to 
take the curve at the highest speed. 

7. A number of amusement parks have rides that make vertical loops like the one shown in Figure 6.33. For safety, the cars are attached to the rails 
in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other 
force acts and what is its direction if: 

(a) The car goes over the top at faster than this speed? 

(b)The car goes over the top at slower than this speed? 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 213 

















// 












i 




1 






















1 






1 









Figure 6.33 Amusement rides with a vertical loop are an example of a form of curved motion. 

8. What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure 6.33 
under the following circumstances: 

(a) The car goes over the top at such a speed that the gravitational force is the only force acting? 

(b) The car goes over the top faster than this speed? 

(c) The car goes over the top slower than this speed? 

9. As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer. 

10. Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax 
paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure 6.34 will the lunch box take when she lets 
go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer. 




Merry-go-round's rotating 
frame of reference 



Figure 6.34 A child riding on a merry-go-round releases her lunch box at point P. This is a view from above the clockwise rotation. Assuming it slides with negligible friction, will 
it follow path A, B, or C, as viewed from Earth's frame of reference? What will be the shape of the path it leaves in the dust on the merry-go-round? 

11. Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car's speed? What is the direction of the 
force exerted on you by the car seat? 

12. Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth's frame of reference, there is no centrifugal 
force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using 
concepts related to centripetal force and Newton's third law, explain what force stretches the string, identifying its physical origin. 



214 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 




Figure 6.35 A mass attached to a nail on a frictionless table moves in a circular path. The force stretching the string is real and not fictional. What is the physical origin of the 
force on the string? 

6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force 

13. When a toilet is flushed or a sink is drained, the water (and other material) begins to rotate about the drain on the way down. Assuming no initial 
rotation and a flow initially directly straight toward the drain, explain what causes the rotation and which direction it has in the northern hemisphere. 
(Note that this is a small effect and in most toilets the rotation is caused by directional water jets.) Would the direction of rotation reverse if water were 
forced up the drain? 

14. Is there a real force that throws water from clothes during the spin cycle of a washing machine? Explain how the water is removed. 

15. In one amusement park ride, riders enter a large vertical barrel and stand against the wall on its horizontal floor. The barrel is spun up and the 
floor drops away. Riders feel as if they are pinned to the wall by a force something like the gravitational force. This is a fictitious force sensed and 
used by the riders to explain events in the rotating frame of reference of the barrel. Explain in an inertial frame of reference (Earth is nearly one) what 
pins the riders to the wall, and identify all of the real forces acting on them. 

16. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the 
truth in physics, and why was this action ultimately accepted? 

17. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a 
satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s . Who do you agree with and why? 

18. A non-rotating frame of reference placed at the center of the Sun is very nearly an inertial one. Why is it not exactly an inertial frame? 

6.5 Newton's Universal Law of Gravitation 

19. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the 
truth in physics, and why was this action ultimately accepted? 

20. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a 
satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s . Who do you agree with and why? 

21. Draw a free body diagram for a satellite in an elliptical orbit showing why its speed increases as it approaches its parent body and decreases as it 
moves away. 

22. Newton's laws of motion and gravity were among the first to convincingly demonstrate the underlying simplicity and unity in nature. Many other 
examples have since been discovered, and we now expect to find such underlying order in complex situations. Is there proof that such order will 
always be found in new explorations? 

6.6 Satellites and Kepler's Laws: An Argument for Simplicity 

23. In what frame(s) of reference are Kepler's laws valid? Are Kepler's laws purely descriptive, or do they contain causal information? 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 215 



Problems & Exercises 



6.1 Rotation Angle and Angular Velocity 

1. Semi-trailer trucks have an odometer on one hub of a trailer wheel. 
The hub is weighted so that it does not rotate, but it contains gears to 
count the number of wheel revolutions — it then calculates the distance 
traveled. If the wheel has a 1.15 m diameter and goes through 200,000 
rotations, how many kilometers should the odometer read? 

2. Microwave ovens rotate at a rate of about 6 rev/min. What is this in 
revolutions per second? What is the angular velocity in radians per 
second? 

3. An automobile with 0.260 m radius tires travels 80,000 km before 
wearing them out. How many revolutions do the tires make, neglecting 
any backing up and any change in radius due to wear? 

4. (a) What is the period of rotation of Earth in seconds? (b) What is the 
angular velocity of Earth? (c) Given that Earth has a radius of 

6.4X 10 m at its equator, what is the linear velocity at Earth's surface? 

5. A baseball pitcher brings his arm forward during a pitch, rotating the 
forearm about the elbow. If the velocity of the ball in the pitcher's hand is 
35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular 
velocity of the forearm? 

6. In lacrosse, a ball is thrown from a net on the end of a stick by rotating 
the stick and forearm about the elbow. If the angular velocity of the ball 
about the elbow joint is 30.0 rad/s and the ball is 1.30 m from the elbow 
joint, what is the velocity of the ball? 

7. A truck with 0.420 m radius tires travels at 32.0 m/s. What is the 
angular velocity of the rotating tires in radians per second? What is this in 
rev/min? 

8. Integrated Concepts When kicking a football, the kicker rotates his 
leg about the hip joint. 

(a) If the velocity of the tip of the kicker's shoe is 35.0 m/s and the hip 
joint is 1.05 m from the tip of the shoe, what is the shoe tip's angular 
velocity? 

(b) The shoe is in contact with the initially nearly stationary 0.500 kg 
football for 20.0 ms. What average force is exerted on the football to give 
it a velocity of 20.0 m/s? 

(c) Find the maximum range of the football, neglecting air resistance. 

9. Construct Your Own Problem 

Consider an amusement park ride in which participants are rotated about 
a vertical axis in a cylinder with vertical walls. Once the angular velocity 
reaches its full value, the floor drops away and friction between the walls 
and the riders prevents them from sliding down. Construct a problem in 
which you calculate the necessary angular velocity that assures the 
riders will not slide down the wall. Include a free body diagram of a single 
rider. Among the variables to consider are the radius of the cylinder and 
the coefficients of friction between the riders' clothing and the wall. 

6.2 Centripetal Acceleration 

10. A fairground ride spins its occupants inside a flying saucer-shaped 
container. If the horizontal circular path the riders follow has an 8.00 m 
radius, at how many revolutions per minute will the riders be subjected to 
a centripetal acceleration 1.50 times that due to gravity? 

11. A runner taking part in the 200 m dash must run around the end of a 
track that has a circular arc with a radius of curvature of 30 m. If he 
completes the 200 m dash in 23.2 s and runs at constant speed 
throughout the race, what is his centripetal acceleration as he runs the 
curved portion of the track? 

12. Taking the age of Earth to be about 4x 10 years and assuming its 

orbital radius of 1.5 XlO has not changed and is circular, calculate 
the approximate total distance Earth has traveled since its birth (in a 
frame of reference stationary with respect to the Sun). 

13. The propeller of a World War II fighter plane is 2.30 m in diameter. 



(a) What is its angular velocity in radians per second if it spins at 1200 
rev/min? 

(b) What is the linear speed of its tip at this angular velocity if the plane is 
stationary on the tarmac? 

(c) What is the centripetal acceleration of the propeller tip under these 
conditions? Calculate it in meters per second squared and convert to 
multiples of g . 

14. An ordinary workshop grindstone has a radius of 7.50 cm and rotates 
at 6500 rev/min. 

(a) Calculate the centripetal acceleration at its edge in meters per second 
squared and convert it to multiples of g . 

(b) What is the linear speed of a point on its edge? 

15. Helicopter blades withstand tremendous stresses. In addition to 
supporting the weight of a helicopter, they are spun at rapid rates and 
experience large centripetal accelerations, especially at the tip. 

(a) Calculate the centripetal acceleration at the tip of a 4.00 m long 
helicopter blade that rotates at 300 rev/min. 

(b) Compare the linear speed of the tip with the speed of sound (taken to 
be 340 m/s). 

16. Olympic ice skaters are able to spin at about 5 rev/s. 

(a) What is their angular velocity in radians per second? 

(b) What is the centripetal acceleration of the skater's nose if it is 0.120 m 
from the axis of rotation? 

(c) An exceptional skater named Dick Button was able to spin much 
faster in the 1950s than anyone since — at about 9 rev/s. What was the 
centripetal acceleration of the tip of his nose, assuming it is at 0.120 m 
radius? 

(d) Comment on the magnitudes of the accelerations found. It is reputed 
that Button ruptured small blood vessels during his spins. 

17. What percentage of the acceleration at Earth's surface is the 
acceleration due to gravity at the position of a satellite located 300 km 
above Earth? 

18. Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, 
and Earth in its orbit is about 30 km/s by calculating: 

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its 
center, rotating at 50,000 rev/min. 

(b) The linear speed of Earth in its orbit about the Sun (use data from the 
text on the radius of Earth's orbit and approximate it as being circular). 

19. A rotating space station is said to create "artificial gravity" — a loosely- 
defined term used for an acceleration that would be crudely similar to 
gravity. The outer wall of the rotating space station would become a floor 
for the astronauts, and centripetal acceleration supplied by the floor 
would allow astronauts to exercise and maintain muscle and bone 
strength more naturally than in non-rotating space environments. If the 
space station is 200 m in diameter, what angular velocity would produce 

an "artificial gravity" of 9.80 m/s at the rim? 

20. At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a 
diameter of 0.850 m. 

(a) At how many rev/min are the tires rotating? 

(b) What is the centripetal acceleration at the edge of the tire? 

(c) With what force must a determined 1.00X 10~ kg bacterium cling 
to the rim? 

(d) Take the ratio of this force to the bacterium's weight. 

21. Integrated Concepts 

Riders in an amusement park ride shaped like a Viking ship hung from a 
large pivot are rotated back and forth like a rigid pendulum. Sometime 
near the middle of the ride, the ship is momentarily motionless at the top 
of its circular arc. The ship then swings down under the influence of 
gravity. 



216 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 



(a) Assuming negligible friction, find the speed of the riders at the bottom 
of its arc, given the system's center of mass travels in an arc having a 
radius of 14.0 m and the riders are near the center of mass. 

(b) What is the centripetal acceleration at the bottom of the arc? 

(c) Draw a free body diagram of the forces acting on a rider at the bottom 
of the arc. 

(d) Find the force exerted by the ride on a 60.0 kg rider and compare it to 
her weight. 

(e) Discuss whether the answer seems reasonable. 

22. Unreasonable Results 

A mother pushes her child on a swing so that his speed is 9.00 m/s at the 
lowest point of his path. The swing is suspended 2.00 m above the child's 
center of mass. 

(a) What is the centripetal acceleration of the child at the low point? 

(b) What force does the child exert on the seat if his mass is 18.0 kg? 

(c) What is unreasonable about these results? 

(d) Which premises are unreasonable or inconsistent? 

6.3 Centripetal Force 

23. (a) A 22.0 kg child is riding a playground merry-go-round that is 
rotating at 40.0 rev/min. What centripetal force must she exert to stay on 
if she is 1.25 m from its center? 

(b) What centripetal force does she need to stay on an amusement park 
merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its 
center? 

(c) Compare each force with her weight. 

24. Calculate the centripetal force on the end of a 100 m (radius) wind 
turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg. 

25. What is the ideal banking angle for a gentle turn of 1.20 km radius on 
a highway with a 105 km/h speed limit (about 65 mi/h), assuming 
everyone travels at the limit? 

26. What is the ideal speed to take a 100 m radius curve banked at a 
20.0° angle? 

27. (a) What is the radius of a bobsled turn banked at 75.0° and taken at 
30.0 m/s, assuming it is ideally banked? 

(b) Calculate the centripetal acceleration. 

(c) Does this acceleration seem large to you? 

28. Part of riding a bicycle involves leaning at the correct angle when 
making a turn, as seen in Figure 6.36. To be stable, the force exerted by 
the ground must be on a line going through the center of gravity. The 
force on the bicycle wheel can be resolved into two perpendicular 
components — friction parallel to the road (this must supply the centripetal 
force), and the vertical normal force (which must equal the system's 
weight). 

(a) Show that 6 (as defined in the figure) is related to the speed v and 
radius of curvature r of the turn in the same way as for an ideally 

banked roadway— that is, 6 = tan" v / rg 



Free-body diagram 
E 




Figure 6.36 A bicyclist negotiating a turn on level ground must lean at the correct 
angle — the ability to do this becomes instinctive. The force of the ground on the wheel 
needs to be on a line through the center of gravity. The net external force on the 
system is the centripetal force. The vertical component of the force on the wheel 
cancels the weight of the system while its horizontal component must supply the 
centripetal force. This process produces a relationship among the angle 6 , the 
speed V , and the radius of curvature r of the turn similar to that for the ideal 
banking of roadways. 

29. A large centrifuge, like the one shown in Figure 6.37(a), is used to 
expose aspiring astronauts to accelerations similar to those experienced 
in rocket launches and atmospheric reentries. 

(a) At what angular velocity is the centripetal acceleration 10 g if the 
rider is 15.0 m from the center of rotation? 

(b) The rider's cage hangs on a pivot at the end of the arm, allowing it to 
swing outward during rotation as shown in Figure 6.37(b). At what angle 
6 below the horizontal will the cage hang when the centripetal 
acceleration is 10 g ? (Hint: The arm supplies centripetal force and 

supports the weight of the cage. Draw a free body diagram of the forces 
to see what the angle 6 should be.) 



(b) Calculate 6 for a 12.0 m/s turn of radius 30.0 m (as in a race). 



CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 217 




1 mm muir 



(a) NASA centrifuge and ride 



Free- 
body 
diagram 




(b) 

Figure 6.37 (a) NASA centrifuge used to subject trainees to accelerations similar to 
those experienced in rocket launches and reentries, (credit: NASA) (b) Rider in cage 
showing how the cage pivots outward during rotation. This allows the total force 
exerted on the rider by the cage to be along its axis at all times. 

30. Integrated Concepts 

If a car takes a banked curve at less than the ideal speed, friction is 
needed to keep it from sliding toward the inside of the curve (a real 
problem on icy mountain roads), (a) Calculate the ideal speed to take a 
100 m radius curve banked at 15.0°. (b) What is the minimum coefficient 
of friction needed for a frightened driver to take the same curve at 20.0 
km/h? 

31. Integrated Concepts 

Modern roller coasters have vertical loops like the one shown in Figure 
6.38. The radius of curvature is smaller at the top than on the sides so 
that the downward centripetal acceleration at the top will be greater than 
the acceleration due to gravity, keeping the passengers pressed firmly 
into their seats. 

(a) What is the speed of the roller coaster at the top of the loop if the 
radius of curvature there is 15.0 m and the downward acceleration of the 
car is 1.50 g? 

(b) How high above the top of the loop must the roller coaster start from 
rest, assuming negligible friction? 

(c) If it actually starts 5.00 m higher than your answer to the previous 
part, how much energy did it lose to friction? Its mass is 1500 kg. 




T 



Figure 6.38 Teardrop-shaped loops are used in the latest roller coasters so that the 
radius of curvature gradually decreases to a minimum at the top. This means that the 
centripetal acceleration builds from zero to a maximum at the top and gradually 
decreases again. A circular loop would cause a jolting change in acceleration at entry, 
a disadvantage discovered long ago in railroad curve design. With a small radius of 
curvature at the top, the centripetal acceleration can more easily be kept greater than 
g so that the passengers do not lose contact with their seats nor do they need seat 

belts to keep them in place. 

32. Unreasonable Results 

(a) Calculate the minimum coefficient of friction needed for a car to 
negotiate an unbanked 50.0 m radius curve at 30.0 m/s. 

(b) What is unreasonable about the result? 

(c) Which premises are unreasonable or inconsistent? 

6.5 Newton's Universal Law of Gravitation 

33. (a) Calculate Earth's mass given the acceleration due to gravity at the 
North Pole is 9.830 m/s 2 and the radius of the Earth is 6371 km from 
pole to pole. 

94 

(b) Compare this with the accepted value of 5.979X 10 kg 

34. (a) Calculate the acceleration due to gravity at Earth due to the 
Moon. 

(b) Calculate the acceleration due to gravity at Earth due to the Sun. 

(c) Take the ratio of the Moon's acceleration to the Sun's and comment 
on why the tides are predominantly due to the Moon in spite of this 
number. 

35. (a) What is the acceleration due to gravity on the surface of the 
Moon? 

(b) On the surface of Mars? The mass of Mars is 6.418x10 kg and 

its radius is 3.38xl0 6 m. 

36. (a) Calculate the acceleration due to gravity on the surface of the 
Sun. 

(b) By what factor would your weight increase if you could stand on the 
Sun? (Never mind that you cannot.) 

37. The Moon and Earth rotate about their common center of mass, 
which is located about 4700 km from the center of Earth. (This is 1690 
km below the surface.) 

(a) Calculate the acceleration due to the Moon's gravity at that point. 

(b) Calculate the centripetal acceleration of the center of Earth as it 
rotates about that point once each lunar month (about 27.3 d) and 
compare it with the acceleration found in part (a). Comment on whether 
or not they are equal and why they should or should not be. 



218 CHAPTER 6 | UNIFORM CIRCULAR MOTION AND GRAVITATION 



38. Solve part (b) of Example 6.6 using a c = v I r . 

39. Astrology, that unlikely and vague pseudoscience, makes much of the 
position of the planets at the moment of one's birth. The only known force 
a planet exerts on Earth is gravitational. 

(a) Calculate the gravitational force exerted on a 4.20 kg baby by a 100 
kg father 0.200 m away at birth (he is assisting, so he is close to the 
child). 

(b) Calculate the force on the baby due to Jupiter if it is at its closest 
distance to Earth, some 6.29x10 m away. How does the force of 
Jupiter on the baby compare to the force of the father on the baby? Other 
objects in the room and the hospital building also exert similar 
gravitational forces. (Of course, there could be an unknown force acting, 
but scientists first need to be convinced that there is even an effect, much 
less that an unknown force causes it.) 

40. The existence of the dwarf planet Pluto was proposed based on 
irregularities in Neptune's orbit. Pluto was subsequently discovered near 
its predicted position. But it now appears that the discovery was 
fortuitous, because Pluto is small and the irregularities in Neptune's orbit 
were not well known. To illustrate that Pluto has a minor effect on the 
orbit of Neptune compared with the closest planet to Neptune: 

(a) Calculate the acceleration due to gravity at Neptune due to Pluto 
when they are 4.50X 10 m apart, as they are at present. The mass of 
Pluto is 1.4xl0 22 kg. 

(b) Calculate the acceleration due to gravity at Neptune due to Uranus, 

1 9 

presently about 2.50X 10 m apart, and compare it with that due to 
Pluto. The mass of Uranus is 8.62X10 25 kg . 

o 

41. (a) The Sun orbits the Milky Way galaxy once each 2.60 x 10 y , 

with a roughly circular orbit averaging 3.00 x 10 light years in radius. 
(A light year is the distance traveled by light in 1 y.) Calculate the 
centripetal acceleration of the Sun in its galactic orbit. Does your result 
support the contention that a nearly inertial frame of reference can be 
located at the Sun? 

(b) Calculate the average speed of the Sun in its galactic orbit. Does the 
answer surprise you? 

42. Unreasonable Result 

A mountain 10.0 km from a person exerts a gravitational force on him 
equal to 2.00% of his weight. 

(a) Calculate the mass of the mountain. 

(b) Compare the mountain's mass with that of Earth. 

(c) What is unreasonable about these results? 

(d) Which premises are unreasonable or inconsistent? (Note that 
accurate gravitational measurements can easily detect the effect of 
nearby mountains and variations in local geology.) 



47. Astronomical observations of our Milky Way galaxy indicate that it has 
a mass of about 8.0X 10 solar masses. A star orbiting on the galaxy's 
periphery is about 6.0x 10 light years from its center, (a) What should 

the orbital period of that star be? (b) If its period is 6.0x10 instead, 
what is the mass of the galaxy? Such calculations are used to imply the 
existence of "dark matter" in the universe and have indicated, for 
example, the existence of very massive black holes at the centers of 
some galaxies. 

48. Integrated Concepts 

Space debris left from old satellites and their launchers is becoming a 
hazard to other satellites, (a) Calculate the speed of a satellite in an orbit 
900 km above Earth's surface, (b) Suppose a loose rivet is in an orbit of 
the same radius that intersects the satellite's orbit at an angle of 90° 
relative to Earth. What is the velocity of the rivet relative to the satellite 
just before striking it? (c) Given the rivet is 3.00 mm in size, how long will 
its collision with the satellite last? (d) If its mass is 0.500 g, what is the 
average force it exerts on the satellite? (e) How much energy in joules is 
generated by the collision? (The satellite's velocity does not change 
appreciably, because its mass is much greater than the rivet's.) 

49. Unreasonable Results 

(a) Based on Kepler's laws and information on the orbital characteristics 
of the Moon, calculate the orbital radius for an Earth satellite having a 
period of 1.00 h. (b) What is unreasonable about this result? (c) What is 
unreasonable or inconsistent about the premise of a 1.00 h orbit? 

50. Construct Your Own Problem 

On February 14, 2000, the NEAR spacecraft was successfully inserted 
into orbit around Eros, becoming the first artificial satellite of an asteroid. 
Construct a problem in which you determine the orbital speed for a 
satellite near Eros. You will need to find the mass of the asteroid and 
consider such things as a safe distance for the orbit. Although Eros is not 
spherical, calculate the acceleration due to gravity on its surface at a 
point an average distance from its center of mass. Your instructor may 
also wish to have you calculate the escape velocity from this point on 
Eros. 



6.6 Satellites and Kepler's Laws: An Argument for 
Simplicity 

43. A geosynchronous Earth satellite is one that has an orbital period of 
precisely 1 day. Such orbits are useful for communication and weather 
observation because the satellite remains above the same point on Earth 
(provided it orbits in the equatorial plane in the same direction as Earth's 
rotation). Calculate the radius of such an orbit based on the data for the 
moon in Table 6.2. 

44. Calculate the mass of the Sun based on data for Earth's orbit and 
compare the value obtained with the Sun's actual mass. 

45. Find the mass of Jupiter based on data for the orbit of one of its 
moons, and compare your result with its actual mass. 

46. Find the ratio of the mass of Jupiter to that of Earth based on data in 
Table 6.2. 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 219 




Figure 7.1 How many forms of energy can you identify in this photograph of a wind farm in Iowa? (credit: Jurgen from Sandesneben, Germany, Wikimedia Commons) 



220 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 



Learning Objectives 



7.1. Work: The Scientific Definition 

• Explain how an object must be displaced for a force on it to do work. 

• Explain how relative directions of force and displacement determine whether the work done is positive, negative, or zero. 

• Calculate the work done by a force on an object given the force and displacement. 

7.2. Kinetic Energy and the Work-Energy Theorem 

• Explain work as a transfer of energy and net work as the work done by the net force. 

• Explain and apply the work-energy theorem. 

7.3. Gravitational Potential Energy 

• Explain gravitational potential energy in terms of work done against gravity. 

• Show that the gravitational potential energy of an object of mass m at height h on Earth is given by PE g = mgh . 

• Show how the work-energy theorem explains the conversion between gravitational potential energy and kinetic energy for an object 
moving under the influence of gravity. 

• Show how knowledge of the potential energy as a function of position can be used to simplify calculations and explain physical 
phenomena. 

7.4. Conservative Forces and Potential Energy 

• Define conservative force, potential energy, and mechanical energy. 

• Explain the potential energy of a spring in terms of its compression when Hooke's law applies. 

• Use the work-energy theorem to show how having only conservative forces implies conservation of mechanical energy. 

• Apply conservation of mechanical energy to solve mechanics problems in terms of potential energies instead of forces. 

7.5. Nonconservative Forces 

• Define nonconservative forces and explain how they affect mechanical energy. 

• Show how the principle of conservation of energy can be applied by treating the conservative forces in terms of their potential energies 
and any nonconservative forces in terms of the work they do. 

7.6. Conservation of Energy 

• Explain the law of the conservation of energy. 

• Express conservation of energy in equation form. 

• Describe some of the many forms of energy. 

• Summarize an effective problem-solving strategy for applying conservation of energy. 

• Examine commonly encountered examples of transformations between forms of energy. 

• Define efficiency of an energy conversion process as the fraction left as useful energy or work, rather than being transformed, for 
example, into thermal energy. 

7.7. Power 

• Define power as the rate of doing work and identify typical examples of power. 

• Calculate power by calculating changes in energy over time. 

• Examine power consumption and calculations of the cost of energy consumed. 

7.8. Work, Energy, and Power in Humans 

• Define the basal metabolic rate. 

• Explain the human body's consumption of energy when at rest vs. when engaged in activities that do useful work. 

• Calculate the conversion of chemical energy in food into useful work. 

7.9. World Energy Use 

• Describe the distinction between renewable and nonrenewable energy sources. 

• Examine the relative consumption of different kinds of energy sources worldwide. 

• Examine the past and projected growth of energy consumption. 

• Examine the correlation between Gross Domestic Product (GDP) and energy consumption for various countries. 

• Explain why the inevitable conversion of energy to less useful forms makes it necessary to conserve energy resources. 



Introduction to Work, Energy, and Energy Resources 

Energy plays an essential role both in everyday events and in scientific phenomena. You can no doubt name many forms of energy, from that 
provided by our foods, to the energy we use to run our cars, to the sunlight that warms us on the beach. You can also cite examples of what people 
call energy that may not be scientific, such as someone having an energetic personality. Not only does energy have many interesting forms, it is 
involved in almost all phenomena, and is one of the most important concepts of physics. What makes it even more important is that the total amount 
of energy in the universe is constant. Energy can change forms, but it cannot appear from nothing or disappear without a trace. Energy is thus one of 
a handful of physical quantities that we say is conserved. 

Conservation of energy (as physicists like to call the principle that energy can neither be created nor destroyed) is based on experiment. Even as 
scientists discovered new forms of energy, conservation of energy has always been found to apply. Perhaps the most dramatic example of this was 

supplied by Einstein when he suggested that mass is equivalent to energy (his famous equation E = mc ). 

From a societal viewpoint, energy is one of the major building blocks of modern civilization. Energy resources are key limiting factors to economic 
growth. The world use of energy resources, especially oil, continues to grow, with ominous consequences economically, socially, politically, and 
environmentally. We will briefly examine the world's energy use patterns at the end of this chapter. 

There is no simple, yet accurate, scientific definition for energy. Energy is characterized by its many forms and the fact that it is conserved. We can 
loosely define energy as the ability to do work, admitting that in some circumstances not all energy is available to do work. Because of the 
association of energy with work, we begin the chapter with a discussion of work. Work is intimately related to energy and how energy moves from one 
system to another or changes form. 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 221 

7.1 Work: The Scientific Definition 

What It Means to Do Work 

The scientific definition of work differs in some ways from its everyday meaning. Certain things we think of as hard work, such as writing an exam or 
carrying a heavy load on level ground, are not work as defined by a scientist. The scientific definition of work reveals its relationship to 
energy — whenever work is done, energy is transferred. 

For work, in the scientific sense, to be done, a force must be exerted and there must be motion or displacement in the direction of the force. 

Formally, the work done on a system by a constant force is defined to be the product of the component of the force in the direction of motion times 
the distance through which the force acts. For one-way motion in one dimension, this is expressed in equation form as 

W= |F| (cosfl) |d| , (7.1) 

where W is work, d is the displacement of the system, and 6 is the angle between the force vector F and the displacement vector d , as in 
Figure 7.2. We can also write this as 

W = Fd cos 6. (7.2) 

To find the work done on a system that undergoes motion that is not one-way or that is in two or three dimensions, we divide the motion into one-way 
one-dimensional segments and add up the work done over each segment. 



What is Work? 


The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through 
which the force acts. For one-way motion in one dimension, this is expressed in equation form as 


W = 


= Fd COS 0, 




(7.3) 


where W is work, F is the magnitude of the force on the system, 


d is the magnitude of the displacement of the system, 


and 


is the angle 


between the force vector F and the displacement vector d . 









222 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 



W= Fd cos 9 



FcosO 




ib} 



(c) 



Electric \, 
generator 






(d) (e) 

Figure 7.2 Examples of work, (a) The work done by the force F on this lawn mower is Fd COS 6 . Note that F COS 6 is the component of the force in the direction of 
motion, (b) A person holding a briefcase does no work on it, because there is no motion. No energy is transferred to or from the briefcase, (c) The person moving the briefcase 
horizontally at a constant speed does no work on it, and transfers no energy to it. (d) Work is done on the briefcase by carrying it up stairs at constant speed, because there is 
necessarily a component of force F in the direction of the motion. Energy is transferred to the briefcase and could in turn be used to do work, (e) When the briefcase is 
lowered, energy is transferred out of the briefcase and into an electric generator. Here the work done on the briefcase by the generator is negative, removing energy from the 
briefcase, because F and d are in opposite directions. 

To examine what the definition of work means, let us consider the other situations shown in Figure 7.2. The person holding the briefcase in Figure 
7.2(b) does no work, for example. Here d = , so W = . Why is it you get tired just holding a load? The answer is that your muscles are doing 
work against one another, but they are doing no work on the system of interest (the "briefcase-Earth system" — see Gravitational Potential Energy 
for more details). There must be motion for work to be done, and there must be a component of the force in the direction of the motion. For example, 
the person carrying the briefcase on level ground in Figure 7.2(c) does no work on it, because the force is perpendicular to the motion. That is, 
cos 90° = , and so W = . 

In contrast, when a force exerted on the system has a component in the direction of motion, such as in Figure 7.2(d), work is done — energy is 
transferred to the briefcase. Finally, in Figure 7.2(e), energy is transferred from the briefcase to a generator. There are two good ways to interpret this 
energy transfer. One interpretation is that the briefcase's weight does work on the generator, giving it energy. The other interpretation is that the 
generator does negative work on the briefcase, thus removing energy from it. The drawing shows the latter, with the force from the generator upward 
on the briefcase, and the displacement downward. This makes 6 = 180° , and cos 180° = -1 ; therefore, W is negative. 



Calculating Work 

Work and energy have the same units. From the definition of work, we see that those units are force times distance. Thus, in SI units, work and 

9 9 

energy are measured in newton-meters. A newton-meter is given the special name joule (J), and 1 J — 1 KT ' Wv M ~ /o 

a large amount of energy; it would lift a small 100-gram apple a distance of about 1 meter. 



1 N • m = 1 kg • m /s . One joule is not 



Example 7.1 Calculating the Work You Do to Push a Lawn Mower Across a Large Lawn 



How much work is done on the lawn mower by the person in Figure 7.2(a) if he exerts a constant force of 75.0 N at an angle 35° below the 
horizontal and pushes the mower 25.0 m on level ground? Convert the amount of work from joules to kilocalories and compare it with this 
person's average daily intake of 10,000 kJ (about 2400 kcal ) of food energy. One calorie (1 cal) of heat is the amount required to warm 1 g of 
water by 1°C , and is equivalent to 4.184 J , while one food calorie (1 kcal) is equivalent to 4184 J . 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 223 

Strategy 

We can solve this problem by substituting the given values into the definition of work done on a system, stated in the equation W = Fd cos 6 . 
The force, angle, and displacement are given, so that only the work W is unknown. 

Solution 

The equation for the work is 

W = Fd cos 0. (7.4) 

Substituting the known values gives 

W = (75.0 N)(25.0m) cos (35.0°) (7.5) 

= 1536 J = 1.54xl0 3 J. 

Converting the work in joules to kilocalories yields W = (1536 J)(l kcal/4184 J) = 0.367 kcal . The ratio of the work done to the daily 
consumption is 

O777n-T=l-53X10- 4 . < 7 - 6 > 

2400 kcal 

Discussion 

This ratio is a tiny fraction of what the person consumes, but it is typical. Very little of the energy released in the consumption of food is used to 
do work. Even when we "work" all day long, less than 10% of our food energy intake is used to do work and more than 90% is converted to 
thermal energy or stored as chemical energy in fat. 

7.2 Kinetic Energy and the Work-Energy Theorem 

Work Transfers Energy 

What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The 
answers depend on the situation. For example, if the lawn mower in Figure 7.2(a) is pushed just hard enough to keep it going at a constant speed, 
then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In 
contrast, work done on the briefcase by the person carrying it up stairs in Figure 7.2(d) is stored in the briefcase-Earth system and can be recovered 
at any time, as shown in Figure 7.2(e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing 
work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth 
system and has the potential to do work. 

In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system 
constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms 
of energy, such as the energy of motion. 

Net Work and the Work-Energy Theorem 

We know from the study of Newton's laws in Dynamics: Force and Newton's Laws of Motion that net force causes acceleration. We will see in this 
section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion. 

Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all external forces — that is, net 
work is the work done by the net external force F net . In equation form, this is W nQt = F nQt d cos 6 where 6 is the angle between the force vector 

and the displacement vector. 

Figure 7.3(a) shows a graph of force versus displacement for the component of the force in the direction of the displacement — that is, an F cos 6 
vs. d graph. In this case, F cos 6 is constant. You can see that the area under the graph is F cos 6 , or the work done. Figure 7.3(b) shows a 
more general process where the force varies. The area under the curve is divided into strips, each having an average force (F cos 6) . The 

work done is (F cos 6) ., .d: for each strip, and the total work done is the sum of the W.- . Thus the total work done is the total area under the 
v y z(ave) l r l 

curve, a useful property to which we shall refer later. 



224 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 

Fcos e 
FqosO 




f Area = FcosG x d 
= Fd cos $ 
= work = W 



(a) 



Fcosd ■■ 



W = Xl/l^ = total area 
under curve 




(b) 

Figure 7.3 (a) A graph of F COS # vs. d , when i* 1 COS # is constant. The area under the curve represents the work done by the force, (b) A graph of F COS 6 vs. d in 
which the force varies. The work done for each interval is the area of each strip; thus, the total area under the curve equals the total work done. 

Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to 
its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in Figure 7.4. 



m = 30,0 kg f=5i00N 



d = 0,800 m 



r\r\i^r\c^i^i^€^ir\f^r^r\r\r\rsr\f^r\f^r\r\r\r\r\r\r\ 



\ 



Figure 7.4 A package on a roller belt is pushed horizontally through a distance d . 

The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal 
in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force F app 

and the horizontal friction force f . Thus, as expected, the net force is parallel to the displacement, so that 6 = 0° and cos 0=1, and the net 
work is given by 



W n 



F nQt d. 



(7.7) 



The effect of the net force F net is to accelerate the package from v to v . The kinetic energy of the package increases, indicating that the net work 

done on the system is positive. (See Example 7.2.) By using Newton's second law, and doing some algebra, we can reach an interesting conclusion. 
Substituting F net = ma from Newton's second law gives 



W nQt = mad. 



(7.8) 



To get a relationship between net work and the speed given to a system by the net force acting on it, we take d = x — Xq and use the equation 

studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance d if the acceleration has the 

2 2 

2 2 ^ r v — v n 

constant value a ; namely, v = v + 2ad (note that a appears in the expression for the net work). Solving for acceleration gives a = — 9 , 

. When a is substituted into the preceding expression for W nQt , we obtain 

(7.9) 



W n 



-M* 



The d cancels, and we rearrange this to obtain 



w 



hnv 2 -hnv^ . 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 225 

(7.10) 



This expression is called the work-energy theorem, and it actually applies in general (even for forces that vary in direction and magnitude), although 

we have derived it for the special case of a constant force parallel to the displacement. The theorem implies that the net work on a system equals the 

1 2 
change in the quantity ^mv . This quantity is our first example of a form of energy. 



The Work-Energy Theorem 




1 2 
The quantity ^-mv in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass m moving at a speed v . 

(Translations! kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the translational kinetic energy, 



Imv 2 , 



is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system 
of objects moving together. 

We are aware that it takes energy to get an object, like a car or the package in Figure 7.4, up to speed, but it may be a bit surprising that kinetic 
energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it 
has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various 
aspects of work and energy. 



Example 7.2 Calculating the Kinetic Energy of a Package 



Suppose a 30.0-kg package on the roller belt conveyor system in Figure 7.4 is moving at 0.500 m/s. What is its kinetic energy? 
Strategy 

1 2 

Because the mass m and speed v are given, the kinetic energy can be calculated from its definition as given in the equation KE = ^-mv . 



Solution 

The kinetic energy is given by 



Entering known values gives 



which yields 



Discussion 



KE = Imv 2 . 



