Kinetic Energy and Work - John Wiley & Sons

226

9In the weight-lifting competi-

tion of the 1996 Olympics,

Andrey Chemerkin lifted a

record-breaking 260.0 kg from

the floor to over his head

(about 2 m). In 1957, Paul

Anderson stooped beneath a

reinforced wood platform,

placed his hands on a short

stool to brace himself, and then

pushed upward on the

platform with his back, lifting

the platform and its load about

a centimeter. On the platform

were auto parts and a safe

filled with lead. The composite

weight of the load was

27 900 N (6270 lb)!

Who did more workon the objects helifted—Chemerkin orAnderson?

The answer is in thischapter.

Kinetic Energy and Work

Introduction 227

FIGURE 9-1 � It is difficult to use New-ton’s Second Law to analyze the motion ofa ski jumper traveling down a curved rampand predict her velocity at the bottom ofthe ramp.

9-1 Introduction

A ski jumper who wants to understand her motion along a curved track is presentedwith a special challenge. If she tries to use Newton’s laws of motion to predict herspeed at each location along that track, she has to account for the fact that the netforce and her acceleration keep changing as the slope of the track changes. The goalof this chapter is to devise a way to simplify the analysis of motions like those of ourski jumper shown in Fig. 9-1.

We can begin by drawing on ideas presented in Chapter 7. There we introducetwo new concepts—momentum and impulse—and use them to derive an alternateform of Newton’s Second Law known as the impulse–momentum theorem. This theo-rem, expressed in Eq. 7-10, tells us that the impulse of a moving particle is equal toits momentum change.

(impulse–momentum theorem). (7-10)

One of the most useful aspects of the impulse–momentum theorem is that we can useit without having to keep track of the particle’s position.

Can we derive another alternate form of Newton’s law to relate a particle’s veloc-ity and position changes without keeping track of time? In this chapter we simplifythe analysis of complex motions like that of the skier by proving an analogous theo-rem called the net work–kinetic energy theorem. But, in order to “derive” our newtheorem we introduce two additional concepts—work and kinetic energy.

We will start our development of the new theorem by introducing the concept ofwork, W, in analogy to the impulse represented by the integral in Eq. 7-10. Initially weconsider a very simple situation in which a net force acts along the line of motion of aparticle. In this case the concept of work as a one-dimensional analogy to impulse in-volving position changes rather than time changes would be

(one-dimensional position analogy to impulse). (9-1)

Here Fxnet and dx are components of force and infinitesimal position change vectors

along an x axis, and x1 is the initial position of the particle and x2 is its position at alater time. In order for the integral to be unique and well defined we will also add therequirement that the component of the force is either constant or only varies with x.That is, we will consider the integral

where the x in parentheses signifies the force on the particle varies with locationalong the x axis. After developing the concept of work we introduce the concept ofkinetic energy and then derive the net work–kinetic energy theorem for one-dimen-sional motions. Next we apply the concept of work and the new theorem to theanalysis of some motions that result from the actions of common one-dimensionalforces.

Although we begin with one-dimensional situations, we will extend the equationswe derive to two (and three) dimensions. As part of this process we will also introducea method for finding a scalar product of two vectors. Then in Section 9-9 toward theend of the chapter, we demonstrate how the two-dimensional form of our new network–kinetic energy theorem enables us to determine the speed of the skier as afunction of her location along a frictionless ramp in a very simple manner.

�x2

x1

F netx (x) dx

W � �x2

x1

F netx dx

J:

� �t2

t1

F: net(t) dt � mv:2 � mv:1

J:

9-2 Introduction to Work and Kinetic Energy

One-Dimensional Relationship for a Net ForceIn order to find an alternative form of Newton’s Second Law that relates position andvelocity, suppose our particle is a bead moving along a wire, as in Fig. 9-2. Suppose thenet force on the bead is a combination of an applied force and a friction force thatacts as the bead slides along. We know that if a net force acts on the bead along its di-rection of motion, its position will change in the direction of the force and its speedwill increase. If the direction of the net force is opposite to that of the bead’s motion,the bead’s speed will decrease.

In our example, the force, , is directed along the wire as shown in Fig. 9-2. This force causes the bead to accelerate in the same direction as the force. Wecan use Newton’s Second Law to relate the force and acceleration components as

, (9-2)

where m is the bead’s mass. As the bead moves through a displacement, the force changes the bead’s velocity from an initial value to an-

other value we will call . Using the definition of acceleration as the rate of velocitychange over a short time interval dt, this gives us

In order to relate the velocity and position, we perform two mathematical operations:First we multiply each term in the equation above by the x-component of velocity, vx.Second, we use the definition of vx as dx /dt to substitute for vx on the left side of ournew equation. This gives us

(9-3)

Since the bead’s mass m is constant, we can use the chain rule of differentiation to seethat the term on the right can be rewritten as

(9-4)

If we substitute the expression on the right side of Eq. 9-4 for the term mvx dvx /dt inEq. 9-3, we get

Now we can eliminate dt from Eq. 9-3 by realizing that during the same infinitesimaltime interval dt the x-component of force times the infinitesimal change in x is equalto the change in the expression . This gives us the following equality betweendifferentials

(9-5)

Because the net force on the bead and its velocity are not necessarily constant over thefull displacement shown in Fig. 9-2, we must integrate both sides of Eq. 9-5 to deter-mine the relationship between position change and velocity change due to our variableforce, so that

F netx (x) dx � d(1

2 mv2

x).

12

mv2x

F netx (x)

dxdt

�d(1

2 mv2x)

dt.

mvx dvx

dt�

d(12

mv2x)

dt.

F netx (x)

dxdt

� mvx dvx

dt.

F netx (x) � max � m

dvx

dt.

v:2

v:1� x: � (x2 � x1) i

F netx (x) � max

F:net � F net

x i

228 CHAPTER 9 Kinetic Energy and Work

FIGURE 9-2 � A simple situation showinga bead on a wire that experiences a netforce that is directed along the wire. Thebead moves in the same direction as theforce.

x

Bead

Wire

Fxnet i

v1 v2

ˆ

ˆ

x1 x2

r = ∆x = (x2 – x1) i

Introduction to Work and Kinetic Energy 229

(9-6)

In summary, by using Newton’s Second Law, the definitions of velocity and accelera-tion, and the rules of calculus, we have derived a new form of the second law that re-lates how a variable force acting over a distance will change the speed of a particle ofmass m.

Note that the expression on the left side of this equation,

is identical to the expression we developed (in Eq. 9-1) as the one-dimensional (1D)integral of force over position that is analogous to the integral of force over time usedin the impulse-momentum theorem.

Initial Definitions of Work and Kinetic EnergyThe left and right sides of Eq. 9-6 are the basis for two new and very important defini-tions. We define the integral as the net work, , done on a particle it moves from aninitial to a final position, so that

(net work definition—1D net force and displacement). (9-7)

If the net force component along a line is made up of the sum of several forcecomponents , we see that the contribution of eachforce to the net work is

If we define the work associated with a single force Fx(x) as

(definition of work—1D single force and displacement), (9-8)

we see that the net work is given by

(9-9)

So there are two ways to calculate the net work. One is to sum the force componentsbefore the net work is calculated. The other is to calculate the work associated witheach of the components separately. The net work is then determined by adding up thework done by each force component.

The right side of Eq. 9-6 tells us how the speed of a particle of mass m is changedby the net work done on it. This is the factor that is changed by the work doneon the bead. Because we are talking about translational motion (as distinct from rota-tional motion), we call this factor translational kinetic energy (or sometimes just “kinetic energy”). In the most general sense, kinetic energy is a quantity associatedwith motion. However, in this chapter, we limit our discussion to the motion of a

12 mv2

x

W net � WA � WB � WC � � � � .

WA � �x2

x1

FA x (x) dx

� �x2

x1

FA x(x) dx � �x2

x1

FB x(x) dx � �x2

x1

FC x(x) dx � ��.

W net � �x2

x1

F netx (x) dx � �x2

x1

(FA x(x) � FB x(x) � FC x(x) ��) dx

FA x(x) � FB x(x) � FC x(x) ��F net

x (x)

W net � �x2

x1

F netx (x) dx

W net

�x2

x1

F netx (x) dx,

�x2

x1

F netx (x) dx � �v2 x

v1 x

d( 12

mv2x) � 1

2 mv22 x � 1

2 mv2

1 x.

single particle, or the center of mass of systems of particles and extended objects.When we turn our attention to the study of thermodynamics later in the book, we willhave to revisit the concept of kinetic energy to take into account the fact that particleswithin objects may well be moving, even if the center of mass is not. So we define thetranslational kinetic energy, K, of a particle-like object in terms of its mass and thesquare of the speed of its center of mass as

(definition of kinetic energy for 1D motion). (9-10)

According to this definition the term on the left side of Eq. 9-6 given byrepresents the change in the object’s translational kinetic energy.

Using our new definitions for net work and translational kinetic energy, Eq. 9-6can be rewritten in streamlined form as

(the net work-energy theorem). (9-11)

Equation 9-11 is known as the net work-kinetic energy theorem. This theorem tells usthat when a net force acts along the direction of motion of a particle that moves fromone position to another, the particle’s kinetic energy changes by .Equation 9-11 is known as a theorem because Eqs. 9-6 and 9-9 were derived mathe-matically from Newton’s Second Law. It is analogous in many ways to the impulse-momentum theorem (also derived from Newton’s Second Law) that relates the actionof a net force over time to the change in momentum of a particle.