KE = 0.5(30.0 kg)(0.500 m/s) 2 , 



KE = 3.75 kg • m 2 /s 2 = 3.75 J. 



(7.13) 



(7.14) 



(7.15) 



Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, 
although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation 
that people can move packages like this without exhausting themselves. 



Example 7.3 Determining the Work to Accelerate a Package 



Suppose that you push on the 30.0-kg package in Figure 7.4 with a constant force of 120 N through a distance of 0.800 m, and that the 
opposing friction force averages 5.00 N. 

(a) Calculate the net work done on the package, (b) Solve the same problem as in part (a), this time by finding the work done by each force that 
contributes to the net force. 

Strategy and Concept for (a) 

This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal 
magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all 
horizontal. (See Figure 7.4.) As expected, the net work is the net force times distance. 

Solution for (a) 

The net force is the push force minus friction, or F net = 120 N - 5.00 N = 1 15 N . Thus the net work is 



226 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 

^net = F net d = (115N)(0.800m) (7.16) 

= 92.0N-m = 92.0J. 

Discussion for (a) 

This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction 
does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the 
work done by each individual force. 

Strategy and Concept for (b) 

The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity 
are each perpendicular to the displacement, and therefore do no work. 

Solution for (b) 

The applied force does work. 

Wapp = i> p dcos(0°) = F app d (7.17) 

= (120N)(0.800m) 
= 96.0 J 

The friction force and displacement are in opposite directions, so that 6 = 180° , and the work done by friction is 

W fY = F fr dcos(l80°) = -F fr d (7.18) 

= -(5.00N)(0.800m) 
= -4.00 J. 

So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively, 

W gI = 0, (7.19) 

W N = 0, 
W app = 96.0 J, 
Wfj. = - 4.00 J. 

The total work done as the sum of the work done by each force is then seen to be 

W to tal = W gr + ^N + W app + W fr = 92.0 J. (7.20) 

Discussion for (b) 

The calculated total work W total as the sum of the work by each force agrees, as expected, with the work W nQt done by the net force. The 
work done by a collection of forces acting on an object can be calculated by either approach. 



Example 7.4 Determining Speed from Work and Energy ^^^^^^| 








Find the speed of the package in Figure 7.4 at the end of the push, using work and energy concepts. 






Strategy 








Here the work-energy theorem can be used, because we have just calculated the net work, 


W nQt , and the initial kinetic energy, 


lmv 2 


. These 


1 2 

calculations allow us to find the final kinetic energy, -^mv , and thus the final speed v . 








Solution 








The work-energy theorem in equation form is 








Wnet = ^mv 2 - lmv 2 . 






(7.21) 


1 2 
Solving for ^-mv gives 








l mv 2 = W nQt + lmv 2 . 






(7.22) 


Thus, 








lmv 2 = 92.0 J+3.75 J = 95.75 J. 






(7.23) 


Solving for the final speed as requested and entering known values gives 









CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 227 





















(7.24) 


jl91.5kg-m 2 /s 2 
1 30.0 kg 


v = f (95 m 75, M 








= 2.53 m/s. 














Discussion 




















Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the 


sum 


of the 


initial kinetic 


energy 


and the net 


work done on 


the package. 


This 


means that the work indeed adds to the 


energy of the package. 













Example 7.5 Work and Energy Can Reveal Distance, Too 



How far does the package in Figure 7.4 coast after the push, assuming friction remains constant? Use work and energy considerations. 

Strategy 

We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has 
removed all of the package's kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the 
angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing. 

Solution 

The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force, and it acts opposite to 
the displacement, so 6 = 180° . To reduce the kinetic energy of the package to zero, the work Wf T by friction must be minus the kinetic energy 

that the package started with plus what the package accumulated due to the pushing. Thus Wf T = —95.75 J . Furthermore, 

Wfj. = fd' cos 6 = -fd' , where d' is the distance it takes to stop. Thus, 

j, = W fr -95.75 J ( 7 - 25 ) 

/ 5.00 N ' 

and so 

d! = 19.2 m. (7.26) 

Discussion 

This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the work done by friction is negative 
(the force is in the opposite direction of motion), so it removes the kinetic energy. 

Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what 
work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics 
and dynamics alone. 

7.3 Gravitational Potential Energy 

Work Done Against Gravity 

Climbing stairs and lifting objects is work in both the scientific and everyday sense — it is work done against the gravitational force. When there is 
work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will 
explore in this section. 

Let us calculate the work done in lifting an object of mass m through a height h , such as in Figure 7.5. If the object is lifted straight up at constant 
speed, then the force needed to lift it is equal to its weight mg . The work done on the mass is then W = Fd = mgh . We define this to be the 
gravitational potential energy (PE g ) put into (or gained by) the object-Earth system. This energy is associated with the state of separation 
between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PE g gained by the object, recognizing 

that this is energy stored in the gravitational field of Earth. Why do we use the word "system"? Potential energy is a property of a system rather than 
of a single object — due to its physical position. An object's gravitational potential is due to its position relative to the surroundings within the Earth- 
object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational 
potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the 
potential energy equal to 0. We usually choose this point to be Earth's surface, but this point is arbitrary; what is important is the difference in 
gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in 
the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs. 

Converting Between Potential Energy and Kinetic Energy 

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an 
amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more 

useful to consider just the conversion of PE g to KE without explicitly considering the intermediate step of work. (See Example 7.7.) This shortcut 

makes it is easier to solve problems using energy (if possible) rather than explicitly using forces. 



228 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 




mgh 



(a) (b) 

Figure 7.5 (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy, (b) As the weight moves downward, this gravitational 
potential energy is transferred to the cuckoo clock. 



More precisely, we define the change in gravitational potential energy APE g to be 

APEg = mgh, 



(7.27) 



where, for simplicity, we denote the change in height by h rather than the usual Ah . Note that h is positive when the final height is greater than the 
initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential 
energy is 

mgh = (0.500 kg)(9.80m/s 2 )(l. 00 m) < 7 - 28 ) 

= 4.90 kg -m 2 /s 2 = 4.90 J. 

Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass 
is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity 
that does the work. 

Using Potential Energy to Simplify Calculations 

The equation APE g = mgh applies for any path that has a change in height of h , not just when the mass is lifted straight up. (See Figure 7.6.) It 

is much easier to calculate mgh (a simple multiplication) than it is to calculate the work done along a complicated path. The idea of gravitational 

potential energy has the double advantage that it is very broadly applicable and it makes calculations easier. From now on, we will consider that any 
change in vertical position h of a mass m is accompanied by a change in gravitational potential energy mgh , and we will avoid the equivalent but 

more difficult task of calculating work done by or against the gravitational force. 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 229 




Figure 7.6 The change in gravitational potential energy (APEg) between points A and B is independent of the path. APEg = mgh for any path between the two points. 
Gravity is one of a small class of forces where the work done by or against the force depends only on the starting and ending points, not on the path between them. 



Example 7.6 The Force to Stop Falling 



A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the 
force on the knee joints. 

Strategy 

This person's energy is brought to zero in this situation by the work done on him by the floor as he stops. The initial PE g is transformed into 

KE as he falls. The work done by the floor reduces this kinetic energy to zero. 

Solution 

The work done on the person by the floor as he stops is given by 

W = Fd cos 6 = -Fd, (7.29) 

with a minus sign because the displacement while stopping and the force from floor are in opposite directions (cos 6 = cos 180° = — 1) . The 

floor removes energy from the system, so it does negative work. 

The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height h : 

KE = -APEg = -mgh, (7.30) 

The distance d that the person's knees bend is much smaller than the height h of the fall, so the additional change in gravitational potential 
energy during the knee bend is ignored. 

The work W done by the floor on the person stops the person and brings the person's kinetic energy to zero: 

W = -KE = mgh. (7.31) 

Combining this equation with the expression for W gives 

-Fd = mgh. (7.32) 

Recalling that h is negative because the person fell down, the force on the knee joints is given by 

ms u (60.0kg)(9.80m/s 2 V-3.00m) , ( 7 - 33 ) 

F = -^p = ^ { = 3.53X10 5 N. 

d 5.00xl0" 3 m 

Discussion 

Such a large force (500 times more than the person's weight) over the short impact time is enough to break bones. A much better way to cushion 
the shock is by bending the legs or rolling on the ground, increasing the time over which the force acts. A bending motion of 0.5 m this way yields 
a force 100 times smaller than in the example. A kangaroo's hopping shows this method in action. The kangaroo is the only large animal to use 
hopping for locomotion, but the shock in hopping is cushioned by the bending of its hind legs in each jump. (See Figure 7.7.) 



230 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 




Figure 7.7 The work done by the ground upon the kangaroo reduces its kinetic energy to zero as it lands. However, by applying the force of the ground on the hind legs over a 
longer distance, the impact on the bones is reduced, (credit: Chris Samuel, Flickr) 



Example 7.7 Finding the Speed of a Roller Coaster from its Height 



(a) What is the final speed of the roller coaster shown in Figure 7.8 if it starts from rest at the top of the 20.0 m hill and work done by frictional 
forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m/s? 




Figure 7.8 The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the roller-coaster-Earth 
system's gravitational potential energy is converted to kinetic energy. If work done by friction is negligible, all APEg is converted to KE . 

Strategy 

The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal 
force, which is perpendicular to the direction of motion and does no work. The net work on the roller coaster is then done by gravity alone. The 
loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy. This can be written in 
equation form as — APE g = AKE . Using the equations for PE g and KE , we can solve for the final speed v , which is the desired quantity. 

Solution for (a) 

1 2 

Here the initial kinetic energy is zero, so that AKE = -^mv . The equation for change in potential energy states that APE g = mgh . Since h 
is negative in this case, we will rewrite this as APE g = — mg \ h \ to show the minus sign clearly. Thus, 



becomes 



Solving for v , we find that mass cancels and that 



-APE g = AKE 



1 2 

mg | h | = -^mv . 



v = i2g\h 



(7.34) 



(7.35) 



(7.36) 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 231 

Substituting known values, 

v = ^2(9.80 m/s 2 )(20.0 m) (737) 

= 19.8 m/s. 
Solution for (b) 

1 2 1 2 

Again — APE g = AKE . In this case there is initial kinetic energy, so AKE = ^mv — -k-mvQ . Thus, 



Rearranging gives 



mg\h\ =±mv 2 -lmv 2 . (7 - 38) 



±mv 2 = mg\h\ + lmv 2 . (7 " 39) 



This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass again cancels, and 



v = l2g\h\W- (7 ' 40) 



This equation is very similar to the kinematics equation v = ]j v + lad , but it is more general — the kinematics equation is valid only for 

constant acceleration, whereas our equation above is valid for any path regardless of whether the object moves with a constant acceleration. 
Now, substituting known values gives 

v = V2(9.80 m/s 2 )(20.0 m) + (5.00 m/s) 2 (741) 

= 20.4 m/s. 

Discussion and Implications 

First, note that mass cancels. This is quite consistent with observations made in Falling Objects that all objects fall at the same rate if friction is 
negligible. Second, only the speed of the roller coaster is considered; there is no information about its direction at any point. This reveals another 
general truth. When friction is negligible, the speed of a falling body depends only on its initial speed and height, and not on its mass or the path 
taken. For example, the roller coaster will have the same final speed whether it falls 20.0 m straight down or takes a more complicated path like 
the one in the figure. Third, and perhaps unexpectedly, the final speed in part (b) is greater than in part (a), but by far less than 5.00 m/s. Finally, 
note that speed can be found at any height along the way by simply using the appropriate value of h at the point of interest. 

We have seen that work done by or against the gravitational force depends only on the starting and ending points, and not on the path between, 
allowing us to define the simplifying concept of gravitational potential energy. We can do the same thing for a few other forces, and we will see that 
this leads to a formal definition of the law of conservation of energy. 

Making Connections: Take-Home Investigation — Converting Potential to Kinetic Energy 

One can study the conversion of gravitational potential energy into kinetic energy in this experiment. On a smooth, level surface, use a ruler of 
the kind that has a groove running along its length and a book to make an incline (see Figure 7.9). Place a marble at the 10-cm position on the 
ruler and let it roll down the ruler. When it hits the level surface, measure the time it takes to roll one meter. Now place the marble at the 20-cm 
and the 30-cm positions and again measure the times it takes to roll 1 m on the level surface. Find the velocity of the marble on the level surface 
for all three positions. Plot velocity squared versus the distance traveled by the marble. What is the shape of each plot? If the shape is a straight 
line, the plot shows that the marble's kinetic energy at the bottom is proportional to its potential energy at the release point. 




Figure 7.9 A marble rolls down a ruler, and its speed on the level surface is measured. 

7.4 Conservative Forces and Potential Energy 

Potential Energy and Conservative Forces 

Work is done by a force, and some forces, such as weight, have special characteristics. A conservative force is one, like the gravitational force, for 
which work done by or against it depends only on the starting and ending points of a motion and not on the path taken. We can define a potential 
energy (PE) for any conservative force, just as we did for the gravitational force. For example, when you wind up a toy, an egg timer, or an old- 



232 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 

fashioned watch, you do work against its spring and store energy in it. (We treat these springs as ideal, in that we assume there is no friction and no 
production of thermal energy.) This stored energy is recoverable as work, and it is useful to think of it as potential energy contained in the spring. 
Indeed, the reason that the spring has this characteristic is that its force is conservative. That is, a conservative force results in stored or potential 
energy. Gravitational potential energy is one example, as is the energy stored in a spring. We will also see how conservative forces are related to the 
conservation of energy. 



Potential Energy and Conservative Forces 

Potential energy is the energy a system has due to position, shape, or configuration. It is stored energy that is completely recoverable. 

A conservative force is one for which work done by or against it depends only on the starting and ending points of a motion and not on the path 
taken. 

We can define a potential energy (PE) for any conservative force. The work done against a conservative force to reach a final configuration 
depends on the configuration, not the path followed, and is the potential energy added. 



Potential Energy of a Spring 

First, let us obtain an expression for the potential energy stored in a spring ( PE S ). We calculate the work done to stretch or compress a spring that 
obeys Hooke's law. (Hooke's law was examined in Elasticity: Stress and Strain, and states that the magnitude of force F on the spring and the 
resulting deformation AL are proportional, F = kAL .) (See Figure 7.10.) For our spring, we will replace AL (the amount of deformation 
produced by a force F) by the distance x that the spring is stretched or compressed along its length. So the force needed to stretch the spring has 
magnitude F = kx , where k is the spring's force constant. The force increases linearly from at the start to kx in the fully stretched position. The 

average force is kx/2 . Thus the work done in stretching or compressing the spring is W s = Fd = y^-\x = -^-kx . Alternatively, we noted in 
Kinetic Energy and the Work-Energy Theorem that the area under a graph of F vs. x is the work done by the force. In Figure 7.10(c) we see 

1 2 

that this area is also -*rkx . We therefore define the potential energy of a spring, PE S , to be 



PE S 



\kx\ 



(7.42) 



where k is the spring's force constant and x is the displacement from its undeformed position. The potential energy represents the work done on 
the spring and the energy stored in it as a result of stretching or compressing it a distance x . The potential energy of the spring PE S does not 
depend on the path taken; it depends only on the stretch or squeeze x in the final configuration. 



#AMMr? 



KAAAA/WW — x 



^x*-l 



(a) 



(b) 




Figure 7.10 (a) An undeformed spring has no PE S stored in it. (b) The force needed to stretch (or compress) the spring a distance X has a magnitude F = kx , and the 

1 2 
work done to stretch (or compress) it is -~kx . Because the force is conservative, this work is stored as potential energy (PE S ) in the spring, and it can be fully recovered. 

1 9 1 9 

(c) A graph of F vs. X has a slope of k , and the area under the graph is -^kx . Thus the work done or potential energy stored is -^kx . 



1 2 

The equation PE S = -k-kx has general validity beyond the special case for which it was derived. Potential energy can be stored in any elastic 
medium by deforming it. Indeed, the general definition of potential energy is energy due to position, shape, or configuration. For shape or position 

1 2 

deformations, stored energy is PE S = ^rkx , where k is the force constant of the particular system and x is its deformation. Another example is 
seen in Figure 7.11 for a guitar string. 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 233 




Figure 7.11 Work is done to deform the guitar string, giving it potential energy. When released, the potential energy is converted to kinetic energy and back to potential as the 
string oscillates back and forth. A very small fraction is dissipated as sound energy, slowly removing energy from the string. 

Conservation of Mechanical Energy 

Let us now consider what form the work-energy theorem takes when only conservative forces are involved. This will lead us to the conservation of 
energy principle. The work-energy theorem states that the net work done by all forces acting on a system equals its change in kinetic energy. In 
equation form, this is 



W n 



If only conservative forces act, then 



\mv 2 - 



W n 



-k mv o 2 : 



W c 



AKE. 



where W c is the total work done by all conservative forces. Thus, 

W c = AKE. 

Now, if the conservative force, such as the gravitational force or a spring force, does work, the system loses potential energy. That is, W c ■ 
Therefore, 

-APE = AKE 



or 



AKE + APE = 0. 



This equation means that the total kinetic and potential energy is constant for any process involving only conservative forces. That is, 

KE + PE = constant ] 



(7.43) 
(7.44) 

(7.45) 
-APE. 

(7.46) 
(7.47) 
(7.48) 



or 



KEi + PEi = KE f + PE f 



^(conservative forces only), 






where i and f denote initial and final values. This equation is a form of the work-energy theorem for conservative forces; it is known as the 
conservation of mechanical energy principle. Remember that this applies to the extent that all the forces are conservative, so that friction is 
negligible. The total kinetic plus potential energy of a system is defined to be its mechanical energy, (KE + PE) . In a system that experiences only 

conservative forces, there is a potential energy associated with each force, and the energy only changes form between KE and the various types of 
PE , with the total energy remaining constant. 



Example 7.8 Using Conservation of Mechanical Energy to Calculate the Speed of a Toy Car 



A 0.100-kg toy car is propelled by a compressed spring, as shown in Figure 7.12. The car follows a track that rises 0.180 m above the starting 
point. The spring is compressed 4.00 cm and has a force constant of 250.0 N/m. Assuming work done by friction to be negligible, find (a) how 
fast the car is going before it starts up the slope and (b) how fast it is going at the top of the slope. 



234 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 





» T>. 


Path of the Car 




WWW-9 


— s- * k * 

/ ft = 18 cm 
/ i 




Alternate path ^i 







Figure 7.12 A toy car is pushed by a compressed spring and coasts up a slope. Assuming negligible friction, the potential energy in the spring is first completely 
converted to kinetic energy, and then to a combination of kinetic and gravitational potential energy as the car rises. The details of the path are unimportant because all 
forces are conservative — the car would have the same final speed if it took the alternate path shown. 

Strategy 

The spring force and the gravitational force are conservative forces, so conservation of mechanical energy can be used. Thus, 

KEi +PEi = KE f + PE f (7.49) 



or 



1 9 1 9 1 9 1 9 

ymvj + mgh x + -*rkx. = ymv f + rnghf + yfor f , 



(7.50) 



where h is the height (vertical position) and x is the compression of the spring. This general statement looks complex but becomes much 
simpler when we start considering specific situations. First, we must identify the initial and final conditions in a problem; then, we enter them into 
the last equation to solve for an unknown. 

Solution for (a) 

This part of the problem is limited to conditions just before the car is released and just after it leaves the spring. Take the initial height to be zero, 
so that both h x and hf are zero. Furthermore, the initial speed v x is zero and the final compression of the spring Xf is zero, and so several 

terms in the conservation of mechanical energy equation are zero and it simplifies to 



^■kx: = ^mv 



f • 



(7.51) 



In other words, the initial potential energy in the spring is converted completely to kinetic energy in the absence of friction. Solving for the final 
speed and entering known values yields 



v f = 



fm A i 



(7.52) 



250.0 N/m (Q 0400 n 
0.100 kg ^• UHUU111 >' 

= 2.00 m/s. 

Solution for (b) 

One method of finding the speed at the top of the slope is to consider conditions just before the car is released and just after it reaches the top of 
the slope, completely ignoring everything in between. Doing the same type of analysis to find which terms are zero, the conservation of 
mechanical energy becomes 

±kx { 2 = ±mv f 2 + mgh f . (7 - 53) 

This form of the equation means that the spring's initial potential energy is converted partly to gravitational potential energy and partly to kinetic 
energy. The final speed at the top of the slope will be less than at the bottom. Solving for v f and substituting known values gives 



kxS 



(7.54) 



-2gh f 



feH" ! 



100 kg 
0.687 m/s. 



2(9.80 m/s z )(0. 180 m) 



Discussion 

Another way to solve this problem is to realize that the car's kinetic energy before it goes up the slope is converted partly to potential 
energy— that is, to take the final conditions in part (a) to be the initial conditions in part (b). 



Note that, for conservative forces, we do not directly calculate the work they do; rather, we consider their effects through their corresponding potential 
energies, just as we did in Example 7.8. Note also that we do not consider details of the path taken — only the starting and ending points are 
important (as long as the path is not impossible). This assumption is usually a tremendous simplification, because the path may be complicated and 
forces may vary along the way. 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 235 



PhET Explorations: Energy Skate Park 



Learn about conservation of energy with a skater dude! Build tracks, ramps and jumps for the skater and view the kinetic energy, potential 
energy and friction as he moves. You can also take the skater to different planets or even space! 



M 



V 



^PhET Interactive Simulation 



Figure 7.13 Energy Skate Park (http://cnx.Org/content/m42149/l.3/energy-skate-park_en.jar) 



7.5 Nonconservative Forces 



Nonconservative Forces and Friction 

Forces are either conservative or nonconservative. Conservative forces were discussed in Conservative Forces and Potential Energy. A 
nonconservative force is one for which work depends on the path taken. Friction is a good example of a nonconservative force. As illustrated in 
Figure 7.14, work done against friction depends on the length of the path between the starting and ending points. Because of this dependence on 
path, there is no potential energy associated with nonconservative forces. An important characteristic is that the work done by a nonconservative 
force adds or removes mechanical energy from a system. Friction, for example, creates thermal energy that dissipates, removing energy from the 
system. Furthermore, even if the thermal energy is retained or captured, it cannot be fully converted back to work, so it is lost or not recoverable in 
that sense as well. 



-? 



•B 






6 



(a) (b) 

Figure 7.14 The amount of the happy face erased depends on the path taken by the eraser between points A and B, as does the work done against friction. Less work is done 
and less of the face is erased for the path in (a) than for the path in (b). The force here is friction, and most of the work goes into thermal energy that subsequently leaves the 
system (the happy face plus the eraser). The energy expended cannot be fully recovered. 



How Nonconservative Forces Affect Mechanical Energy 

Mechanical energy may not be conserved when nonconservative forces act. For example, when a car is brought to a stop by friction on level ground, 
it loses kinetic energy, which is dissipated as thermal energy, reducing its mechanical energy. Figure 7.15 compares the effects of conservative and 
nonconservative forces. We often choose to understand simpler systems such as that described in Figure 7.15(a) first before studying more 
complicated systems as in Figure 7.15(b). 



236 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 



System 



System 




PE g = PE S = KE 



(a) 



(b) 



Figure 7.15 Comparison of the effects of conservative and nonconservative forces on the mechanical energy of a system, (a) A system with only conservative forces. When a 
rock is dropped onto a spring, its mechanical energy remains constant (neglecting air resistance) because the force in the spring is conservative. The spring can propel the 
rock back to its original height, where it once again has only potential energy due to gravity, (b) A system with nonconservative forces. When the same rock is dropped onto the 
ground, it is stopped by nonconservative forces that dissipate its mechanical energy as thermal energy, sound, and surface distortion. The rock has lost mechanical energy. 

How the Work-Energy Theorem Applies 

Now let us consider what form the work-energy theorem takes when both conservative and nonconservative forces act. We will see that the work 
done by nonconservative forces equals the change in the mechanical energy of a system. As noted in Kinetic Energy and the Work-Energy 
Theorem, the work-energy theorem states that the net work on a system equals the change in its kinetic energy, or W net = AKE . The net work is 

the sum of the work by nonconservative forces plus the work by conservative forces. That is, 



W n 



W nc + W c 



(7.55) 



so that 



W nc + W C = AKE, 



(7.56) 



where W nc is the total work done by all nonconservative forces and W c is the total work done by all conservative forces. 




Figure 7.16 A person pushes a crate up a ramp, doing work on the crate. Friction and gravitational force (not shown) also do work on the crate; both forces oppose the 
person's push. As the crate is pushed up the ramp, it gains mechanical energy, implying that the work done by the person is greater than the work done by friction. 

Consider Figure 7.16, in which a person pushes a crate up a ramp and is opposed by friction. As in the previous section, we note that work done by 
a conservative force comes from a loss of gravitational potential energy, so that W c = —APE . Substituting this equation into the previous one and 

solving for W nc gives 



W n 



AKE + APE. 



(7.57) 



This equation means that the total mechanical energy (KE + PE) changes by exactly the amount of work done by nonconservative forces. In 

Figure 7.16, this is the work done by the person minus the work done by friction. So even if energy is not conserved for the system of interest (such 
as the crate), we know that an equal amount of work was done to cause the change in total mechanical energy. 

We rearrange W nc = AKE + APE to obtain 



KE + PEi + W nc = KE f + PE f . 



(7.58) 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 237 

This means that the amount of work done by nonconservative forces adds to the mechanical energy of a system. If W nc is positive, then mechanical 

energy is increased, such as when the person pushes the crate up the ramp in Figure 7.16. If W nc is negative, then mechanical energy is 

decreased, such as when the rock hits the ground in Figure 7.15(b). If W nc is zero, then mechanical energy is conserved, and nonconservative 

forces are balanced. For example, when you push a lawn mower at constant speed on level ground, your work done is removed by the work of 
friction, and the mower has a constant energy. 

Applying Energy Conservation with Nonconservative Forces 

When no change in potential energy occurs, applying KE + PEj + W nc = KE f + PE f amounts to applying the work-energy theorem by setting 

the change in kinetic energy to be equal to the net work done on the system, which in the most general case includes both conservative and 
nonconservative forces. But when seeking instead to find a change in total mechanical energy in situations that involve changes in both potential and 
kinetic energy, the previous equation KEj + PEj + W nc = KE f + PE f says that you can start by finding the change in mechanical energy that 

would have resulted from just the conservative forces, including the potential energy changes, and add to it the work done, with the proper sign, by 
any nonconservative forces involved. 



Example 7.9 Calculating Distance Traveled: How Far a Baseball Player Slides 



Consider the situation shown in Figure 7.17, where a baseball player slides to a stop on level ground. Using energy considerations, calculate the 
distance the 65.0-kg baseball player slides, given that his initial speed is 6.00 m/s and the force of friction against him is a constant 450 N. 




Figure 7.17 The baseball player slides to a stop in a distance d . In the process, friction removes the player's kinetic energy by doing an amount of work fd equal to 
the initial kinetic energy. 

Strategy 

Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the work-energy theorem, the 
work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The work done by friction is negative, because 
f is in the opposite direction of the motion (that is, 6 = 180° , and so cos 6 = — 1 ). Thus W nc = —fd . The equation simplifies to 



mVi z -fd = 



(7.59) 



or 



fd: 



\mv 2 . 

2 i 



(7.60) 



This equation can now be solved for the distance d . 

Solution 

Solving the previous equation for d and substituting known values yields 

2 



d = 



mv- x 



(7.61) 



(65.0 kg)(6.00 m/s) z 
(2)(450 N) 
= 2.60 m. 

Discussion 

The most important point of this example is that the amount of nonconservative work equals the change in mechanical energy. For example, you 
must work harder to stop a truck, with its large mechanical energy, than to stop a mosquito. 



238 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 



Example 7.10 Calculating Distance Traveled: Sliding Up an Incline 



Suppose that the player from Example 7.9 is running up a hill having a 5.00° incline upward with a surface similar to that in the baseball 
stadium. The player slides with the same initial speed. Determine how far he slides. 




Figure 7.18 The same baseball player slides to a stop on a 5.00° slope. 

Strategy 

In this case, the work done by the nonconservative friction force on the player reduces the mechanical energy he has from his kinetic energy at 
zero height, to the final mechanical energy he has by moving through distance d to reach height h along the hill, with h = d sin 5.00° . This is 
expressed by the equation 

KE + PEi + W nc = KE f + PE f . (7.62) 



1 2 

fd ; initially the potential energy is PEj = mg -0 = and the kinetic energy is KEj = ^-mv. ; 



Solution 

The work done by friction is again W nc 

the final energy contributions are KE f = for the kinetic energy and PE f = mgh = mgd sin 6 for the potential energy. 

Substituting these values gives 

Imvj 2 + + (-fd) = + mgd sin 6. 



Solve this for d to obtain 



f + mg sin 6 



(7.63) 



(7.64) 



(0.5X65.0 kg)(6.00m/s)" 



450 N+(65.0 kg)(9.80 m/s 2 ) sin (5.00°) 
= 2.31m. 



Discussion 

As might have been expected, the player slides a shorter distance by sliding uphill. Note that the problem could also have been solved in terms 
of the forces directly and the work energy theorem, instead of using the potential energy. This method would have required combining the normal 
force and force of gravity vectors, which no longer cancel each other because they point in different directions, and friction, to find the net force. 
You could then use the net force and the net work to find the distance d that reduces the kinetic energy to zero. By applying conservation of 
energy and using the potential energy instead, we need only consider the gravitational potential energy mgh , without combining and resolving 
force vectors. This simplifies the solution considerably. 



Making Connections: Take-Home Investigation — Determining Friction from the Stopping Distance 



This experiment involves the conversion of gravitational potential energy into thermal energy. Use the ruler, book, and marble from Take-Home 
Investigation — Converting Potential to Kinetic Energy. In addition, you will need a foam cup with a small hole in the side, as shown in Figure 
7.19. From the 10-cm position on the ruler, let the marble roll into the cup positioned at the bottom of the ruler. Measure the distance d the cup 
moves before stopping. What forces caused it to stop? What happened to the kinetic energy of the marble at the bottom of the ruler? Next, place 
the marble at the 20-cm and the 30-cm positions and again measure the distance the cup moves after the marble enters it. Plot the distance the 
cup moves versus the initial marble position on the ruler. Is this relationship linear? 

With some simple assumptions, you can use these data to find the coefficient of kinetic friction /i k of the cup on the table. The force of friction 
/ on the cup is ju^N , where the normal force N is just the weight of the cup plus the marble. The normal force and force of gravity do no 
work because they are perpendicular to the displacement of the cup, which moves horizontally. The work done by friction is fd . You will need 
the mass of the marble as well to calculate its initial kinetic energy. 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 239 

It is interesting to do the above experiment also with a steel marble (or ball bearing). Releasing it from the same positions on the ruler as you did 
with the glass marble, is the velocity of this steel marble the same as the velocity of the marble at the bottom of the ruler? Is the distance the cup 
moves proportional to the mass of the steel and glass marbles? 




Figure 7.19 Rolling a marble down a ruler into a foam cup. 
PhET Explorations: The Ramp 



Explore forces, energy and work as you push household objects up and down a ramp. Lower and raise the ramp to see how the angle of 
inclination affects the parallel forces acting on the file cabinet. Graphs show forces, energy and work. 

Pit 

Jj' p>PhET Interactive Simulation 



^ 



Figure 7.20 The Ramp (http://cnx.Org/content/m42150/l.4/the-ramp_en.jar) 

7.6 Conservation of Energy 

Law of Conservation of Energy 

Energy, as we have noted, is conserved, making it one of the most important physical quantities in nature. The law of conservation of energy can 
be stated as follows: 

Total energy is constant in any process. It may change in form or be transferred from one system to another, but the total remains the same. 

We have explored some forms of energy and some ways it can be transferred from one system to another. This exploration led to the definition of two 
major types of energy — mechanical energy (KE + PE) and energy transferred via work done by nonconservative forces ( W nc ) . But energy takes 

many other forms, manifesting itself in many different ways, and we need to be able to deal with all of these before we can write an equation for the 
above general statement of the conservation of energy. 

Other Forms of Energy than Mechanical Energy 

At this point, we deal with all other forms of energy by lumping them into a single group called other energy ( OE ). Then we can state the 
conservation of energy in equation form as 

KEi + PEi + W nc + OEi = KE f + PE f + 0E f . (7.65) 

All types of energy and work can be included in this very general statement of conservation of energy. Kinetic energy is KE , work done by a 
conservative force is represented by PE , work done by nonconservative forces is W nc , and all other energies are included as OE . This equation 
applies to all previous examples; in those situations OE was constant, and so it subtracted out and was not directly considered. 

Making Connections: Usefulness of the Energy Conservation Principle 

The fact that energy is conserved and has many forms makes it very important. You will find that energy is discussed in many contexts, because 
it is involved in all processes. It will also become apparent that many situations are best understood in terms of energy and that problems are 
often most easily conceptualized and solved by considering energy. 

When does OE play a role? One example occurs when a person eats. Food is oxidized with the release of carbon dioxide, water, and energy. Some 
of this chemical energy is converted to kinetic energy when the person moves, to potential energy when the person changes altitude, and to thermal 
energy (another form of OE ). 

Some of the Many Forms of Energy 

What are some other forms of energy? You can probably name a number of forms of energy not yet discussed. Many of these will be covered in later 
chapters, but let us detail a few here. Electrical energy is a common form that is converted to many other forms and does work in a wide range of 
practical situations. Fuels, such as gasoline and food, carry chemical energy that can be transferred to a system through oxidation. Chemical fuel 



240 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 

can also produce electrical energy, such as in batteries. Batteries can in turn produce light, which is a very pure form of energy. Most energy sources 
on Earth are in fact stored energy from the energy we receive from the Sun. We sometimes refer to this as radiant energy, or electromagnetic 
radiation, which includes visible light, infrared, and ultraviolet radiation. Nuclear energy comes from processes that convert measurable amounts of 
mass into energy. Nuclear energy is transformed into the energy of sunlight, into electrical energy in power plants, and into the energy of the heat 
transfer and blast in weapons. Atoms and molecules inside all objects are in random motion. This internal mechanical energy from the random 
motions is called thermal energy, because it is related to the temperature of the object. These and all other forms of energy can be converted into 
one another and can do work. 

Table 7.1 gives the amount of energy stored, used, or released from various objects and in various phenomena. The range of energies and the 
variety of types and situations is impressive. 

Problem-Solving Strategies for Energy 



You will find the following problem-solving strategies useful whenever you deal with energy. The strategies help in organizing and reinforcing 
energy concepts. In fact, they are used in the examples presented in this chapter. The familiar general problem-solving strategies presented 
earlier — involving identifying physical principles, knowns, and unknowns, checking units, and so on — continue to be relevant here. 

Step 1. Determine the system of interest and identify what information is given and what quantity is to be calculated. A sketch will help. 

Step 2. Examine all the forces involved and determine whether you know or are given the potential energy from the work done by the forces. 
Then use step 3 or step 4. 

Step 3. If you know the potential energies for the forces that enter into the problem, then forces are all conservative, and you can apply 
conservation of mechanical energy simply in terms of potential and kinetic energy. The equation expressing conservation of energy is 

KEi + PEi = KE f + PE f . (7.66) 

Step 4. If you know the potential energy for only some of the forces, possibly because some of them are nonconservative and do not have a 
potential energy, or if there are other energies that are not easily treated in terms of force and work, then the conservation of energy law in its 
most general form must be used. 

KEi + PE i + ^nc + OEi = KE f + PE f + OE f . (7.67) 

In most problems, one or more of the terms is zero, simplifying its solution. Do not calculate W c , the work done by conservative forces; it is 
already incorporated in the PE terms. 

Step 5. You have already identified the types of work and energy involved (in step 2). Before solving for the unknown, eliminate terms wherever 
possible to simplify the algebra. For example, choose h = at either the initial or final point, so that PE g is zero there. Then solve for the 

unknown in the customary manner. 

Step 6. Check the answer to see if it is reasonable. Once you have solved a problem, reexamine the forms of work and energy to see if you have 
set up the conservation of energy equation correctly. For example, work done against friction should be negative, potential energy at the bottom 
of a hill should be less than that at the top, and so on. Also check to see that the numerical value obtained is reasonable. For example, the final 
speed of a skateboarder who coasts down a 3-m-high ramp could reasonably be 20 km/h, but not 80 km/h. 