Units of Work and EnergyThe SI unit for both work and kinetic energy (and every other type of energy) is thejoule (J). It is defined directly from (Eq. 9-10) in terms of the units for massand velocity as 1 kg � m2/s2. It is easy to show that the units for work, which is theproduct of a force in newtons (N) and a distance in meters (m), are N � m, which alsoreduce to 1 kg � m2/s2. In summary,

1 joule = 1 J = 1 kg � m2/s2 = 1 N � m (SI unit for energy).

Other units of energy which you may encounter are the erg (or g � cm2/s2) and thefoot-pound.

Generalizing Work and Kinetic Energy ConceptsSo far we have only related position and velocity for the very special case of a netforce acting along a line of motion. What if the forces on a particle that make up thenet force have components that do not lie along the direction of motion of our bodyof interest? If forces do not act parallel to a particle’s displacement, how can we mul-tiply and then integrate two vectors such as force and displacement to calculate work?

The remainder of this chapter is devoted to understanding how to calculate thework done by single forces in some common situations. This will enable us to applythe net work-kinetic energy theorem to relate how changes in a particle’s position dueto a net force are related to changes in its speed.

READI NG EXERC IS E 9-1 : A particle moves along an x axis. Does the kinetic energyof the particle increase, decrease, or remain the same if the particle’s velocity changes (a) from�3 m/s to �2 m/s and (b) from �2 m/s to 2 m/s? (c) In each situation, is the net work done onthe particle positive, negative, or zero? �

K � 12 mv2

12 mv2

2 x � 12mv2

1 x

W net � K2 � K1 � �K

12 mv2

2 x � 12 mv2

1 x

K � 12 mv2

x


The Concept of Physical Work 231

FIGURE 9-3 � Pushing one and then twotextbooks across a tabletop with a smallbut steady force.

9-3 The Concept of Physical Work

So far we have only discussed the work done on small particles with no internalstructure. Now we would like to apply the concept of work to changes in motion offamiliar extended objects. If we push on an object that deforms and changes itsshape as it moves, then it’s impossible to describe its motion in terms of a single dis-placement. For this reason, when we apply the concept of work to extended objects,we are assuming that these objects are particle-like as defined in Section 2-1. Thus,when we refer to doing work on an object, we assume the object is rigid enough thatthe work done distorting it is negligible compared to the work that displaces its cen-ter of mass.

Now let’s get back to work. In casual conversation, most of us think of work as anexpenditure of effort. It takes effort to push a rigid box down the hallway or to lift it.But you also expend effort to hold a heavy object steady in midair or to shove againsta massive object that won’t budge. If we examine our expression for work (Eq. 9-8), wesee that at least for one-dimensional motion, work is given by the product of the com-ponents of a force along the line of motion of a particle and the particle’s displacementalong that line. This means that even though shoving really hard on a massive objectthat is at rest takes a lot of effort, no physical work is done on it unless it starts to movein the direction of the force. As we examine ways to define work in more general situa-tions, we will find that the definition of work in physics requires that

No work is done on a rigid object by a force unless there is a component of the force alongthe object’s line of motion.

We have defined work in such a way that it requires a force and a displacement.How do we know how much work is done? Let’s consider how much effort it takes topush a heavy, very rigid box down a hallway with a steady force. In this special casethe force and the displacement of the box are in the same direction. In the next sec-tion we will consider what happens when the force acting on an object is in the oppo-site direction as its displacement.

Suppose you push the box to the right so the x-component of its displacement is�x � x2 � x1. Now imagine that you use the same steady force to push the same boxthrough twice the distance so its displacement component is 2�x. How much more ef-fort did this take? How much effort would it take to just watch the box? The answersto these questions give us important insights into the nature of work. Namely, for agiven force the magnitude of physical work should be proportional to the distancethat an object is moved. This is consistent with the way, we have defined work— sothat it is proportional to the distance an object moves under the influence of forces.

Imagine pushing another larger box through a rightward displacement of �x us-ing twice the force. How much more effort would you guess it takes to push the largerbox than it takes to push the smaller box? You can get a feel for this by pushing oneand then two larger textbooks of identical mass across a tabletop as shown in Fig. 9-3.

If you took a moment to do this experiment, you found that it takes about twicethe effort to push two books through a displacement of �x using a force of as itdid to push one book through the same displacement with a force of . We can con-clude that the amount (that is, the absolute value) of work done is not just propor-tional to the distance an object is moved— it is also proportional to the magnitude ofthe force acting on the object. The concepts of proportionality between displacementforce and the amount of physical work done can be summarized by the equation

(amount of work done by a steady force along a line of motion).� W � � � Fx �x �

F:

x

2F:

x

INTRODUCTIONT O P H Y S I C S FI

FTH

EDIT

ION

AJA

X

AP


FTH

EDIT

ION

AJA

X

AP


FTH

EDIT

ION

AJA

X

AP

(a)

(b)


In the next section we will use the mathematical definition of work to considerhow much work is done under the influence of a steady force and also the circum-stances under which work is positive or negative.

READI NG EXERC IS E 9-2 : The figure shows four situations in which a force acts ona box while the box either slides to the right with a displacement � or doesn’t budge as indi-cated. The force on each box is shown. Rank the situations according to the amount of the workdone by the force on the box during its displacement from greatest to least.

�

9-4 Calculating Work for Constant Forces

One-Dimensional Forces and Motions Along the Same LineLet’s start by reconsidering the formal definition of the work associated with one-dimensional motion (presented in Eq. 9-8),

(definition of work for a 1D motion and force). (9-12)

As we established in Section 9-1, Fx(x) denotes the x-component of a force that canvary with x. However, if the force does not vary with x, then we can simply denoteFx(x) as Fx and take it out of the integral. This allows us to write

so that W � Fx �x (work done by a constant force along a line of motion). (9-13)

Positive vs. Negative Work At the end of the last section we presented the equationto represent the amount of work done on a rigid object. This is an in-

formal expression we developed by imagining the effort needed to slide rigid objectson a table or down a hall. Equation 9-13 that we just derived is very similar except ithas no absolute value signs. Since both Fx and �x represent components of vectorsalong an axis, either component can be positive or negative. If this is the case, then thesign of the work, W, calculated as the products of these components can also be posi-tive and negative. This raises some questions. How can we tell when the work done bya force will be positive? Negative? Does this mean that work is a vector component?To answer these questions let’s consider the work done on a puck that is free to movealong a line on a sheet of ice with no friction forces on it.

Imagine that the puck is initially at rest at your chosen origin. When you push itto the right with a steady horizontal force component of Fx � �50 N, it speeds upuntil its x-position is �1 m (Fig. 9-4a). The work you do on the puck is positive since itis given by

(speeding up).W � Fx �x � Fx(x2 � x1) � (�50 N)(�1 m � 0 m) � �50 J

� W � � � Fx �x �

W � �x2

x1

Fx dx � Fx(x2 � x1) � Fx �x,

W � �x2

x1

Fx(x) dx

x:

Fx = (1 N) i

ˆx = (1 m) i∆(a)

Fx = (2 N) i

ˆx = (1 m) i∆(c)

Fy = (2 N) j

ˆx = (2 m) i∆(d)

Fx = (3 N) i

ˆx = (0 m) i∆(b)

FIGURE 9-4 � (a) A puck is pushed fromrest with a positive force component. (b) Apuck that is already moving in a positivedirection is pushed with a force that has anegative force component along a chosenx axis.

Origin

Fx = +50 N

x1 = 0 m

x1 = +1.0 m

(a)

x

v1x = (0 m/s) i

Origin

Fx = –50 N0 m

(b)

x

v1x

Calculating Work for Constant Forces 233

Suppose that now as the puck is moving away from the origin along your chosenpositive x axis, you suddenly take your other hand and push on it in the opposite di-rection with a steady horizontal force component of Fx � �50 N. Since you are push-ing the puck in a direction opposite to its motion, it starts slowing down and reaches azero velocity at a distance of 2.0 m from the origin (Fig. 9-4b). In this case the workyou do on the puck while slowing it down is negative since it is given by

(slowing down).

If we consider many similar situations with different types of forces acting, we canmake the following general statement about the sign of the work done by a singleforce or net force acting on the center of mass of a rigid object:

POSITIVE VS. NEGATIVE WORK: If a single (or net) force has a component that acts in thedirection of the object’s displacement, the work that the force does is positive. If a single (ornet) force has a component that acts in a direction opposite to the object’s displacement,the work it does is negative.

Work Is a Scalar Quantity Recall that in Section 2-2 we defined a scalar (unlike thecomponent of a vector) to be a quantity that is independent of coordinate systems.Based on this definition, we can see that work is a scalar quantity even though it canbe positive or negative. If we rotate our x axis by 180° so that all the components offorce and displacement in our puck example above change sign, the sign of the work(which is the product of components) would not change sign. So even though thereare directions associated with both force and displacement, there is no direction asso-ciated with the positive or negative work done by a force on an object. Therefore,work is a scalar quantity.

Work Done by a Gravitational Force We next examine the work done on an object by a particular type of constant force—namely, the gravitational force. Suppose a particle-like object of mass m, such asa tomato, is thrown upward with initial speed of v1 y as in Fig. 9–5a. As it rises, it isslowed by a gravitational force that acts downward in the direction opposite thetomato’s motion. We expect that does negative work on the tomato as it risesbecause the force is in the direction opposite the motion.

To verify this, let’s choose our y axis to be positive in the upward direction, so thatthe y-component of the gravitational force (acts downward) is given by We calculate the work done on the tomato as where the y-componentof displacement is given by .

Since during the rise y2 is greater than y1, �y is positive. The gravitational forcecomponent is negative and so we can write

(rising object). (9-14)

After the object has reached its maximum height it begins falling back down. Weexpect the work done by the gravitational force to be positive in this case because theforce and motion are in the same direction. Here y2 is less than y1 (Fig. 9-5b). Henceboth the y-component of the gravitational force and displacement �y � y2 � y1

are negative. This gives us an expression for positive work of

(falling object). (9-15)

Thus, as we saw for the puck on the ice, the work done by the gravitational forceis positive when the force and the displacement of the tomato are in the same direc-tion and the work done by the force is negative when the force and displacement arein opposite directions.

W grav � mg � �y �

F gravy

W grav � �mg(y2 � y1) � �mg � �y �

�y � y2 � y1

W grav � F gravy �y

F gravy � �mg.

F: grav

F:grav

W � Fx �x � Fx(x2 � x1) � (�50 N)(�2 m � 1 m) � �50 J

F grav

y∆

Tomato

(a)

F grav

y∆

Tomato

(b)

FIGURE 9-5 � If the only force acting on atomato is gravitational: (a) As the tomatorises, the gravitational force does negativework on the object. (b) As the tomato fallsdownward, the gravitational force doespositive work on it.

9-5 Work Done by a Spring Force

So far, we have limited our discussion to the work done by constant forces that do notchange with position or time. Our goal in this section and the next is to explore howto calculate the work done by variable forces. A very common one-dimensional vari-able force is the spring force. The spring force is of great interest because many forcesin our natural and man-made surroundings have the same mathematical form as thespring force. Examples include the interaction between atoms bound in a solid, theflexing of a bridge under the weight of vehicles, and the sway of a building during anearthquake—as long as displacements remain small. Thus, by examining this one ide-alized force, you can gain an understanding of many phenomena.

As you may have experienced, the magnitude of the force exerted by a springincreases when it is stretched or compressed more. Figure 9-7a shows a spring in itsrelaxed state— that is, neither compressed nor extended. One end is fixed, and a rigidblock is attached to the other, free end. If we stretch the spring by pulling the block tothe right, as in Fig. 9-7b, the spring pulls back on the block toward the left. (Because aspring’s force acts to restore the relaxed state, it is sometimes said to be a restoringforce.) If we compress the spring by pushing the block to the left, as in Fig. 9-7c, thespring now pushes on the block back toward the right.

To a good approximation for many springs, the force exerted by it isproportional to the displacement of the free end from its relaxed position. Asusual, the fact that depends on x is symbolized by writing it as a function of x,

. If its displacement is not too large, many spring-like objects have a springforce given by

(Hooke’s law for a 1D ideal spring). (9-16)

This “law” is named after Robert Hooke, an English scientist of the late 1600s. SinceHooke’s law is based on the measured behavior of specific objects, it does not havethe same status as Newton’s laws.

F:

xspring(x) � �k �x:

F:

xspring(x)

F:

xspring

�x:F:

xspring


During a storm, a crate of crepe is sliding across a slick, oily parkinglot through a displacement while a steady windpushes against the crate with a force � . The situationand coordinate axes are shown in Fig. 9-6.

(a) How much work does this force from the wind do on the crateduring the displacement?

S O L U T I O N � The Ke y I d e a here is that, because we can treatthe crate as a particle and because the wind force is constant(“steady”) in both magnitude and direction during the displace-ment, we can use Eq. 9-13 (W � Fx �x) to calculate the work,

W � Fx �x

� (2.0 N)(�3.0 m) (Answer)

� �6.0 J.

So, the wind’s force does negative 6.0 J of work on the crate.

(2.0 N)iF:

�x: � (�3.0 m)i

(b) If the crate has a kinetic energy of 10 J at the beginning ofdisplacement , what is its kinetic energy at the end of ?

S O L U T I O N � The Key I dea here is that, because the forcedoes negative work on the crate, it reduces the crate’s kinetic energy.Using the work-kinetic energy theorem in the form of Eq. 9-11,we have

K2 � K1 � W � 10 J � (�6.0 J) � 4.0 J. (Answer)

Because the kinetic energy is decreased to 4.0 J, the crate has beenslowed.

�x:�x:

TOUCHSTONE EXAMPLE 9-1: Crepe Crate

FIGURE 9-6 � A constantnet force net slows a cratedown as it undergoes a dis-placement .�x:

F:

y

xF

x∆

FIGURE 9-7 � One end of a spring is at-tached to a fixed wall and the other end toa block that is free to slide. (a) The originof an x axis is located at the point wherethe relaxed spring is connected to theblock. (b) The block and spring are given apositive displacement. Note the directionof the restoring force exerted bythe spring. (c) The spring is compressed bya negative amount x. Again, note the direc-tion of the restoring force.

F: spring

x (x)

Blockattachedto spring

x0

x0

x

x

0x

x = 0Fx

spring = 0

x positive∆x positiveF negative

x negative∆x negativeF positive

(a)

(b)

(c)

∆

Fxspring (x)

x

∆xFx

spring (x)

Work Done by a Spring Force 235

The minus sign in (Eq. 9-16) indicates that the spring force isalways opposite in direction from the displacement of the free end so the force is“restoring.” The constant of proportionality k is called the spring constant. It is alwayspositive and is a measure of the stiffness of the spring. The larger k is, the stiffer thespring— that is, the stronger will be its pull or push for a given displacement. The SIunit for k is the N/m.

In Fig. 9-7, an x axis has been placed parallel to the length of a spring, with theorigin (x � 0) at the position of the free end when the spring is in its relaxed state. Forthis coordinate system and arrangement, we can write in compo-nent form as

(Hooke’s law for x � 0 in the relaxed state). (9-17)

The equation correctly describes ideal spring behavior. It tells us that if x is positive(the spring is stretched toward the right on the x axis), then the component

is negative (it is a pull toward the left). If x is negative (the spring is com-pressed toward the left), then the component is positive (it is a pushtoward the right). Also note that Hooke’s law gives us a linear relationship betweenFx and x.

Work Done by a Spring ForceIn the situation shown in Fig. 9-7, the spring force components and displacements liealong the same line, so we can substitute the spring force in the more generalexpression presented in Eq. 9-8 to determine the work done by a one-dimensionalvariable force. We get

(Eq. 9-8)

To apply this equation to the work done by the spring force as the block in Fig. 9-7amoves, let us make two simplifying assumptions about the spring and block. (1) Thespring is massless; that is, its mass is negligible compared to the block’s mass. (2) Thespring is ideal so it obeys Hooke’s law exactly. Making these simplifying assumptionsmight seem to make the results unreal. But for many interesting cases, these simplifi-cations give us results that agree fairly well with experimental findings.

Back to the integral. We use Hooke’s law (Eq. 9-16) to substitute �kx forthe component .We also pull k out of the integral since it is constant. Thus,we get

(9-18)

so the work done on the block by the spring force as the block moves is

(work by a spring force). (9-19)

This work W spring, done by the spring force, can have a positive or negative value,depending on whether the block is moving toward or away from its zero position. Thisis quite similar to the way the gravitational work done on the tomato in the previoussection changes as the tomato rises and falls.

Note that the final position x2 appears in the second term on the right side ofEq. 9-19. Therefore,

W spring � �12

kx21 � 1

2 k x2

2

� (�12 k)[x2]x1

x2 � �12 k(x2

2 � x12),

W spring � �x2

x1

(�kx)dx � �k �x2

x1

x dx

Fxspring(x)

W spring � �x2

x1

Fxspring

(x) dx.

Fxspring(x)

Fxspring(x)

Fxspring(x) � �kx

F:

xspring(x) � �k�x:

F:

xspring(x) � �k �x:


The work done by the spring force on the block W spring is positive if the block moves closerto the relaxed position (x � 0). The work done by the spring force on the block is negativeif the block moves farther away from x � 0. It is zero if the block ends up at the same dis-tance from x � 0.

READI NG EXERC IS E 9-3 : For three situations, the initial and final positions, alongthe x axis for the block in Fig. 9-7 are, respectively, (a) �3 cm, 2 cm; (b) 2 cm, 3 cm; and (c)�2 cm, 2 cm. In each situation, is the work done by the spring force on the block positive, nega-tive, or zero? �

A package of spicy Cajun pralines lies on a frictionless floor, at-tached to the free end of a spring in the arrangement of Fig. 9-7a.An applied force of magnitude F app � 4.9 N would be needed tohold the package stationary at x2 � 12 mm (Fig. 9-7b).

(a) How much work does the spring force do on the package if thepackage is pulled rightward from x1 � 0 to x3 � 17 mm?

S O L U T I O N � A Ke y I d e a here is that as the package movesfrom one position to another, the spring force does work on it asgiven by Eq. 9-19. We know that the initial position x1 is 0 andthe final position x3 is 17 mm, but we do not know the spring con-stant k.