Transformation of Energy 

The transformation of energy from one form into others is happening all the time. The chemical energy in food is converted into thermal energy 
through metabolism; light energy is converted into chemical energy through photosynthesis. In a larger example, the chemical energy contained in 
coal is converted into thermal energy as it burns to turn water into steam in a boiler. This thermal energy in the steam in turn is converted to 
mechanical energy as it spins a turbine, which is connected to a generator to produce electrical energy. (In all of these examples, not all of the initial 
energy is converted into the forms mentioned. This important point is discussed later in this section.) 

Another example of energy conversion occurs in a solar cell. Sunlight impinging on a solar cell (see Figure 7.21) produces electricity, which in turn 
can be used to run an electric motor. Energy is converted from the primary source of solar energy into electrical energy and then into mechanical 
energy. 




Figure 7.21 Solar energy is converted into electrical energy by solar cells, which is used to run a motor in this solar-power aircraft, (credit: NASA) 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 241 



Table 7.1 Energy of Various Objects and Phenomena 







Big Bang 


10 68 


Energy released in a supernova 


10 44 


Fusion of all the hydrogen in Earth's oceans 


10 34 


Annual world energy use 


4xl0 20 


Large fusion bomb (9 megaton) 


3.8xl0 16 


1 kg hydrogen (fusion to helium) 


6.4xl0 14 


1 kg uranium (nuclear fission) 


8.0xl0 13 


Hiroshima-size fission bomb (10 kiloton) 


4.2xl0 13 


90,000-ton aircraft carrier at 30 knots 


l.lxlO 10 


1 barrel crude oil 


5.9xl0 9 


1 ton TNT 


4.2xl0 9 


1 gallon of gasoline 


1.2xl0 8 


Daily home electricity use (developed countries) 


7xl0 7 


Daily adult food intake (recommended) 


1.2xl0 7 


1000-kg car at 90 km/h 


3.1xl0 5 


1 g fat (9.3 kcal) 


3.9xl0 4 


ATP hydrolysis reaction 


3.2xl0 4 


1 g carbohydrate (4.1 kcal) 


1.7xl0 4 


1 g protein (4.1 kcal) 


1.7xl0 4 


Tennis ball at 100 km/h 


22 


Mosquito (l0~ 2 g at 0.5 m/s) 


1.3xl0" 6 


Single electron in a TV tube beam 


4.0xl0" 15 


Energy to break one DNA strand 


lO" 19 



Efficiency 

Even though energy is conserved in an energy conversion process, the output of useful energy or work will be less than the energy input. The 
efficiency Eff of an energy conversion process is defined as 

Efficiency (Elf) = useful ener gY or work Qut P ut = ^out 
total energy input E m ' 



(7.68) 



Table 7.2 lists some efficiencies of mechanical devices and human activities. In a coal-fired power plant, for example, about 40% of the chemical 
energy in the coal becomes useful electrical energy. The other 60% transforms into other (perhaps less useful) energy forms, such as thermal energy, 
which is then released to the environment through combustion gases and cooling towers. 



242 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 



Table 7.2 Efficiency of the Human Body and 
Mechanical Devices 







Cycling and climbing 


20 


Swimming, surface 


2 


Swimming, submerged 


4 


Shoveling 


3 


Weightlifting 


9 


Steam engine 


17 


Gasoline engine 


30 


Diesel engine 


35 


Nuclear power plant 


35 


Coal power plant 


42 


Electric motor 


98 


Compact fluorescent light 


20 


Gas heater (residential) 


90 


Solar cell 


10 



PhET Explorations: Masses and Springs 



A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. 
Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energies for each spring. 



/% 
^ 



PhET Interactive Simulation 



Figure 7.22 Masses and Springs (http://cnx.Org/content/m42151/l.4/mass-spring-lab_en.jar) 



7.7 Power 



What is Power? 

Power — the word conjures up many images: a professional football player muscling aside his opponent, a dragster roaring away from the starting 
line, a volcano blowing its lava into the atmosphere, or a rocket blasting off, as in Figure 7.23. 




Figure 7.23 This powerful rocket on the Space Shuttle Endeavor did work and consumed energy at a very high rate, (credit: NASA) 

These images of power have in common the rapid performance of work, consistent with the scientific definition of power ( P ) as the rate at which 
work is done. 



1. Representative values 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 243 



Power 


Power is the rate at which work is done. 






P _ W 

t 




(7.69) 


The SI unit for power is the watt ( W ), where 1 watt equals 1 joule/second (1 W = 


1 J/s). 





Because work is energy transfer, power is also the rate at which energy is expended. A 60-W light bulb, for example, expends 60 J of energy per 
second. Great power means a large amount of work or energy developed in a short time. For example, when a powerful car accelerates rapidly, it 
does a large amount of work and consumes a large amount of fuel in a short time. 

Calculating Power from Energy 



Example 7.11 Calculating the Power to Climb Stairs 



What is the power output for a 60.0-kg woman who runs up a 3.00 m high flight of stairs in 3.50 s, starting from rest but having a final speed of 
2.00 m/s? (See Figure 7.24.) 

KE + PE a 





w - mg 



Figure 7.24 When this woman runs upstairs starting from rest, she converts the chemical energy originally from food into kinetic energy and gravitational potential energy. 
Her power output depends on how fast she does this. 

Strategy and Concept 

The work going into mechanical energy is W= KE + PE . At the bottom of the stairs, we take both KE and PE g as initially zero; thus, 

1 2 

W = KE f + PEg = -kmVf + mgh , where h is the vertical height of the stairs. Because all terms are given, we can calculate W and then 

divide it by time to get power. 

Solution 

Substituting the expression for W into the definition of power given in the previous equation, P = W 1 1 yields 



ymv f + mgh 



Entering known values yields 



P-W_l 

t 



0.5(60.0 kg)(2.00 m/s) 2 + (60.0 kg)(9.80 m/s 2 )(3.00 m) 
P = 3^0l 

= 120 J +1764 J 
3.50 s 

= 538 W. 



(7.70) 



(7.71) 



Discussion 



The woman does 1764 J of work to move up the stairs compared with only 120 J to increase her kinetic energy; thus, most of her power output is 
required for climbing rather than accelerating. 



It is impressive that this woman's useful power output is slightly less than 1 horsepower (1 hp = 746 W) ! People can generate more than a 

horsepower with their leg muscles for short periods of time by rapidly converting available blood sugar and oxygen into work output. (A horse can put 
out 1 hp for hours on end.) Once oxygen is depleted, power output decreases and the person begins to breathe rapidly to obtain oxygen to 



244 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 

metabolize more food — this is known as the aerobic stage of exercise. If the woman climbed the stairs slowly, then her power output would be much 
less, although the amount of work done would be the same. 



Making Connections: Take-Home Investigation — Measure Your Power Rating 

Determine your own power rating by measuring the time it takes you to climb a flight of stairs. We will ignore the gain in kinetic energy, as the 
above example showed that it was a small portion of the energy gain. Don't expect that your output will be more than about 0.5 hp. 



Examples of Power 

Examples of power are limited only by the imagination, because there are as many types as there are forms of work and energy. (See Table 7.3 for 

some examples.) Sunlight reaching Earth's surface carries a maximum power of about 1.3 kilowatts per square meter (kW/m ). A tiny fraction of 

this is retained by Earth over the long term. Our consumption rate of fossil fuels is far greater than the rate at which they are stored, so it is inevitable 
that they will be depleted. Power implies that energy is transferred, perhaps changing form. It is never possible to change one form completely into 
another without losing some of it as thermal energy. For example, a 60-W incandescent bulb converts only 5 W of electrical power to light, with 55 W 
dissipating into thermal energy. Furthermore, the typical electric power plant converts only 35 to 40% of its fuel into electricity. The remainder 
becomes a huge amount of thermal energy that must be dispersed as heat transfer, as rapidly as it is created. A coal-fired power plant may produce 

1000 megawatts; 1 megawatt (MW) is 10 W of electric power. But the power plant consumes chemical energy at a rate of about 2500 MW, 
creating heat transfer to the surroundings at a rate of 1500 MW. (See Figure 7.25.] 




Figure 7.25 Tremendous amounts of electric power are generated by coal-fired power plants such as this one in China, but an even larger amount of power goes into heat 
transfer to the surroundings. The large cooling towers here are needed to transfer heat as rapidly as it is produced. The transfer of heat is not unique to coal plants but is an 
unavoidable consequence of generating electric power from any fuel — nuclear, coal, oil, natural gas, or the like, (credit: Kleinolive, Wikimedia Commons) 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 245 



Table 7.3 Power Output or Consumption 







Supernova (at peak) 


5xl0 37 


Milky Way galaxy 


10 37 


Crab Nebula pulsar 


10 28 


The Sun 


4xl0 26 


Volcanic eruption (maximum) 


4xl0 15 


Lightning bolt 


2xl0 12 


Nuclear power plant (total electric and heat transfer) 


3xl0 9 


Aircraft carrier (total useful and heat transfer) 


10 8 


Dragster (total useful and heat transfer) 


2xl0 6 


Car (total useful and heat transfer) 


8xl0 4 


Football player (total useful and heat transfer) 


5xl0 3 


Clothes dryer 


4xl0 3 


Person at rest (all heat transfer) 


100 


Typical incandescent light bulb (total useful and heat transfer) 


60 


Heart, person at rest (total useful and heat transfer) 


8 


Electric clock 


3 


Pocket calculator 


10" 3 



Power and Energy Consumption 

We usually have to pay for the energy we use. It is interesting and easy to estimate the cost of energy for an electrical appliance if its power 
consumption rate and time used are known. The higher the power consumption rate and the longer the appliance is used, the greater the cost of that 
appliance. The power consumption rate is P = W 1 t = E 1 t , where E is the energy supplied by the electricity company. So the energy consumed 
over a time t is 

E = Pt. (7.72) 

Electricity bills state the energy used in units of kilowatt- hours (kW • h), which is the product of power in kilowatts and time in hours. This unit is 
convenient because electrical power consumption at the kilowatt level for hours at a time is typical. 



Example 7.12 Calculating Energy Costs 



What is the cost of running a 0.200-kW computer 6.00 h per day for 30.0 d if the cost of electricity is $0,120 per kW • h ? 

Strategy 

Cost is based on energy consumed; thus, we must find E from E = Pt and then calculate the cost. Because electrical energy is expressed in 

kW • h , at the start of a problem such as this it is convenient to convert the units into kW and hours. 

Solution 

The energy consumed in kW • h is 

E = Pt = (0.200 kW)(6.00h/d)(30.0d) (7.73) 

= 36.0 kW-h, 



and the cost is simply given by 



Discussion 



cost = (36.0 kW • h)($0.120 per kW • h) = $4.32 per month. 



(7.74) 



The cost of using the computer in this example is neither exorbitant nor negligible. It is clear that the cost is a combination of power and time. 
When both are high, such as for an air conditioner in the summer, the cost is high. 



246 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 

The motivation to save energy has become more compelling with its ever-increasing price. Armed with the knowledge that energy consumed is the 
product of power and time, you can estimate costs for yourself and make the necessary value judgments about where to save energy. Either power 
or time must be reduced. It is most cost-effective to limit the use of high-power devices that normally operate for long periods of time, such as water 
heaters and air conditioners. This would not include relatively high power devices like toasters, because they are on only a few minutes per day. It 
would also not include electric clocks, in spite of their 24-hour-per-day usage, because they are very low power devices. It is sometimes possible to 
use devices that have greater efficiencies — that is, devices that consume less power to accomplish the same task. One example is the compact 
fluorescent light bulb, which produces over four times more light per watt of power consumed than its incandescent cousin. 

Modern civilization depends on energy, but current levels of energy consumption and production are not sustainable. The likelihood of a link between 
global warming and fossil fuel use (with its concomitant production of carbon dioxide), has made reduction in energy use as well as a shift to non- 
fossil fuels of the utmost importance. Even though energy in an isolated system is a conserved quantity, the final result of most energy 
transformations is waste heat transfer to the environment, which is no longer useful for doing work. As we will discuss in more detail in 
Thermodynamics, the potential for energy to produce useful work has been "degraded" in the energy transformation. 

7.8 Work, Energy, and Power in Humans 

Energy Conversion in Humans 

Our own bodies, like all living organisms, are energy conversion machines. Conservation of energy implies that the chemical energy stored in food is 
converted into work, thermal energy, and/or stored as chemical energy in fatty tissue. (See Figure 7.26.) The fraction going into each form depends 
both on how much we eat and on our level of physical activity. If we eat more than is needed to do work and stay warm, the remainder goes into body 
fat. 



OEi 



Food 
energy 



Work Wnc (negative) 




^OE f 



OE; + W nc = OE f 



Figure 7.26 Energy consumed by humans is converted to work, thermal energy, and stored fat. By far the largest fraction goes to thermal energy, although the fraction varies 
depending on the type of physical activity. 

Power Consumed at Rest 

The rate at which the body uses food energy to sustain life and to do different activities is called the metabolic rate. The total energy conversion rate 
of a person at rest is called the basal metabolic rate (BMR) and is divided among various systems in the body, as shown in Table 7.4. The largest 
fraction goes to the liver and spleen, with the brain coming next. Of course, during vigorous exercise, the energy consumption of the skeletal muscles 
and heart increase markedly. About 75% of the calories burned in a day go into these basic functions. The BMR is a function of age, gender, total 
body weight, and amount of muscle mass (which burns more calories than body fat). Athletes have a greater BMR due to this last factor. 



Table 7.4 Basal Metabolic Rates (BMR) 





1 1 1 


Power consumed at rest (W) 


Oxygen consumption (mL/mm) 


Percent of BMR 


Liver & spleen 


23 


67 


27 


Brain 


16 


47 


19 


Skeletal muscle 


15 


45 


18 


Kidney 


9 


26 


10 


Heart 


6 


17 


7 


Other 


16 


48 


19 


Totals 


85 W 


250 mL/min 


100% 



Energy consumption is directly proportional to oxygen consumption because the digestive process is basically one of oxidizing food. We can measure 
the energy people use during various activities by measuring their oxygen use. (See Figure 7.27.) Approximately 20 kJ of energy are produced for 
each liter of oxygen consumed, independent of the type of food. Table 7.5 shows energy and oxygen consumption rates (power expended) for a 
variety of activities. 

Power of Doing Useful Work 

Work done by a person is sometimes called useful work, which is work done on the outside world, such as lifting weights. Useful work requires a 
force exerted through a distance on the outside world, and so it excludes internal work, such as that done by the heart when pumping blood. Useful 
work does include that done in climbing stairs or accelerating to a full run, because these are accomplished by exerting forces on the outside world. 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 247 

Forces exerted by the body are nonconservative, so that they can change the mechanical energy ( KE + PE ) of the system worked upon, and this is 
often the goal. A baseball player throwing a ball, for example, increases both the ball's kinetic and potential energy. 

If a person needs more energy than they consume, such as when doing vigorous work, the body must draw upon the chemical energy stored in fat. 
So exercise can be helpful in losing fat. However, the amount of exercise needed to produce a loss in fat, or to burn off extra calories consumed that 
day, can be large, as Example 7.13 illustrates. 



Example 7.13 Calculating Weight Loss from Exercising 



If a person who normally requires an average of 12,000 kJ (3000 kcal) of food energy per day consumes 13,000 kJ per day, he will steadily gain 
weight. How much bicycling per day is required to work off this extra 1000 kJ? 

Solution 

Table 7.5 states that 400 W are used when cycling at a moderate speed. The time required to work off 1000 kJ at this rate is then 

Time = .S2L = MM = 2500 s = 42 min. < 7 - 75 > 

energy 
time 

Discussion 



/energy \ 4Q0 W 
V time / 



If this person uses more energy than he or she consumes, the person's body will obtain the needed energy by metabolizing body fat. If the 
person uses 13,000 kJ but consumes only 12,000 kJ, then the amount of fat loss will be 



Fat loss = (1000 W 1 '^^ ) = 26 g> 



(7.76) 



assuming the energy content of fat to be 39 kJ/g. 




Figure 7.27 A pulse oxymeter is an apparatus that measures the amount of oxygen in blood. Oxymeters can be used to determine a person's metabolic rate, which is the rate 
at which food energy is converted to another form. Such measurements can indicate the level of athletic conditioning as well as certain medical problems, (credit: UusiAjaja, 
Wikimedia Commons) 



Table 7.5 Energy and Oxygei 






i 


Energy consumption in watts 


Oxygen consumption in liters 02/min 


Sleeping 


83 


0.24 


Sitting at rest 


120 


0.34 


Standing relaxed 


125 


0.36 


Sitting in class 


210 


0.60 


Walking (5 km/h) 


280 


0.80 


Cycling (13-18 km/h) 


400 


1.14 


Shivering 


425 


1.21 


Playing tennis 


440 


1.26 


Swimming breaststroke 


475 


1.36 


Ice skating (14.5 km/h) 


545 


1.56 


Climbing stairs (116/min) 


685 


1.96 


Cycling (21 km/h) 


700 


2.00 


Running cross-country 


740 


2.12 


Playing basketball 


800 


2.28 


Cycling, professional racer 


1855 


5.30 


Sprinting 


2415 


6.90 



2. for an average 76-kg male 



248 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 

All bodily functions, from thinking to lifting weights, require energy. (See Figure 7.28.) The many small muscle actions accompanying all quiet activity, 
from sleeping to head scratching, ultimately become thermal energy, as do less visible muscle actions by the heart, lungs, and digestive tract. 
Shivering, in fact, is an involuntary response to low body temperature that pits muscles against one another to produce thermal energy in the body 
(and do no work). The kidneys and liver consume a surprising amount of energy, but the biggest surprise of all it that a full 25% of all energy 
consumed by the body is used to maintain electrical potentials in all living cells. (Nerve cells use this electrical potential in nerve impulses.) This 
bioelectrical energy ultimately becomes mostly thermal energy, but some is utilized to power chemical processes such as in the kidneys and liver, and 
in fat production. 









^^^^9^ 


1 


MfW^^^^ ^"^ 


ittss!^ 




$L,. 


■ s 




atSo^'tf 


J»X( 


4 


w& 


M 


f 


+ 













Figure 7.28 This fMRI scan shows an increased level of energy consumption in the vision center of the brain. Here, the patient was being asked to recognize faces, (credit: 
NIH via Wikimedia Commons) 

7.9 World Energy Use 

Energy is an important ingredient in all phases of society. We live in a very interdependent world, and access to adequate and reliable energy 
resources is crucial for economic growth and for maintaining the quality of our lives. But current levels of energy consumption and production are not 
sustainable. About 40% of the world's energy comes from oil, and much of that goes to transportation uses. Oil prices are dependent as much upon 
new (or foreseen) discoveries as they are upon political events and situations around the world. The U.S., with 4.5% of the world's population, 
consumes 24% of the world's oil production per year; 66% of that oil is imported! 

Renewable and Nonrenewable Energy Sources 

The principal energy resources used in the world are shown in Figure 7.29. The fuel mix has changed over the years but now is dominated by oil, 
although natural gas and solar contributions are increasing. Renewable forms of energy are those sources that cannot be used up, such as water, 
wind, solar, and biomass. About 85% of our energy comes from nonrenewable fossil fuels — oil, natural gas, coal. The likelihood of a link between 
global warming and fossil fuel use, with its production of carbon dioxide through combustion, has made, in the eyes of many scientists, a shift to non- 
fossil fuels of utmost importance — but it will not be easy. 

Petroleum: 3527 - 35.43% 

Coal: 29C2 - 23,15% 

Dry natural gas: 233S - 23.46% 

Hydro-electricity: 624- 6.27% 

Nuclear-electricity; 576 - 5.79% 




Geothermal, wind< 
solar, biomass: 



86 - O.B6% 



Geothermal, bin mass, _ - „., 

solar not used 
for electricity: 

Total! 9955 

Figure 7.29 World energy consumption by source, in billions of kilowatt- hours: 2006. (credit: KVDP) 

The World's Growing Energy Needs 

World energy consumption continues to rise, especially in the developing countries. (See Figure 7.30.) Global demand for energy has tripled in the 
past 50 years and might triple again in the next 30 years. While much of this growth will come from the rapidly booming economies of China and 
India, many of the developed countries, especially those in Europe, are hoping to meet their energy needs by expanding the use of renewable 
sources. Although presently only a small percentage, renewable energy is growing very fast, especially wind energy. For example, Germany plans to 
meet 20% of its electricity and 10% of its overall energy needs with renewable resources by the year 2020. (See Figure 7.31.) Energy is a key 
constraint in the rapid economic growth of China and India. In 2003, China surpassed Japan as the world's second largest consumer of oil. However, 
over 1/3 of this is imported. Unlike most Western countries, coal dominates the commercial energy resources of China, accounting for 2/3 of its 
energy consumption. In 2009 China surpassed the United States as the largest generator of C0 2 . In India, the main energy resources are biomass 

(wood and dung) and coal. Half of India's oil is imported. About 70% of India's electricity is generated by highly polluting coal. Yet there are sizeable 
strides being made in renewable energy. India has a rapidly growing wind energy base, and it has the largest solar cooking program in the world. 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 249 



1000 

800 

1 600 

i 

! 400 

200 





World Energy Consumption 








761 


812 


654 ™ 




605 




532 


373 A2S 









1990 2000 2008 2015 2020 2025 2030 2035 
Year 



Figure 7.30 Past and projected world energy use (source: Based on data from U.S. Energy Information Administration, 2011) 




Figure 7.31 Solar cell arrays at a power plant in Steindorf, Germany (credit: Michael Betke, Flickr) 

Table 7.6 displays the 2006 commercial energy mix by country for some of the prime energy users in the world. While non-renewable sources 
dominate, some countries get a sizeable percentage of their electricity from renewable resources. For example, about 67% of New Zealand's 
electricity demand is met by hydroelectric. Only 10% of the U.S. electricity is generated by renewable resources, primarily hydroelectric. It is difficult 
to determine total contributions of renewable energy in some countries with a large rural population, so these percentages in this table are left blank. 



Table 7.6 Energy Consumption- 


— Selectei 


















Donsumotion 
















. Electricity Use Energy Use 








Nuclear Hydro ~ per capita (kWh/ per capita (GJ/ 

y) y) 


Australia 


5.4 


34% 


17% 


44% 


0% 


3% 


1% 


10000 


260 


Brazil 


9.6 


48% 


7% 


5% 


1% 


35% 


2% 


2000 


50 


China 


63 


22% 


3% 


69% 


1% 


6% 




1500 


35 


Egypt 


2.4 


50% 


41% 


1% 


0% 


6% 




990 


32 


Germany 


16 


37% 


24% 


24% 


11% 


1% 


3% 


6400 


173 


India 


15 


34% 


7% 


52% 


1% 


5% 




470 


13 


Indonesia 


4.9 


51% 


26% 


16% 


0% 


2% 


3% 


420 


22 


Japan 


24 


48% 


14% 


21% 


12% 


4% 


1% 


7100 


176 


New 
Zealand 


0.44 


32% 


26% 


6% 


0% 


11% 


19% 


8500 


102 


Russia 


31 


19% 


53% 


16% 


5% 


6% 




5700 


202 


U.S. 


105 


40% 


23% 


22% 


8% 


3% 


1% 


12500 


340 


World 


432 


39% 


23% 


24% 


6% 


6% 


2% 


2600 


71 



Energy and Economic Well-being 

The last two columns in this table examine the energy and electricity use per capita. Economic well-being is dependent upon energy use, and in most 
countries higher standards of living, as measured by GDP (gross domestic product) per capita, are matched by higher levels of energy consumption 
per capita. This is borne out in Figure 7.32. Increased efficiency of energy use will change this dependency. A global problem is balancing energy 
resource development against the harmful effects upon the environment in its extraction and use. 



250 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 



$45,000 
$40,OD0 
535,000 
$30,OD0 

| $26,000 

p 

q $20,000 

$15,000 

$10,000 

55,000 

3- 











Japan* 






* USA 




. U< 






Canada 


Germany * ^ Francfl 


# Australia 






Italy* 

■ f Spain 


t Korea 


Argentina 

+ ^Wsrtd average 


• Saucf 

Arabia 






* . * •" South AMM R|lss | a 

i Ti — ^ T ■ 1 1 1 1 



6 
kW/cnpita 



10 



V2 



Figure 7.32 Power consumption per capita versus GDP per capita for various countries. Note the increase in energy usage with increasing GDP. (2007, credit: Frank van 
Mierlo, Wikimedia Commons) 

Conserving Energy 

As we finish this chapter on energy and work, it is relevant to draw some distinctions between two sometimes misunderstood terms in the area of 
energy use. As has been mentioned elsewhere, the "law of the conservation of energy" is a very useful principle in analyzing physical processes. It is 
a statement that cannot be proven from basic principles, but is a very good bookkeeping device, and no exceptions have ever been found. It states 
that the total amount of energy in an isolated system will always remain constant. Related to this principle, but remarkably different from it, is the 
important philosophy of energy conservation. This concept has to do with seeking to decrease the amount of energy used by an individual or group 
through (1) reduced activities (e.g., turning down thermostats, driving fewer kilometers) and/or (2) increasing conversion efficiencies in the 
performance of a particular task — such as developing and using more efficient room heaters, cars that have greater miles-per-gallon ratings, energy- 
efficient compact fluorescent lights, etc. 

Since energy in an isolated system is not destroyed or created or generated, one might wonder why we need to be concerned about our energy 
resources, since energy is a conserved quantity. The problem is that the final result of most energy transformations is waste heat transfer to the 
environment and conversion to energy forms no longer useful for doing work. To state it in another way, the potential for energy to produce useful 
work has been "degraded" in the energy transformation. (This will be discussed in more detail in Thermodynamics.) 



Glossary 



basal metabolic rate: the total energy conversion rate of a person at rest 

chemical energy: the energy in a substance stored in the bonds between atoms and molecules that can be released in a chemical reaction 

conservation of mechanical energy: the rule that the sum of the kinetic energies and potential energies remains constant if only conservative 
forces act on and within a system 

conservative force: a force that does the same work for any given initial and final configuration, regardless of the path followed 

efficiency: a measure of the effectiveness of the input of energy to do work; useful energy or work divided by the total input of energy 

electrical energy: the energy carried by a flow of charge 

energy: the ability to do work 

fossil fuels: oil, natural gas, and coal 

friction: the force between surfaces that opposes one sliding on the other; friction changes mechanical energy into thermal energy 

gravitational potential energy: the energy an object has due to its position in a gravitational field 

horsepower: an older non-SI unit of power, with 1 hp = 746 W 

joule: SI unit of work and energy, equal to one newton-meter 

kilowatt-hour: (kW • h) unit used primarily for electrical energy provided by electric utility companies 

kinetic eneruv 1 2 

My " the energy an object has by reason of its motion, equal to ^-mv for the translational (i.e., non-rotational) motion of an object of 

mass m moving at speed v 

law of conservation of energy: the general law that total energy is constant in any process; energy may change in form or be transferred from 
one system to another, but the total remains the same 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 251 

mechanical energy: the sum of kinetic energy and potential energy 

metabolic rate: the rate at which the body uses food energy to sustain life and to do different activities 

net work: work done by the net force, or vector sum of all the forces, acting on an object 

nonconservative force: a force whose work depends on the path followed between the given initial and final configurations 

nuclear energy: energy released by changes within atomic nuclei, such as the fusion of two light nuclei or the fission of a heavy nucleus 

potential energy of a spring: the stored energy of a spring as a function of its displacement; when Hooke's law applies, it is given by the 

1 2 

expression ^kx where x is the distance the spring is compressed or extended and k is the spring constant 

potential energy: energy due to position, shape, or configuration 

power: the rate at which work is done 

radiant energy: the energy carried by electromagnetic waves 

renewable forms of energy: those sources that cannot be used up, such as water, wind, solar, and biomass 

thermal energy: the energy within an object due to the random motion of its atoms and molecules that accounts for the object's temperature 

useful work: work done on an external system 

watt: (W) SI unit of power, with 1 W = 1 J/s 

work-energy theorem: the result, based on Newton's laws, that the net work done on an object is equal to its change in kinetic energy 

work: the transfer of energy by a force that causes an object to be displaced; the product of the component of the force in the direction of the 
displacement and the magnitude of the displacement 



Section Summary 



7.1 Work: The Scientific Definition 

• Work is the transfer of energy by a force acting on an object as it is displaced. 

• The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, 
times the cosine of the angle 6 between them. In symbols, 

W = Fd cos 0. 

9 9 

• The SI unit for work and energy is the joule (J), where lJ=lN-m=lkg- m /s . 

• The work done by a force is zero if the displacement is either zero or perpendicular to the force. 

• The work done is positive if the force and displacement have the same direction, and negative if they have opposite direction. 

7.2 Kinetic Energy and the Work-Energy Theorem 

• The net work W nQt is the work done by the net force acting on an object. 

• Work done on an object transfers energy to the object. 

1 2 

• The translational kinetic energy of an object of mass m moving at speed v is KE = ^mv . 

1 2 1 2 

• The work-energy theorem states that the net work W nQt on a system changes its kinetic energy, W nQt = ^mv — ^rnv$ . 

7.3 Gravitational Potential Energy 

• Work done against gravity in lifting an object becomes potential energy of the object-Earth system. 

• The change in gravitational potential energy, APE g , is APE g = mgh , with h being the increase in height and g the acceleration due to 

gravity. 

• The gravitational potential energy of an object near Earth's surface is due to its position in the mass-Earth system. Only differences in 
gravitational potential energy, APE g , have physical significance. 

• As an object descends without friction, its gravitational potential energy changes into kinetic energy corresponding to increasing speed, so that 

AKE=-APE g . 

7.4 Conservative Forces and Potential Energy 

• A conservative force is one for which work depends only on the starting and ending points of a motion, not on the path taken. 

• We can define potential energy (PE) for any conservative force, just as we defined PE g for the gravitational force. 

1 2 

• The potential energy of a spring is PE S = -^kx , where k is the spring's force constant and x is the displacement from its undeformed 

position. 

• Mechanical energy is defined to be KE + PE for a conservative force. 



252 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 

• When only conservative forces act on and within a system, the total mechanical energy is constant. In equation form, 

KE + PE = constant ] 

or I 

KEi + PEi = KE f + PE f J 

where i and f denote initial and final values. This is known as the conservation of mechanical energy. 

7.5 Nonconservative Forces 

• A nonconservative force is one for which work depends on the path. 

• Friction is an example of a nonconservative force that changes mechanical energy into thermal energy. 

• Work W nc done by a nonconservative force changes the mechanical energy of a system. In equation form, W nc = AKE + APE or, 

equivalents KE^ + PEi + W nc = KE f + PE f . 

• When both conservative and nonconservative forces act, energy conservation can be applied and used to calculate motion in terms of the 
known potential energies of the conservative forces and the work done by nonconservative forces, instead of finding the net work from the net 
force, or having to directly apply Newton's laws. 

7.6 Conservation of Energy 

• The law of conservation of energy states that the total energy is constant in any process. Energy may change in form or be transferred from one 
system to another, but the total remains the same. 

• When all forms of energy are considered, conservation of energy is written in equation form as 

KEj + PEj + W nc + OEj = KE f + PE f + OE f , where OE is all other forms of energy besides mechanical energy. 

• Commonly encountered forms of energy include electric energy, chemical energy, radiant energy, nuclear energy, and thermal energy. 

• Energy is often utilized to do work, but it is not possible to convert all the energy of a system to work. 

W 

• The efficiency Eff of a machine or human is defined to be Eff = out , where W out is useful work output and E m is the energy 

consumed. 

7.7 Power 

• Power is the rate at which work is done, or in equation form, for the average power P for work W done over a time t , P = W 1 1 . 

• The SI unit for power is the watt (W), where 1 W = 1 J/s . 

• The power of many devices such as electric motors is also often expressed in horsepower (hp), where 1 hp = 746 W . 

7.8 Work, Energy, and Power in Humans 

• The human body converts energy stored in food into work, thermal energy, and/or chemical energy that is stored in fatty tissue. 

• The rate at which the body uses food energy to sustain life and to do different activities is called the metabolic rate, and the corresponding rate 
when at rest is called the basal metabolic rate (BMR) 

• The energy included in the basal metabolic rate is divided among various systems in the body, with the largest fraction going to the liver and 
spleen, and the brain coming next. 

• About 75% of food calories are used to sustain basic body functions included in the basal metabolic rate. 

• The energy consumption of people during various activities can be determined by measuring their oxygen use, because the digestive process is 
basically one of oxidizing food. 

7.9 World Energy Use 

• The relative use of different fuels to provide energy has changed over the years, but fuel use is currently dominated by oil, although natural gas 
and solar contributions are increasing. 

• Although non-renewable sources dominate, some countries meet a sizeable percentage of their electricity needs from renewable resources. 

• The United States obtains only about 10% of its energy from renewable sources, mostly hydroelectric power. 

• Economic well-being is dependent upon energy use, and in most countries higher standards of living, as measured by GDP (Gross Domestic 
Product) per capita, are matched by higher levels of energy consumption per capita. 

• Even though, in accordance with the law of conservation of energy, energy can never be created or destroyed, energy that can be used to do 
work is always partly converted to less useful forms, such as waste heat to the environment, in all of our uses of energy for practical purposes. 



Conceptual Questions 



7.1 Work: The Scientific Definition 

1. Give an example of something we think of as work in everyday circumstances that is not work in the scientific sense. Is energy transferred or 
changed in form in your example? If so, explain how this is accomplished without doing work. 

2. Give an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does no work. 

3. Describe a situation in which a force is exerted for a long time but does no work. Explain. 

7.2 Kinetic Energy and the Work-Energy Theorem 

4. The person in Figure 7.33 does work on the lawn mower. Under what conditions would the mower gain energy? Under what conditions would it 
lose energy? 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 253 



W=Fd cose 




Figure 7.33 

5. Work done on a system puts energy into it. Work done by a system removes energy from it. Give an example for each statement. 

6. When solving for speed in Example 7.4, we kept only the positive root. Why? 

7.3 Gravitational Potential Energy 

7. In Example 7.7, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. 
Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 
20 m below the start. We would find in that case that it had the same final speed. Explain in terms of conservation of energy. 

8. Does the work you do on a book when you lift it onto a shelf depend on the path taken? On the time taken? On the height of the shelf? On the 
mass of the book? 

7.4 Conservative Forces and Potential Energy 

9. What is a conservative force? 

10. The force exerted by a diving board is conservative, provided the internal friction is negligible. Assuming friction is negligible, describe changes in 
the potential energy of a diving board as a swimmer dives from it, starting just before the swimmer steps on the board until just after his feet leave it. 

11. Define mechanical energy. What is the relationship of mechanical energy to nonconservative forces? What happens to mechanical energy if only 
conservative forces act? 

12. What is the relationship of potential energy to conservative force? 

7.6 Conservation of Energy 

13. Consider the following scenario. A car for which friction is not negligible accelerates from rest down a hill, running out of gasoline after a short 
distance. The driver lets the car coast farther down the hill, then up and over a small crest. He then coasts down that hill into a gas station, where he 
brakes to a stop and fills the tank with gasoline. Identify the forms of energy the car has, and how they are changed and transferred in this series of 
events. (See Figure 7.34.) 

Coasts Down 
Hill 

Coasts Up 
Over Crest 

Coasts Down 
Hill 

Stops for 
Gasoline 



Hili 

I 




Figure 7.34 A car experiencing non-negligible friction coasts down a hill, over a small crest, then downhill again, and comes to a stop at a gas station. 

14. Describe the energy transfers and transformations for a javelin, starting from the point at which an athlete picks up the javelin and ending when 
the javelin is stuck into the ground after being thrown. 

15. Do devices with efficiencies of less than one violate the law of conservation of energy? Explain. 

16. List four different forms or types of energy. Give one example of a conversion from each of these forms to another form. 

17. List the energy conversions that occur when riding a bicycle. 

7.7 Power 

18. Most electrical appliances are rated in watts. Does this rating depend on how long the appliance is on? (When off, it is a zero-watt device.) 
Explain in terms of the definition of power. 

19. Explain, in terms of the definition of power, why energy consumption is sometimes listed in kilowatt- hours rather than joules. What is the 
relationship between these two energy units? 

20. A spark of static electricity, such as that you might receive from a doorknob on a cold dry day, may carry a few hundred watts of power. Explain 
why you are not injured by such a spark. 



7.8 Work, Energy, and Power in Humans 



254 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 

21. Explain why it is easier to climb a mountain on a zigzag path rather than one straight up the side. Is your increase in gravitational potential energy 
the same in both cases? Is your energy consumption the same in both? 

22. Do you do work on the outside world when you rub your hands together to warm them? What is the efficiency of this activity? 

23. Shivering is an involuntary response to lowered body temperature. What is the efficiency of the body when shivering, and is this a desirable 
value? 

24. Discuss the relative effectiveness of dieting and exercise in losing weight, noting that most athletic activities consume food energy at a rate of 400 
to 500 W, while a single cup of yogurt can contain 1360 kJ (325 kcal). Specifically, is it likely that exercise alone will be sufficient to lose weight? You 
may wish to consider that regular exercise may increase the metabolic rate, whereas protracted dieting may reduce it. 

7.9 World Energy Use 

25. What is the difference between energy conservation and the law of conservation of energy? Give some examples of each. 

26. If the efficiency of a coal-fired electrical generating plant is 35%, then what do we mean when we say that energy is a conserved quantity? 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 255 



Problems & Exercises 



7.1 Work: The Scientific Definition 

1. How much work does a supermarket checkout attendant do on a can 
of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express 
your answer in joules and kilocalories. 

2. A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the 
work done to accomplish this task. 

3. (a) Calculate the work done on a 1500-kg elevator car by its cable to 
lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) 
What is the work done on the lift by the gravitational force in this 
process? (c) What is the total work done on the lift? 

4. Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal 
of gasoline. Only 30% of the gasoline goes into useful work by the force 
that keeps the car moving at constant speed despite friction. (See Table 
7.1 for the energy content of gasoline.) (a) What is the force exerted to 
keep the car moving at constant speed? (b) If the required force is 
directly proportional to speed, how many gallons will be used to drive 108 
km at a speed of 28.0 m/s? 

5. Calculate the work done by an 85.0-kg man who pushes a crate 4.00 
m up along a ramp that makes an angle of 20.0° with the horizontal. 
(See Figure 7.35.) He exerts a force of 500 N on the crate parallel to the 
ramp and moves at a constant speed. Be certain to include the work he 
does on the crate and on his body to get up the ramp. 




Figure 7.35 A man pushes a crate up a ramp. 

6. How much work is done by the boy pulling his sister 30.0 m in a wagon 
as shown in Figure 7.36? Assume no friction acts on the wagon. 




Figure 7.36 The boy does work on the system of the wagon and the child when he 
pulls them as shown. 

7. A shopper pushes a grocery cart 20.0 m at constant speed on level 
ground, against a 35.0 N frictional force. He pushes in a direction 25.0° 
below the horizontal, (a) What is the work done on the cart by friction? (b) 
What is the work done on the cart by the gravitational force? (c) What is 
the work done on the cart by the shopper? (d) Find the force the shopper 
exerts, using energy considerations, (e) What is the total work done on 
the cart? 

8. Suppose the ski patrol lowers a rescue sled and victim, having a total 
mass of 90.0 kg, down a 60.0° slope at constant speed, as shown in 
Figure 7.37. The coefficient of friction between the sled and the snow is 
0.100. (a) How much work is done by friction as the sled moves 30.0 m 
along the hill? (b) How much work is done by the rope on the sled in this 
distance? (c) What is the work done by the gravitational force on the 
sled? (d) What is the total work done? 




A 60 ° 



Figure 7.37 A rescue sled and victim are lowered down a steep slope. 

7.2 Kinetic Energy and the Work-Energy Theorem 

9. Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h 
with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h. 

10. (a) How fast must a 3000-kg elephant move to have the same kinetic 
energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the 
larger energies needed for the movement of larger animals would relate 
to metabolic rates. 

11. Confirm the value given for the kinetic energy of an aircraft carrier in 
Table 7.1. You will need to look up the definition of a nautical mile (1 knot 
= 1 nautical mile/h). 

12. (a) Calculate the force needed to bring a 950-kg car to rest from a 
speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a 
non-panic stop), (b) Suppose instead the car hits a concrete abutment at 
full speed and is brought to a stop in 2.00 m. Calculate the force exerted 
on the car and compare it with the force found in part (a). 

13. A car's bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision 
with an immovable object without damage to the body of the car. The 
bumper cushions the shock by absorbing the force over a distance. 
Calculate the magnitude of the average force on a bumper that collapses 
0.200 m while bringing a 900-kg car to rest from an initial speed of 1.1 
m/s. 

14. Boxing gloves are padded to lessen the force of a blow, (a) Calculate 
the force exerted by a boxing glove on an opponent's face, if the glove 
and face compress 7.50 cm during a blow in which the 7.00-kg arm and 
glove are brought to rest from an initial speed of 10.0 m/s. (b) Calculate 
the force exerted by an identical blow in the gory old days when no 
gloves were used and the knuckles and face would compress only 2.00 
cm. (c) Discuss the magnitude of the force with glove on. Does it seem 
high enough to cause damage even though it is lower than the force with 
no glove? 

15. Using energy considerations, calculate the average force a 60.0-kg 
sprinter exerts backward on the track to accelerate from 2.00 to 8.00 m/s 
in a distance of 25.0 m, if he encounters a headwind that exerts an 
average force of 30.0 N against him. 

7.3 Gravitational Potential Energy 

16. A hydroelectric power facility (see Figure 7.38) converts the 
gravitational potential energy of water behind a dam to electric energy, 
(a) What is the gravitational potential energy relative to the generators of 

a lake of volume 50.0 km ( mass = 5.00X 10 kg) , given that the 

lake has an average height of 40.0 m above the generators? (b) 
Compare this with the energy stored in a 9-megaton fusion bomb. 



256 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 




Figure 7.38 Hydroelectric facility (credit: Denis Belevich, Wikimedia Commons) 

17. (a) How much gravitational potential energy (relative to the ground on 
which it is built) is stored in the Great Pyramid of Cheops, given that its 

mass is about 7x10 kg and its center of mass is 36.5 m above the 

surrounding ground? (b) How does this energy compare with the daily 
food intake of a person? 

18. Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g 
snake and raises it 2.5 m from the ground to a branch, (a) How much 
work did the bird do on the snake? (b) How much work did it do to raise 
its own center of mass to the branch? 

19. In Example 7.7, we found that the speed of a roller coaster that had 
descended 20.0 m was only slightly greater when it had an initial speed 
of 5.00 m/s than when it started from rest. This implies that 

APE >> KEj . Confirm this statement by taking the ratio of APE to 

KEj . (Note that mass cancels.) 

20. A 100-g toy car is propelled by a compressed spring that starts it 
moving. The car follows the curved track in Figure 7.39. Show that the 
final speed of the toy car is 0.687 m/s if its initial speed is 2.00 m/s and it 
coasts up the frictionless slope, gaining 0.180 m in altitude. 




Figure 7.39 A toy car moves up a sloped track, (credit: Leszek Leszczynski, Flickr) 

21. In a downhill ski race, surprisingly, little advantage is gained by 
getting a running start. (This is because the initial kinetic energy is small 
compared with the gain in gravitational potential energy on even small 
hills.) To demonstrate this, find the final speed and the time taken for a 
skier who skies 70.0 m along a 30° slope neglecting friction: (a) Starting 
from rest, (b) Starting with an initial speed of 2.50 m/s. (c) Does the 
answer surprise you? Discuss why it is still advantageous to get a 
running start in very competitive events. 

7.4 Conservative Forces and Potential Energy 

22. A 5.00x 10 -kg subway train is brought to a stop from a speed of 

0.500 m/s in 0.400 m by a large spring bumper at the end of its track. 
What is the force constant k of the spring? 

23. A pogo stick has a spring with a force constant of 2.50x 10 N/m , 
which can be compressed 12.0 cm. To what maximum height can a child 
jump on the stick using only the energy in the spring, if the child and stick 



have a total mass of 40.0 kg? Explicitly show how you follow the steps in 
the Problem-Solving Strategies for Energy. 

7.5 Nonconservative Forces 

24. A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m- 
high rise as shown in Figure 7.40. Find her final speed at the top, given 
that the coefficient of friction between her skis and the snow is 0.0800. 
(Hint: Find the distance traveled up the incline assuming a straight-line 
path as shown in the figure.) 

v< = ? 




Figure 7.40 The skier's initial kinetic energy is partially used in coasting to the top of a 
rise. 

25. (a) How high a hill can a car coast up (engine disengaged) if work 
done by friction is negligible and its initial speed is 110 km/h? (b) If, in 
actuality, a 750-kg car with an initial speed of 110 km/h is observed to 
coast up a hill to a height 22.0 m above its starting point, how much 
thermal energy was generated by friction? (c) What is the average force 
of friction if the hill has a slope 2.5° above the horizontal? 

7.6 Conservation of Energy 

26. Using values from Table 7.1, how many DNA molecules could be 
broken by the energy carried by a single electron in the beam of an old- 
fashioned TV tube? (These electrons were not dangerous in themselves, 
but they did create dangerous x rays. Later model tube TVs had shielding 
that absorbed x rays before they escaped and exposed viewers.) 

27. Using energy considerations and assuming negligible air resistance, 
show that a rock thrown from a bridge 20.0 m above water with an initial 
speed of 15.0 m/s strikes the water with a speed of 24.8 m/s independent 
of the direction thrown. 

28. If the energy in fusion bombs were used to supply the energy needs 
of the world, how many of the 9-megaton variety would be needed for a 
year's supply of energy (using data from Table 7.1)? This is not as far- 
fetched as it may sound — there are thousands of nuclear bombs, and 
their energy can be trapped in underground explosions and converted to 
electricity, as natural geothermal energy is. 

29. (a) Use of hydrogen fusion to supply energy is a dream that may be 
realized in the next century. Fusion would be a relatively clean and 
almost limitless supply of energy, as can be seen from Table 7.1. To 
illustrate this, calculate how many years the present energy needs of the 
world could be supplied by one millionth of the oceans' hydrogen fusion 
energy, (b) How does this time compare with historically significant 
events, such as the duration of stable economic systems? 

7.7 Power 

30. The Crab Nebula (see Figure 7.41) pulsar is the remnant of a 
supernova that occurred in A.D. 1054. Using data from Table 7.3, 
calculate the approximate factor by which the power output of this 
astronomical object has declined since its explosion. 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 257 




Figure 7.41 Crab Nebula (credit: ESO, via Wiki media Commons) 

31. Suppose a star 1000 times brighter than our Sun (that is, emitting 
1000 times the power) suddenly goes supernova. Using data from Table 
7.3: (a) By what factor does its power output increase? (b) How many 
times brighter than our entire Milky Way galaxy is the supernova? (c) 
Based on your answers, discuss whether it should be possible to observe 

supernovas in distant galaxies. Note that there are on the order of 10 
observable galaxies, the average brightness of which is somewhat less 
than our own galaxy. 

32. A person in good physical condition can put out 100 W of useful 
power for several hours at a stretch, perhaps by pedaling a mechanism 
that drives an electric generator. Neglecting any problems of generator 
efficiency and practical considerations such as resting time: (a) How 
many people would it take to run a 4.00-kW electric clothes dryer? (b) 
How many people would it take to replace a large electric power plant 
that generates 800 MW? 

33. What is the cost of operating a 3.00-W electric clock for a year if the 
cost of electricity is $0.0900 per kW • h ? 

34. A large household air conditioner may consume 15.0 kW of power. 
What is the cost of operating this air conditioner 3.00 h per day for 30.0 d 
if the cost of electricity is $0,110 per kW • h ? 

35. (a) What is the average power consumption in watts of an appliance 
that uses 5.00 kW • h of energy per day? (b) How many joules of 
energy does this appliance consume in a year? 

36. (a) What is the average useful power output of a person who does 
6.00xl0 6 J of useful work in 8.00 h? (b) Working at this rate, how long 

will it take this person to lift 2000 kg of bricks 1.50 m to a platform? (Work 
done to lift his body can be omitted because it is not considered useful 
output here.) 

37. A 500-kg dragster accelerates from rest to a final speed of 110 m/s in 
400 m (about a quarter of a mile) and encounters an average frictional 
force of 1200 N. What is its average power output in watts and 
horsepower if this takes 7.30 s? 

38. (a) How long will it take an 850-kg car with a useful power output of 
40.0 hp (1 hp = 746 W) to reach a speed of 15.0 m/s, neglecting friction? 
(b) How long will this acceleration take if the car also climbs a 3.00-m- 
high hill in the process? 

39. (a) Find the useful power output of an elevator motor that lifts a 
2500-kg load a height of 35.0 m in 12.0 s, if it also increases the speed 
from rest to 4.00 m/s. Note that the total mass of the counterbalanced 
system is 10,000 kg — so that only 2500 kg is raised in height, but the full 
10,000 kg is accelerated, (b) What does it cost, if electricity is $0.0900 
per kW • h ? 



40. (a) What is the available energy content, in joules, of a battery that 
operates a 2.00-W electric clock for 18 months? (b) How long can a 

battery that can supply 8.00x 10 J run a pocket calculator that 
consumes energy at the rate of 1.00x10 W? 

41. (a) How long would it take a 1.50x10 -kg airplane with engines that 
produce 100 MW of power to reach a speed of 250 m/s and an altitude of 
12.0 km if air resistance were negligible? (b) If it actually takes 900 s, 
what is the power? (c) Given this power, what is the average force of air 
resistance if the airplane takes 1200 s? (Hint: You must find the distance 
the plane travels in 1200 s assuming constant acceleration.) 

42. Calculate the power output needed for a 950-kg car to climb a 2.00° 
slope at a constant 30.0 m/s while encountering wind resistance and 
friction totaling 600 N. Explicitly show how you follow the steps in the 
Problem-Solving Strategies for Energy. 

43. (a) Calculate the power per square meter reaching Earth's upper 
atmosphere from the Sun. (Take the power output of the Sun to be 

4.00X 10 26 W.) (b) Part of this is absorbed and reflected by the 
atmosphere, so that a maximum of 1.30 kW/m reaches Earth's 

surface. Calculate the area in km of solar energy collectors needed to 
replace an electric power plant that generates 750 MW if the collectors 
convert an average of 2.00% of the maximum power into electricity. (This 
small conversion efficiency is due to the devices themselves, and the fact 
that the sun is directly overhead only briefly.) With the same 
assumptions, what area would be needed to meet the United States' 

energy needs (1.05x10 J)? Australia's energy needs 

(5.4X10 18 J)? China's energy needs (6.3xl0 19 J)? (These energy 
consumption values are from 2006.) 

7.8 Work, Energy, and Power in Humans 

44. (a) How long can you rapidly climb stairs (116/min) on the 93.0 kcal of 
energy in a 10.0-g pat of butter? (b) How many flights is this if each flight 
has 16 stairs? 

45. (a) What is the power output in watts and horsepower of a 70.0-kg 
sprinter who accelerates from rest to 10.0 m/s in 3.00 s? (b) Considering 
the amount of power generated, do you think a well-trained athlete could 
do this repetitively for long periods of time? 

46. Calculate the power output in watts and horsepower of a shot-putter 
who takes 1.20 s to accelerate the 7.27-kg shot from rest to 14.0 m/s, 
while raising it 0.800 m. (Do not include the power produced to 
accelerate his body.) 




Figure 7.42 Shot putter at the Dornoch Highland Gathering in 2007. (credit: John 
Haslam, Flickr) 

47. (a) What is the efficiency of an out-of-condition professor who does 
2.10x10 J of useful work while metabolizing 500 kcal of food energy? 
(b) How many food calories would a well-conditioned athlete metabolize 
in doing the same work with an efficiency of 20%? 



258 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 



48. Energy that is not utilized for work or heat transfer is converted to the 
chemical energy of body fat containing about 39 kJ/g. How many grams 
of fat will you gain if you eat 10,000 kJ (about 2500 kcal) one day and do 
nothing but sit relaxed for 16.0 h and sleep for the other 8.00 h? Use data 
from Table 7.5 for the energy consumption rates of these activities. 

49. Using data from Table 7.5, calculate the daily energy needs of a 
person who sleeps for 7.00 h, walks for 2.00 h, attends classes for 4.00 
h, cycles for 2.00 h, sits relaxed for 3.00 h, and studies for 6.00 h. 
(Studying consumes energy at the same rate as sitting in class.) 

50. What is the efficiency of a subject on a treadmill who puts out work at 
the rate of 100 W while consuming oxygen at the rate of 2.00 L/min? 
(Hint: See Table 7.5.) 

51. Shoveling snow can be extremely taxing because the arms have 
such a low efficiency in this activity. Suppose a person shoveling a 
footpath metabolizes food at the rate of 800 W. (a) What is her useful 
power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? 
(This could be the amount of heavy snow on 20 m of footpath.) (c) How 
much waste heat transfer in kilojoules will she generate in the process? 

52. Very large forces are produced in joints when a person jumps from 
some height to the ground, (a) Calculate the force produced if an 80.0-kg 
person jumps from a 0.600-m-high ledge and lands stiffly, compressing 
joint material 1.50 cm as a result. (Be certain to include the weight of the 
person.) (b) In practice the knees bend almost involuntarily to help extend 
the distance over which you stop. Calculate the force produced if the 
stopping distance is 0.300 m. (c) Compare both forces with the weight of 
the person. 

53. Jogging on hard surfaces with insufficiently padded shoes produces 
large forces in the feet and legs, (a) Calculate the force needed to stop 
the downward motion of a jogger's leg, if his leg has a mass of 13.0 kg, a 
speed of 6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to 
include the weight of the 75.0-kg jogger's body.) (b) Compare this force 
with the weight of the jogger. 

54. (a) Calculate the energy in kJ used by a 55.0-kg woman who does 50 
deep knee bends in which her center of mass is lowered and raised 
0.400 m. (She does work in both directions.) You may assume her 
efficiency is 20%. (b) What is the average power consumption rate in 
watts if she does this in 3.00 min? 

55. Kanellos Kanellopoulos flew 119 km from Crete to Santorini, Greece, 
on April 23, 1988, in the Daedalus 88, an aircraft powered by a bicycle- 
type drive mechanism (see Figure 7.43). His useful power output for the 
234-min trip was about 350 W. Using the efficiency for cycling from Table 
7.2, calculate the food energy in kilojoules he metabolized during the 
flight. 




Figure 7.43 The Daedalus 88 in flight, (credit: NASA photo by Beasley) 

56. The swimmer shown in Figure 7.44 exerts an average horizontal 
backward force of 80.0 N with his arm during each 1.80 m long stroke, (a) 
What is his work output in each stroke? (b) Calculate the power output of 
his arms if he does 120 strokes per minute. 




Figure 7.44 

57. Mountain climbers carry bottled oxygen when at very high altitudes, 
(a) Assuming that a mountain climber uses oxygen at twice the rate for 
climbing 116 stairs per minute (because of low air temperature and 
winds), calculate how many liters of oxygen a climber would need for 
10.0 h of climbing. (These are liters at sea level.) Note that only 40% of 
the inhaled oxygen is utilized; the rest is exhaled, (b) How much useful 
work does the climber do if he and his equipment have a mass of 90.0 kg 
and he gains 1000 m of altitude? (c) What is his efficiency for the 10.0-h 
climb? 

58. The awe-inspiring Great Pyramid of Cheops was built more than 
4500 years ago. Its square base, originally 230 m on a side, covered 13.1 

acres, and it was 146 m high, with a mass of about 7x 10 kg . (The 

pyramid's dimensions are slightly different today due to quarrying and 
some sagging.) Historians estimate that 20,000 workers spent 20 years 
to construct it, working 12-hour days, 330 days per year, (a) Calculate the 
gravitational potential energy stored in the pyramid, given its center of 
mass is at one-fourth its height, (b) Only a fraction of the workers lifted 
blocks; most were involved in support services such as building ramps 
(see Figure 7.45), bringing food and water, and hauling blocks to the site. 
Calculate the efficiency of the workers who did the lifting, assuming there 
were 1000 of them and they consumed food energy at the rate of 300 
kcal/h. What does your answer imply about how much of their work went 
into block-lifting, versus how much work went into friction and lifting and 
lowering their own bodies? (c) Calculate the mass of food that had to be 
supplied each day, assuming that the average worker required 3600 kcal 
per day and that their diet was 5% protein, 60% carbohydrate, and 35% 
fat. (These proportions neglect the mass of bulk and nondigestible 
materials consumed.) 




Figure 7.45 Ancient pyramids were probably constructed using ramps as simple 
machines, (credit: Franck Monnier, Wikimedia Commons) 

59. (a) How long can you play tennis on the 800 kJ (about 200 kcal) of 
energy in a candy bar? (b) Does this seem like a long time? Discuss why 
exercise is necessary but may not be sufficient to cause a person to lose 
weight. 

7.9 World Energy Use 

60. Integrated Concepts 

(a) Calculate the force the woman in Figure 7.46 exerts to do a push-up 
at constant speed, taking all data to be known to three digits, (b) How 
much work does she do if her center of mass rises 0.240 m? (c) What is 
her useful power output if she does 25 push-ups in 1 min? (Should work 



CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 259 



done lowering her body be included? See the discussion of useful work in 
Work, Energy, and Power in Humans. 



m = 50 kg 

CG 



0,90 m 



■1.50 m 



w 


^"^ 


V__^ 




^reaction 









Figure 7.46 Forces involved in doing push-ups. The woman's weight acts as a force 
exerted downward on her center of gravity (CG). 

61. Integrated Concepts 

A 75.0-kg cross-country skier is climbing a 3.0° slope at a constant 
speed of 2.00 m/s and encounters air resistance of 25.0 N. Find his 
power output for work done against the gravitational force and air 
resistance, (b) What average force does he exert backward on the snow 
to accomplish this? (c) If he continues to exert this force and to 
experience the same air resistance when he reaches a level area, how 
long will it take him to reach a velocity of 10.0 m/s? 

62. Integrated Concepts 

The 70.0-kg swimmer in Figure 7.44 starts a race with an initial velocity 
of 1.25 m/s and exerts an average force of 80.0 N backward with his 
arms during each 1.80 m long stroke, (a) What is his initial acceleration if 
water resistance is 45.0 N? (b) What is the subsequent average 
resistance force from the water during the 5.00 s it takes him to reach his 
top velocity of 2.50 m/s? (c) Discuss whether water resistance seems to 
increase linearly with velocity. 

63. Integrated Concepts 

A toy gun uses a spring with a force constant of 300 N/m to propel a 
10.0-g steel ball. If the spring is compressed 7.00 cm and friction is 
negligible: (a) How much force is needed to compress the spring? (b) To 
what maximum height can the ball be shot? (c) At what angles above the 
horizontal may a child aim to hit a target 3.00 m away at the same height 
as the gun? (d) What is the gun's maximum range on level ground? 

64. Integrated Concepts 

(a) What force must be supplied by an elevator cable to produce an 
acceleration of 0.800 m/s against a 200-N frictional force, if the mass 
of the loaded elevator is 1500 kg? (b) How much work is done by the 
cable in lifting the elevator 20.0 m? (c) What is the final speed of the 
elevator if it starts from rest? (d) How much work went into thermal 
energy? 

65. Unreasonable Results 

A car advertisement claims that its 900-kg car accelerated from rest to 
30.0 m/s and drove 100 km, gaining 3.00 km in altitude, on 1.0 gal of 
gasoline. The average force of friction including air resistance was 700 N. 
Assume all values are known to three significant figures, (a) Calculate the 
car's efficiency, (b) What is unreasonable about the result? (c) Which 
premise is unreasonable, or which premises are inconsistent? 

66. Unreasonable Results 

Body fat is metabolized, supplying 9.30 kcal/g, when dietary intake is less 
than needed to fuel metabolism. The manufacturers of an exercise 
bicycle claim that you can lose 0.500 kg of fat per day by vigorously 
exercising for 2.00 h per day on their machine, (a) How many kcal are 
supplied by the metabolization of 0.500 kg of fat? (b) Calculate the kcal/ 
min that you would have to utilize to metabolize fat at the rate of 0.500 kg 
in 2.00 h. (c) What is unreasonable about the results? (d) Which premise 
is unreasonable, or which premises are inconsistent? 

67. Construct Your Own Problem 

Consider a person climbing and descending stairs. Construct a problem 
in which you calculate the long-term rate at which stairs can be climbed 
considering the mass of the person, his ability to generate power with his 



legs, and the height of a single stair step. Also consider why the same 
person can descend stairs at a faster rate for a nearly unlimited time in 
spite of the fact that very similar forces are exerted going down as going 
up. (This points to a fundamentally different process for descending 
versus climbing stairs.) 

68. Construct Your Own Problem 

Consider humans generating electricity by pedaling a device similar to a 
stationary bicycle. Construct a problem in which you determine the 
number of people it would take to replace a large electrical generation 
facility. Among the things to consider are the power output that is 
reasonable using the legs, rest time, and the need for electricity 24 hours 
per day. Discuss the practical implications of your results. 

69. Integrated Concepts 

A 105-kg basketball player crouches down 0.400 m while waiting to jump. 
After exerting a force on the floor through this 0.400 m, his feet leave the 
floor and his center of gravity rises 0.950 m above its normal standing 
erect position, (a) Using energy considerations, calculate his velocity 
when he leaves the floor, (b) What average force did he exert on the 
floor? (Do not neglect the force to support his weight as well as that to 
accelerate him.) (c) What was his power output during the acceleration 
phase? 



260 CHAPTER 7 | WORK, ENERGY, AND ENERGY RESOURCES 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 261 




Figure 8.1 Each rugby player has great momentum, which will affect the outcome of their collisions with each other and the ground, (credit: ozzzie, Flickr) 



Learning Objectives 



8.1. Linear Momentum and Force 

Define linear momentum. 

Explain the relationship between momentum and force. 
State Newton's second law of motion in terms of momentum. 
Calculate momentum given mass and velocity. 

8.2. Impulse 
Define impulse. 

Describe effects of impulses in everyday life. 

Determine the average effective force using graphical representation. 
Calculate average force and impulse given mass, velocity, and time. 

8.3. Conservation of Momentum 
Describe the principle of conservation of momentum. 
Derive an expression for the conservation of momentum. 
Explain conservation of momentum with examples. 
Explain the principle of conservation of momentum as it relates to atomic and subatomic particles. 

8.4. Elastic Collisions in One Dimension 
Describe an elastic collision of two objects in one dimension. 
Define internal kinetic energy. 

Derive an expression for conservation of internal kinetic energy in a one dimensional collision. 
Determine the final velocities in an elastic collision given masses and initial velocities. 

8.5. Inelastic Collisions in One Dimension 
Define inelastic collision. 
Explain perfectly inelastic collision. 
Apply an understanding of collisions to sports. 
Determine recoil velocity and loss in kinetic energy given mass and initial velocity. 

8.6. Collisions of Point Masses in Two Dimensions 
Discuss two dimensional collisions as an extension of one dimensional analysis. 
Define point masses. 

Derive an expression for conservation of momentum along x-axis and y-axis. 
Describe elastic collisions of two objects with equal mass. 
Determine the magnitude and direction of the final velocity given initial velocity, and scattering angle 

8.7. Introduction to Rocket Propulsion 
State Newton's third law of motion. 

Explain the principle involved in propulsion of rockets and jet engines. 
Derive an expression for the acceleration of the rocket. 
Discuss the factors that affect the rocket's acceleration. 
Describe the function of a space shuttle. 



262 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 



Introduction to Linear Momentum and Collisions 



We use the term momentum in various ways in everyday language, and most of these ways are consistent with its precise scientific definition. We 
speak of sports teams or politicians gaining and maintaining the momentum to win. We also recognize that momentum has something to do with 
collisions. For example, looking at the rugby players in the photograph colliding and falling to the ground, we expect their momenta to have great 
effects in the resulting collisions. Generally, momentum implies a tendency to continue on course — to move in the same direction — and is associated 
with great mass and speed. 

Momentum, like energy, is important because it is conserved. Only a few physical quantities are conserved in nature, and studying them yields 
fundamental insight into how nature works, as we shall see in our study of momentum. 

8.1 Linear Momentum and Force 

Linear Momentum 

The scientific definition of linear momentum is consistent with most people's intuitive understanding of momentum: a large, fast-moving object has 
greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system's mass multiplied by its velocity. In 
symbols, linear momentum is expressed as 



p = my. 



(8.1) 



Momentum is directly proportional to the object's mass and also its velocity. Thus the greater an object's mass or the greater its velocity, the greater 
its momentum. Momentum p is a vector having the same direction as the velocity v . The SI unit for momentum is kg • m/s . 



Linear Momentum 



Linear momentum is defined as the product of a system's mass multiplied by its velocity: 

p = my. 



(8.2) 



Example 8.1 Calculating Momentum: A Football Player and a Football 



(a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player's momentum with the momentum of a hard- 
thrown 0.410-kg football that has a speed of 25.0 m/s. 

Strategy 

No information is given regarding direction, and so we can calculate only the magnitude of the momentum, p . (As usual, a symbol that is in 

italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this example, the magnitude of 
momentum can be calculated directly from the definition of momentum given in the equation, which becomes 

p = mv (8.3) 

when only magnitudes are considered. 
Solution for (a) 

To determine the momentum of the player, substitute the known values for the player's mass and speed into the equation. 

^player = ( U0 k gX 8 ' 00 ^ = 88 ° k S ' m/s 
Solution for (b) 

To determine the momentum of the ball, substitute the known values for the ball's mass and speed into the equation. 

p ban = (0.410 kg)(25.0 m/s) = 10.3 kg • m/s 

The ratio of the player's momentum to that of the ball is 

^player _ 880 



/'ball 



10.3 



85.9. 



(8.4) 



(8.5) 



(8.6) 



Discussion 



Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the 
momentum of the football, as you might guess. As a result, the player's motion is only slightly affected if he catches the ball. We shall quantify 
what happens in such collisions in terms of momentum in later sections. 



Momentum and Newton's Second Law 

The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics. Momentum was 
deemed so important that it was called the "quantity of motion." Newton actually stated his second law of motion in terms of momentum: The net 
external force equals the change in momentum of a system divided by the time over which it changes. Using symbols, this law is 



F -*B. 

'net- At > 
where F net is the net external force, Ap is the change in momentum, and At is the change in time. 



(8.7) 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 263 



Newton's Second Law of Motion in Ten 



urn 



The net external force equals the change in momentum of a system divided by the time over which it changes. 

F -AE 

net ~ A? 



(8.8) 



Making Connections: Force and Momentum 

Force and momentum are intimately related. Force acting over time can change momentum, and Newton's second law of motion, can be stated 
in its most broadly applicable form in terms of momentum. Momentum continues to be a key concept in the study of atomic and subatomic 
particles in quantum mechanics. 

This statement of Newton's second law of motion includes the more familiar F net =ma as a special case. We can derive this form as follows. First, 
note that the change in momentum Ap is given by 

Ap = A(mv). (8.9) 

If the mass of the system is constant, then 

A(mv) = mAv. (8.10) 

So that for constant mass, Newton's second law of motion becomes 



Because ^- = a , we get the familiar equation 



F _ Ap _ mAv 

'net- A , - At ■ 



F npt =raa 



(8.11) 



(8.12) 



when the mass of the system is constant. 

Newton's second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems where the mass is 
changing, such as rockets, as well as to systems of constant mass. We will consider systems with varying mass in some detail; however, the 
relationship between momentum and force remains useful when mass is constant, such as in the following example. 



Example 8.2 Calculating Force: Venus Williams' Racquet 



During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women's match, reaching a speed of 58 m/s (209 km/ 
h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams' racquet, assuming that the ball's speed just after impact is 
58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 
5.0 ms (milliseconds)? 

Strategy 

This problem involves only one dimension because the ball starts from having no horizontal velocity component before impact. Newton's second 
law stated in terms of momentum is then written as 

(8.13) 



f -AE 

net ~ At' 



As noted above, when mass is constant, the change in momentum is given by 

Ap = mAv = m(v f — Vj). 



(8.14) 



_Ap 



In this example, the velocity just after impact and the change in time are given; thus, once Ap is calculated, F nQt = —^- can be used to find 

the force. 
Solution 

To determine the change in momentum, substitute the values for the initial and final velocities into the equation above. 

Ap = m(v f -v { ) (8.15) 

= (0.057 kg)(58 m/s - m/s) 
= 3.306 kg -m/s « 3.3 kg -m/s 



Now the magnitude of the net external force can determined by using F n 



At 



_ Ap_ _ 3.306 kg • m/s 
net " AF" 5.0xl0" 3 s 
= 661N«660N, 



(8.16) 



264 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 

where we have retained only two significant figures in the final step. 

Discussion 

This quantity was the average force exerted by Venus Williams' racquet on the tennis ball during its brief impact (note that the ball also 
experienced the 0.56-N force of gravity, but that force was not due to the racquet). This problem could also be solved by first finding the 
acceleration and then using F net =ma , but one additional step would be required compared with the strategy used in this example. 



8.2 Impulse 

The effect of a force on an object depends on how long it acts, as well as how great the force is. In Example 8.1, a very large force acting for a short 
time had a great effect on the momentum of the tennis ball. A small force could cause the same change in momentum, but it would have to act for a 
much longer time. For example, if the ball were thrown upward, the gravitational force (which is much smaller than the tennis racquet's force) would 
eventually reverse the momentum of the ball. Quantitatively, the effect we are talking about is the change in momentum Ap . 



By rearranging the equation F E 



Ap 

A; 



to be 



Ap = F net Af, 



(8.17) 



we can see how the change in momentum equals the average net external force multiplied by the time this force acts. The quantity ¥ nQt At is given 
the name impulse. Impulse is the same as the change in momentum. 

Impulse: Change in Momentum 

Change in momentum equals the average net external force multiplied by the time this force acts. 



The quantity ¥ nQt At is given the name impulse. 



Ap = ¥ nQt At 



(8.18) 



There are many ways in which an understanding of impulse can save lives, or at least limbs. The dashboard padding in a car, and certainly the 
airbags, allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is 
the same for an occupant, whether an air bag is deployed or not, but the force (to bring the occupant to a stop) will be much less if it acts over a 
larger time. Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. 
Another advantage is that a car will crumple in a collision, especially in the event of a head-on collision. A longer collision time means the force 
on the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could 
crumple or collapse in the event of an accident. 

Bones in a body will fracture if the force on them is too large. If you jump onto the floor from a table, the force on your legs can be immense if you 
land stiff-legged on a hard surface. Rolling on the ground after jumping from the table, or landing with a parachute, extends the time over which 
the force (on you from the ground) acts. 



Example 8.3 Calculating Magnitudes of Impulses: Two Billiard Balls Striking a Rigid Wall 



Two identical billiard balls strike a rigid wall with the same speed, and are reflected without any change of speed. The first ball strikes 
perpendicular to the wall. The second ball strikes the wall at an angle of 30° from the perpendicular, and bounces off at an angle of 30° from 
perpendicular to the wall. 

(a) Determine the direction of the force on the wall due to each ball. 

(b) Calculate the ratio of the magnitudes of impulses on the two balls by the wall. 

Strategy for (a) 

In order to determine the force on the wall, consider the force on the ball due to the wall using Newton's second law and then apply Newton's 
third law to determine the direction. Assume the x -axis to be normal to the wall and to be positive in the initial direction of motion. Choose the y 

-axis to be along the wall in the plane of the second ball's motion. The momentum direction and the velocity direction are the same. 

Solution for (a) 

The first ball bounces directly into the wall and exerts a force on it in the +x direction. Therefore the wall exerts a force on the ball in the -x 

direction. The second ball continues with the same momentum component in the y direction, but reverses its x -component of momentum, as 

seen by sketching a diagram of the angles involved and keeping in mind the proportionality between velocity and momentum. 

These changes mean the change in momentum for both balls is in the -x direction, so the force of the wall on each ball is along the -x 

direction. 

Strategy for (b) 

Calculate the change in momentum for each ball, which is equal to the impulse imparted to the ball. 

Solution for (b) 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 265 

Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball. Choose the x -axis and y -axis as 
previously described, and consider the change in momentum of the first ball which strikes perpendicular to the wall. 

p xi = mu\ p yi = (8.19) 

p xf = -mu; p yf = (8.20) 

Impulse is the change in momentum vector. Therefore the x -component of impulse is equal to —2mu and the y -component of impulse is 

equal to zero. 