We can probably find k with Eq. 9-16 (Hooke’s law), but weneed a second Ke y I d e a to use it: if the package were held sta-tionary at x2 � 12 mm, the spring force would have to balance theapplied force (by Newton’s Second Law). Thus, the x-component ofthe spring force Fx

spring would have to be �4.9 N (toward the left inFig. 9-7b), so Eq. 9-16 gives us

k � � � Fxspring

x 2� � � � �4.9 N

12 � 10�3 m � � �408 N/m.

Now, with the package at x3 � 17 mm, Eq. 9-19 yields

� �0.059 J � �59 mJ (Answer)

(b) Next, the package is moved leftward from x3 � 17 mm tox4 � �12 mm. How much work does the spring force do on thepackage during this displacement? Explain the sign of this work.

S O L U T I O N � The Ke y I d e a here is the first one we noted inpart (a). Now x3 � �17 mm and x4 � �12 mm where x3 and x4 aretwo positions of the spring relative to its equilibrium position. Theydo not represent displacements. Eq. 9-19 yields

�

� 0.030 J � 30 mJ. (Answer)

This work done on the block by the spring force is positive becausethe block ends up closer to the spring’s relaxed position.

12 (408 N/m)[(17 � 10�3 m)2 � (�12 � 10�3 m)2]

Ws � 12 k x2

3 � 12 k x 2

4 � 12 k(x2

3 � x24)

W spring � �12 k x2

3 � � 12 (408 N/m)(17 � 10�3 m)2

TOUCHSTONE EXAMPLE 9-2: Pralines and a Spring

In Fig. 9-8, a cumin canister of mass m � 0.40 kg slides across a hor-izontal frictionless counter with velocity . It thenruns into and compresses a spring of spring constant k � 750 N/m.When the canister is momentarily stopped by the spring, by whatamount �x is the spring compressed?

S O L U T I O N � There are three Ke y I d e a s here:

1. The work W spring done on the canister by the spring force is re-lated to the requested displacement �x � x2 � x1 by Eq. 9-19( ).

2. The work W spring is also related to the kinetic energy of thecanister by Eq. 9-11 .(W net � K2 � K1)

W spring � 12 kx2

1 � 12 kx2

2

v: � (�0.50 m/s)i

3. The canister’s kinetic energy has an initial value of and a value of zero when the canister is momentarily at rest.

K1 � 12 mv2

x

TOUCHSTONE EXAMPLE 9-3: Cumin Canister

kmFrictionless

v

FIGURE 9-8 � A canister of mass m moves at velocity toward aspring with spring constant k.

v:

Work for a One-Dimensional Variable force—General Considerations 237

Putting the first two of these ideas together, and noting thatx1 � 0 here since the spring is initially uncompressed, we write thework-kinetic energy theorem for the canister as

Substituting according to the third idea makes this

Simplifying, solving for x, and substituting known data then give us

0 � 12 mv2

x � �12 kx2

2.

K2 � K1 � �12 kx2

2.� �1.2 � 10�2 m

� �1.2 cm.

We reject x2 � �1.2 cm as a solution, since clearly the mass movesto the left as it compresses the spring. So

�x � x2 � x1

� �1.2 cm � 0 cm

� �1.2 cm. (Answer)

x2 � �√ mvx2

k� �√ (0.40 kg)(�0.50 m/s)2

750 N/m

9-6 Work for a One-Dimensional Variable Force—General Considerations

Calculating Work for Well-Behaved ForcesIn the previous section, we were able to find the work done by our spring force usingcalculus to perform the integration called for in Eq. 9-8. This is because the springforce is a “well-behaved,” continuous mathematical function that can be integrated. Ifyou know the function , you can substitute it into Eq. 9-8, introduce the properlimits of integration, carry out the integration, and thus find the work. (Appendix Econtains a list of common integrals.) In summary, whenever a one-dimensional vari-able force is a function that can be integrated using the rules of calculus, the use ofEq. 9-8 is the preferred way to find the work done by the force on an object thatmoves along that same line.

Calculating Work Using Numerical IntegrationSuppose that instead of a spring force, our variable force on an object is caused bysomeone pushing and pulling erratically on the bead sliding along a wire depicted inFig. 9-2. In that case, the force will probably not vary with x the way a familiar mathe-matical function does, so we cannot use the rules of calculus to perform our integra-tion. Whenever this is the case, we can use numerical methods to examine the variableforce during small displacements where the force is approximately constant. We canthen calculate the work done during each small displacement, and we can add eachcontribution to the work together to determine the total work. In this situation we aredoing a numerical integration.

Let’s start our exploration of numerical integration by considering the x-componentof a one-dimensional force that varies as a particle moves. A general plot of such a one-dimensional variable force is shown in Fig. 9-9a. One method for finding the work doneon the particle is to divide the distance between the initial location of a particle, x1,and its final location, xN, into N small steps of width �x. We can choose a large N so thatthe values of �x are small enough so the force component along the x axis is rea-sonably constant over that interval. Let be the component representing the av-erage value of within the nth interval. As shown in Fig. 9-9b or c, is theheight of the nth strip. The value of x for the nth strip is given by , where�x � (xN � x1)/N.

xn � (n � 12) �x

�Fx n(x)�Fx(x)�Fx n(x)�

F:

x(x)

F:

x(x)

With taken to be constant, the small increment of work �Wn done by theforce in the nth interval is approximately given by Eq. 9-13 as

. (9-20)

Referring to the most darkly shaded region in Fig. 9-9b or c, we see that �Wn is thenequal to the area of the nth rectangular strip.

To approximate the total work W done by the force as the particle moves from x0

to xN, we add the areas of all the strips between x1 and xN in Fig. 9-9c,

. (9-21)

This is not an exact calculation of the actual work done because the broken “skyline”formed by the tops of the rectangular strips in Fig. 9-9b (representing the �Fx n� s asconstants) only approximates the actual curve of .

If needed in a particular situation we can make the approximation better by reduc-ing the strip width �x and using more strips, as in Fig. 9-9c. Once the strip width is suffi-ciently small, Eq. 9-21 can be used to compute the total work done by the variable force.

Defining the IntegralIt is interesting to note that in the limit where the strip width approaches zero, thenumber of strips then becomes infinitely large and we approach an exact result,

(9-22)

This limit is precisely what we mean by the integral of the function Fx(x) betweenthe limits x0 and xN. Thus, Eq. 9-22 becomes

(work done by a variable force in one dimension). (Eq. 9-8)

Geometrically, the work is equal to the area between the curve and the x axis,taken between the limits x0 and xN (shaded in Fig. 9-9d). Remember that whenever Fx

is negative, the area between the graph of Fx and the x axis is also negative.

F:

x(x)

W � � xN

x1

Fx(x) dx

W � lim�x : 0

N

n�1�Fn x(x)� �x.

F:

x(x)

W N

n�1�Wn �

N

n�1�Fx n(x)� �x

�Wn �Fx n� �x

�Fx n�


FIGURE 9-9 � A particle only moves inone dimension. (a) A one-dimensionalforce component is plotted againstthe displacement of the particle.(b) Same as (a) but with the area underthe curve divided into narrow strips. (c)Same as (b) but with the area divided intonarrower strips. (d) The limiting case. Thework done by the force is given in Eq. 9-8and is represented by the shaded areabetween the curve and the x axis andbetween and .xNx1

xN � x1

Fx(x)

Fx(x)

xx0 xN0

(a)

Fx(x)

xx0 xN

< Fx n>

∆ x0

(b)

∆Wj = < Fx n> ∆x

12

n

N

Fx(x)

xx0 xN0∆ x

(c)

12

n

N Fx(x)

xx0 xN0

W

(d)

Force and Displacement in More Than One Dimension 239

A 2.0 kg stone moves along an x axis on a horizontal frictionlesssurface, accelerated by a force Fx(x) that varies with the stone’s po-sition as shown in Fig. 9-10.

(a) How much work is done on the stone by the force as the stonemoves from its initial point at x1 � 0 to x2 � 5 m?

S O L U T I O N � A Ke y I d e a is that the work done by a singleone-dimensional force is given by Eq. 9-8:

.

Here the limits are x1 � 0 and x2 � 5 m, and Fx(x) is given by Fig. 9-10. A second Ke y I d e a is that we can easily evaluate theintegral graphically from Fig. 9-10. To do so, we find the area be-tween the plot of Fx(x) and the x axis, between the limits x1 � 0 andx2 � 5 m. Note that we can split that area into three parts: a righttriangle at the left (from x � 0 to x � 2 m), a central rectangle(from x � 2 m to x � 4 m), and a triangle at the right (fromx � 4 m to x � 5 m).

W � �x2

x1

Fx(x) dx

Recall that the area of a triangle is . The workthat was done on the stone from x1 � 0 and x2 � 5 m is then

� 17.5 J. (Answer)

(b) The stone starts from rest at x � 0. What is its speed at x � 8 m?

S O L U T I O N � A Ke y I d e a here is that the stone’s speed is re-lated to its kinetic energy, and its kinetic energy is changed becauseof the work done on the stone by the force. Because the stone isinitially at rest, its initial kinetic energy K1 is 0. If we write its finalkinetic energy at x � 8 m as , then we can write thework-kinetic energy theorem of Eq. 9-11 (K3 � K1 � W) as

(9-23)

where is the work done on the stone from x1 � 0 to x3 � 8 m.A second Ke y I d e a is that, as in part (a), we can find the

work graphically from Fig. 9-10 by finding the area between theplotted curve and the x axis. However, we must be careful aboutsigns. We must take an area to be positive when the plotted curve isabove the x axis and negative when it is below the x axis. We al-ready know that work � 17.5 J, so completing the calculationof the area gives us

� 9.5 J.