Now consider the change in momentum of the second ball. 

p xi = mu cos 30°; p { = -mu sin 30° (8.21) 

p xf = -mu cos 30°; /? yf = —mu sin 30° (8.22) 

It should be noted here that while p x changes sign after the collision, p y does not. Therefore the x -component of impulse is equal to 
—2mu cos 30° and the y -component of impulse is equal to zero. 



The ratio of the magnitudes of the impulse imparted to the balls is 

2mu 



= — = 1 155 
2mwcos30° V3 



(8.23) 



Discussion 

The direction of impulse and force is the same as in the case of (a); it is normal to the wall and along the negative x -direction. Making use of 
Newton's third law, the force on the wall due to each ball is normal to the wall along the positive x -direction. 

Our definition of impulse includes an assumption that the force is constant over the time interval At . Forces are usually not constant. Forces vary 
considerably even during the brief time intervals considered. It is, however, possible to find an average effective force F eff that produces the same 

result as the corresponding time-varying force. Figure 8.2 shows a graph of what an actual force looks like as a function of time for a ball bouncing 
off the floor. The area under the curve has units of momentum and is equal to the impulse or change in momentum between times t± and t 2 ■ That 

area is equal to the area inside the rectangle bounded by F eff , t^ , and t 2 ■ Thus the impulses and their effects are the same for both the actual 

and effective forces. 



actual 



eff 




ti h t 

Figure 8.2 A graph of force versus time with time along the X -axis and force along the y -axis for an actual force and an equivalent effective force. The areas under the two 
curves are equal. 



Making Connections: Take-Home Investigation — Hand Movement and Impulse 

Try catching a ball while "giving" with the ball, pulling your hands toward your body. Then, try catching a ball while keeping your hands still. Hit 
water in a tub with your full palm. After the water has settled, hit the water again by diving your hand with your fingers first into the water. (Your 
full palm represents a swimmer doing a belly flop and your diving hand represents a swimmer doing a dive.) Explain what happens in each case 
and why. Which orientations would you advise people to avoid and why? 



Making Connections: Constant Force and Constant Acceleration 

The assumption of a constant force in the definition of impulse is analogous to the assumption of a constant acceleration in kinematics. In both 
cases, nature is adequately described without the use of calculus. 



266 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 



8.3 Conservation of Momentum 

Momentum is an important quantity because it is conserved. Yet it was not conserved in the examples in Impulse and Linear Momentum and 
Force, where large changes in momentum were produced by forces acting on the system of interest. Under what circumstances is momentum 
conserved? 

The answer to this question entails considering a sufficiently large system. It is always possible to find a larger system in which total momentum is 
constant, even if momentum changes for components of the system. If a football player runs into the goalpost in the end zone, there will be a force on 
him that causes him to bounce backward. However, the Earth also recoils — conserving momentum — because of the force applied to it through the 
goalpost. Because Earth is many orders of magnitude more massive than the player, its recoil is immeasurably small and can be neglected in any 
practical sense, but it is real nevertheless. 

Consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth — for example, one car 
bumping into another, as shown in Figure 8.3. Both cars are coasting in the same direction when the lead car (labeled m 2 ) is bumped by the trailing 

car (labeled m{). The only unbalanced force on each car is the force of the collision. (Assume that the effects due to friction are negligible.) Car 1 

slows down as a result of the collision, losing some momentum, while car 2 speeds up and gains some momentum. We shall now show that the total 
momentum of the two-car system remains constant. 



Before 



net F = 



System 

of interest 




System 
of interest 



Figure 8.3 A car of mass m j moving with a velocity of V j bumps into another car of mass m 2 and velocity V 2 that it is following. As a result, the first car slows down to 
a velocity of v' ^ and the second speeds up to a velocity of v' 2 ■ The momentum of each car is changed, but the total momentum /? tot of the two cars is the same before 
and after the collision (if you assume friction is negligible). 

Using the definition of impulse, the change in momentum of car 1 is given by 

Ap 1 =F 1 At, (8.24) 

where Fj is the force on car 1 due to car 2, and At is the time the force acts (the duration of the collision). Intuitively, it seems obvious that the 

collision time is the same for both cars, but it is only true for objects traveling at ordinary speeds. This assumption must be modified for objects 
travelling near the speed of light, without affecting the result that momentum is conserved. 

Similarly, the change in momentum of car 2 is 

Ap 2 = F 2 At, (8.25) 

where F 2 is the force on car 2 due to car 1, and we assume the duration of the collision At is the same for both cars. We know from Newton's third 
law that F 2 = - F± , and so 

Ap 2 = -F x At = -Ap x . (8.26) 

Thus, the changes in momentum are equal and opposite, and 

Ap l + Ap 2 = 0. (8.27) 

Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is, 

Pi + Pi — constant, (8.28) 

P\+Pl = P'i+P f 2> (8 - 29) 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 267 



where p' 1 and p' \ are the momenta of cars 1 and 2 after the collision. (We often use primes to denote the final state.) 

This result — that momentum is conserved — has validity far beyond the preceding one-dimensional case. It can be similarly shown that total 
momentum is conserved for any isolated system, with any number of objects in it. In equation form, the conservation of momentum principle for an 
isolated system is written 

Ptot — constant, (8.30) 



or 



Ptot — P tOt' 



(8.31) 



where p tot is the total momentum (the sum of the momenta of the individual objects in the system) and p' tot is the total momentum some time 
later. (The total momentum can be shown to be the momentum of the center of mass of the system.) An isolated system is defined to be one for 



which the net external force is zero (F r 



0). 



Conservation of Momentum Principle 



p tot = constant 

Ptot — P tot (isolated system) 



(8.32) 



Isolated System 



An isolated system is defined to be one for which the net external force is zero (F net = 0). 



Perhaps an easier way to see that momentum is conserved for an isolated system is to consider Newton's second law in terms of momentum, 



_ Ap t0 



For an isolated system, (F net = 0) ; thus, Ap tot = , and p tot is constant. 



We have noted that the three length dimensions in nature — x , y , and z — are independent, and it is interesting to note that momentum can be 

conserved in different ways along each dimension. For example, during projectile motion and where air resistance is negligible, momentum is 
conserved in the horizontal direction because horizontal forces are zero and momentum is unchanged. But along the vertical direction, the net vertical 
force is not zero and the momentum of the projectile is not conserved. (See Figure 8.4.) However, if the momentum of the projectile-Earth system is 
considered in the vertical direction, we find that the total momentum is conserved. 



After 



Before 




Px = Px 



Py*Py\ 



Figure 8.4 The horizontal component of a projectile's momentum is conserved if air resistance is negligible, even in this case where a space probe separates. The forces 
causing the separation are internal to the system, so that the net external horizontal force F x _ net is still zero. The vertical component of the momentum is not conserved, 

because the net vertical force Fy _ net is not zero. In the vertical direction, the space probe-Earth system needs to be considered and we find that the total momentum is 

conserved. The center of mass of the space probe takes the same path it would if the separation did not occur. 

The conservation of momentum principle can be applied to systems as different as a comet striking Earth and a gas containing huge numbers of 
atoms and molecules. Conservation of momentum is violated only when the net external force is not zero. But another larger system can always be 
considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered 
above, the two-car system conserves momentum while each one-car system does not. 



268 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 



Making Connections: Take-Home Investigation — Drop of Tennis Ball and a Basketball 



Hold a tennis ball side by side and in contact with a basketball. Drop the balls together. (Be careful!) What happens? Explain your observations. 
Now hold the tennis ball above and in contact with the basketball. What happened? Explain your observations. What do you think will happen if 
the basketball ball is held above and in contact with the tennis ball? 

Making Connections: Take-Home Investigation — Two Tennis Balls in a Ballistic Trajectory 



Tie two tennis balls together with a string about a foot long. Hold one ball and let the other hang down and throw it in a ballistic trajectory. Explain 
your observations. Now mark the center of the string with bright ink or attach a brightly colored sticker to it and throw again. What happened? 
Explain your observations. 

Some aquatic animals such as jellyfish move around based on the principles of conservation of momentum. A jellyfish fills its umbrella section 
with water and then pushes the water out resulting in motion in the opposite direction to that of the jet of water. Squids propel themselves in a 
similar manner but, in contrast with jellyfish, are able to control the direction in which they move by aiming their nozzle forward or backward. 
Typical squids can move at speeds of 8 to 12 km/h. 

The ballistocardiograph (BCG) was a diagnostic tool used in the second half of the 20th century to study the strength of the heart. About once a 
second, your heart beats, forcing blood into the aorta. A force in the opposite direction is exerted on the rest of your body (recall Newton's third 
law). A ballistocardiograph is a device that can measure this reaction force. This measurement is done by using a sensor (resting on the person) 
or by using a moving table suspended from the ceiling. This technique can gather information on the strength of the heart beat and the volume of 
blood passing from the heart. However, the electrocardiogram (ECG or EKG) and the echocardiogram (cardiac ECHO or ECHO; a technique that 
uses ultrasound to see an image of the heart) are more widely used in the practice of cardiology. 

Making Connections: Conservation of Momentum and Collision 

Conservation of momentum is quite useful in describing collisions. Momentum is crucial to our understanding of atomic and subatomic particles 
because much of what we know about these particles comes from collision experiments. 



Subatomic Collisions and Momentum 

The conservation of momentum principle not only applies to the macroscopic objects, it is also essential to our explorations of atomic and subatomic 
particles. Giant machines hurl subatomic particles at one another, and researchers evaluate the results by assuming conservation of momentum 
(among other things). 

On the small scale, we find that particles and their properties are invisible to the naked eye but can be measured with our instruments, and models of 
these subatomic particles can be constructed to describe the results. Momentum is found to be a property of all subatomic particles including 
massless particles such as photons that compose light. Momentum being a property of particles hints that momentum may have an identity beyond 
the description of an object's mass multiplied by the object's velocity. Indeed, momentum relates to wave properties and plays a fundamental role in 
what measurements are taken and how we take these measurements. Furthermore, we find that the conservation of momentum principle is valid 
when considering systems of particles. We use this principle to analyze the masses and other properties of previously undetected particles, such as 
the nucleus of an atom and the existence of quarks that make up particles of nuclei. Figure 8.5 below illustrates how a particle scattering backward 
from another implies that its target is massive and dense. Experiments seeking evidence that quarks make up protons (one type of particle that 
makes up nuclei) scattered high-energy electrons off of protons (nuclei of hydrogen atoms). Electrons occasionally scattered straight backward in a 
manner that implied a very small and very dense particle makes up the proton — this observation is considered nearly direct evidence of quarks. The 
analysis was based partly on the same conservation of momentum principle that works so well on the large scale. 

Macroscopic 
target 





Proton 

Figure 8.5 A subatomic particle scatters straight backward from a target particle. In experiments seeking evidence for quarks, electrons were observed to occasionally scatter 
straight backward from a proton. 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 269 

8.4 Elastic Collisions in One Dimension 

Let us consider various types of two-object collisions. These collisions are the easiest to analyze, and they illustrate many of the physical principles 
involved in collisions. The conservation of momentum principle is very useful here, and it can be used whenever the net external force on a system is 
zero. 

We start with the elastic collision of two objects moving along the same line — a one-dimensional problem. An elastic collision is one that also 
conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Figure 8.6 illustrates an 
elastic collision in which internal kinetic energy and momentum are conserved. 

Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but 
not quite, elastic — some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. One 
macroscopic collision that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring 
bumpers on an air track. Icy surfaces and air tracks are nearly frictionless, more readily allowing nearly elastic collisions on them. 

Elastic Collision 



An elastic collision is one that conserves internal kinetic energy. 



Internal Kinetic Energy 



Internal kinetic energy is the sum of the kinetic energies of the objects in the system. 



System of interest 
net F m 



Before 




System of interest 
Elastic =* KE; + KE£ = KE 1 + KE 2 
After 




Figure 8.6 An elastic one-dimensional two-object collision. Momentum and internal kinetic energy are conserved. 

Now, to solve problems involving one-dimensional elastic collisions between two objects we can use the equations for conservation of momentum 
and conservation of internal kinetic energy. First, the equation for conservation of momentum for two objects in a one-dimensional collision is 

Pl+P2 = P'l+P'2 (Fnct = 0) (8-33) 



m l v l + m 2 v 2 = m x v\ + m 2 v' 2 (^net = 0), 



(8.34) 



where the primes (') indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic 
energies before the collision equals the sum after the collision. Thus, 

^-m 1 v 1 +ym 2 V2 =^m 1 v / j +^-m 2 v / 2 (two-object elastic collision) ^' ' 

expresses the equation for conservation of internal kinetic energy in a one-dimensional collision. 



Example 8.4 Calculating Velocities Following an Elastic Collision 



Calculate the velocities of two objects following an elastic collision, given that 

m 1 = 0.500 kg, m 2 = 3.50 kg, v x = 4.00 m/s, and v 2 = 0. 



(8.36) 



270 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 

Strategy and Concept 

First, visualize what the initial conditions mean — a small object strikes a larger object that is initially at rest. This situation is slightly simpler than 
the situation shown in Figure 8.6 where both objects are initially moving. We are asked to find two unknowns (the final velocities v' i and v' 2 ). 

To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can 
be simplified by the fact that object 2 is initially at rest, and thus v 2 = . Once we simplify these equations, we combine them algebraically to 

solve for the unknowns. 

Solution 

For this problem, note that v 2 = and use conservation of momentum. Thus, 

p l =p' l +p' 2 ( 8 - 37 ) 

or 

m^ Vi = mjv'i + ^2 V 2- (8.38) 

Using conservation of internal kinetic energy and that v 2 = , 

l m v 2 _ l m v / 2 , 1 , 2 (8.39) 



Solving the first equation (momentum equation) for v' 2 , we obtain 



-Ulll^-WA (8-40) 



v 2 = mti v l- v l 



Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable v' 2 , leaving only v\ as an 
unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are 

v' 1= 4.00 m/s (8.41) 

and 

v\ = -3.00m/s. (8.42) 

As noted when quadratic equations were encountered in earlier chapters, both solutions may or may not be meaningful. In this case, the first 
solution is the same as the initial condition. The first solution thus represents the situation before the collision and is discarded. The second 
solution (v' i = —3.00 m/s) is negative, meaning that the first object bounces backward. When this negative value of v' 1 is used to find the 

velocity of the second object after the collision, we get 

m 1( M O500kg r/inn , Qnml . (8.43) 

v 2 = Wp\ ~ V i) = 3> 5o kg t 4 - 00 " (-3.00)] m/s 

or 

v' 2 = 1.00 m/s. (8.44) 

Discussion 

The result of this example is intuitively reasonable. A small object strikes a larger one at rest and bounces backward. The larger one is knocked 
forward, but with a low speed. (This is like a compact car bouncing backward off a full-size SUV that is initially at rest.) As a check, try calculating 
the internal kinetic energy before and after the collision. You will see that the internal kinetic energy is unchanged at 4.00 J. Also check the total 
momentum before and after the collision; you will find it, too, is unchanged. 

The equations for conservation of momentum and internal kinetic energy as written above can be used to describe any one-dimensional elastic 
collision of two objects. These equations can be extended to more objects if needed. 

Making Connections: Take-Ho me Investigation — Ice Cubes and Elastic Collision 

Find a few ice cubes which are about the same size and a smooth kitchen tabletop or a table with a glass top. Place the ice cubes on the surface 
several centimeters away from each other. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes 
after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. Have you created approximately elastic collisions? 
Explain the speeds and directions of the ice cubes using momentum. 

PhET Explorations: Collision Lab 

Investigate collisions on an air hockey table. Set up your own experiments: vary the number of discs, masses and initial conditions. Is momentum 
conserved? Is kinetic energy conserved? Vary the elasticity and see what happens. 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 271 



tZ t^PhET Interactive Simulation 

Figure 8.7 Collision Lab (http://cnx.Org/content/m42163/l.3/collision-lab_en.jar) 

8.5 Inelastic Collisions in One Dimension 

We have seen that in an elastic collision, internal kinetic energy is conserved. An inelastic collision is one in which the internal kinetic energy 
changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add internal kinetic energy. 
Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, 
this internal work may transform some internal kinetic energy into heat transfer. Or it may convert stored energy into internal kinetic energy, such as 
when exploding bolts separate a satellite from its launch vehicle. 

Inelastic Collision 



An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). 

Figure 8.8 shows an example of an inelastic collision. Two objects that have equal masses head toward one another at equal speeds and then stick 
together. Their total internal kinetic energy is initially mv f^-mv + -^mv I . The two objects come to rest after sticking together, conserving 

momentum. But the internal kinetic energy is zero after the collision. A collision in which the objects stick together is sometimes called a perfectly 
inelastic collision because it reduces internal kinetic energy more than does any other type of inelastic collision. In fact, such a collision reduces 
internal kinetic energy to the minimum it can have while still conserving momentum. 

Perfectly Inelastic Collision 



A collision in which the objects stick together is sometimes called "perfectly inelastic' 

System of interest 
net F = 



Before 



System of interest 




After 




(a) 

Figure 8.8 An inelastic one-dimensional two-object collision. Momentum is conserved, but internal kinetic energy is not conserved, (a) Two objects of equal mass initially head 
directly toward one another at the same speed, (b) The objects stick together (a perfectly inelastic collision), and so their final velocity is zero. The internal kinetic energy of the 
system changes in any inelastic collision and is reduced to zero in this example. 



Example 8.5 Calculating Velocity and Change in Kinetic Energy: Inelastic Collision of a Puck and a Goalie 



(a) Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 
35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. (See 
Figure 8.9 ) 



272 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 



System of interest 



After System of interest 

KE' jnt < KE int 
Pi + Pi = Ptot 




Frictionless ice surface 



Frictionless ice surface 



Figure 8.9 An ice hockey goalie catches a hockey puck and recoils backward. The initial kinetic energy of the puck is almost entirely converted to thermal energy and 
sound in this inelastic collision. 

Strategy 

Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find 
the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie 
are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested. 

Solution for (a) 

Momentum is conserved because the net external force on the puck-goalie system is zero. 
Conservation of momentum is 

Pl+P2 = P'l+P'2 ( 8 - 45 ) 



or 



(8.46) 



m 1 v 1 + m 2 v 2 = M\v' i + m 2 v' ' 2 . 

Because the goalie is initially at rest, we know v 2 = . Because the goalie catches the puck, the final velocities are equal, or V i = v' 2 = v' . 
Thus, the conservation of momentum equation simplifies to 

m 1 v 1 = (m 1 + m 2 )v'. (8.47) 

Solving for v' yields 



wii + m 9 1" 



Entering known values in this equation, we get 



V = ( TO.OkgTo^Okg ) 35 - ""»> = 7 - 48Xl0 " 2 »*• 



(8.48) 



(8.49) 



Discussion for (a) 

This recoil velocity is small and in the same direction as the puck's original velocity, as we might expect. 

Solution for (b) 

Before the collision, the internal kinetic energy KE int of the system is that of the hockey puck, because the goalie is initially at rest. Therefore, 

KE int is initially 



KE: n 



= l mv 2 = 1(0.150 kg)(35.0 m/s) 2 



= 91.9 J. 

After the collision, the internal kinetic energy is 

2 
KE' int = l(m + M)v 2 = 1(70.15 kg)(7.48xl0" 2 m/s) 



The change in internal kinetic energy is thus 



2 V 
0.196 J. 



KE/ int _KE int 



= 0.196 J -91.9 J 
= -91.7 J 



(8.50) 



(8.51) 



(8.52) 



where the minus sign indicates that the energy was lost. 
Discussion for (b) 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 273 



Nearly all of the initial internal kinetic energy is lost in this perfectly inelastic collision. KE int is mostly converted to thermal energy and sound. 

During some collisions, the objects do not stick together and less of the internal kinetic energy is removed — such as happens in most automobile 
accidents. Alternatively, stored energy may be converted into internal kinetic energy during a collision. Figure 8.10 shows a one-dimensional 
example in which two carts on an air track collide, releasing potential energy from a compressed spring. Example 8.6 deals with data from such 
a collision. 



Before 



net F = 



System of interest 




Frictionless 
surface 



After 



KEL > KE 



System of interest 




Uncoiled spring 



-J} 



V2 



Frictionless surface 



Figure 8.10 An air track is nearly frictionless, so that momentum is conserved. Motion is one-dimensional. In this collision, examined in Example 8.6, the potential energy 
of a compressed spring is released during the collision and is converted to internal kinetic energy. 

Collisions are particularly important in sports and the sporting and leisure industry utilizes elastic and inelastic collisions. Let us look briefly at 
tennis. Recall that in a collision, it is momentum and not force that is important. So, a heavier tennis racquet will have the advantage over a 
lighter one. This conclusion also holds true for other sports — a lightweight bat (such as a softball bat) cannot hit a hardball very far. 

The location of the impact of the tennis ball on the racquet is also important, as is the part of the stroke during which the impact occurs. A smooth 
motion results in the maximizing of the velocity of the ball after impact and reduces sports injuries such as tennis elbow. A tennis player tries to 
hit the ball on the "sweet spot" on the racquet, where the vibration and impact are minimized and the ball is able to be given more velocity. Sports 
science and technologies also use physics concepts such as momentum and rotational motion and vibrations. 

Take-Home Experiment — Bouncing of Tennis Ball 

1. Find a racquet (a tennis, badminton, or other racquet will do). Place the racquet on the floor and stand on the handle. Drop a tennis ball on 
the strings from a measured height. Measure how high the ball bounces. Now ask a friend to hold the racquet firmly by the handle and drop 
a tennis ball from the same measured height above the racquet. Measure how high the ball bounces and observe what happens to your 
friend's hand during the collision. Explain your observations and measurements. 

2. The coefficient of restitution (c) is a measure of the elasticity of a collision between a ball and an object, and is defined as the ratio of the 

speeds after and before the collision. A perfectly elastic collision has a c of 1. For a ball bouncing off the floor (or a racquet on the floor), 

I/O 

c can be shown to be c = (hi H) where h is the height to which the ball bounces and H is the height from which the ball is 

dropped. Determine c for the cases in Part 1 and for the case of a tennis ball bouncing off a concrete or wooden floor ( c = 0.85 for new 
tennis balls used on a tennis court). 



Example 8.6 Calculating Final Velocity and Energy Release: Two Carts Collide 



In the collision pictured in Figure 8.10, two carts collide inelastically. Cart 1 (denoted m 1 carries a spring which is initially compressed. During 

the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and 
the cart and the spring together have an initial velocity of 2.00 m/s . Cart 2 (denoted ra 2 in Figure 8.10) has a mass of 0.500 kg and an initial 

velocity of —0.500 m/s . After the collision, cart 1 is observed to recoil with a velocity of —4.00 m/s . (a) What is the final velocity of cart 2? (b) 

How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)? 

Strategy 



274 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 

We can use conservation of momentum to find the final velocity of cart 2, because F net = (the track is frictionless and the force of the spring 

is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy 
was released by the spring. 

Solution for (a) 

As before, the equation for conservation of momentum in a two-object system is 

m 1 v 1 + m 2 v 2 = m\v' i + ^2 V 2- (8.53) 

The only unknown in this equation is v\ . Solving for v' 2 and substituting known values into the previous equation yields 

/ _ m l v l + m 2 v 2 — m l v l (8.54) 

v 2 - nT 2 

(0.350 kg)(2.QQ m/s) + (0.500 kg)(-Q.5QQ m/s) (0.350 kg)(-4.QQ m/s) 
0.500 kg 0.500 kg 

= 3.70 m/s. 
Solution for (b) 

The internal kinetic energy before the collision is 

KE int = l mi v? + lm 2 v2 ^ 

= 1(0.350 kg)(2.00 m/s) 2 + 1(0.500 kg)( - 0.500 m/s) 2 
= 0.763 J. 
After the collision, the internal kinetic energy is 

KF / _ l m v /2 , 1 ,2 (8.56) 

K±1 int - 2 l 1 2 2 2 

= 1(0.350 kg)(-4.00 m/s) 2 + 1(0.500 kg)(3.70 m/s) 2 
= 6.22 J. 
The change in internal kinetic energy is thus 

KE' int -KE int = 6.22 J -0.763 J (8.57) 

= 5.46 J. 
Discussion 

The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision 
increases by 5.46 J. That energy was released by the spring. 

8.6 Collisions of Point Masses in Two Dimensions 

In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along 
the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional 
collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the 
approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into 
components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously. 

One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters 
hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so 
that no rotation is possible. To avoid rotation, we consider only the scattering of point masses — that is, structureless particles that cannot rotate or 
spin. 

We start by assuming that F net = , so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. 

(See Figure 8.11.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 
8.11. Because momentum is conserved, the components of momentum along the x - and y -axes (p x and p y ) will also be conserved, but with the 

chosen coordinate system, p y is initially zero and p x is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the 

simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from 
the analysis of two-dimensional collisions.) 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 275 



Before 

Pi = Ptoi 




After 

P'l + P2 = Plot 



System of interest 
(before collision) 



System of interest 
{after collision) 



Figure 8.11 A two-dimensional collision with the coordinate system chosen so that m 2 is initially at rest and Vj is parallel to the X -axis. This coordinate system is 

sometimes called the laboratory coordinate system, because many scattering experiments have a target that is stationary in the laboratory, while particles are scattered from it 
to determine the particles that make-up the target and how they are bound together. The particles may not be observed directly, but their initial and final velocities are. 



Along the x -axis, the equation for conservation of momentum is 

Plx + P2x = P'ix + P'2x- 



(8.58) 



Where the subscripts denote the particles and axes and the primes denote the situation after the collision. In terms of masses and velocities, this 
equation is 

m l v lx + m 2 v 2x — m \ v * \x + m 2 v/ 2r (8.59) 

But because particle 2 is initially at rest, this equation becomes 

m \ v \x = m \ V \x + m 2 V 2r (8 - 60) 

The components of the velocities along the x -axis have the form v cos 6 . Because particle 1 initially moves along the x -axis, we find v lx = v^ . 
Conservation of momentum along the x -axis gives the following equation: 

m 1 v 1 = m 1 v / 1 cos Q x +m 2 v' 2 cos 2 , (8.61) 

where 6^ and 6 2 are as shown in Figure 8.11. 

Conservation of Momentum along the x -axis 



i l v l = m\v\ cos Q x +m 2 v' 2 cos 6 2 



(8.62) 



(8.63) 



(8.64) 



Along the y -axis, the equation for conservation of momentum is 

Ply + P2y = P'\y + P'2y 

or 

m l v ly + m 2 v 2y = m l v ' \y + m 2 v ' 2y 
But v ly is zero, because particle 1 initially moves along the x -axis. Because particle 2 is initially at rest, v 2 is also zero. The equation for 
conservation of momentum along the y -axis becomes 

= mi v' ly + m 2 v' 2y . (8.65) 

The components of the velocities along the y -axis have the form v sin 6 . 
Thus, conservation of momentum along the y -axis gives the following equation: 

= m 1 v / 1 sin 6 ) 1 +m 2 v / 2 sin 6 2 . (8.66) 



Conservation of Momentum along the y -axis 



= miv' x sin 6 1 + m 2 v' ' 2 sin 6 2 



(8.67) 



276 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 

The equations of conservation of momentum along the x -axis and y -axis are very useful in analyzing two-dimensional collisions of particles, where 

one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be 
necessary when collision experiments are used to explore nature at the subatomic level. 



Example 8.7 Determining the Final Velocity of an Unseen Object from the Scattering of Another Object 



Suppose the following experiment is performed. A 0.250-kg object (m^ is slid on a frictionless surface into a dark room, where it strikes an 

initially stationary object with mass of 0.400 kg (m 2 ) . The 0.250-kg object emerges from the room at an angle of 45.0° with its incoming 

direction. 

The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity 
(v' 2 and 6 2 ) °f tne 0.400-kg object after the collision. 

Strategy 

Momentum is conserved because the surface is frictionless. The coordinate system shown in Figure 8.12 is one in which m 2 is originally at rest 
and the initial velocity is parallel to the x -axis, so that conservation of momentum along the x - and y -axes is applicable. 

Everything is known in these equations except v' 2 and 6 2 , which are precisely the quantities we wish to find. We can find two unknowns 
because we have two independent equations: the equations describing the conservation of momentum in the x - and y -directions. 

Solution 

Solving m x v x =m x V \ cos 6 l +m 2 v' 2 cos # 2 and = m l v f l sm6 l +m 2 v' 2 sin # 2 for v' 2 sin 6 2 and taking the ratio yields an 

equation (because (tan 6 = sm ^ J in which all but one quantity is known: 

v\ sinfli (8.68) 

tan 6 ? = — — l —^ — l - — . 
z v 1 cos 6 1 — Vj 



(1.50 m/s)(0.7071) = _i 129 (869) 



Entering known values into the previous equation gives 

f n _ yi.JU 111/SA U - /u/ -U 

2 " (1.50 m/s)(0.7071) - 2.00 m/s 
Thus, 

6 2 = tan _1 (-1.129) = 311.5° « 312°. ( 87 °) 

Angles are defined as positive in the counter clockwise direction, so this angle indicates that m 2 is scattered to the right in Figure 8.12, as 

expected (this angle is in the fourth quadrant). Either equation for the x - or y -axis can now be used to solve for v' 2 , but the latter equation is 

easiest because it has fewer terms. 

, _ _m ± , sin 6^ (8.71) 

V 2" rn 2 v l sin 2 

Entering known values into this equation gives 



_ / 0.250kg V /q.7 

V 2" Vo.400kg/ L5Um/S) Uo. 



7485/ 



Thus, 



Discussion 



v' 2 = 0.886 m/s. (8.73) 



It is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of- 
chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. 
This type of result makes a physicist want to explore the system further. 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 277 



y* 



net F = 

Before After 

Pi = Ptot Pi + P2 = Plot 



m z = 0.400 mg v 2 = ,\ ^ = t 




= 45 c 



©s_~ 



7 



14= ? © 2 = ? 

Figure 8.12 A collision taking place in a dark room is explored in Example 8.7. The incoming object m^ is scattered by an initially stationary object. Only the stationary 
object's mass m 2 is known. By measuring the angle and speed at which m j emerges from the room, it is possible to calculate the magnitude and direction of the initially 
stationary object's velocity after the collision. 

Elastic Collisions of Two Objects with Equal Mass 

Some interesting situations arise when the two colliding objects have equal mass and the collision is elastic. This situation is nearly the case with 
colliding billiard balls, and precisely the case with some subatomic particle collisions. We can thus get a mental image of a collision of subatomic 
particles by thinking about billiards (or pool). (Refer to Figure 8.11 for masses and angles.) First, an elastic collision conserves internal kinetic energy. 
Again, let us assume object 2 (m 2 ) is initially at rest. Then, the internal kinetic energy before and after the collision of two objects that have equal 



masses is 



l mVl 2 = hnv f 



2 1 / 2 



Because the masses are equal, m 1 = m 2 = m . Algebraic manipulation (left to the reader) of conservation of momentum in the x - and y - 
directions can show that 



ymvj = ±mv\ -\-^mV 2 + mv\V 2 cos(6 , 1 - 6 2 ). 



(8.74) 



(8.75) 



(Remember that 6 2 is negative here.) The two preceding equations can both be true only if 

mv\v' 2 cos{6 l -6 2 ) = 0. (8-76) 

There are three ways that this term can be zero. They are 

• v'j = : head-on collision; incoming ball stops 

• V 2 — : no collision; incoming ball continues unaffected 

• cos^j — 6 2 ) = : angle of separation (^ 1 — 2 ) is 90° after the collision 

All three of these ways are familiar occurrences in billiards and pool, although most of us try to avoid the second. If you play enough pool, you will 
notice that the angle between the balls is very close to 90° after the collision, although it will vary from this value if a great deal of spin is placed on 
the ball. (Large spin carries in extra energy and a quantity called angular momentum, which must also be conserved.) The assumption that the 
scattering of billiard balls is elastic is reasonable based on the correctness of the three results it produces. This assumption also implies that, to a 
good approximation, momentum is conserved for the two-ball system in billiards and pool. The problems below explore these and other 
characteristics of two-dimensional collisions. 

Connections to Nuclear and Particle Physics 

Two-dimensional collision experiments have revealed much of what we know about subatomic particles, as we shall see in Medical 
Applications of Nuclear Physics and Particle Physics. Ernest Rutherford, for example, discovered the nature of the atomic nucleus from such 
experiments. 



8.7 Introduction to Rocket Propulsion 

Rockets range in size from fireworks so small that ordinary people use them to immense Saturn Vs that once propelled massive payloads toward the 
Moon. The propulsion of all rockets, jet engines, deflating balloons, and even squids and octopuses is explained by the same physical 



278 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 

principle — Newton's third law of motion. Matter is forcefully ejected from a system, producing an equal and opposite reaction on what remains. 
Another common example is the recoil of a gun. The gun exerts a force on a bullet to accelerate it and consequently experiences an equal and 
opposite force, causing the gun's recoil or kick. 

Making Connections: Take-Home Experiment — Propulsion of a Balloon 

Hold a balloon and fill it with air. Then, let the balloon go. In which direction does the air come out of the balloon and in which direction does the 
balloon get propelled? If you fill the balloon with water and then let the balloon go, does the balloon's direction change? Explain your answer. 

Figure 8.13 shows a rocket accelerating straight up. In part (a), the rocket has a mass m and a velocity v relative to Earth, and hence a momentum 

mv . In part (b), a time At has elapsed in which the rocket has ejected a mass Am of hot gas at a velocity v e relative to the rocket. The 

remainder of the mass (m — Am) now has a greater velocity (v + Av) . The momentum of the entire system (rocket plus expelled gas) has 

actually decreased because the force of gravity has acted for a time At , producing a negative impulse Ap = —mgAt . (Remember that impulse is 

the net external force on a system multiplied by the time it acts, and it equals the change in momentum of the system.) So, the center of mass of the 
system is in free fall but, by rapidly expelling mass, part of the system can accelerate upward. It is a commonly held misconception that the rocket 
exhaust pushes on the ground. If we consider thrust; that is, the force exerted on the rocket by the exhaust gases, then a rocket's thrust is greater in 
outer space than in the atmosphere or on the launch pad. In fact, gases are easier to expel into a vacuum. 

By calculating the change in momentum for the entire system over At , and equating this change to the impulse, the following expression can be 
shown to be a good approximation for the acceleration of the rocket. 

n = XeAm_ Q (8.77) 

m a? g 

"The rocket" is that part of the system remaining after the gas is ejected, and g is the acceleration due to gravity. 



Acceleration of a Rocket 


Acceleration of a rocket is 








a = — ^■ 
u m 


Am 

At ' 


-g> 
















(8.78) 


where a is the acceleration of the rocket, 


v e 


is the 


escape velocity, 


m is 


the 


mass 


of the rocket, 


Am 


is the 


mass 


of the 


ejected gas, 


and A^ 


is the time in which the gas 


is ejected. 






























System of 

interest at 

t + At 



Rocket 
free-body 
diagram 



w 



(a) (b) 

Figure 8.13 (a) This rocket has a mass m and an upward velocity V . The net external force on the system is —mg , if air resistance is neglected, (b) A time A^ later the 

system has two main parts, the ejected gas and the remainder of the rocket. The reaction force on the rocket is what overcomes the gravitational force and accelerates it 
upward. 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 279 

A rocket's acceleration depends on three major factors, consistent with the equation for acceleration of a rocket . First, the greater the exhaust 
velocity of the gases relative to the rocket, v e , the greater the acceleration is. The practical limit for v e is about 2.5x10 m/s for conventional 
(non-nuclear) hot-gas propulsion systems. The second factor is the rate at which mass is ejected from the rocket. This is the factor Am/ At in the 
equation. The quantity (Am/ At)v e , with units of newtons, is called "thrust." The faster the rocket burns its fuel, the greater its thrust, and the 
greater its acceleration. The third factor is the mass m of the rocket. The smaller the mass is (all other factors being the same), the greater the 
acceleration. The rocket mass m decreases dramatically during flight because most of the rocket is fuel to begin with, so that acceleration increases 
continuously, reaching a maximum just before the fuel is exhausted. 

Factors Affecting a Rocket's Acceleration 



The greater the exhaust velocity v e of the gases relative to the rocket, the greater the acceleration. 