Substituting this and m � 2.0 kg into Eq. 9-23 and solving for v3, wefind

v3 � 3.1 m/s. (Answer)

� 17.5 J � 12 (1 m)(4 N) � (1 m)(4 N) � 1

2 (1 m)(4 N)

W0:8 � W0:5 � W5:8

W0:5

W0:8

12 mv2

3 � 0 � W0:8,

K3 � 12 mv2

3

W0:5 � 12(2 m)(5 N) � (2 m)(5 N) � 1

2(1 m)(5 N)

W0:5

12 (base)(height)

TOUCHSTONE EXAMPLE 9-4: Work on a Stone

–2

–4

1

3

5

21 3 4 5876 x (m)

Fx (N)

Fx (x)

FIGURE 9-10 � A graphshowing the variation of aone-dimensional forcecomponent with a stone’sposition.

9-7 Force and Displacement in More Than One Dimension

In this section we will explore a quite general situation in which a particle moves in acurved three-dimensional path while acted upon by a three-dimensional force thatcould vary with the position of the particle and might not act in the same direction asthe particle’s displacement. How can we calculate the work done on the particle bythe force in this much more complex situation?

Before undertaking this more general treatment of work, we will actually start ourexploration with a simple example of the work done by a constant two-dimensionalforce acting on a sled that is moving in only one dimension. Our simple example willlead us to conclude that we need to devise a general method for finding work as theproduct of two vectors.

Work Done by a Force Applied at an AngleImagine that you are pulling a loaded sled with no friction present Fig. 9-11a. Youhold the rope handle of the sled at some angle relative to the ground. You pull ashard as you can and the sled starts to move. However, you find that you are gettingtired quickly and still have a significant distance to go. What would you do? Is the sit-uation hopeless? One thing that you could try is to change the angle at which you pullon the handle of the sled. Should you make the angle larger or smaller?

As you probably know from your everyday experiences, you must pull more orless horizontally on a heavy object to pull it along. If you make the angle smaller,then you will pull the sled more efficiently. As discussed in Chapter 6, is because theperpendicular force component can only change the direction of the motion (and inthis situation we assume the sled glides on top of packed snow that prevents it frommoving down). Only the component of a force along the line of motion is effective inchanging an object’s speed. So, by the work-kinetic energy theorem, it must be thatonly the component of a force along the line of motion contributes to the work doneby the force. Saying this more formally:

To calculate the work done on an object by a force during a displacement, we use only thecomponent of force along the line of the object’s displacement. The component of forceperpendicular to the displacement does zero work.

From Fig. 9-11b, we see that we can write the x-component of the force Fx in terms ofthe magnitude of the force and the angle between the force and the positive x axis.That is,

(9-24)

To find the work done, we can use Eq. 9-13 (W � Fx�x) for a constant force to get

(work for displacement parallel to an x axis), (9-25)

where �x is the sled’s displacement. Since the sled is moving to the left in the direc-tion of the pull, both Fx and �x are positive in our coordinate system and so the workdone by the force is positive.

We can derive a similar but more general expression for the work done by a two-dimensional force along any line of displacement (that does not necessarily liealong a chosen axis). To do this we must always take the angle between the force andthe direction of the displacement (rather than the direction of a positive axis). In thiscase the expression for the work becomes

(work in terms of angle between and ). (9-26)

As shown in Fig. 9-12, using the angle between the force and displacement andabsolute values for both, the sign of the work comes out correctly.

Why the Sign of the Work Is Correct in Eq. 9-26 As shown in Fig. 9-12, if we set theangle in (Eq. 9-26) to any value less than 90°, then cos is posi-tive and so is the work. If is greater than 90° (up to 180°), then cos is negative andso is the work. Referring to Fig. 9-11, we see that this way of determining the sign ofthe work done by an applied force is equivalent to determining the sign based onwhether there is a component of force in the same or opposite direction as the mo-tion. (Can you see why the work is zero when � 90°?)

W � � F:

� ��r: � cos

� F:

��r:W � � F:

� � �r: � cos

�r:

W � Fx �x � (� F:

� cos ) �x

Fx � � F:

� cos .


FIGURE 9-11 � (a) A sled is pulled by arope that makes an angle with theground as it moves toward the left. (b) Thecomponents of the pulling force along apositive x axis and perpendicular to it.

F:

(a)

(b)

φ

–x

+y

F (Fy) j = F sin θ jˆ ˆ

(Fx) i = F cos θ iˆ ˆ

θφ

180°

90°

270°

0°∆r axis

∆r axis

FA

FA'

φ'φ

(a)

180°

90°

270°

0°FB

FB'

φ' φ

(b)

r∆

r∆

FIGURE 9-12 � (a) If force or hasan angle or with respectto the displacement vector of an ob-ject, its component relative to the displace-ment will be positive. (b) If a force or

has an angle with re-spect to the displacement of an object,its components relative to the displace-ment will be negative.

�r:90 � � 270F

:�B

F:

B

�r: 270 � 90

F:

�AF:

A

Force and Displacement in More Than One Dimension 241

You can use similar consideration to determine the sign of work for 180� � 270and for 270 � � 360. Notice that once again it is the relative directions of the forceand displacement vectors that determine the work done. As we already stated, no mat-ter which way you choose to have the coordinate system pointing, the work for a partic-ular process (including its sign) stays the same.Thus, work is indeed a scalar quantity.

Cautions: There are two restrictions to using the equations above to calculatework done on an object by a force. First, the force must be a constant force; that is, itmust not change in magnitude or direction as the object moves through its displace-ment . Second, the object must be particle-like. This means that the object must berigid and not change shape as its center of mass moves.

Net Work Done by Several Forces Suppose the net force on a rigid object is given byand we want to calculate the net work done by these

forces. As we discussed earlier, it is simple to prove mathematically that the net workdone on the object is the sum of the work done by the individual forces. We can calcu-late the net work in two ways: (1) We can use WA � (Eq. 9-26) where is the angle between the direction of and the object’s displacement to find thework done by each force and then sum those works. Work is a scalar quantity, so sum-ming the work done by the forces is as simple as adding up positive and negativenumbers. (2) Alternatively, we can first find the net force by finding the vectorsum of the individual forces. Then we can use (Eq. 9-26), substi-tuting the magnitude of for the magnitude of , and the angle between the di-rections of the net force and the displacement for .

Work Done by a Three-Dimensional Variable ForceIn general, even if a force varies with position, a particle can move through an infini-tesimal displacement while being acted on by a three-dimensional force .The displacement can be expressed in rectangular coordinates as

(9-27)

If we restrict ourselves to considering forces with rectangular components that de-pend only on the position component of the particle along a given axis, then

(9-28)

Given the fact that no work is done unless there is a force component along the lineof displacement, we can write the infinitesimal amount of work dW done on the parti-cle by the force as

(9-29)

The work W done by while the particle moves from an initial position with coor-dinates (x1, y1, z1) to a final position with coordinates (x2, y2, z2) is then

(9-30)

Note that if has only an x-component, then the y and z terms in the equa-tion above are zero, so this equation reduces to

(Eq. 9-8)

which is known as a line integral.

W � �x2

x1

Fx(x) dx.

F:

( r:)

W � �r2:

r1:

dW � �x2

x1

Fx (x) dx � �y2

y1

Fy (y) dy � �z2

z1

Fz (z) dz.

r:2

r:1F:

dW � Fx(x) dx � Fy(y) dy � Fz(z) dz.

F:

( r:)

F:

( r:) � Fx (x)i � Fy(y)j � Fz(z)k.

dr: � dx i � dy j � dzk.

F:

( r:)dr:

F:

F: net

W � � F:

� ��r: � cos F: net

F:

A

� F:

A �� r: �cos

F: net � F

:

A � F:

B � F:

C � � � ,

�r:


Figure 9-13a shows two industrial spies sliding an initially stationary225 kg floor safe a displacement of magnitude 8.50 m, alonga straight line toward their truck.The push of Spy 001 is 12.0 N, di-rected at an angle of 30° downward from the horizontal; the pull of Spy 002 is 10.0 N, directed at 40° above the horizontal. The magni-tudes and directions of these forces do not change as the safe moves,and the floor and safe make frictionless contact.

(a) What is the work done on the safe by forces and duringthe displacement ?

S O L U T I O N � We use two Ke y I d e a s here. First, the work Wdone on the safe by the two forces is the sum of the works they doindividually. Second, because we can treat the safe as a particle andthe forces are constant in both magnitude and direction, we can useEq. 9-26,

,

to calculate those works. Note: cos(�) � cos(�) From this and thefree-body diagram for the safe in Fig. 9-13b, the work done by isF

:

1

(W � � F:

� � �r: � cos )

�r:F:

2F:

1

F:

2

F:

1

�r:� 88.33 J,

and the work done by is

� 65.11 J.

Thus, the work done by both forces is

W � W1 � W2 � 88.33 J � 65.11 J

� 153.4 J � 153 J. (Answer)

During the 8.50 m displacement, therefore, the spies transfer 153 Jof energy to the kinetic energy of the safe.

(b) During the displacement, what is the work W grav done on thesafe by the gravitational force and what is the work W Normal

done on the safe by the normal force from the floor?

S O L U T I O N � The Ke y I d e a is that, because these forces areconstant in both magnitude and direction, we can find the workthey do with Eq. 9-26. Thus, with mg as the magnitude of the gravi-tational force, we write

(Answer)

and (Answer)

We should have known this result. Because these forces are perpen-dicular to the displacement of the safe, they do zero work on thesafe and do not transfer any energy to or from it.

W Normal � N ��r: � cos 90 � N ��r: �(0) � 0.

W grav � mg ��r: � cos 90 � mg ��r: �(0) � 0,

N:

F: grav

W2 � � F2:

� � �r: � cos 2 � (10.0 N)(8.50 m)(cos 40)

F:

2

W1 � � F:

1 � � �r: � cos 1 � (12.0 N)(8.50 m)(cos 30)

TOUCHSTONE EXAMPLE 9-5: Sliding a Safe

READI NG EXERC IS E 9-4 : The figure shows four situations in which a force acts on abox while the box slides rightward a distance across a frictionless floor. The magnitudesof the forces are identical; their orientations are as shown. Rank the situations accordingto the work done on the box by the force during the displacement, from most positive to mostnegative.

�

� �x: �

(a) (b) (c) (d)

(a)

Safe

(b)

40°30°

Spy 001Spy 002

F grav F1

F2N

r∆

FIGURE 9-13 � (a) Two spies move a floor safe through displace-ment . (b) A free-body diagram showing the forces on the safe.�r:

Multplying a Vector by a Vector: The Dot Product 243

9-8 Multiplying a Vector by a Vector: The Dot Product

In the previous section, we discussed how to calculate the work done by a force thatacts at some angle to the direction of an object’s motion. We saw that in one dimen-sion work is defined as the scalar product of two vector components (force and dis-placement). This may seem strange, but it is because only the component of the forcealong a line relative to the direction of displacement contributes to the work done bythe force. This type of relationship between two vector quantities is so common thatmathematicians have defined an operation to represent it. That operation is called thedot (or scalar) product. Learning about how to represent and calculate this productwill lead us to a more general mathematical definition of work for three-dimensionalsituations. Application of the dot product will make the key equations we derived inthe previous section easier to represent.

The Dot Product of Two VectorsThe scalar or dot product of the vectors and in Fig. 9-14a is written as anddefined to be

(definition of scalar product), (9-31)

where is the magnitude of , is the magnitude of , and is the angle betweenand (or, more properly, between the directions of and ). There are actu-

ally two such angles, and 360° � . Either can be used in � , be-cause their cosines are the same.

Note that there are only scalars on the right side of � (includ-ing the value of cos ). Thus on the left side represents a scalar quantity. Beingscalars, the values of these quantities do not change, no matter how we choose to de-fine our coordinate system. Because of the dot placed between the two vectors todenote this product, the name usually used for it is “dot product” and is spokenas “a dot b.”

As in the case of work, the dot product can be regarded as the product of twoquantities: (1) the magnitude of one of the vectors and (2) the component of the sec-ond vector along the direction of the first vector. For example, in Fig. 9-14b, has acomponent ( cos ) along the direction of . Note that a perpendicular droppedfrom the head of to determines that component. Alternatively, has a compo-nent along the direction of .

If the angle between two vectors is 0°, the component of one vector along the other ismaximum, and so also is the dot product of the vectors. If the angle between two vectorsis 180, the component of one vector along the other is a minimum. If, instead, is 90°or270°, the component of one vector along the other is zero, and so is the dot product.

Equation 9-31 ( � ) is sometimes rewritten as follows to em-phasize the components:

� (9-32)

Here, ( ) is the component of along , and ( ) is the component ofalong . The commutative law applies to a scalar product, so we can write

� b:

� a:.a: � b

:

b:

a:� a: � cos a:b

:� b

:� cos

� a: � (� b:

� cos ) � (� a: � cos )� b:

�.a: � b

:

� a: � � b:

� cos a: � b

:

a:� b:

� cos b:

b:

a:b:

� a: �a:

a: � b

:

a: � b

:� a: � � b

:� cos a:

� b

:

� a: � � b:

� cos a: � b

:b:

a:b:

a:b:

� b:

�a:� a: �

a: � b

:

� � a: � � b

:�

cos

a: � b

:b:

a:

FIGURE 9-14 � (a) Two vectors and with an angle between them.Since each vector has a component alongthe direction of the other vector, the samedot product results from: (b) multiplyingthe component of on by or (c)multiplying the component of on by

.� a: �a:b

:� b

:�b

:a:

b:

a:

a

a

b

bφ

(a)

(b)

Component of aalong direction of

b is a cos φ

φ

a

bφ

(c)

Component of balong direction of

a is b cos φ

When two vectors are in unit-vector notation in one, two, or three dimensions, it canbe shown mathematically that we will get the same result shown in Eq. 9-32 by writingthe dot product as

(9-33)

which we can expand according to the distributive law: Each component of the firstvector is to be “dotted” with each component of the second vector. For example, thefirst step is

Since is perpendicular to both and , there is no component of along the othertwo unit vectors, the angle between them is 90o, and so . On theother hand, is completely along , the angle here is 0o, and so . Therefore,

If we continue along these lines, we find that

(9-34)

Defining the Work Done as a Dot Product If the force is constant over a displacement , we can use the definition of a dotproduct above and the relationship cos in Eq. 9-26 to produce an al-ternative mathematical definition for work,

(definition of work done by a constant force). (9-35)

If the force is variable we can still use the definition of a dot product above along withthe relationship presented in Eq. 9-30, where we integrated over infinitesimal dis-placements:

to produce a more general alternative mathematical definition for work:

(definition of work done by a variable force). (9-36)

This dot product representation of work has some advantages. For one, the notationis more compact. It is also especially useful for calculating work when and or are given in unit-vector notation because we can exploit the fact that

� axbx � ayby � azbz (Eq. 9-34).

9-9 Net Work and Translational Kinetic Energy

Generalizing the Net Work-Kinetic Energy TheoremWe know from Newton’s laws that if you apply a force to an object in the same direc-tion as the object’s motion, the object’s speed will increase. From our discussion of

a: � b

:�r:

dr:F:

W � �r2:

r1:

F:

( r: ) � dr:

W � �r2:

r1:

dW � �x2

x1

Fx (x) dx � � y2

y1

Fy (y) dy � � z2

z1

Fz (z) dz,

W � F:

� �r:

W � � F:

� � � r: �� r:

a: � b

:� ax bx � ay by � az bz.

ax i � (bx i � by j � bz k) � axbx.

i � i � 1iii � j � i � k � 0

ikji

ax i � (bx i � by j � bz k) � axbx (i � i) � ax by (i � j) � ax bz (i � k).

a: � b

:� (ax i � ay j � az k) � (bx i � by j � bz k),


work, we also know that the force does positive work on the object. If you apply theforce in the direction opposite the direction of the object’s motion, the object’s speedwill decrease, and we know that in that case the force will do negative work on the ob-ject. This suggests that work done by forces correlates with changes in speed. We usedthese considerations to relate work and kinetic energy for the special case of a beadon a wire that experiences a single force in the direction of the wire. In doing so, wedeveloped a net work-kinetic energy theorem for one-dimensional forces and mo-tions given by

(the 1D net work-kinetic energy theorem), (Eq. 9-11)

where (Eq. 9-7) and (Eq. 9-10)

Can we extend this to our more general three-dimensional situation? Fortunatelywe can combine

(Eq. 9-6)

and our general expression for work in three dimensions (Eq. 9-30) and rearrangeterms to get

(9-37)

Since the speed of a particle moving in three dimensions is given by and

this reduces to

(3D net work-kinetic energy theorem for variable force), (9-38)

where represents a more general definition of the kinetic energy of a parti-cle of mass m with its center of mass moving with speed v. In words, Eq. 9-38 tellsus that

Net work done on the particle � Change in its translational kinetic energy.

This relationship is valid in one, two, or three dimensions.

Experimental Verification of the Net Work-Kinetic Energy TheoremExperimental verification of the one-dimensional net work-kinetic energy theorem isshown in Figs. 9-15 and 9-16. A low-friction cart with a force sensor attached to it ispulled along a smooth track from x1 � 0.6 m to x2 � 1.2 m with a variable appliedforce. The applied force is measured with a force sensor. The distance along the trackis measured with a motion detector. Both measurements are fed to a computer for

K � 12mv2

W net � K2 � K1

W � �r2:

r1:

F:

( r:)�dr: � �x2

x1

Fx (x) dx � �y2

y1

Fy (y) dy � �z2

z1

Fz (z) dz,

v2 � v2x � v2

y � v2z

� 12 m(v2

2 x � v22 y � v2

2 z) � 12 m(v2

1 x � v21 y � v2

1 z).

W net � �x2

x1

F netx (x) dx � �y2

y1

F nety (y) dy � �z2

z1

F netz (z) dz

�x2

x1

F netx (x) dx � 1

2 mv22 x � 1

2mv21 x,

K � 12 mv2

x.W net � �x2

x1

F netx (x) dx

W net � K2 � K1

Net Work andTranslational Kinetic Energy 245

display. If we ignore friction, then W net � W app. The net work is given by the area un-der the curve obtained when data for the net force vs. distance is graphed. This area(determined by numerical integration as in Fig. 9-9) gives us

The distance and time data are used to determine the velocity of the cart at each loca-tion. The cart mass (m � 1.5 kg) and velocity are then used to determine the transla-tional kinetic energy of the cart as a function of its location along the track. Thechange in kinetic energy between x1 � 0.6 m and x2 � 1.6 m is

as expected according to the net work-kinetic energy theorem.