The faster the rocket burns its fuel, the greater its acceleration. 

The smaller the rocket's mass (all other factors being the same), the greater the acceleration. 



Example 8.8 Calculating Acceleration: Initial Acceleration of a Moon Launch 



A Saturn V's mass at liftoff was 2.80X 10 kg , its fuel-burn rate was 1.40X 10 kg/s , and the exhaust velocity was 2.40X 10 m/s . 

Calculate its initial acceleration. 

Strategy 

This problem is a straightforward application of the expression for acceleration because a is the unknown and all of the terms on the right side 

of the equation are given. 

Solution 

Substituting the given values into the equation for acceleration yields 

n = ZeAm_ (8.79) 

" m At 

= 2.40X10 3 m/s ^ 4QxlQ 4 k j \ _ 9 80 ^2 
2.80xl0 6 kg v } 

= 2.20 m/s 2 . 

Discussion 

This value is fairly small, even for an initial acceleration. The acceleration does increase steadily as the rocket burns fuel, because m decreases 

while v Q and -^^ remain constant. Knowing this acceleration and the mass of the rocket, you can show that the thrust of the engines was 
3.36xl0 7 N. 



To achieve the high speeds needed to hop continents, obtain orbit, or escape Earth's gravity altogether, the mass of the rocket other than fuel must 
be as small as possible. It can be shown that, in the absence of air resistance and neglecting gravity, the final velocity of a one-stage rocket initially at 
rest is 



i m 



(8.80) 



where ln(mQ/m r ) is the natural logarithm of the ratio of the initial mass of the rocket (ra ) to what is left (m r ) after all of the fuel is exhausted. 

(Note that v is actually the change in velocity, so the equation can be used for any segment of the flight. If we start from rest, the change in velocity 
equals the final velocity.) For example, let us calculate the mass ratio needed to escape Earth's gravity starting from rest, given that the escape 
velocity from Earth is about 11.2x10 m/s , and assuming an exhaust velocity v e = 2.5x10 m/s . 



l n ^0 = JL = 11.2X10^ m/s = 4 48 
mr Ve 2.5xl0 3 m/s 



Solving for m /m r gives 



Thus, the mass of the rocket is 



m _ ,4.48 _ 
m r ~ e ~ 



m r = 



_ m 



(8.81) 



(8.82) 



(8.83) 



This result means that only 1/88 of the mass is left when the fuel is burnt, and 87 / 88 of the initial mass was fuel. Expressed as percentages, 
98.9% of the rocket is fuel, while payload, engines, fuel tanks, and other components make up only 1.10%. Taking air resistance and gravitational 
force into account, the mass m r remaining can only be about m / 180 . It is difficult to build a rocket in which the fuel has a mass 180 times 



280 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 

everything else. The solution is multistage rockets. Each stage only needs to achieve part of the final velocity and is discarded after it burns its fuel. 
The result is that each successive stage can have smaller engines and more payload relative to its fuel. Once out of the atmosphere, the ratio of 
payload to fuel becomes more favorable, too. 

The space shuttle was an attempt at an economical vehicle with some reusable parts, such as the solid fuel boosters and the craft itself. (See Figure 
8.14) The shuttle's need to be operated by humans, however, made it at least as costly for launching satellites as expendable, unmanned rockets. 
Ideally, the shuttle would only have been used when human activities were required for the success of a mission, such as the repair of the Hubble 
space telescope. Rockets with satellites can also be launched from airplanes. Using airplanes has the double advantage that the initial velocity is 
significantly above zero and a rocket can avoid most of the atmosphere's resistance. 




Figure 8.14 The space shuttle had a number of reusable parts. Solid fuel boosters on either side were recovered and refueled after each flight, and the entire orbiter returned 
to Earth for use in subsequent flights. The large liquid fuel tank was expended. The space shuttle was a complex assemblage of technologies, employing both solid and liquid 
fuel and pioneering ceramic tiles as reentry heat shields. As a result, it permitted multiple launches as opposed to single-use rockets, (credit: NASA) 

PhET Explorations: Lunar Lander 

Can you avoid the boulder field and land safely, just before your fuel runs out, as Neil Armstrong did in 1969? Our version of this classic video 
game accurately simulates the real motion of the lunar lander with the correct mass, thrust, fuel consumption rate, and lunar gravity. The real 
lunar lander is very hard to control. 



cij f^PhET Interactive Simulation 



Figure 8.15 Lunar Lander (http://cnx.Org/content/m42166/l.3/lunar-lander_en.jar) 



Glossary 



change in momentum: the difference between the final and initial momentum; the mass times the change in velocity 

conservation of momentum principle: when the net external force is zero, the total momentum of the system is conserved or constant 

elastic collision: a collision that also conserves internal kinetic energy 

impulse: the average net external force times the time it acts; equal to the change in momentum 

inelastic collision: a collision in which internal kinetic energy is not conserved 

internal kinetic energy: the sum of the kinetic energies of the objects in a system 

isolated system: a system in which the net external force is zero 

linear momentum: the product of mass and velocity 

perfectly inelastic collision: a collision in which the colliding objects stick together 

point masses: structureless particles with no rotation or spin 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 281 



quark: fundamental constituent of matter and an elementary particle 



second law of motion: physical law that states that the net external force equals the change in momentum of a system divided by the time over 
which it changes 



Section Summary 



8.1 Linear Momentum and Force 

• Linear momentum (momentum for brevity) is defined as the product of a system's mass multiplied by its velocity. 

• In symbols, linear momentum p is defined to be 

p = mv, 
where m is the mass of the system and v is its velocity. 

• The SI unit for momentum is kg • m/s . 

• Newton's second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by 
the time over which it changes. 

• In symbols, Newton's second law of motion is defined to be 

F -AE 

net ~ At 9 

F net is the net external force, Ap is the change in momentum, and At is the change time. 

8.2 Impulse 

• Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts: 

Ap = F net Af. 

• Forces are usually not constant over a period of time. 

8.3 Conservation of Momentum 

• The conservation of momentum principle is written 



= constant 



or 



Ptot — P tot (isolated system), 
p tot is the initial total momentum and p' tot is the total momentum some time later. 

• An isolated system is defined to be one for which the net external force is zero (F net = 0). 

• During projectile motion and where air resistance is negligible, momentum is conserved in the horizontal direction because horizontal forces are 
zero. 

• Conservation of momentum applies only when the net external force is zero. 

• The conservation of momentum principle is valid when considering systems of particles. 

8.4 Elastic Collisions in One Dimension 

• An elastic collision is one that conserves internal kinetic energy. 

• Conservation of kinetic energy and momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one 
dimensional two-body collisions. 

8.5 Inelastic Collisions in One Dimension 

• An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). 

• A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does 
any other type of inelastic collision. 

• Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations. 

8.6 Collisions of Point Masses in Two Dimensions 

• The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along 
perpendicular axes. Choose a coordinate system with the x -axis parallel to the velocity of the incoming particle. 

• Two-dimensional collisions of point masses where mass 2 is initially at rest conserve momentum along the initial direction of mass 1 (the x - 
axis), stated by m 1 v 1 = m^v\ cos 0^ + m 2 v y ' 2 cos 6 2 and along the direction perpendicular to the initial direction (the y -axis) stated by 

= m l v , ly +m 2 v , 2 y 

• The internal kinetic before and after the collision of two objects that have equal masses is 

l/m^ 2 = lmv / 1 2 + lmv / 2 2 + mv / 1 v / 2 cos^ - 2 ). 

• Point masses are structureless particles that cannot spin. 

8.7 Introduction to Rocket Propulsion 

• Newton's third law of motion states that to every action, there is an equal and opposite reaction. 

• Acceleration of a rocket is a = jjf ■^ LL — g . 

• A rocket's acceleration depends on three main factors. They are 

1. The greater the exhaust velocity of the gases, the greater the acceleration. 



282 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 

2. The faster the rocket burns its fuel, the greater its acceleration. 

3. The smaller the rocket's mass, the greater the acceleration. 



Conceptual Questions 



8.1 Linear Momentum and Force 

1. An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy? 

2. An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum? 

3. Professional Application 

Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of 
momentum, work, and energy, explain how a football player can be more effective with his feet on the ground. 

4. How can a small force impart the same momentum to an object as a large force? 

8.2 Impulse 

5. Professional Application 

Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advantages of a carpeted vs. 
tile floor for a day care center. 

6. While jumping on a trampoline, sometimes you land on your back and other times on your feet. In which case can you reach a greater height and 
why? 

7. Professional Application 

Tennis racquets have "sweet spots." If the ball hits a sweet spot then the player's arm is not jarred as much as it would be otherwise. Explain why this 
is the case. 

8.3 Conservation of Momentum 

8. Professional Application 

If you dive into water, you reach greater depths than if you do a belly flop. Explain this difference in depth using the concept of conservation of 
energy. Explain this difference in depth using what you have learned in this chapter. 

9. Under what circumstances is momentum conserved? 

10. Can momentum be conserved for a system if there are external forces acting on the system? If so, under what conditions? If not, why not? 

11. Momentum for a system can be conserved in one direction while not being conserved in another. What is the angle between the directions? Give 
an example. 

12. Professional Application 

Explain in terms of momentum and Newton's laws how a car's air resistance is due in part to the fact that it pushes air in its direction of motion. 

13. Can objects in a system have momentum while the momentum of the system is zero? Explain your answer. 

14. Must the total energy of a system be conserved whenever its momentum is conserved? Explain why or why not. 

8.4 Elastic Collisions in One Dimension 

15. What is an elastic collision? 

8.5 Inelastic Collisions in One Dimension 

16. What is an inelastic collision? What is a perfectly inelastic collision? 

17. Mixed-pair ice skaters performing in a show are standing motionless at arms length just before starting a routine. They reach out, clasp hands, 
and pull themselves together by only using their arms. Assuming there is no friction between the blades of their skates and the ice, what is their 
velocity after their bodies meet? 

18. A small pickup truck that has a camper shell slowly coasts toward a red light with negligible friction. Two dogs in the back of the truck are moving 
and making various inelastic collisions with each other and the walls. What is the effect of the dogs on the motion of the center of mass of the system 
(truck plus entire load)? What is their effect on the motion of the truck? 

8.6 Collisions of Point Masses in Two Dimensions 

19. Figure 8.16 shows a cube at rest and a small object heading toward it. (a) Describe the directions (angle Qy ) at which the small object can 

emerge after colliding elastically with the cube. How does Qy depend on b , the so-called impact parameter? Ignore any effects that might be due to 

rotation after the collision, and assume that the cube is much more massive than the small object, (b) Answer the same questions if the small object 
instead collides with a massive sphere. 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 283 



After 



Before 




n 




v 2 = 



m 



:i< 




h = ? 



v^-0 



Figure 8.16 A small object approaches a collision with a much more massive cube, after which its velocity has the direction 0\ . The angles at which the small object can be 
scattered are determined by the shape of the object it strikes and the impact parameter b . 

8.7 Introduction to Rocket Propulsion 

20. Professional Application 

Suppose a fireworks shell explodes, breaking into three large pieces for which air resistance is negligible. How is the motion of the center of mass 
affected by the explosion? How would it be affected if the pieces experienced significantly more air resistance than the intact shell? 

21. Professional Application 

During a visit to the International Space Station, an astronaut was positioned motionless in the center of the station, out of reach of any solid object 
on which he could exert a force. Suggest a method by which he could move himself away from this position, and explain the physics involved. 

22. Professional Application 

It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and gas 
momentum are in the same direction as that of the rocket. How is the rocket still able to obtain thrust by ejecting the gases? 



284 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 



Problems & Exercises 



8.1 Linear Momentum and Force 

1. (a) Calculate the momentum of a 2000-kg elephant charging a hunter 
at a speed of 7.50 m/s . (b) Compare the elephant's momentum with the 
momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s . 
(c) What is the momentum of the 90.0-kg hunter running at 7.40 m/s 
after missing the elephant? 

2. (a) What is the mass of a large ship that has a momentum of 
1 .60x 10 kg • m/s , when the ship is moving at a speed of 

48.0 km/h? (b) Compare the ship's momentum to the momentum of a 
1100-kg artillery shell fired at a speed of 1200 m/s . 

3. (a) At what speed would a 2.00X 10 -kg airplane have to fly to have 

a momentum of 1.60X 10 kg • m/s (the same as the ship's momentum 

in the problem above)? (b) What is the plane's momentum when it is 
taking off at a speed of 60.0 m/s ? (c) If the ship is an aircraft carrier that 
launches these airplanes with a catapult, discuss the implications of your 
answer to (b) as it relates to recoil effects of the catapult on the ship. 

4. (a) What is the momentum of a garbage truck that is 1 .20x 10 kg 

and is moving at 10.0 m/s ? (b) At what speed would an 8.00-kg trash 
can have the same momentum as the truck? 

5. A runaway train car that has a mass of 15,000 kg travels at a speed of 
5.4 m/s down a track. Compute the time required for a force of 1500 N 

to bring the car to rest. 

6. The mass of Earth is 5.972x10 kg and its orbital radius is an 
average of 1.496x10 m . Calculate its linear momentum. 

8.2 Impulse 

7. A bullet is accelerated down the barrel of a gun by hot gases produced 
in the combustion of gun powder. What is the average force exerted on a 
0.0300-kg bullet to accelerate it to a speed of 600 m/s in a time of 2.00 
ms (milliseconds)? 

8. Professional Application 

A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate 
the force the seat belt exerts on a passenger in the car to bring him to a 
halt. The mass of the passenger is 70 kg. 

9. A person slaps her leg with her hand, bringing her hand to rest in 2.50 
milliseconds from an initial speed of 4.00 m/s. (a) What is the average 
force exerted on the leg, taking the effective mass of the hand and 
forearm to be 1.50 kg? (b) Would the force be any different if the woman 
clapped her hands together at the same speed and brought them to rest 
in the same time? Explain why or why not. 

10. Professional Application 

A professional boxer hits his opponent with a 1000-N horizontal blow that 
lasts for 0.150 s. (a) Calculate the impulse imparted by this blow, (b) 
What is the opponent's final velocity, if his mass is 105 kg and he is 
motionless in midair when struck near his center of mass? (c) Calculate 
the recoil velocity of the opponent's 10.0-kg head if hit in this manner, 
assuming the head does not initially transfer significant momentum to the 
boxer's body, (d) Discuss the implications of your answers for parts (b) 
and (c). 

11. Professional Application 

Suppose a child drives a bumper car head on into the side rail, which 
exerts a force of 4000 N on the car for 0.200 s. (a) What impulse is 
imparted by this force? (b) Find the final velocity of the bumper car if its 
initial velocity was 2.80 m/s and the car plus driver have a mass of 200 
kg. You may neglect friction between the car and floor. 

12. Professional Application 



One hazard of space travel is debris left by previous missions. There are 
several thousand objects orbiting Earth that are large enough to be 
detected by radar, but there are far greater numbers of very small 
objects, such as flakes of paint. Calculate the force exerted by a 
0.100-mg chip of paint that strikes a spacecraft window at a relative 

speed of 4.00X10 3 m/s , given the collision lasts 6.00x10 " 8 s . 

13. Professional Application 

A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs 
into a bridge abutment, (a) Calculate the average force on the person if 
he is stopped by a padded dashboard that compresses an average of 
1.00 cm. (b) Calculate the average force on the person if he is stopped 
by an air bag that compresses an average of 15.0 cm. 

14. Professional Application 

Military rifles have a mechanism for reducing the recoil forces of the gun 
on the person firing it. An internal part recoils over a relatively large 
distance and is stopped by damping mechanisms in the gun. The larger 
distance reduces the average force needed to stop the internal part, (a) 
Calculate the recoil velocity of a 1.00-kg plunger that directly interacts 
with a 0.0200-kg bullet fired at 600 m/s from the gun. (b) If this part is 
stopped over a distance of 20.0 cm, what average force is exerted upon it 
by the gun? (c) Compare this to the force exerted on the gun if the bullet 
is accelerated to its velocity in 10.0 ms (milliseconds). 

n 

15. A cruise ship with a mass of 1 .00x 10 kg strikes a pier at a speed 

of 0.750 m/s. It comes to rest 6.00 m later, damaging the ship, the pier, 
and the tugboat captain's finances. Calculate the average force exerted 
on the pier using the concept of impulse. (Hint: First calculate the time it 
took to bring the ship to rest.) 

16. Calculate the final speed of a 110-kg rugby player who is initially 
running at 8.00 m/s but collides head-on with a padded goalpost and 

experiences a backward force of 1.76x10 N for 5.50x10" s . 

17. Water from a fire hose is directed horizontally against a wall at a rate 
of 50.0 kg/s and a speed of 42.0 m/s. Calculate the force exerted on the 
wall, assuming the water's horizontal momentum is reduced to zero. 

18. A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes 
a nail and comes to rest after driving the nail 1.00 cm into a board, (a) 
Calculate the duration of the impact, (b) What was the average force 
exerted on the nail? 

19. Starting with the definitions of momentum and kinetic energy, derive 
an equation for the kinetic energy of a particle expressed as a function of 
its momentum 

20. A ball with an initial velocity of 10 m/s moves at an angle 60° above 
the +x -direction. The ball hits a vertical wall and bounces off so that it is 
moving 60° above the — x -direction with the same speed. What is the 
impulse delivered by the wall? 

21. When serving a tennis ball, a player hits the ball when its velocity is 
zero (at the highest point of a vertical toss). The racquet exerts a force of 
540 N on the ball for 5.00 ms, giving it a final velocity of 45.0 m/s. Using 
these data, find the mass of the ball. 

22. A punter drops a ball from rest vertically 1 meter down onto his foot. 
The ball leaves the foot with a speed of 18 m/s at an angle 55° above 
the horizontal. What is the impulse delivered by the foot (magnitude and 
direction)? 

8.3 Conservation of Momentum 

23. Professional Application 

Train cars are coupled together by being bumped into one another. 
Suppose two loaded train cars are moving toward one another, the first 
having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second 
having a mass of 110,000 kg and a velocity of —0.120 m/s . (The minus 
indicates direction of motion.) What is their final velocity? 

24. Suppose a clay model of a koala bear has a mass of 0.200 kg and 
slides on ice at a speed of 0.750 m/s. It runs into another clay model, 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 285 



which is initially motionless and has a mass of 0.350 kg. Both being soft 
clay, they naturally stick together. What is their final velocity? 

25. Professional Application 

Consider the following question: A car moving at 10 m/s crashes into a 
tree and stops in 0.26 s. Calculate the force the seatbelt exerts on a 
passenger in the car to bring him to a halt. The mass of the passenger is 
70 kg. Would the answer to this question be different if the car with the 
70-kg passenger had collided with a car that has a mass equal to and is 
traveling in the opposite direction and at the same speed? Explain your 
answer. 

26. What is the velocity of a 900-kg car initially moving at 30.0 m/s, just 
after it hits a 150-kg deer initially running at 12.0 m/s in the same 
direction? Assume the deer remains on the car. 

27. A 1.80-kg falcon catches a 0.650-kg dove from behind in midair. What 
is their velocity after impact if the falcon's velocity is initially 28.0 m/s and 
the dove's velocity is 7.00 m/s in the same direction? 

8.4 Elastic Collisions in One Dimension 

28. Two identical objects (such as billiard balls) have a one-dimensional 
collision in which one is initially motionless. After the collision, the moving 
object is stationary and the other moves with the same speed as the 
other originally had. Show that both momentum and kinetic energy are 
conserved. 

29. Professional Application 

Two manned satellites approach one another at a relative speed of 0.250 
m/s, intending to dock. The first has a mass of 4.00x10 kg , and the 

second a mass of 7.50x10 kg . If the two satellites collide elastically 
rather than dock, what is their final relative velocity? 

30. A 70.0-kg ice hockey goalie, originally at rest, catches a 0. 150-kg 
hockey puck slapped at him at a velocity of 35.0 m/s. Suppose the goalie 
and the ice puck have an elastic collision and the puck is reflected back 
in the direction from which it came. What would their final velocities be in 
this case? 

8.5 Inelastic Collisions in One Dimension 

31. A 0.240-kg billiard ball that is moving at 3.00 m/s strikes the bumper 
of a pool table and bounces straight back at 2.40 m/s (80% of its original 
speed). The collision lasts 0.0150 s. (a) Calculate the average force 
exerted on the ball by the bumper, (b) How much kinetic energy in joules 
is lost during the collision? (c) What percent of the original energy is left? 

32. During an ice show, a 60.0-kg skater leaps into the air and is caught 
by an initially stationary 75.0-kg skater, (a) What is their final velocity 
assuming negligible friction and that the 60.0-kg skater's original 
horizontal velocity is 4.00 m/s? (b) How much kinetic energy is lost? 

33. Professional Application 

Using mass and speed data from Example 8.1 and assuming that the 
football player catches the ball with his feet off the ground with both of 
them moving horizontally, calculate: (a) the final velocity if the ball and 
player are going in the same direction and (b) the loss of kinetic energy in 
this case, (c) Repeat parts (a) and (b) for the situation in which the ball 
and the player are going in opposite directions. Might the loss of kinetic 
energy be related to how much it hurts to catch the pass? 

n 

34. A battleship that is 6.00X 10 kg and is originally at rest fires a 

1100-kg artillery shell horizontally with a velocity of 575 m/s. (a) If the 
shell is fired straight aft (toward the rear of the ship), there will be 
negligible friction opposing the ship's recoil. Calculate its recoil velocity, 
(b) Calculate the increase in internal kinetic energy (that is, for the ship 
and the shell). This energy is less than the energy released by the gun 
powder — significant heat transfer occurs. 

35. Professional Application 



the second a mass of 7.50X 10 kg . (a) Calculate the final velocity 

(after docking) by using the frame of reference in which the first satellite 
was originally at rest, (b) What is the loss of kinetic energy in this inelastic 
collision? (c) Repeat both parts by using the frame of reference in which 
the second satellite was originally at rest. Explain why the change in 
velocity is different in the two frames, whereas the change in kinetic 
energy is the same in both. 

36. Professional Application 

A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction 
under a hopper that dumps 110,000 kg of scrap metal into it. (a) What is 
the final velocity of the loaded freight car? (b) How much kinetic energy is 
lost? 

37. Professional Application 

Space probes may be separated from their launchers by exploding bolts. 
(They bolt away from one another.) Suppose a 4800-kg satellite uses this 
method to separate from the 1500-kg remains of its launcher, and that 
5000 J of kinetic energy is supplied to the two parts. What are their 
subsequent velocities using the frame of reference in which they were at 
rest before separation? 

38. A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 
3.00-kg rifle. The pain of the rifle's kick is much worse if you hold the gun 
loosely a few centimeters from your shoulder rather than holding it tightly 
against your shoulder, (a) Calculate the recoil velocity of the rifle if it is 
held loosely away from the shoulder, (b) How much kinetic energy does 
the rifle gain? (c) What is the recoil velocity if the rifle is held tightly 
against the shoulder, making the effective mass 28.0 kg? (d) How much 
kinetic energy is transferred to the rifle-shoulder combination? The pain is 
related to the amount of kinetic energy, which is significantly less in this 
latter situation, (e) Calculate the momentum of a 110-kg football player 
running at 8.00 m/s. Compare the player's momentum with the 
momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 
m/s. Discuss its relationship to this problem. 

39. Professional Application 

One of the waste products of a nuclear reactor is plutonium-239 
f Pu) . This nucleus is radioactive and decays by splitting into a 

helium-4 nucleus and a uranium-235 nucleus ( He + Uj , the latter 

of which is also radioactive and will itself decay some time later. The 
energy emitted in the plutonium decay is 8.40x10 J and is entirely 
converted to kinetic energy of the helium and uranium nuclei. The mass 

— 97 

of the helium nucleus is 6.68X 10 kg , while that of the uranium is 

3.92X 10 " 25 kg (note that the ratio of the masses is 4 to 235). (a) 

Calculate the velocities of the two nuclei, assuming the plutonium 
nucleus is originally at rest, (b) How much kinetic energy does each 
nucleus carry away? Note that the data given here are accurate to three 
digits only. 

40. Professional Application 

The Moon's craters are remnants of meteorite collisions. Suppose a fairly 

1 9 

large asteroid that has a mass of 5.00x10 kg (about a kilometer 

across) strikes the Moon at a speed of 15.0 km/s. (a) At what speed does 
the Moon recoil after the perfectly inelastic collision (the mass of the 

99 

Moon is 7.36X 10 kg ) ? (b) How much kinetic energy is lost in the 

collision? Such an event may have been observed by medieval English 
monks who reported observing a red glow and subsequent haze about 
the Moon, (c) In October 2009, NASA crashed a rocket into the Moon, 
and analyzed the plume produced by the impact. (Significant amounts of 
water were detected.) Answer part (a) and (b) for this real-life experiment. 
The mass of the rocket was 2000 kg and its speed upon impact was 
9000 km/h. How does the plume produced alter these results? 



Two manned satellites approaching one another, at a relative speed of 41. Professional Application 



0.250 m/s, intending to dock. The first has a mass of 4.00x10 kg , and 



Two football players collide head-on in midair while trying to catch a 
thrown football. The first player is 95.0 kg and has an initial velocity of 



286 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 



6.00 m/s, while the second player is 115 kg and has an initial velocity of 
-3.50 m/s. What is their velocity just after impact if they cling together? 

42. What is the speed of a garbage truck that is 1 .20x 10 kg and is 

initially moving at 25.0 m/s just after it hits and adheres to a trash can 
that is 80.0 kg and is initially at rest? 

43. During a circus act, an elderly performer thrills the crowd by catching 
a cannon ball shot at him. The cannon ball has a mass of 10.0 kg and the 
horizontal component of its velocity is 8.00 m/s when the 65.0-kg 
performer catches it. If the performer is on nearly frictionless roller 
skates, what is his recoil velocity? 

44. (a) During an ice skating performance, an initially motionless 80.0-kg 
clown throws a fake barbell away. The clown's ice skates allow her to 
recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and 
the barbell is thrown with a velocity of 10.0 m/s, what is the mass of the 
barbell? (b) How much kinetic energy is gained by this maneuver? (c) 
Where does the kinetic energy come from? 

8.6 Collisions of Point Masses in Two Dimensions 

45. Two identical pucks collide on an air hockey table. One puck was 
originally at rest, (a) If the incoming puck has a speed of 6.00 m/s and 
scatters to an angle of 30.0° ,what is the velocity (magnitude and 
direction) of the second puck? (You may use the result that 

61 — 62 = 90° for elastic collisions of objects that have identical 

masses.) (b) Confirm that the collision is elastic. 

46. Confirm that the results of the example Example 8.7 do conserve 
momentum in both the x - and y -directions. 

47. A 3000-kg cannon is mounted so that it can recoil only in the 
horizontal direction, (a) Calculate its recoil velocity when it fires a 15.0-kg 
shell at 480 m/s at an angle of 20.0° above the horizontal, (b) What is 
the kinetic energy of the cannon? This energy is dissipated as heat 
transfer in shock absorbers that stop its recoil, (c) What happens to the 
vertical component of momentum that is imparted to the cannon when it 
is fired? 

48. Professional Application 

A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg 
bowling pin, which is scattered at an angle of 85.0° to the initial direction 
of the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final 
velocity (magnitude and direction) of the bowling ball, (b) Is the collision 
elastic? (c) Linear kinetic energy is greater after the collision. Discuss 
how spin on the ball might be converted to linear kinetic energy in the 
collision. 

49. Professional Application 

Ernest Rutherford (the first New Zealander to be awarded the Nobel 
Prize in Chemistry) demonstrated that nuclei were very small and dense 

by scattering helium-4 nuclei ( He) from gold-197 nuclei ( Auj . 
The energy of the incoming helium nucleus was 8.00x10 J , and the 

— 97 

masses of the helium and gold nuclei were 6.68x10 kg and 

3.29X 10 kg , respectively (note that their mass ratio is 4 to 197). (a) 

If a helium nucleus scatters to an angle of 120° during an elastic 
collision with a gold nucleus, calculate the helium nucleus's final speed 
and the final velocity (magnitude and direction) of the gold nucleus, (b) 
What is the final kinetic energy of the helium nucleus? 

50. Professional Application 

Two cars collide at an icy intersection and stick together afterward. The 
first car has a mass of 1200 kg and is approaching at 8.00 m/s due 
south. The second car has a mass of 850 kg and is approaching at 
17.0 m/s due west, (a) Calculate the final velocity (magnitude and 
direction) of the cars, (b) How much kinetic energy is lost in the collision? 
(This energy goes into deformation of the cars.) Note that because both 
cars have an initial velocity, you cannot use the equations for 



conservation of momentum along the x -axis and y -axis; instead, you 
must look for other simplifying aspects. 

51. Starting with equations m 1 v 1 =m 1 v / 1 cos^ 1 + m 2 v / 2 cos 6 2 
and = m^v\ sin 6^ + ^2 V 2 sm ^2 for conservation of momentum 
in the x - and y -directions and assuming that one object is originally 
stationary, prove that for an elastic collision of two objects of equal 
masses, ^-mv 1 = ^mv' l +^-mv' 2 +mv' 1 v' 2 cos (6 , 1 - 6 2 ) 

as discussed in the text. 

52. Integrated Concepts 

A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a 
velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, 
how far does the player recoil in the time it takes the puck to reach the 
goal 15.0 m away? 

8.7 Introduction to Rocket Propulsion 

53. Professional Application 

Antiballistic missiles (ABMs) are designed to have very large 
accelerations so that they may intercept fast-moving incoming missiles in 
the short time available. What is the takeoff acceleration of a 10,000-kg 
ABM that expels 196 kg of gas per second at an exhaust velocity of 

2.50xl0 3 m/s? 

54. Professional Application 

What is the acceleration of a 5000-kg rocket taking off from the Moon, 
where the acceleration due to gravity is only 1.6 m/s , if the rocket 
expels 8.00 kg of gas per second at an exhaust velocity of 
2.20xl0 3 m/s? 

55. Professional Application 

Calculate the increase in velocity of a 4000-kg space probe that expels 
3500 kg of its mass at an exhaust velocity of 2.00X 10 m/s . You may 
assume the gravitational force is negligible at the probe's location. 

56. Professional Application 

Ion-propulsion rockets have been proposed for use in space. They 
employ atomic ionization techniques and nuclear energy sources to 
produce extremely high exhaust velocities, perhaps as great as 

8.00X 10 m/s . These techniques allow a much more favorable 
payload-to-fuel ratio. To illustrate this fact: (a) Calculate the increase in 
velocity of a 20,000-kg space probe that expels only 40.0-kg of its mass 
at the given exhaust velocity, (b) These engines are usually designed to 
produce a very small thrust for a very long time — the type of engine that 
might be useful on a trip to the outer planets, for example. Calculate the 

acceleration of such an engine if it expels 4.50X 10~ kg/s at the given 
velocity, assuming the acceleration due to gravity is negligible. 

57. Derive the equation for the vertical acceleration of a rocket. 

58. Professional Application 

(a) Calculate the maximum rate at which a rocket can expel gases if its 
acceleration cannot exceed seven times that of gravity. The mass of the 
rocket just as it runs out of fuel is 75,000-kg, and its exhaust velocity is 

2.40X 10 m/s . Assume that the acceleration of gravity is the same as 
on Earth's surface (9.80 m/s . (b) Why might it be necessary to limit 
the acceleration of a rocket? 

59. Given the following data for a fire extinguisher-toy wagon rocket 
experiment, calculate the average exhaust velocity of the gases expelled 
from the extinguisher. Starting from rest, the final velocity is 10.0 m/s. 
The total mass is initially 75.0 kg and is 70.0 kg after the extinguisher is 
fired. 



CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 287 



60. How much of a single-stage rocket that is 100,000 kg can be anything 
but fuel if the rocket is to have a final speed of 8.00 km/s , given that it 

expels gases at an exhaust velocity of 2.20X 10 m/s? 

61. Professional Application 

(a) A 5.00-kg squid initially at rest ejects 0.250-kg of fluid with a velocity 
of 10.0 m/s. What is the recoil velocity of the squid if the ejection is done 
in 0.100 s and there is a 5.00-N frictional force opposing the squid's 
movement, (b) How much energy is lost to work done against friction? 

62. Unreasonable Results 

Squids have been reported to jump from the ocean and travel 30.0 m 
(measured horizontally) before re-entering the water, (a) Calculate the 
initial speed of the squid if it leaves the water at an angle of 20.0° , 
assuming negligible lift from the air and negligible air resistance, (b) The 
squid propels itself by squirting water. What fraction of its mass would it 
have to eject in order to achieve the speed found in the previous part? 
The water is ejected at 12.0 m/s ; gravitational force and friction are 
neglected, (c) What is unreasonable about the results? (d) Which 
premise is unreasonable, or which premises are inconsistent? 

63. Professional Application 

Consider an astronaut in deep space cut free from her space ship and 
needing to get back to it. The astronaut has a few packages that she can 
throw away to move herself toward the ship. Construct a problem in 
which you calculate the time it takes her to get back by throwing all the 
packages at one time compared to throwing them one at a time. Among 
the things to be considered are the masses involved, the force she can 
exert on the packages through some distance, and the distance to the 
ship. 

64. Consider an artillery projectile striking armor plating. Construct a 
problem in which you find the force exerted by the projectile on the plate. 
Among the things to be considered are the mass and speed of the 
projectile and the distance over which its speed is reduced. Your 
instructor may also wish for you to consider the relative merits of 
depleted uranium versus lead projectiles based on the greater density of 
uranium. 



288 CHAPTER 8 | LINEAR MOMENTUM AND COLLISIONS 



CHAPTER 9 | STATICS AND TORQUE 289 



STATICS AND TORQUE 




Figure 9.1 On a short time scale, rocks like these in Australia's Kings Canyon are static, or motionless relative to the Earth, (credit: freeaussiestock.com) 



Learning Objectives 



9.1. The First Condition for Equilibrium 

• State the first condition of equilibrium. 

• Explain static equilibrium. 

• Explain dynamic equilibrium. 

9.2. The Second Condition for Equilibrium 

• State the second condition that is necessary to achieve equilibrium. 

• Explain torque and the factors on which it depends. 

• Describe the role of torque in rotational mechanics. 

9.3. Stability 

• State the types of equilibrium. 

• Describe stable and unstable equilibriums. 

• Describe neutral equilibrium. 

9.4. Applications of Statics, Including Problem-Solving Strategies 

• Discuss the applications of Statics in real life. 

• State and discuss various problem-solving strategies in Statics. 

9.5. Simple Machines 

• Describe different simple machines. 

• Calculate the mechanical advantage. 

9.6. Forces and Torques in Muscles and Joints 

• Explain the forces exerted by muscles. 

• State how a bad posture causes back strain. 

• Discuss the benefits of skeletal muscles attached close to joints. 

• Discuss various complexities in the real system of muscles, bones, and joints. 



Introduction to Statics and Torque 

What might desks, bridges, buildings, trees, and mountains have in common — at least in the eyes of a physicist? The answer is that they are 
ordinarily motionless relative to the Earth. Furthermore, their acceleration is zero because they remain motionless. That means they also have 



290 CHAPTER 9 | STATICS AND TORQUE 

something in common with a car moving at a constant velocity, because anything with a constant velocity also has an acceleration of zero. Now, the 
important part — Newton's second law states that net F = ma , and so the net external force is zero for all stationary objects and for all objects 
moving at constant velocity. There are forces acting, but they are balanced. That is, they are in equilibrium. 

Statics 



Statics is the study of forces in equilibrium, a large group of situations that makes up a special case of Newton's second law. We have already 
considered a few such situations; in this chapter, we cover the topic more thoroughly, including consideration of such possible effects as the 
rotation and deformation of an object by the forces acting on it. 

How can we guarantee that a body is in equilibrium and what can we learn from systems that are in equilibrium? There are actually two conditions 
that must be satisfied to achieve equilibrium. These conditions are the topics of the first two sections of this chapter. 



9.1 The First Condition for Equilibrium 

The first condition necessary to achieve equilibrium is the one already mentioned: the net external force on the system must be zero. Expressed as 
an equation, this is simply 



net F = 



(9.1) 



Note that if net F is zero, then the net external force in any direction is zero. For example, the net external forces along the typical x- and y-axes are 
zero. This is written as 

(9.2) 



net F x = and F y = 



Figure 9.2 and Figure 9.3 illustrate situations where net F = for both static equilibrium (motionless), and dynamic equilibrium (constant 
velocity). 



net F = 




Free-body 
diagram 



Stationary t 

■ 



Figure 9.2 This motionless person is in static equilibrium. The forces acting on him add up to zero. Both forces are vertical in this case. 