Lifting and Lowering—Net Work and Kinetic Energy Suppose we lift a particle-like object by applying a vertical force to it as shown inFig. 9-17. During the upward displacement, our applied force does positive work W app

on the object while the gravitational force does negative work W grav on it. Our forceadds energy to (or transfers energy to) the object while the gravitational force re-moves energy from (or transfers energy from) it. By �K � K2 � K1 (Eq. 9-38), thechange �K in the translational kinetic energy of the object due to these two energytransfers is

(9-39)

This equation also applies if we lower the object. However, then the gravitationalforce tends to transfer energy to the object whereas our force tends to transfer energyfrom it.

A common situation involves an object that is stationary before and after beinglifted. For example, suppose you lift a book from the floor to a shelf. Then K2 and K1

are both zero, and reduces to

or (if object starts are ends at rest). (9-40)W app � �W grav

W net � W app � W grav � 0 N,

�K � K2 � K1 � W net � W app � W grav

�K � K2 � K1 � W net � W app � W grav.

F:

y

�K � K2 � K1 � 1.4 J � 0.1 J � 1.3 J,

W net � �x2

x1

Fxnet (x) dx � 1.3 J.


FIGURE 9-16 � Experimental verification of the net work-kinetic energy theorem for a cartundergoing one-dimensional horizontal motion under the influence of a variable applied forceand a kinetic friction force.

FIGURE 9-15 � A variable force is appliedto a force sensor attached to a cart on ahorizontal track. The cart’s position andvelocity are recorded using a motion sen-sor. The friction force is negligible.

0.7 0.8 0.9x (m)

1.0 1.1 1.20.60.00.4

0.8

1.2

1.6

2.0

K1 = 0.1 J

K2 = 1.4 J

K (

J)

FIGURE 9-17 � An upward force is ap-plied to an object in the presence of adownward gravitational force: (a) As theobject rises, the applied force does positivework while the gravitational force doesnegative work. (b) As the object is loweredthe applied force does negative work onthe object while the gravitational forcedoes negative work.

(a) (b)

F grav

F grav

F

F

y

Object

Object

∆

y∆

0.7 0.8 0.9x (m)

1.0 1.1 1.2

F xn

et (

N) Fx

net (x)dx = 1.3 J1.2 m

0.6 m

0.60.00.51.01.52.02.53.0

Net Work andTranslational Kinetic Energy 247

Note that we get the same result if K2 and K1 are not zero but are still equal. This re-sult means that the work done by the applied force is the negative of the work doneby the gravitational force. That is, the applied force transfers the same amount of en-ergy to the object as the gravitational force takes away from it (whenever the initialand final speeds of an object are the same).

Falling on an Incline—The Skier on a Curved RampFinally, we are ready to return to the question of how to find the speed of a skier(shown in Fig. 9-1) as a function of how far she has descended on a frictionless ramp.Suppose we would like to know the speed of the skier shown in Fig. 9-1 as she leavesthe end of a long curved ramp in order to predict how far she can jump. Let us revisitour initial claim that the net work-kinetic energy theorem is much more useful thanNewton’s Second Law for this calculation. Actually the net work-kinetic energy theo-rem is only useful in this particular case if we make the simplifying assumptions that:(1) we can neglect frictional forces; (2) the skier doesn’t push with her poles as sheslides down the ramp; and (3) she holds her body rigid.

Given these assumptions, we can determine the net force on the skier’s center ofmass when she is at an arbitrary location on the ramp (Fig. 9-18a). This net force is thesum of forces shown in the free-body diagram in Fig. 9-18b. Figure 9-18c shows thecomponents of the normal force and the gravitational force parallel and perpendicu-lar to the ramp. Since there is no motion perpendicular to the ramp at a given loca-tion, the force components perpendicular to the ramp cancel out, so the net force actsparallel to the ramp, and its component down the ramp is given by where � is the angle between the horizontal and the ramp.

Our problem now is to take into account the fact that � keeps changing along thecurved ramp. To do this we can divide the ramp into a whole series of tiny ramps hav-ing sides dx, dy, and a length dr as shown in Fig. 9-19a. A greatly enlarged picture ofone of these infinitesimal ramps is shown in Fig. 9-19b. The infinitesimal work done intraveling a distance dr down any one of the tiny ramps is given by

(9-41)

but since sin � � dy/dr, we see that dW becomes simply (�mg) dy. This is a very pro-found result because it tells us that the work done by a rigid object as it falls down africtionless ramp does not depend on the angle of the ramp but only on the constantfactor �mg and the vertical distance through which the object’s center of mass falls.We will return to this idea in Chapter 10.

If we integrate the net force over the collection of tiny ramps we get

(9-42)

where �y is a negative quantity because y2 � y1.Now that we have obtained a simple expression for the net work done by the

gravitational force as the skier goes down the ramp, we can use the net work-kineticenergy theorem (Eq. 9-38) to find the skier’s speed at the bottom of the ramp. If theskier starts from rest so that her initial speed is v1 � 0 m/s, then

so that (9-43)

Solving for the final speed gives us

(9-44)v2 � √2g �y.

W net � mg�y � 12m(v2).W net � K2 � K1

� �y2

y1

(�mg) dy � �mg(y2 � y1) � �mg �y � Fygrav �y,

W net � �r2:

r1:

F:net

( r:) � dr: � �r2

r1

F�� dr

dW � F��dr � (�mg sin � dr),

F��net � mg sin �

FIGURE 9-18 � (a) A curved ramp makesan angle with respect to the horizontal atthe location of a skier. (b) A free-body dia-gram showing the forces on the skier. (c) Adiagram showing the resolution of into the components parallel and perpen-dicular to the ramp.

F: grav

�

θ

y1

y2

v

(a)

(b)

(c)

F grav

N

F grav

Fgrav = mg sin θ

F grav = mg cos θ

N

drAdrB

drn

(a)

(b)

comdr

dx

dyF = mg sin θ

θ

FIGURE 9-19 � (a) The ramp can be di-vided into many smaller ramps, each possi-bly having a different �. (b) A ramp of infin-itesimal length dr with a vertical componentdy and a horizontal component dx.


TOUCHSTONE EXAMPLE 9-7: Crate on a Ramp

An initially stationary 15.0 kg crate of cheese wheels is pulled, via acable, a distance d � 5.70 m up a frictionless ramp to a height h of2.50 m, where it stops (Fig. 9-20a).

(a) How much work W grav is done on the crate by the gravitationalforce during the lift?

S O L U T I O N � A Ke y I d e a is that we can treat the crate as aparticle and thus use Eq. 9-26 to find thework W grav done by . However, we do not know the angle between the directions of and displacement . From thecrate’s free-body diagram in Fig. 9-20b, we find that is � � 90°,where � is the (unknown) angle of the ramp. Equation 9-26 thengives us

(9-45)W grav � mgd cos(� � 90) � �mgd sin �,

�r:F: grav

F: grav

(W � � F:

� � �r: � cos )

F: grav

where we have used a trigonometric identity to simplify the expres-sion. The result seems to be useless because � is unknown. But(continuing with physics courage) we see from Fig. 9-20a that

, where h is a known quantity. With this substitution,Eq. 9-45 becomes

(9-46)

(Answers)

Note that Eq. 9-46 tells us that the work W grav done by the gravita-tional force depends on the vertical displacement but (perhaps sur-prisingly) not on the horizontal displacement. (Again, we return tothis point in Chapter 10.)

� �368 J.

� �(15.0 kg)(9.8 N/kg)(2.50 m)

W grav � �mgh

sin � � h/d

Let us return to the lifting feats of Andrey Chemerkin shown onthe opening page of this chapter.

(a) Chemerkin made his record-breaking lift with rigidly connectedobjects (a barbell and disk weights) having a total mass m � 260.0 kg.He lifted them a distance of 2.0 m. During the lift, how much workwas done on the objects by the gravitational force acting onthem?

S O L U T I O N � The Ke y I d e a here is that we can treat therigidly connected objects as a single particle and thus use Eq. 9-14,

to find the work W grav done on them by . The total weight mgwas 2548 N, and . Thus,

� �5100 J.(Answer)

(b) How much work was done on the objects by Chemerkin’s forceduring the lift?

S O L U T I O N � We do not have an expression for Chemerkin’sforce on the object, and even if we did, his force was certainlynot constant. Thus, one Ke y I d e a here is that we cannot just sub-stitute his force into Eq. 9-12 to find his work. However, we knowthat the objects were stationary at the start and end of the lift, so

W grav � �mg�y � �(2548 N)(2.0 m)

�y � �2.0 mF: grav

W grav � �mg�y,

F: grav

TOUCHSTONE EXAMPLE 9-6: Weight Lifting

that K2 � K1 � 0. Therefore, as a second Ke y I d e a , we know bythe net work-kinetic energy theorem that the work W app done byChemerkin’s applied force was the negative of the work W grav doneby the gravitational force . Equation 9-40 expresses this factand gives us

(Answer)

(c) While Chemerkin held the objects stationary above his head,how much work was done on them by his force?