Free-body 

diagram 

Ns 

tttt 



v constant 




f 



w 



Figure 9.3 This car is in dynamic equilibrium because it is moving at constant velocity. There are horizontal and vertical forces, but the net external force in any direction is 
zero. The applied force ^app between the tires and the road is balanced by air friction, and the weight of the car is supported by the normal forces, here shown to be equal 

for all four tires. 

However, it is not sufficient for the net external force of a system to be zero for a system to be in equilibrium. Consider the two situations illustrated in 
Figure 9.4 and Figure 9.5 where forces are applied to an ice hockey stick lying flat on ice. The net external force is zero in both situations shown in 
the figure; but in one case, equilibrium is achieved, whereas in the other, it is not. In Figure 9.4, the ice hockey stick remains motionless. But in 



CHAPTER 9 I STATICS AND TORQUE 291 

Figure 9.5, with the same forces applied in different places, the stick experiences accelerated rotation. Therefore, we know that the point at which a 
force is applied is another factor in determining whether or not equilibrium is achieved. This will be explored further in the next section. 



Equilibrium: remains stationary 



net F = 




Free-body diagram 



Figure 9.4 An ice hockey stick lying flat on ice with two equal and opposite horizontal forces applied to it. Friction is negligible, and the gravitational force is balanced by the 
support of the ice (a normal force). Thus, net F = . Equilibrium is achieved, which is static equilibrium in this case. 

Nonequilibrium: rotation accelerates 



net F = 




Free-body diagram 



Figure 9.5 The same forces are applied at other points and the stick rotates — in fact, it experiences an accelerated rotation. Here net F = but the system is not at 
equilibrium. Hence, the net F = is a necessary — but not sufficient — condition for achieving equilibrium. 

PhET Explorations: Torque 

Investigate how torque causes an object to rotate. Discover the relationships between angular acceleration, moment of inertia, angular 
momentum and torque. 

ct p PhET Interactive Simulation 

Figure 9.6 Torque (http://cnx.Org/content/m42170/l.4/torque_en.jar) 



9.2 The Second Condition for Equilibrium 



Torque 



The second condition necessary to achieve equilibrium involves avoiding accelerated rotation (maintaining a constant angular velocity. A rotating 
body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what 
factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges. 



Several familiar factors determine how effective you are in opening the door. See Figure 9.7. First of all, the larger the force, the more effective it is in 
opening the door — obviously, the harder you push, the more rapidly the door opens. Also, the point at which you push is crucial. If you apply your 
force too close to the hinges, the door will open slowly, if at all. Most people have been embarrassed by making this mistake and bumping up against 



292 CHAPTER 9 | STATICS AND TORQUE 



a door when it did not open as quickly as expected. Finally, the direction in which you push is also important. The most effective direction is 
perpendicular to the door — we push in this direction almost instinctively. 




r x = rslne 




6 = 0° 



Figure 9.7 Torque is the turning or twisting effectiveness of a force, illustrated here for door rotation on its hinges (as viewed from overhead). Torque has both magnitude and 
direction, (a) Counterclockwise torque is produced by this force, which means that the door will rotate in a counterclockwise due to F . Note that r_j_ is the perpendicular 

distance of the pivot from the line of action of the force, (b) A smaller counterclockwise torque is produced by a smaller force F acting at the same distance from the hinges 
(the pivot point), (c) The same force as in (a) produces a smaller counterclockwise torque when applied at a smaller distance from the hinges, (d) The same force as in (a), but 
acting in the opposite direction, produces a clockwise torque, (e) A smaller counterclockwise torque is produced by the same magnitude force acting at the same point but in a 
different direction. Here, 6 is less than 90° . (f) Torque is zero here since the force just pulls on the hinges, producing no rotation. In this case, 6 = 0° . 

The magnitude, direction, and point of application of the force are incorporated into the definition of the physical quantity called torque. Torque is the 
rotational equivalent of a force. It is a measure of the effectiveness of a force in changing or accelerating a rotation (changing the angular velocity 
over a period of time). In equation form, the magnitude of torque is defined to be 

t = rF sin 6 (9.3) 

where r (the Greek letter tau) is the symbol for torque, r is the distance from the pivot point to the point where the force is applied, F is the 
magnitude of the force, and 6 is the angle between the force and the vector directed from the point of application to the pivot point, as seen in 
Figure 9.7 and Figure 9.8. An alternative expression for torque is given in terms of the perpendicular lever arm r j_ as shown in Figure 9.7 and 

Figure 9.8, which is defined as 

r j_ = r sin 6 (9.4) 

so that 

r = r ± F. (9.5) 

Nail 





el 



Nail / / r 



(a) 

Figure 9.8 A force applied to an object can produce a torque, which depends on the location of the pivot point, (a) The three factors T , F , and 6 for pivot point A on a body 
are shown here — r is the distance from the chosen pivot point to the point where the force F is applied, and 6 is the angle between F and the vector directed from the 
point of application to the pivot point. If the object can rotate around point A, it will rotate counterclockwise. This means that torque is counterclockwise relative to pivot A. (b) In 
this case, point B is the pivot point. The torque from the applied force will cause a clockwise rotation around point B, and so it is a clockwise torque relative to B. 




CHAPTER 9 | STATICS AND TORQUE 293 

The perpendicular lever arm r j_ is the shortest distance from the pivot point to the line along which F acts; it is shown as a dashed line in Figure 
9.7 and Figure 9.8. Note that the line segment that defines the distance r j_ is perpendicular to F , as its name implies. It is sometimes easier to 
find or visualize r j_ than to find both r and 6 . In such cases, it may be more convenient to use t = r j_ F rather than t = rF sin 6 for torque, 
but both are equally valid. 

The SI unit of torque is newtons times meters, usually written as N • m . For example, if you push perpendicular to the door with a force of 40 N at a 
distance of 0.800 m from the hinges, you exert a torque of 32 N-m(0.800 mx40 Nxsin 90°) relative to the hinges. If you reduce the force to 20 N, 
the torque is reduced to 16 N-m , and so on. 

The torque is always calculated with reference to some chosen pivot point. For the same applied force, a different choice for the location of the pivot 
will give you a different value for the torque, since both r and 6 depend on the location of the pivot. Any point in any object can be chosen to 
calculate the torque about that point. The object may not actually pivot about the chosen "pivot point." 

Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point, 
as illustrated for points B and A, respectively, in Figure 9.8. If the object can rotate about point A, it will rotate counterclockwise, which means that the 
torque for the force is shown as counterclockwise relative to A. But if the object can rotate about point B, it will rotate clockwise, which means the 
torque for the force shown is clockwise relative to B. Also, the magnitude of the torque is greater when the lever arm is longer. 

Now, the second condition necessary to achieve equilibrium is that the net external torque on a system must be zero. An external torque is one that is 
created by an external force. You can choose the point around which the torque is calculated. The point can be the physical pivot point of a system or 
any other point in space — but it must be the same point for all torques. If the second condition (net external torque on a system is zero) is satisfied for 
one choice of pivot point, it will also hold true for any other choice of pivot point in or out of the system of interest. (This is true only in an inertial frame 
of reference.) The second condition necessary to achieve equilibrium is stated in equation form as 

net t = (9.6) 

where net means total. Torques, which are in opposite directions are assigned opposite signs. A common convention is to call counterclockwise (ccw) 
torques positive and clockwise (cw) torques negative. 

When two children balance a seesaw as shown in Figure 9.9, they satisfy the two conditions for equilibrium. Most people have perfect intuition about 
seesaws, knowing that the lighter child must sit farther from the pivot and that a heavier child can keep a lighter one off the ground indefinitely. 




Figure 9.9 Two children balancing a seesaw satisfy both conditions for equilibrium. The lighter child sits farther from the pivot to create a torque equal in magnitude to that of 
the heavier child. 



Example 9.1 She Saw Torques On A Seesaw 



The two children shown in Figure 9.9 are balanced on a seesaw of negligible mass. (This assumption is made to keep the example 

simple — more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot.(a) If the second child has a 

mass of 32.0 kg, how far is she from the pivot? (b) What is Fp , the supporting force exerted by the pivot? 

Strategy 

Both conditions for equilibrium must be satisfied. In part (a), we are asked for a distance; thus, the second condition (regarding torques) must be 
used, since the first (regarding only forces) has no distances in it. To apply the second condition for equilibrium, we first identify the system of 
interest to be the seesaw plus the two children. We take the supporting pivot to be the point about which the torques are calculated. We then 
identify all external forces acting on the system. 

Solution (a) 

The three external forces acting on the system are the weights of the two children and the supporting force of the pivot. Let us examine the 
torque produced by each. Torque is defined to be 

r = rF sin 6. (9.7) 



Here 6 = 90° , so that sin 6 = 1 for all three forces. That means r 



r for all three. The torques exerted by the three forces are first, 



294 CHAPTER 9 | STATICS AND TORQUE 

r 1 = r 1 w 1 (9.8) 

second, 

r 2 = - r 2 w 2 (9.9) 

and third, 

r p = r v F v (9.10) 

= 0-F p 

= 0. 

Note that a minus sign has been inserted into the second equation because this torque is clockwise and is therefore negative by convention. 
Since Fp acts directly on the pivot point, the distance r p is zero. A force acting on the pivot cannot cause a rotation, just as pushing directly on 

the hinges of a door will not cause it to rotate. Now, the second condition for equilibrium is that the sum of the torques on both children is zero. 
Therefore 

t 2 = -z h (9.11) 

or 

r 2 w 2 — r l w l- (9.12) 

Weight is mass times the acceleration due to gravity. Entering mg for w , we get 

r 2 m 2 g = r l m l g. (9.13) 

Solve this for the unknown r 2 : 

r - r 01± (9-14) 

r 2- r lm 2 

The quantities on the right side of the equation are known; thus, r 2 is 

n m , 26.0 kg (9.15) 

r 2 = (1.60 m) 32Qk * = 1.30 m. 

As expected, the heavier child must sit closer to the pivot (1.30 m versus 1.60 m) to balance the seesaw. 

Solution (b) 

This part asks for a force F p . The easiest way to find it is to use the first condition for equilibrium, which is 

netF = 0. (9.16) 

The forces are all vertical, so that we are dealing with a one-dimensional problem along the vertical axis; hence, the condition can be written as 

netF.y = (9.17) 

where we again call the vertical axis the y-axis. Choosing upward to be the positive direction, and using plus and minus signs to indicate the 
directions of the forces, we see that 

F v -w l -w 2 = 0. (9.18) 

This equation yields what might have been guessed at the beginning: 

F p = w l +w 2 - (9.19) 

So, the pivot supplies a supporting force equal to the total weight of the system: 

^p = M\g + M2g- ( 9 - 2 °) 

Entering known values gives 

F v = (26.0 kg)(9.80 m/s 2 ) + (32.0 kg)(9.80 m/s 2 ) < 9 - 21 ) 

= 568 N. 
Discussion 

The two results make intuitive sense. The heavier child sits closer to the pivot. The pivot supports the weight of the two children. Part (b) can also 
be solved using the second condition for equilibrium, since both distances are known, but only if the pivot point is chosen to be somewhere other 
than the location of the seesaw's actual pivot! 

Several aspects of the preceding example have broad implications. First, the choice of the pivot as the point around which torques are calculated 
simplified the problem. Since Fp is exerted on the pivot point, its lever arm is zero. Hence, the torque exerted by the supporting force F p is zero 

relative to that pivot point. The second condition for equilibrium holds for any choice of pivot point, and so we choose the pivot point to simplify the 
solution of the problem. 



CHAPTER 9 | STATICS AND TORQUE 295 

Second, the acceleration due to gravity canceled in this problem, and we were left with a ratio of masses. This will not always be the case. Always 
enter the correct forces — do not jump ahead to enter some ratio of masses. 

Third, the weight of each child is distributed over an area of the seesaw, yet we treated the weights as if each force were exerted at a single point. 
This is not an approximation — the distances r^ and r 2 are the distances to points directly below the center of gravity of each child. As we shall 

see in the next section, the mass and weight of a system can act as if they are located at a single point. 

Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays 
in linear motion. We will examine this in the next chapter. 

Take-Home Experiment 

Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder 
while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies? 

9.3 Stability 

It is one thing to have a system in equilibrium; it is quite another for it to be stable. The toy doll perched on the man's hand in Figure 9.10, for 
example, is not in stable equilibrium. There are three types of equilibrium: stable, unstable, and neutral. Figures throughout this module illustrate 
various examples. 

Figure 9.10 presents a balanced system, such as the toy doll on the man's hand, which has its center of gravity (eg) directly over the pivot, so that 
the torque of the total weight is zero. This is equivalent to having the torques of the individual parts balanced about the pivot point, in this case the 
hand. The cgs of the arms, legs, head, and torso are labeled with smaller type. 




Figure 9.10 A man balances a toy doll on one hand. 

A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite to the 
direction of the displacement. For example, a marble at the bottom of a bowl will experience a restoring force when displaced from its equilibrium 
position. This force moves it back toward the equilibrium position. Most systems are in stable equilibrium, especially for small displacements. For 
another example of stable equilibrium, see the pencil in Figure 9.11. 




w 



Pivot point 



Figure 9.11 This pencil is in the condition of equilibrium. The net force on the pencil is zero and the total torque about any pivot is zero. 

A system is in unstable equilibrium if, when displaced, it experiences a net force or torque in the same direction as the displacement from 
equilibrium. A system in unstable equilibrium accelerates away from its equilibrium position if displaced even slightly. An obvious example is a ball 
resting on top of a hill. Once displaced, it accelerates away from the crest. See the next several figures for examples of unstable equilibrium. 



296 CHAPTER 9 | STATICS AND TORQUE 



Pivot point 




■t 



Figure 9.12 If the pencil is displaced slightly to the side (counterclockwise), it is no longer in equilibrium. Its weight produces a clockwise torque that returns the pencil to its 
equilibrium position. 




Figure 9.13 If the pencil is displaced too far, the torque caused by its weight changes direction to counterclockwise and causes the displacement to increase. 



CG 

w 



■f 



Free-body 
diagram 

In 
|« 

Pivot point 



Figure 9.14 This figure shows unstable equilibrium, although both conditions for equilibrium are satisfied. 



CHAPTER 9 | STATICS AND TORQUE 297 




Figure 9.15 If the pencil is displaced even slightly, a torque is created by its weight that is in the same direction as the displacement, causing the displacement to increase. 

A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. A marble on a flat horizontal surface is 
an example. Combinations of these situations are possible. For example, a marble on a saddle is stable for displacements toward the front or back of 
the saddle and unstable for displacements to the side. Figure 9.16 shows another example of neutral equilibrium. 



T = 


Free-body 
diagram 

•n 


kc 


/, Pivot poin 


.♦ 




Neutral 
(a) 






Free-body 
diagram 


w r = 


In 

t- 



• CG 



Limited neutral 
(b) 

Figure 9.16 (a) Here we see neutral equilibrium. The eg of a sphere on a flat surface lies directly above the point of support, independent of the position on the surface. The 
sphere is therefore in equilibrium in any location, and if displaced, it will remain put. (b) Because it has a circular cross section, the pencil is in neutral equilibrium for 
displacements perpendicular to its length. 

When we consider how far a system in stable equilibrium can be displaced before it becomes unstable, we find that some systems in stable 
equilibrium are more stable than others. The pencil in Figure 9.11 and the person in Figure 9.17(a) are in stable equilibrium, but become unstable for 
relatively small displacements to the side. The critical point is reached when the eg is no longer above the base of support. Additionally, since the eg 
of a person's body is above the pivots in the hips, displacements must be quickly controlled. This control is a central nervous system function that is 
developed when we learn to hold our bodies erect as infants. For increased stability while standing, the feet should be spread apart, giving a larger 
base of support. Stability is also increased by lowering one's center of gravity by bending the knees, as when a football player prepares to receive a 
ball or braces themselves for a tackle. A cane, a crutch, or a walker increases the stability of the user, even more as the base of support widens. 
Usually, the eg of a female is lower (closer to the ground) than a male. Young children have their center of gravity between their shoulders, which 
increases the challenge of learning to walk. 



298 CHAPTER 9 | STATICS AND TORQUE 



Free-body 
diagram 

N||N 




(a) (b) 

Figure 9.17 (a) The center of gravity of an adult is above the hip joints (one of the main pivots in the body) and lies between two narrowly-separated feet. Like a pencil 
standing on its eraser, this person is in stable equilibrium in relation to sideways displacements, but relatively small displacements take his eg outside the base of support and 
make him unstable. Humans are less stable relative to forward and backward displacements because the feet are not very long. Muscles are used extensively to balance the 
body in the front-to-back direction, (b) While bending in the manner shown, stability is increased by lowering the center of gravity. Stability is also increased if the base is 
expanded by placing the feet farther apart. 

Animals such as chickens have easier systems to control. Figure 9.18 shows that the eg of a chicken lies below its hip joints and between its widely 
separated and broad feet. Even relatively large displacements of the chicken's eg are stable and result in restoring forces and torques that return the 
eg to its equilibrium position with little effort on the chicken's part. Not all birds are like chickens, of course. Some birds, such as the flamingo, have 
balance systems that are almost as sophisticated as that of humans. 

Figure 9.18 shows that the eg of a chicken is below the hip joints and lies above a broad base of support formed by widely-separated and large feet. 
Hence, the chicken is in very stable equilibrium, since a relatively large displacement is needed to render it unstable. The body of the chicken is 
supported from above by the hips and acts as a pendulum between the hips. Therefore, the chicken is stable for front-to-back displacements as well 
as for side-to-side displacements. 




Figure 9.18 The center of gravity of a chicken is below the hip joints. The chicken is in stable equilibrium. The body of the chicken is supported from above by the hips and 
acts as a pendulum between them. 

Engineers and architects strive to achieve extremely stable equilibriums for buildings and other systems that must withstand wind, earthquakes, and 
other forces that displace them from equilibrium. Although the examples in this section emphasize gravitational forces, the basic conditions for 
equilibrium are the same for all types of forces. The net external force must be zero, and the net torque must also be zero. 



Take-Home Experiment 

Stand straight with your heels, back, and head against a wall. Bend forward from your waist, keeping your heels and bottom against the wall, to 
touch your toes. Can you do this without toppling over? Explain why and what you need to do to be able to touch your toes without losing your 
balance. Is it easier for a woman to do this? 



9.4 Applications of Statics, Including Problem-Solving Strategies 

Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of 
problem-solving strategies specifically used for statics. Since statics is a special case of Newton's laws, both the general problem-solving strategies 
and the special strategies for Newton's laws, discussed in Problem-Solving Strategies, still apply. 



Problem-Solving Strategy: Static Equilibrium Situations 

1. The first step is to determine whether or not the system is in static equilibrium. This condition is always the case when the acceleration of 
the system is zero and accelerated rotation does not occur. 

2. It is particularly important to draw a free body diagram for the system of interest. Carefully label all forces, and note their relative 
magnitudes, directions, and points of application whenever these are known. 

3. Solve the problem by applying either or both of the conditions for equilibrium (represented by the equations net F = and net r = , 
depending on the list of known and unknown factors. If the second condition is involved, choose the pivot point to simplify the solution. Any 
pivot point can be chosen, but the most useful ones cause torques by unknown forces to be zero. (Torque is zero if the force is applied at 
the pivot (then r = ), or along a line through the pivot point (then 6 = )). Always choose a convenient coordinate system for projecting 
forces. 



CHAPTER 9 | STATICS AND TORQUE 299 

4. Check the solution to see if it is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step 
never diminishes, although in unfamiliar applications, it is usually more difficult to judge reasonableness. These judgments become 
progressively easier with experience. 

Now let us apply this problem-solving strategy for the pole vaulter shown in the three figures below. The pole is uniform and has a mass of 5.00 kg. In 
Figure 9.19, the pole's eg lies halfway between the vaulter's hands. It seems reasonable that the force exerted by each hand is equal to half the 
weight of the pole, or 24.5 N. This obviously satisfies the first condition for equilibrium (net F = 0) . The second condition (net r = 0) is also 

satisfied, as we can see by choosing the eg to be the pivot point. The weight exerts no torque about a pivot point located at the eg, since it is applied 
at that point and its lever arm is zero. The equal forces exerted by the hands are equidistant from the chosen pivot, and so they exert equal and 
opposite torques. Similar arguments hold for other systems where supporting forces are exerted symmetrically about the eg. For example, the four 
legs of a uniform table each support one-fourth of its weight. 

In Figure 9.19, a pole vaulter holding a pole with its eg halfway between his hands is shown. Each hand exerts a force equal to half the weight of the 
pole, F R = F L = w/2.(b) The pole vaulter moves the pole to his left, and the forces that the hands exert are no longer equal. See Figure 9.19. If 

the pole is held with its eg to the left of the person, then he must push down with his right hand and up with his left. The forces he exerts are larger 
here because they are in opposite directions and the eg is at a long distance from either hand. 

Similar observations can be made using a meter stick held at different locations along its length. 




F R 



F L 



Free-body 
diagram 



Figure 9.19 A pole vaulter holds a pole horizontally with both hands. 



F„ 



k 

• Free-body 
diagram 



Figure 9.20 A pole vaulter is holding a pole horizontally with both hands. The center of gravity is near his right hand 




Free-body 
diagram 



Figure 9.21 A pole vaulter is holding a pole horizontally with both hands. The center of gravity is to the left side of the vaulter. 

If the pole vaulter holds the pole as shown in Figure 9.19, the situation is not as simple. The total force he exerts is still equal to the weight of the 
pole, but it is not evenly divided between his hands. (If F L = F R , then the torques about the eg would not be equal since the lever arms are 



300 CHAPTER 9 | STATICS AND TORQUE 

different.) Logically, the right hand should support more weight, since it is closer to the eg. In fact, if the right hand is moved directly under the eg, it 
will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the eg carries more of its weight. Finding 
the forces F L and F R is straightforward, as the next example shows. 

If the pole vaulter holds the pole from near the end of the pole (Figure 9.21), the direction of the force applied by the right hand of the vaulter 
reverses its direction. 



Example 9.2 What Force Is Needed to Support a Weight Held Near Its CG? 



For the situation shown in Figure 9.19, calculate: (a) F R , the force exerted by the right hand, and (b) F L , the force exerted by the left hand. 

The hands are 0.900 m apart, and the eg of the pole is 0.600 m from the left hand. 

Strategy 

Figure 9.19 includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for 
equilibrium (net F = ), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is 

enough information to use the second condition for equilibrium (net r = 0) if the pivot point is chosen to be at either hand, thereby making the 

torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand. 

Solution for (a) 

There are now only two nonzero torques, those from the gravitational force ( t w ) and from the push or pull of the right hand ( t r ). Stating the 

second condition in terms of clockwise and counterclockwise torques, 

net r cw = -net r ccw . (9.22) 

or the algebraic sum of the torques is zero. 
Here this is 

r R = -t w (9.23) 

since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise toque. Using the definition of torque, 
i = rF sin 6 , noting that 6 = 90° , and substituting known values, we obtain 

(0.900 m)(F R ) = (0.600 m)(mg). (9- 24 ) 

Thus, 

F R = (0.667)(5.00kg)(9.80m/s 2 ) < 9 - 25 ) 

= 32.7 N. 
Solution for (b) 
The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton's second law: 

F L + F R - mg = (9.26) 

From this we can conclude: 

F L + F R = w = mg (9.27) 

Solving for F L , we obtain 

F L = mg-F R (9.28) 

= mg-32.7N 

= (5.00 kg)(9.80 m/s 2 ) - 32.7 N 

= 16.3 N 
Discussion 
F L is seen to be exactly half of F R , as we might have guessed, since F L is applied twice as far from the eg as F R . 

If the pole vaulter holds the pole as he might at the start of a run, shown in Figure 9.21, the forces change again. Both are considerably greater, and 
one force reverses direction. 

Take-Home Experiment 

This is an experiment to perform while standing in a bus or a train. Stand facing sideways. How do you move your body to readjust the 
distribution of your mass as the bus accelerates and decelerates? Now stand facing forward. How do you move your body to readjust the 
distribution of your mass as the bus accelerates and decelerates? Why is it easier and safer to stand facing sideways rather than forward? Note: 
For your safety (and those around you), make sure you are holding onto something while you carry out this activity! 



CHAPTER 9 | STATICS AND TORQUE 301 



PhET Explorations: Balancing Act 



Play with objects on a teeter totter to learn about balance. Test what you've learned by trying the Balance Challenge game. 



Jj r^PhET Interactive Simulation 



9 



'j 



Figure 9.22 Balancing Act (http://phet.colorado.edu/en/simulation/balancing-act) 



9.5 Simple Machines 

Simple machines are devices that can be used to multiply or augment a force that we apply - often at the expense of a distance through which we 
apply the force. The word for "machine" comes from the Greek word meaning "to help make things easier." Levers, gears, pulleys, wedges, and 
screws are some examples of machines. Energy is still conserved for these devices because a machine cannot do more work than the energy put 
into it. However, machines can reduce the input force that is needed to perform the job. The ratio of output to input force magnitudes for any simple 
machine is called its mechanical advantage (MA). 






(9.29) 



One of the simplest machines is the lever, which is a rigid bar pivoted at a fixed place called the fulcrum. Torques are involved in levers, since there is 
rotation about a pivot point. Distances from the physical pivot of the lever are crucial, and we can obtain a useful expression for the MA in terms of 
these distances. 



Free-body diagram 




Figure 9.23 A nail puller is a lever with a large mechanical advantage. The external forces on the nail puller are represented by solid arrows. The force that the nail puller 
applies to the nail ( F ) is not a force on the nail puller. The reaction force the nail exerts back on the puller ( F n ) is an external force and is equal and opposite to F . The 

perpendicular lever arms of the input and output forces are l x and /q . 

Figure 9.23 shows a lever type that is used as a nail puller. Crowbars, seesaws, and other such levers are all analogous to this one. Fj is the input 
force and F is the output force. There are three vertical forces acting on the nail puller (the system of interest) - these are Fj, F , and N . F n is 
the reaction force back on the system, equal and opposite to F . (Note that F is not a force on the system.) N is the normal force upon the lever, 
and its torque is zero since it is exerted at the pivot. The torques due to Fj and F n must be equal to each other if the nail is not moving, to satisfy 
the second condition for equilibrium (net t = 0) . (In order for the nail to actually move, the torque due to Fj must be ever-so-slightly greater than 
torque due to F n .) Hence, 



l x b j — l r 



(9.30) 



where l x and l are the distances from where the input and output forces are applied to the pivot, as shown in the figure. Rearranging the last 
equation gives 

Fo h (9-31) 

What interests us most here is that the magnitude of the force exerted by the nail puller, F , is much greater than the magnitude of the input force 
applied to the puller at the other end, F x . For the nail puller, 



302 CHAPTER 9 | STATICS AND TORQUE 



F I- 
MA = ^ = f . 



(9.32) 



This equation is true for levers in general. For the nail puller, the MA is certainly greater than one. The longer the handle on the nail puller, the greater 
the force you can exert with it. 

Two other types of levers that differ slightly from the nail puller are a wheelbarrow and a shovel, shown in Figure 9.24. All these lever types are 
similar in that only three forces are involved - the input force, the output force, and the force on the pivot - and thus their MAs are given by 

F di 

MA = -=?■ and MA = -f- , with distances being measured relative to the physical pivot. The wheelbarrow and shovel differ from the nail puller 
r i «2 

because both the input and output forces are on the same side of the pivot. 

In the case of the wheelbarrow, the output force or load is between the pivot (the wheel's axle) and the input or applied force. In the case of the 
shovel, the input force is between the pivot (at the end of the handle) and the load, but the input lever arm is shorter than the output lever arm. In this 
case, the MA is less than one. 



Free -body 
diagram 




Free -body 
F ft diagram 



Figure 9.24 (a) In the case of the wheelbarrow, the output force or load is between the pivot and the input force. The pivot is the wheel's axle. Here, the output force is greater 
than the input force. Thus, a wheelbarrow enables you to lift much heavier loads than you could with your body alone, (b) In the case of the shovel, the input force is between 
the pivot and the load, but the input lever arm is shorter than the output lever arm. The pivot is at the handle held by the right hand. Here, the output force (supporting the 
shovel's load) is less than the input force (from the hand nearest the load), because the input is exerted closer to the pivot than is the output. 



Example 9.3 What is the Advantage for the Wheelbarrow': 



In the wheelbarrow of Figure 9.24, the load has a perpendicular lever arm of 7.50 cm, while the hands have a perpendicular lever arm of 1.02 m. 
(a) What upward force must you exert to support the wheelbarrow and its load if their combined mass is 45.0 kg? (b) What force does the 
wheelbarrow exert on the ground? 

Strategy 

Here, we use the concept of mechanical advantage. 
Solution 



(a) In this case, 



becomes 



F=F { 



^o 



Adding values into this equation yields 



F: = (45.0 kg)(9.80 mis 2 ) :™ m = 32.4 N. 
1 v 0/ v / 1.02 m 



(9.33) 



(9.34) 



The free-body diagram (see Figure 9.24) gives the following normal force: F^ + N = W . Therefore, 

TV = (45.0 kg)(9.80 m/s J - 32.4 N = 409 N . N is the normal force acting on the wheel; by Newton's third law, the force the wheel exerts 

on the ground is 409 N . 



CHAPTER 9 | STATICS AND TORQUE 303 



Discussion 

An even longer handle would reduce the force needed to lift the load. The MA here is MA = 1.02/0.0750 = 13.6 . 



Another very simple machine is the inclined plane. Pushing a cart up a plane is easier than lifting the same cart straight up to the top using a ladder, 
because the applied force is less. However, the work done in both cases (assuming the work done by friction is negligible) is the same. Inclined lanes 
or ramps were probably used during the construction of the Egyptian pyramids to move large blocks of stone to the top. 

A crank is a lever that can be rotated 360° about its pivot, as shown in Figure 9.25. Such a machine may not look like a lever, but the physics of its 
actions remain the same. The MA for a crank is simply the ratio of the radii rj r . Wheels and gears have this simple expression for their MAs too. 

The MA can be greater than 1, as it is for the crank, or less than 1, as it is for the simplified car axle driving the wheels, as shown. If the axle's radius 
is 2.0 cm and the wheel's radius is 24.0 cm , then MA = 2.0/24.0 = 0.083 and the axle would have to exert a force of 12,000 N on the 

wheel to enable it to exert a force of 1000 N on the ground. 



o 




(a) 




(b) ^F, 




(c) 

Figure 9.25 (a) A crank is a type of lever that can be rotated 360° about its pivot. Cranks are usually designed to have a large MA. (b) A simplified automobile axle drives a 
wheel, which has a much larger diameter than the axle. The MA is less than 1. (c) An ordinary pulley is used to lift a heavy load. The pulley changes the direction of the force 
T exerted by the cord without changing its magnitude. Hence, this machine has an MA of 1. 

An ordinary pulley has an MA of 1; it only changes the direction of the force and not its magnitude. Combinations of pulleys, such as those illustrated 
in Figure 9.26, are used to multiply force. If the pulleys are friction-free, then the force output is approximately an integral multiple of the tension in 
the cable. The number of cables pulling directly upward on the system of interest, as illustrated in the figures given below, is approximately the MA of 
the pulley system. Since each attachment applies an external force in approximately the same direction as the others, they add, producing a total 
force that is nearly an integral multiple of the input force T . 



304 CHAPTER 9 | STATICS AND TORQUE 





MA= 3 



TTT 



. is in 




(a) (b) (c) 

Figure 9.26 (a) The combination of pulleys is used to multiply force. The force is an integral multiple of tension if the pulleys are frictionless. This pulley system has two cables 
attached to its load, thus applying a force of approximately 2T . This machine has MA « 2 . (b) Three pulleys are used to lift a load in such a way that the mechanical 
advantage is about 3. Effectively, there are three cables attached to the load, (c) This pulley system applies a force of AT , so that it has MA « 4 . Effectively, four cables 
are pulling on the system of interest. 



9.6 Forces and Torques in Muscles and Joints 

Muscles, bones, and joints are some of the most interesting applications of statics. There are some surprises. Muscles, for example, exert far greater 
forces than we might think. Figure 9.27 shows a forearm holding a book and a schematic diagram of an analogous lever system. The schematic is a 
good approximation for the forearm, which looks more complicated than it is, and we can get some insight into the way typical muscle systems 
function by analyzing it. 

Muscles can only contract, so they occur in pairs. In the arm, the biceps muscle is a flexor — that is, it closes the limb. The triceps muscle is an 
extensor that opens the limb. This configuration is typical of skeletal muscles, bones, and joints in humans and other vertebrates. Most skeletal 
muscles exert much larger forces within the body than the limbs apply to the outside world. The reason is clear once we realize that most muscles 
are attached to bones via tendons close to joints, causing these systems to have mechanical advantages much less than one. Viewing them as 
simple machines, the input force is much greater than the output force, as seen in Figure 9.27. 



CHAPTER 9 | STATICS AND TORQUE 305 



Triceps 
muscle 



CG 



CG 



-J" 



W b 



Free -body 

diagram 




(b) 

Figure 9.27 (a) The figure shows the forearm of a person holding a book. The biceps exert a force Fg to support the weight of the forearm and the book. The triceps are 
assumed to be relaxed, (b) Here, you can view an approximately equivalent mechanical system with the pivot at the elbow joint as seen in Example 9.4. 



Example 9.4 Muscles Exert Bigger Forces Than You Might Think 



Calculate the force the biceps muscle must exert to hold the forearm and its load as shown in Figure 9.27, and compare this force with the 
weight of the forearm plus its load. You may take the data in the figure to be accurate to three significant figures. 

Strategy 

There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is F B ; that of the 
elbow joint is F E ; that of the weights of the forearm is w a , and its load is w b . Two of these are unknown ( F B and F E ), so that the first 
condition for equilibrium cannot by itself yield F B . But if we use the second condition and choose the pivot to be at the elbow, then the torque 
due to F E is zero, and the only unknown becomes F B . 

Solution 

The torques created by the weights are clockwise relative to the pivot, while the torque created by the biceps is counterclockwise; thus, the 
second condition for equilibrium (net t = 0) becomes 



r 2 w 3L + r 3 w h = r 1 F B . 

Note that sin 6 = 1 for all forces, since 6 = 90° for all forces. This equation can easily be solved for F B in terms of known quantities, 
yielding 



_ r 2 w a + r 3 w b 

^b- r x • 



Entering the known values gives 



^B = 



(0.160 m)(2.50 kg)(9.80 m/s 2 ) + (0.380 m)(4.00 kg)(9.80 m/s 2 ) 



(9.35) 



(9.36) 



(9.37) 



0.0400 m 



306 CHAPTER 9 | STATICS AND TORQUE 



which yields 




F B = 470 N. 












(9.38) 


Now, the combined weight of the arm 


and its load is (6.50 kg)(9.80 m/s 2 ) = 


:63.7N, 


so that the ratio of the force exerted 


by 


the 


biceps 


to the 


total weight is 




F B _ 470 _ 


7.38. 










(9.39) 


Wa + ^b 63.7 


Discussion 


















This means that the biceps muscle is 


exerting 


a force 7.38 times the weight supported. 













In the above example of the biceps muscle, the angle between the forearm and upper arm is 90°. If this angle changes, the force exerted by the 
biceps muscle also changes. In addition, the length of the biceps muscle changes. The force the biceps muscle can exert depends upon its length; it 
is smaller when it is shorter than when it is stretched. 

Very large forces are also created in the joints. In the previous example, the downward force F E exerted by the humerus at the elbow joint equals 

407 N, or 6.38 times the total weight supported. (The calculation of F E is straightforward and is left as an end-of-chapter problem.) Because 

muscles can contract, but not expand beyond their resting length, joints and muscles often exert forces that act in opposite directions and thus 
subtract. (In the above example, the upward force of the muscle minus the downward force of the joint equals the weight supported — that is, 
470 N - 407 N = 63 N , approximately equal to the weight supported.) Forces in muscles and joints are largest when their load is a long distance 
from the joint, as the book is in the previous example. 

In racquet sports such as tennis the constant extension of the arm during game play creates large forces in this way. The mass times the lever arm of 
a tennis racquet is an important factor, and many players use the heaviest racquet they can handle. It is no wonder that joint deterioration and 
damage to the tendons in the elbow, such as "tennis elbow," can result from repetitive motion, undue torques, and possibly poor racquet selection in 
such sports. Various tried techniques for holding and using a racquet or bat or stick not only increases sporting prowess but can minimize fatigue and 
long-term damage to the body. For example, tennis balls correctly hit at the "sweet spot" on the racquet will result in little vibration or impact force 
being felt in the racquet and the body — less torque as explained in Collisions of Extended Bodies in Two Dimensions. Twisting the hand to 
provide top spin on the ball or using an extended rigid elbow in a backhand stroke can also aggravate the tendons in the elbow. 

Training coaches and physical therapists use the knowledge of relationships between forces and torques in the treatment of muscles and joints. In 
physical therapy, an exercise routine can apply a particular force and torque which can, over a period of time, revive muscles and joints. Some 
exercises are designed to be carried out under water, because this requires greater forces to be exerted, further strengthening muscles. However, 
connecting tissues in the limbs, such as tendons and cartilage as well as joints are sometimes damaged by the large forces they carry. Often, this is 
due to accidents, but heavily muscled athletes, such as weightlifters, can tear muscles and connecting tissue through effort alone. 

The back is considerably more complicated than the arm or leg, with various muscles and joints between vertebrae, all having mechanical 
advantages less than 1. Back muscles must, therefore, exert very large forces, which are borne by the spinal column. Discs crushed by mere exertion 
are very common. The jaw is somewhat exceptional — the masseter muscles that close the jaw have a mechanical advantage greater than 1 for the 
back teeth, allowing us to exert very large forces with them. A cause of stress headaches is persistent clenching of teeth where the sustained large 
force translates into fatigue in muscles around the skull. 

Figure 9.28 shows how bad posture causes back strain. In part (a), we see a person with good posture. Note that her upper body's eg is directly 
above the pivot point in the hips, which in turn is directly above the base of support at her feet. Because of this, her upper body's weight exerts no 
torque about the hips. The only force needed is a vertical force at the hips equal to the weight supported. No muscle action is required, since the 
bones are rigid and transmit this force from the floor. This is a position of unstable equilibrium, but only small forces are needed to bring the upper 
body back to vertical if it is slightly displaced. Bad posture is shown in part (b); we see that the upper body's eg is in front of the pivot in the hips. This 
creates a clockwise torque around the hips that is counteracted by muscles in the lower back. These muscles must exert large forces, since they 
have typically small mechanical advantages. (In other words, the perpendicular lever arm for the muscles is much smaller than for the eg.) Poor 
posture can also cause muscle strain for people sitting at their desks using computers. Special chairs are available that allow the body's CG to be 
more easily situated above the seat, to reduce back pain. Prolonged muscle action produces muscle strain. Note that the eg of the entire body is still 
directly above the base of support in part (b) of Figure 9.28. This is compulsory; otherwise the person would not be in equilibrium. We lean forward 
for the same reason when carrying a load on our backs, to the side when carrying a load in one arm, and backward when carrying a load in front of 
us, as seen in Figure 9.29. 



CHAPTER 9 | STATICS AND TORQUE 307 



Upper body 
pivot point" 







(a) (b) 

Figure 9.28 (a) Good posture places the upper body's eg over the pivots in the hips, eliminating the need for muscle action to balance the body, (b) Poor posture requires 
exertion by the back muscles to counteract the clockwise torque produced around the pivot by the upper body's weight. The back muscles have a small effective perpendicular 
lever arm, r^ j_ , and must therefore exert a large force F^ . Note that the legs lean backward to keep the eg of the entire body above the base of support in the feet. 

You have probably been warned against lifting objects with your back. This action, even more than bad posture, can cause muscle strain and 
damage discs and vertebrae, since abnormally large forces are created in the back muscles and spine. 




(a) 



(b) 



(c) 



Figure 9.29 People adjust their stance to maintain balance, (a) A father carrying his son piggyback leans forward to position their overall eg above the base of support at his 
feet, (b) A student carrying a shoulder bag leans to the side to keep the overall eg over his feet, (c) Another student carrying a load of books in her arms leans backward for the 
same reason. 



Example 9.5 Do Not Lift with Your Back 



Consider the person lifting a heavy box with his back, shown in Figure 9.30. (a) Calculate the magnitude of the force F B - in the back muscles 

that is needed to support the upper body plus the box and compare this with his weight. The mass of the upper body is 55.0 kg and the mass of 
the box is 30.0 kg. (b) Calculate the magnitude and direction of the force Fy - exerted by the vertebrae on the spine at the indicated pivot 

point. Again, data in the figure may be taken to be accurate to three significant figures. 

Strategy 

By now, we sense that the second condition for equilibrium is a good place to start, and inspection of the known values confirms that it can be 
used to solve for F B - if the pivot is chosen to be at the hips. The torques created by w ub and w box - are clockwise, while that created by 



1 B 



is counterclockwise. 



Solution for (a) 

Using the perpendicular lever arms given in the figure, the second condition for equilibrium (net r = 0) becomes 

(0.350 m)(55.0 kg)(9.80 m/s 2 ) + (0.500 m)(30.0 kg)(9.80 m/s 2 ) = (0.0800 m)F B . 
Solving for F B yields 

F B = 4.20x10 s N. 
The ratio of the force the back muscles exert to the weight of the upper body plus its load is 



(9.40) 



(9.41) 



308 CHAPTER 9 | STATICS AND TORQUE 



1b = 4200 N = 5 04 



(9.42) 



w ub + w box 833 N 
This force is considerably larger than it would be if the load were not present. 
Solution for (b) 

More important in terms of its damage potential is the force on the vertebrae Fy . The first condition for equilibrium ( net F = ) can be used to 
find its magnitude and direction. Using y for vertical and x for horizontal, the condition for the net external forces along those axes to be zero 

net F y = and net F x = 0. (9.43) 

Starting with the vertical ( y ) components, this yields 

^Vy " ^ub " "W " ^B sin 29.0° = 0. (9.44) 



Thus, 



yielding 



Similarly, for the horizontal (x) components, 



yielding 



Fy y = w ub + w b0X + F B sin 29.0° (9.45) 

= 833 N + (4200 N) sin 29.0° 



F V}7 = 2.87xl0 3 N. < 9 - 46 ) 



F Yx - F B cos 29.0° = (9.47) 



F Vjc = 3.67xl0 3 N. ( 9 - 48 > 



The magnitude of Fy is given by the Pythagorean theorem: 



F V = i F lx + F ly = 4 ' 66 >< 103 N - (9 ' 49) 



The direction of Fy is 

6 = tan 
Note that the ratio of Fy to the weight supported is 



■(£)-»* 



^V = 4660 N = c eg 
^ub + ^box 833 N J ' Jy ' 



(9.50) 



(9.51) 



Discussion 

This force is about 5.6 times greater than it would be if the person were standing erect. The trouble with the back is not so much that the forces 
are large — because similar forces are created in our hips, knees, and ankles — but that our spines are relatively weak. Proper lifting, performed 
with the back erect and using the legs to raise the body and load, creates much smaller forces in the back — in this case, about 5.6 times smaller. 



CHAPTER 9 | STATICS AND TORQUE 309 



8 cm 




Pivot 
point 



Figure 9.30 This figure shows that large forces are exerted by the back muscles and experienced in the vertebrae when a person lifts with their back, since these muscles 
have small effective perpendicular lever arms. The data shown here are analyzed in the preceding example, Example 9.5. 

What are the benefits of having most skeletal muscles attached so close to joints? One advantage is speed because small muscle contractions can 
produce large movements of limbs in a short period of time. Other advantages are flexibility and agility, made possible by the large numbers of joints 
and the ranges over which they function. For example, it is difficult to imagine a system with biceps muscles attached at the wrist that would be 
capable of the broad range of movement we vertebrates possess. 

There are some interesting complexities in real systems of muscles, bones, and joints. For instance, the pivot point in many joints changes location 
as the joint is flexed, so that the perpendicular lever arms and the mechanical advantage of the system change, too. Thus the force the biceps 
muscle must exert to hold up a book varies as the forearm is flexed. Similar mechanisms operate in the legs, which explain, for example, why there is 
less leg strain when a bicycle seat is set at the proper height. The methods employed in this section give a reasonable description of real systems 
provided enough is known about the dimensions of the system. There are many other interesting examples of force and torque in the body — a few of 
these are the subject of end-of-chapter problems. 



Glossary 



center of gravity: the point where the total weight of the body is assumed to be concentrated 

dynamic equilibrium: a state of equilibrium in which the net external force and torque on a system moving with constant velocity are zero 

mechanical advantage: the ratio of output to input forces for any simple machine 

neutral equilibrium: a state of equilibrium that is independent of a system's displacements from its original position 

perpendicular lever arm: the shortest distance from the pivot point to the line along which F lies 

SI units of torque: newton times meters, usually written as N-m 

stable equilibrium: a system, when displaced, experiences a net force or torque in a direction opposite to the direction of the displacement 

static equilibrium: a state of equilibrium in which the net external force and torque acting on a system is zero 

static equilibrium: equilibrium in which the acceleration of the system is zero and accelerated rotation does not occur 

torque: turning or twisting effectiveness of a force 

unstable equilibrium: a system, when displaced, experiences a net force or torque in the same direction as the displacement from equilibrium 



Section Summary 



9.1 The First Condition for Equilibrium 

• Statics is the study of forces in equilibrium. 

• Two conditions must be met to achieve equilibrium, which is defined to be motion without linear or rotational acceleration. 

• The first condition necessary to achieve equilibrium is that the net external force on the system must be zero, so that net F = . 

9.2 The Second Condition for Equilibrium 

• The second condition assures those torques are also balanced. Torque is the rotational equivalent of a force in producing a rotation and is 
defined to be 

t = rF sin 6 



310 CHAPTER 9 | STATICS AND TORQUE 

where i is torque, r is the distance from the pivot point to the point where the force is applied, F is the magnitude of the force, and 6 is the 
angle between F and the vector directed from the point where the force acts to the pivot point. The perpendicular lever arm r_|_ is defined to 

be 

r_|_ = r sin 6 

so that 

r = r_|_ F. 

• The perpendicular lever arm r_|_ is the shortest distance from the pivot point to the line along which F acts. The SI unit for torque is newton- 

meter (N-m) . The second condition necessary to achieve equilibrium is that the net external torque on a system must be zero: 

net t = 
By convention, clockwise torques are positive, and counterclockwise torques are negative. 

9.3 Stability 

• A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite the 
direction of the displacement. 

• A system is in unstable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in the same direction as the 
displacement from equilibrium. 

• A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. 

9.4 Applications of Statics, Including Problem-Solving Strategies 

• Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We have discussed the 
problem-solving strategies specifically useful for statics. Statics is a special case of Newton's laws, both the general problem-solving strategies 
and the special strategies for Newton's laws, discussed in Problem-Solving Strategies, still apply. 

9.5 Simple Machines 

• Simple machines are devices that can be used to multiply or augment a force that we apply - often at the expense of a distance through which 
we have to apply the force. 

• The ratio of output to input forces for any simple machine is called its mechanical advantage 

• A few simple machines are the lever, nail puller, wheelbarrow, crank, etc. 

9.6 Forces and Torques in Muscles and Joints 

• Statics plays an important part in understanding everyday strains in our muscles and bones. 

• Many lever systems in the body have a mechanical advantage of significantly less than one, as many of our muscles are attached close to 
joints. 

• Someone with good posture stands or sits in such as way that their center of gravity lies directly above the pivot point in their hips, thereby 
avoiding back strain and damage to disks. 



Conceptual Questions 



9.1 The First Condition for Equilibrium 

1. What can you say about the velocity of a moving body that is in dynamic equilibrium? Draw a sketch of such a body using clearly labeled arrows to 
represent all external forces on the body. 

2. Under what conditions can a rotating body be in equilibrium? Give an example. 

9.2 The Second Condition for Equilibrium 

3. What three factors affect the torque created by a force relative to a specific pivot point? 

4. A wrecking ball is being used to knock down a building. One tall unsupported concrete wall remains standing. If the wrecking ball hits the wall near 
the top, is the wall more likely to fall over by rotating at its base or by falling straight down? Explain your answer. How is it most likely to fall if it is 
struck with the same force at its base? Note that this depends on how firmly the wall is attached at its base. 

5. Mechanics sometimes put a length of pipe over the handle of a wrench when trying to remove a very tight bolt. How does this help? (It is also 
hazardous since it can break the bolt.) 

9.3 Stability 

6. A round pencil lying on its side as in Figure 9.13 is in neutral equilibrium relative to displacements perpendicular to its length. What is its stability 
relative to displacements parallel to its length? 

7. Explain the need for tall towers on a suspension bridge to ensure stable equilibrium. 

9.4 Applications of Statics, Including Problem-Solving Strategies 

8. When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load needs to be directly 
above the person's neck vertebrae. 

9.5 Simple Machines 

9. Scissors are like a double-lever system. Which of the simple machines in Figure 9.23 and Figure 9.24 is most analogous to scissors? 

10. Suppose you pull a nail at a constant rate using a nail puller as shown in Figure 9.23. Is the nail puller in equilibrium? What if you pull the nail 
with some acceleration - is the nail puller in equilibrium then? In which case is the force applied to the nail puller larger and why? 



CHAPTER 9 | STATICS AND TORQUE 311 

11. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the 
body? 

12. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be 
even greater than muscle forces (see previous Question)? 

9.6 Forces and Torques in Muscles and Joints 

13. Why are the forces exerted on the outside world by the limbs of our bodies usually much smaller than the forces exerted by muscles inside the 
body? 

14. Explain why the forces in our joints are several times larger than the forces we exert on the outside world with our limbs. Can these forces be 
even greater than muscle forces? 

15. Certain types of dinosaurs were bipedal (walked on two legs). What is a good reason that these creatures invariably had long tails if they had long 
necks? 

16. Swimmers and athletes during competition need to go through certain postures at the beginning of the race. Consider the balance of the person 
and why start-offs are so important for races. 

17. If the maximum force the biceps muscle can exert is 1000 N, can we pick up an object that weighs 1000 N? Explain your answer. 

18. Suppose the biceps muscle was attached through tendons to the upper arm close to the elbow and the forearm near the wrist. What would be the 
advantages and disadvantages of this type of construction for the motion of the arm? 

19. Explain one of the reasons why pregnant women often suffer from back strain late in their pregnancy. 



312 CHAPTER 9 | STATICS AND TORQUE 



Problems & Exercises 



9.2 The Second Condition for Equilibrium 

1. (a) When opening a door, you push on it perpendicularly with a force of 
55.0 N at a distance of 0.850m from the hinges. What torque are you 
exerting relative to the hinges? (b) Does it matter if you push at the same 
height as the hinges? 

2. When tightening a bolt, you push perpendicularly on a wrench with a 
force of 165 N at a distance of 0.140 m from the center of the bolt, (a) 
How much torque are you exerting in newton x meters (relative to the 
center of the bolt)? (b) Convert this torque to footpounds. 

3. Two children push on opposite sides of a door during play. Both push 
horizontally and perpendicular to the door. One child pushes with a force 
of 17.5 N at a distance of 0.600 m from the hinges, and the second child 
pushes at a distance of 0.450 m. What force must the second child exert 
to keep the door from moving? Assume friction is negligible. 

4. Use the second condition for equilibrium (net x = 0) to calculate F p Figure 9.32 



10. A 17.0-m-high and 11.0-m-long wall under construction and its 
bracing are shown in Figure 9.32. The wall is in stable equilibrium 
without the bracing but can pivot at its base. Calculate the force exerted 
by each of the 10 braces if a strong wind exerts a horizontal force of 650 
N on each square meter of the wall. Assume that the net force from the 
wind acts at a height halfway up the wall and that all braces exert equal 
forces parallel to their lengths. Neglect the thickness of the wall. 




17.0 m 



10 braces 



in Example 9.1, employing any data given or solved for in part (a) of the 
example. 

5. Repeat the seesaw problem in Example 9.1 with the center of mass of 
the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) 
and assuming a mass of 12.0 kg for the seesaw. The other data given in 
the example remain unchanged. Explicitly show how you follow the steps 
in the Problem-Solving Strategy for static equilibrium. 

9.3 Stability 

6. Suppose a horse leans against a wall as in Figure 9.31. Calculate the 
force exerted on the wall assuming that force is horizontal while using the 
data in the schematic representation of the situation. Note that the force 
exerted on the wall is equal and opposite to the force exerted on the 
horse, keeping it in equilibrium. The total mass of the horse and rider is 
500 kg. Take the data to be accurate to three digits. 




1.4 m 



Figure 9.31 

7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with 
the pivot point located at the center of the seesaw. If the children are 
separated by a distance of 3 m, at what distance from the pivot point is 
the small child sitting in order to maintain the balance? 

8. (a) Calculate the magnitude and direction of the force on each foot of 
the horse in Figure 9.31 (two are on the ground), assuming the center of 
mass of the horse is midway between the feet. The total mass of the 
horse and rider is 500kg. (b) What is the minimum coefficient of friction 
between the hooves and ground? Note that the force exerted by the wall 
is horizontal. 

9. A person carries a plank of wood 2 m long with one hand pushing 
down on it at one end with a force F 1 and the other hand holding it up at 

50 cm from the end of the plank with force F 2 . If the plank has a mass 

of 20 kg and its center of gravity is at the middle of the plank, what are 
the forces F 1 and F 2 ? 



11. (a) What force must be exerted by the wind to support a 2.50-kg 
chicken in the position shown in Figure 9.33? (b) What is the ratio of this 
force to the chicken's weight? (c) Does this support the contention that 
the chicken has a relatively stable construction? 




9.0 cm 

Figure 9.33 

12. Suppose the weight of the drawbridge in Figure 9.34 is supported 
entirely by its hinges and the opposite shore, so that its cables are slack, 
(a) What fraction of the weight is supported by the opposite shore if the 
point of support is directly beneath the cable attachments? (b) What is 
the direction and magnitude of the force the hinges exert on the bridge 
under these circumstances? The mass of the bridge is 2500 kg. 




Figure 9.34 A small drawbridge, showing the forces on the hinges ( F ), its weight ( 
W ), and the tension in its wires ( T ). 

13. Suppose a 900-kg car is on the bridge in Figure 9.34 with its center 
of mass halfway between the hinges and the cable attachments. (The 
bridge is supported by the cables and hinges only.) (a) Find the force in 
the cables, (b) Find the direction and magnitude of the force exerted by 
the hinges on the bridge. 

14. A sandwich board advertising sign is constructed as shown in Figure 
9.35. The sign's mass is 8.00 kg. (a) Calculate the tension in the chain 



CHAPTER 9 | STATICS AND TORQUE 313 



assuming no friction between the legs and the sidewalk, (b) What force is 
exerted by each side on the hinge? 

Hinge -^— H^- - - - --r r 



0.50 m 



Chain 



i • 



i 



1.30 m 



CG CG 



* Uniform board 

(CG at center) 

I* 1.10 m Z *T 

Figure 9.35 A sandwich board advertising sign demonstrates tension. 

15. (a) What minimum coefficient of friction is needed between the legs 
and the ground to keep the sign in Figure 9.35 in the position shown if 
the chain breaks? (b) What force is exerted by each side on the hinge? 

16. A gymnast is attempting to perform splits. From the information given 
in Figure 9.36, calculate the magnitude and direction of the force exerted 
on each foot by the floor. 



Hip pivot 




i.30 m 



Figure 9.36 A gymnast performs full split. The center of gravity and the various 
distances from it are shown. 

9.4 Applications of Statics, Including Problem-Solving 
Strategies 

17. To get up on the roof, a person (mass 70.0 kg) places a 6.00-m 
aluminum ladder (mass 10.0 kg) against the house on a concrete pad 
with the base of the ladder 2.00 m from the house. The ladder rests 
against a plastic rain gutter, which we can assume to be frictionless. The 
center of mass of the ladder is 2 m from the bottom. The person is 
standing 3 m from the bottom. What are the magnitudes of the forces on 
the ladder at the top and bottom? 

18. In Figure 9.21, the eg of the pole held by the pole vaulter is 2.00 m 
from the left hand, and the hands are 0.700 m apart. Calculate the force 
exerted by (a) his right hand and (b) his left hand, (c) If each hand 
supports half the weight of the pole in Figure 9.19, show that the second 
condition for equilibrium (net r = 0) is satisfied for a pivot other than 

the one located at the center of gravity of the pole. Explicitly show how 
you follow the steps in the Problem-Solving Strategy for static equilibrium 
described above. 

9.5 Simple Machines 

19. What is the mechanical advantage of a nail puller — similar to the one 
shown in Figure 9.23 — where you exert a force 45 cm from the pivot 



and the nail is 1.8 cm on the other side? What minimum force must you 
exert to apply a force of 1250 N to the nail? 

20. Suppose you needed to raise a 250-kg mower a distance of 6.0 cm 
above the ground to change a tire. If you had a 2.0-m long lever, where 
would you place the fulcrum if your force was limited to 300 N? 

21. a) What is the mechanical advantage of a wheelbarrow, such as the 
one in Figure 9.24, if the center of gravity of the wheelbarrow and its load 
has a perpendicular lever arm of 5.50 cm, while the hands have a 
perpendicular lever arm of 1.02 m? (b) What upward force should you 
exert to support the wheelbarrow and its load if their combined mass is 
55.0 kg? (c) What force does the wheel exert on the ground? 

22. A typical car has an axle with 1.10 cm radius driving a tire with a 
radius of 27.5 cm . What is its mechanical advantage assuming the very 
simplified model in Figure 9.25(b)? 

23. What force does the nail puller in Exercise 9.19 exert on the 
supporting surface? The nail puller has a mass of 2.10 kg. 

24. If you used an ideal pulley of the type shown in Figure 9.26(a) to 
support a car engine of mass 1 15 kg , (a) What would be the tension in 

the rope? (b) What force must the ceiling supply, assuming you pull 
straight down on the rope? Neglect the pulley system's mass. 

25. Repeat Exercise 9.24 for the pulley shown in Figure 9.26(c), 
assuming you pull straight up on the rope. The pulley system's mass is 
7.00 kg . 

9.6 Forces and Torques in Muscles and Joints 

26. Verify that the force in the elbow joint in Example 9.4 is 407 N, as 
stated in the text. 

27. Two muscles in the back of the leg pull on the Achilles tendon as 
shown in Figure 9.37. What total force do they exert? 

F 2 (200 N) F/200 N) 




Figure 9.37 The Achilles tendon of the posterior leg serves to attach plantaris, 
gastrocnemius, and soleus muscles to calcaneus bone. 

28. The upper leg muscle (quadriceps) exerts a force of 1250 N, which is 
carried by a tendon over the kneecap (the patella) at the angles shown in 
Figure 9.38. Find the direction and magnitude of the force exerted by the 
kneecap on the upper leg bone (the femur). 



314 CHAPTER 9 | STATICS AND TORQUE 




F a = 1250 N 



Figure 9.38 The knee joint works like a hinge to bend and straighten the lower leg. It 
permits a person to sit, stand, and pivot. 

29. A device for exercising the upper leg muscle is shown in Figure 9.39, 
together with a schematic representation of an equivalent lever system. 
Calculate the force exerted by the upper leg muscle to lift the mass at a 
constant speed. Explicitly show how you follow the steps in the Problem- 
Solving Strategy for static equilibrium in Applications of Statistics, 
Including Problem-Solving Strategies. 




Figure 9.39 A mass is connected by pulleys and wires to the ankle in this exercise 
device. 

30. A person working at a drafting board may hold her head as shown in 
Figure 9.40, requiring muscle action to support the head. The three 
major acting forces are shown. Calculate the direction and magnitude of 
the force supplied by the upper vertebrae F v to hold the head 

stationary, assuming that this force acts along a line through the center of 
mass as do the weight and muscle force. 




F, = 60N 



Figure 9.40 

31. We analyzed the biceps muscle example with the angle between 
forearm and upper arm set at 90° . Using the same numbers as in 
Example 9.4, find the force exerted by the biceps muscle when the angle 
is 120° and the forearm is in a downward position. 

32. Even when the head is held erect, as in Figure 9.41, its center of 
mass is not directly over the principal point of support (the atlanto- 
occipital joint). The muscles at the back of the neck should therefore 
exert a force to keep the head erect. That is why your head falls forward 
when you fall asleep in the class, (a) Calculate the force exerted by these 
muscles using the information in the figure, (b) What is the force exerted 
by the pivot on the head? 




Figure 9.41 The center of mass of the head lies in front of its major point of support, 
requiring muscle action to hold the head erect. A simplified lever system is shown. 

33. A 75-kg man stands on his toes by exerting an upward force through 
the Achilles tendon, as in Figure 9.42. (a) What is the force in the 
Achilles tendon if he stands on one foot? (b) Calculate the force at the 
pivot of the simplified lever system shown — that force is representative of 
forces in the ankle joint. 



CHAPTER 9 | STATICS AND TORQUE 315 



CQ 




Figure 9.42 The muscles in the back of the leg pull the Achilles tendon when one 
stands on one's toes. A simplified lever system is shown. 

34. A father lifts his child as shown in Figure 9.43. What force should the 
upper leg muscle exert to lift the child at a constant speed? 



^ 



r' = 20.0 cm 




Figure 9.43 A child being lifted by a father's lower leg. 

35. Unlike most of the other muscles in our bodies, the masseter muscle 
in the jaw, as illustrated in Figure 9.44, is attached relatively far from the 
joint, enabling large forces to be exerted by the back teeth, (a) Using the 
information in the figure, calculate the force exerted by the teeth on the 
bullet, (b) Calculate the force on the joint. 






r u = 2.9 cm 



Masseter. 
muscle 




36. Integrated Concepts 

Suppose we replace the 4.0-kg book in Exercise 9.31 of the biceps 
muscle with an elastic exercise rope that obeys Hooke's Law. Assume its 
force constant k = 600 N/m . (a) How much is the rope stretched (past 
equilibrium) to provide the same force F B as in this example? Assume 

the rope is held in the hand at the same location as the book, (b) What 
force is on the biceps muscle if the exercise rope is pulled straight up so 
that the forearm makes an angle of 25° with the horizontal? Assume the 
biceps muscle is still perpendicular to the forearm. 

37. (a) What force should the woman in Figure 9.45 exert on the floor 
with each hand to do a push-up? Assume that she moves up at a 
constant speed, (b) The triceps muscle at the back of her upper arm has 
an effective lever arm of 1.75 cm, and she exerts force on the floor at a 
horizontal distance of 20.0 cm from the elbow joint. Calculate the 
magnitude of the force in each triceps muscle, and compare it to her 
weight, (c) How much work does she do if her center of mass rises 0.240 
m? (d) What is her useful power output if she does 25 pushups in one 
minute? 

m=50-0 kg 




reaction 



Figure 9.45 A woman doing pushups. 

38. You have just planted a sturdy 2-m-tall palm tree in your front lawn for 
your mother's birthday. Your brother kicks a 500 g ball, which hits the top 
of the tree at a speed of 5 m/s and stays in contact with it for 10 ms. The 
ball falls to the ground near the base of the tree and the recoil of the tree 
is minimal, (a) What is the force on the tree? (b) The length of the sturdy 
section of the root is only 20 cm. Furthermore, the soil around the roots is 
loose and we can assume that an effective force is applied at the tip of 
the 20 cm length. What is the effective force exerted by the end of the tip 
of the root to keep the tree from toppling? Assume the tree will be 
uprooted rather than bend, (c) What could you have done to ensure that 
the tree does not uproot easily? 

39. Unreasonable Results 

Suppose two children are using a uniform seesaw that is 3.00 m long and 
has its center of mass over the pivot. The first child has a mass of 30.0 
kg and sits 1.40 m from the pivot, (a) Calculate where the second 18.0 kg 
child must sit to balance the seesaw, (b) What is unreasonable about the 
result? (c) Which premise is unreasonable, or which premises are 
inconsistent? 

40. Construct Your Own Problem 

Consider a method for measuring the mass of a person's arm in 
anatomical studies. The subject lies on her back, extends her relaxed 
arm to the side and two scales are placed below the arm. One is placed 
under the elbow and the other under the back of her hand. Construct a 
problem in which you calculate the mass of the arm and find its center of 
mass based on the scale readings and the distances of the scales from 
the shoulder joint. You must include a free body diagram of the arm to 
direct the analysis. Consider changing the position of the scale under the 
hand to provide more information, if needed. You may wish to consult 
references to obtain reasonable mass values. 



Figure 9.44 A person clenching a bullet between his teeth. 



316 CHAPTER 9 | STATICS AND TORQUE 



CHAPTER 10 | ROTATIONAL MOTION AND ANGULAR MOMENTUM 317 



ROTATIONAL MOTION AND ANGULAR 
'MENTUM 




Figure 10.1 The mention of a tornado conjures up images of raw destructive power. Tornadoes blow houses away as if they were made of paper and have been known to 
pierce tree trunks with pieces of straw. They descend from clouds in funnel-like shapes that spin violently, particularly at the bottom where they are most narrow, producing 
winds as high as 500 km/h. (credit: Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) 



Learning Objectives 



10.1. 



10.2 



10.3 



10.4. 



10.5. 



10.6 



10.7 



Angular Acceleration 

• Describe uniform circular motion. 

• Explain non-uniform circular motion. 

• Calculate angular acceleration of an object. 

• Observe the link between linear and angular acceleration. 
Kinematics of Rotational Motion 

• Observe the kinematics of rotational motion. 

• Derive rotational kinematic equations. 

• Evaluate problem solving strategies for rotational kinematics. 
Dynamics of Rotational Motion: Rotational Inertia 

• Understand the relationship between force, mass and acceleration. 

• Study the turning effect of force. 

• Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration. 
Rotational Kinetic Energy: Work and Energy Revisited 

• Derive the equation for rotational work. 

• Calculate rotational kinetic energy. 

• Demonstrate the Law of Conservation of Energy. 
Angular Momentum and Its Conservation 

• Understand the analogy between angular momentum and linear momentum. 

• Observe the relationship between torque and angular momentum. 

• Apply the law of conservation of angular momentum. 
Collisions of Extended Bodies in Two Dimensions 

• Observe collisions of extended bodies in two dimensions. 

• Examine collision at the point of percussion. 
Gyroscopic Effects: Vector Aspects of Angular Momentum 

• Describe the right-hand rule to find the direction of angular velocity, momentum, and torque. 

• Explain the gyroscopic effect. 

• Study how Earth acts like a gigantic gyroscope. 



318 CHAPTER 10 | ROTATIONAL MOTION AND ANGULAR MOMENTUM 



Introduction to Rotational Motion and Angular Momentum 



Why do tornadoes spin at all? And why do tornados spin so rapidly? The answer is that air masses that produce tornadoes are themselves rotating, 
and when the radii of the air masses decrease, their rate of rotation increases. An ice skater increases her spin in an exactly analogous manner as 
seen in Figure 10.2. The skater starts her rotation with outstretched limbs and increases her spin by pulling them in toward her body. The same 
physics describes the exhilarating spin of a skater and the wrenching force of a tornado. 

Clearly, force, energy, and power are associated with rotational motion. These and other aspects of rotational motion are covered in this chapter. We 
shall see that all important aspects of rotational motion either have already been defined for linear motion or have exact analogs in linear motion. 
First, we look at angular acceleration — the rotational analog of linear acceleration. 




Figure 10.2 This figure skater increases her rate of spin by pulling her arms and her extended leg closer to her axis of rotation, (credit: Luu, Wikimedia Commons) 

10.1 Angular Acceleration 

Uniform Circular Motion and Gravitation discussed only uniform circular motion, which is motion in a circle at constant speed and, hence, constant 
angular velocity. Recall that angular velocity co was defined as the time rate of change of angle 6 : 



<» = ¥-, 

At 



(10.1) 



where 6 is the angle of rotation as seen in Figure 10.3. The relationship between angular velocity co and linear velocity v was also defined in 
Rotation Angle and Angular Velocity as 

v = rco (10.2) 

or 

CO = £. ( 10 - 3 ) 



where r is the radius of curvature, also seen in Figure 10.3. According to the sign convention, the counter clockwise direction is considered as 
positive direction and clockwise direction as negative 

v 




Figure 10.3 This figure shows uniform circular motion and some of its defined quantities. 

Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer's hard disk 
slows to a halt when switched off. In all these cases, there is an angular acceleration, in which co changes. The faster the change occurs, the 
greater the angular acceleration. Angular acceleration a is defined as the rate of change of angular velocity. In equation form, angular acceleration is 
expressed as follows: 



a ■ 



Acq 

At' 



(10.4) 



CHAPTER 10 | ROTATIONAL MOTION AND ANGULAR MOMENTUM 319 

where Aw is the change in angular velocity and At is the change in time. The units of angular acceleration are (rad/s)/s , or rad/s . If w 
increases, then a is positive. If co decreases, then a is negative. 



Example 10.1 Calculating the Angular Acceleration and Deceleration of a Bike Wheel 



Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) 
Calculate the angular acceleration in rad/s . (b) If she now slams on the brakes, causing an angular acceleration of - 87.3 rad/s , how long 
does it take the wheel to stop? 
Strategy for (a) 

The angular acceleration can be found directly from its definition in a = ^@- because the final angular velocity and time are given. We see that 

Am is 250 rpm and At is 5.00 s. 
Solution for (a) 

Entering known information into the definition of angular acceleration, we get 

a = Aoi (10-5) 

A^ 

250 rpm 
5.00 s ' 

Because Acq is in revolutions per minute (rpm) and we want the standard units of rad/s for angular acceleration, we need to convert Acq 
from rpm to rad/s: 



Aco = 250- 



2jtrad . 1 min (10.6) 



min rev 60 sec 



26.2^. 



Entering this quantity into the expression for a , we get 



a = 4^ < ia7 > 

A? 

= 26.2 rad/s 
5.00 s 

= 5.24 rad/s 2 . 
Strategy for (b) 

In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular 
acceleration and solving for At , yielding 



A '" a 



Aw (10.8) 



Solution for (b) 

Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that Am is - 26.2 rad/s , and a is given to be - 87.3 rad/s . 



Thus, 

At = - 26.2 rad/s (10.9) 

- 87.3 rad/s 2 
= 0.300 s. 

Discussion 

Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When 
she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are 
analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall — the velocity change 
is large in a short time interval. 

If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then 
come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear 
and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in Figure 10.4. 
Thus, linear acceleration is called tangential acceleration a t . 



320 CHAPTER 10 | ROTATIONAL MOTION AND ANGULAR MOMENTUM 




Figure 10.4 In circular motion, linear acceleration a , occurs as the magnitude of the velocity changes: a is tangent to the motion. In the context of circular motion, linear 
acceleration is also called tangential acceleration a t . 

Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know from Uniform Circular Motion and 
Gravitation that in circular motion centripetal acceleration, a c , refers to changes in the direction of the velocity but not its magnitude. An object 

undergoing circular motion experiences centripetal acceleration, as seen in Figure 10.5. Thus, a t and a c are perpendicular and independent of 

one another. Tangential acceleration a t is directly related to the angular acceleration a and is linked to an increase or decrease in the velocity, but 

not its direction. 




a t affects magnitude 
a c affects direction 

Figure 10.5 Centripetal acceleration a c occurs as the direction of velocity changes; it is perpendicular to the circular motion. Centripetal and tangential acceleration are thus 
perpendicular to each other. 

Now we can find the exact relationship between linear acceleration a t and angular acceleration a . Because linear acceleration is proportional to a 
change in the magnitude of the velocity, it is defined (as it was in One-Dimensional Kinematics) to be 

„ _ Av (10.10) 



At' 



For circular motion, note that v = roo , so that 



_ Ajrm) 

The radius r is constant for circular motion, and so A{rco) = r(Aco) . Thus, 

a t- r At . 

By definition, a = ^p- . Thus, 

y At 



a t = ra, 



(10.11) 



(10.12) 



(10.13) 



(10.14) 



These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger 
the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car's drive wheels, the greater the 
acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration 
a . 



CHAPTER 10 | ROTATIONAL MOTION AND ANGULAR MOMENTUM 321