S O L U T I O N � The Ke y I d e a is that when he supported theobjects, they were stationary. Thus, their displacement � 0 and,by Eq. 9-36, the work done on them was zero (even though sup-porting them was a very tiring task).

(d) How much work was done by the force Paul Anderson appliedto lift objects with a total weight of 27 900 N a distance of 1.0 cm?

S O L U T I O N � Following the argument of parts (a) and (b) butnow with and , we find

� (27 900 N)(0.010 m) � 280 J. (Answer)

Anderson’s lift required a tremendous upward force but only asmall energy transfer of 280 J because of the short displacement in-volved.

W app � �W grav � �(�mg �y) � �mg �y

�y � 1.0 cmmg � 27 900 N

� r:

W app � �W grav � �5100 J.

F: grav

Note that using Newton’s Second Law to find this speed would be extremely diffi-cult because it requires us to keep track of the angle of the ramp at each location. Wewill further explore the advantages of the net work-kinetic energy theorem for othersituations in the next chapter.

Power 249

(b) How much work W rope is done on the crate by the force fromthe cable during the lift?

S O L U T I O N � We cannot just substitute the force magnitude T for in Eq. 9-26 because we do not knowthe value of T. However, a Ke y I d e a to get us going is that wecan treat the crate as a particle and then apply the work-kinetic en-ergy theorem to it. Because the crate is stationary be-fore and after the lift, the change �K in its kinetic energy is zero.For the net work W net done on the crate, we must sum the worksdone by all three forces acting on the crate. From (a), the work

(W net � �K)

(W � � F:

�� r: �cos )� F:

�

T:

W grav done by the gravitational force is �368 J. The workW Normal done by the normal force on the crate from the ramp is zero because is perpendicular to the displacement. We wantthe work W rope done by . Thus, the work-kinetic energy theoremgives us

or

and so (Answer)W rope � 368 J.

0 � W rope � 368 J � 0,

�K � W rope � W grav � W Normal,

T:

N:

N:

F: grav

Crate

d

Cable

(a)

θ

hFrictionless

T θ

(b)

φ

r

T

N

F grav

∆

FIGURE 9-20 � (a) A crate is pulled up a frictionless ramp by a force parallel to the ramp.(b) A free-body diagram for the crate, showing all the forces on it. Its displacement is alsoshown.

�r:T:

9-10 Power

A contractor wishes to lift a load of bricks from the sidewalk to the top of a buildingusing a winch. We can now calculate how much work the force applied by the winchmust do on the load to make the lift. The contractor, however, is much more inter-ested in the rate at which that work is done. Will the job take 5 minutes (acceptable)or a week (unacceptable)?

The rate at which work is done by a force is called the power. If an amount ofwork W is done in an amount of time by a force, we define the average power dueto the work done by a force during that time interval as

(definition of average power). (9-47)

We define the instantaneous power P as the instantaneous rate of doing work, so that

(definition of instantaneous power), (9-48)

where dW is the infinitesimal amount of work done in an infinitesimal time intervaldt. Suppose we know the work W(t) done by a force as a continuous well-behaved

P � dWdt

�P� � W�t

�t


function of time. Then to get the instantaneous power P at, say, time duringthe work, we would first take the time derivative of W(t), and then evaluate the resultfor .

The SI unit of power is the joule per second. This unit is used so often that it has aspecial name, the watt (W), after James Watt (who greatly improved the rate at whichsteam engines could do work). In the British system, the unit of power is the foot-pound per second. Often the horsepower is used. Some relations among these unitsare

(9-49)

and . (9-50)

Inspection of Eq. 9-47 shows that we can express work as power multiplied bytime, . When we do this, we commonly use the unit of kilowatt-hour. Thus,

(9-51)

Perhaps because the unit of kilowatt-hour appears on our utility bills, it has be-come identified as an electrical unit. However, the kilowatt-hour can be usedequally well as a unit for other examples of work (or energy). Thus, if you pick upthis book from the floor and put it on a tabletop, you are free to report the workthat you have done as (or alternatively converting to milliwatts toget as ).

We can also express the rate at which a force does work on a particle (or particle-like object) in terms of that force and the body’s velocity. For a particle that is movingalong a straight line (say, the x axis) and acted on by a constant force directed atsome angle to that line, (Eq. 9-48) becomes

,

but since we get

(9-52)

Reorganizing the right side of this equation as the dot product we may rewriteEq. 9-52 as

(instantaneous power). (9-53)

For example, the truck in Fig. 9-21 exerts a force on the trailing load, which has ve-locity at some instant. The instantaneous power due to is the rate at which does work on the load at that instant and is given by Eq. 9-52 and (Eq. 9-53). Saying that this power is “the power of the truck” is often acceptable, butwe should keep in mind what is meant: Power is the rate at which the applied forcedoes work.

READI NG EXERC IS E 9-5 : A block moves with uniform circular motion because acord tied to the block is anchored at the center of a circle. Is the power due to the force on theblock from the cord positive, negative, or zero? �

P � F:

� v:

F:

F:

v:F:

P � F:

� v:

F:

� v:

P � � F:

�� vx �cos .

vx � dx/dt,

dxdt

P �dWdt

� (� F

:�cos ) � dx �

dt� � F

:�

cos

P � dW/dtF:

4 mW � h4 � 10�6 kW � h

� 3.6 � 106 J � 3.6 MJ.

1 kilowatt-hour � 1 kW � h � (103 W)(3600 s)

W � �P� �t

1 horsepower � 1 hp � 550 ft � lb/s � 746 W

1 watt � 1 W � 1 J/s � 0.738 ft � lb/s

t � 3.0 s

t � 3.0 s

FIGURE 9-21 � The power due to thetruck’s applied force on the trailing load isthe rate at which that force does work onthe load.

SEC. 9-2 � INTRODUCTION TO WORK AND

KINETIC ENERGY

1. Electron in Copper If an electron (mass m � 9.11 � l0�31 kg) incopper near the lowest possible temperature has a kinetic energy of6.7 � l0�19 J, what is the speed of the electron?

2. Large Meteorite vs. TNTOn August 10, l972, a largemeteorite skipped acrossthe atmosphere above west-ern United States andCanada, much like a stoneskipped across water. Theaccompanying fireball wasso bright that it could beseen in the daytime sky (seeFig. 9-22 for a similarevent). The meteorite’smass was about 4 � 106 kg;its speed was about 15 km/s.Had it entered the atmosphere vertically, it would have hit Earth’ssurface with about the same speed. (a) Calculate the meteorite’sloss of kinetic energy (in joules) that would have been associatedwith the vertical impact. (b) Express the energy as a multiple of theexplosive energy of 1 megaton of TNT, which is 4.2 � 1015 J. (c) Theenergy associated with the atomic bomb explosion over Hiroshimawas equivalent to 13 kilotons of TNT. To how many Hiroshimabombs would the meteorite impact have been equivalent?

Problems 251

3. Calculate Kinetic Energy Calculate the kinetic energies of thefollowing objects moving at the given speeds: (a) a 110 kg footballlinebacker running at 8.1 m/s; (b) a 4.2 g bullet at 950 m/s; (c) theaircraft carrier Nimitz, 40.2 � 108 kg at 32 knots.

4. Father Racing Son A father racing his son has half the kineticenergy of the son, who has half the mass of the father. The fatherspeeds up by 1.0 m/s and then has the same kinetic energy as theson. What are the original speeds of (a) the father and (b) the son?

5. A Proton is Accelerated A proton (mass m � 1.67 � 10�27 kg) isbeing accelerated along a straight line at 3.6 � 1013 m/s2 in a machine.If the proton has an initial speed of 2.4 � 107 m/s and travels 3.5 cm,what then is (a) its speed and (b) the increase in its kinetic energy?

6. Vehicle’s Kinetic Energy If a vehicle with a mass of 1200 kg hasa speed of 120 km/h, what is the vehicle’s kinetic energy as deter-mined by someone at rest alongside the vehicle’s road?

7. Truck Traveling North A 2100 kg truck traveling north at 41 km/hturns east and accelerates to 51 km/h. (a) What is the change in thekinetic energy of the truck? What are the (b) magnitude and (c) direc-tion of the change in the translational momentum of the truck?

8. Two Pieces from One An object, with mass m and speed v rela-tive to an observer, explodes into two pieces, one three times as mas-sive as the other; the explosion takes place in deep space. The lessmassive piece stops relative to the observer. How much kinetic energyis added to the system in the explosion, as measured in the observer’sreference frame? Hint:Translational momentum is conserved.

9. Freight Car A railroad freight car of mass 3.18 � l04 kg collideswith a stationary caboose car. They couple together, and 27.0% of

TOUCHSTONE EXAMPLE 9-8: Average and Instantaneous Power

A horizontal cable accelerates a suspicious package across a fric-tionless horizontal floor. The amount of work that has been done bythe cable’s force on the package is given by .

(a) What is the average power due to the cable’s force in thetime interval to ?

S O L U T I O N � The Key Idea here is that the average power is the ratio of the amount of work W done in the given time intervalto that time interval (Eq. 9-47). To find the work W, we evaluate theamount of work that has been done, W(t), at and . Atthose times, the cable has done work W1 and W2, respectively:

and

Therefore, in the 10 s interval, the work done is Equation 9-47 then gives us

(Answer)�P� �W�t

�20 J10 s

� 2.0 W.

W2 � W1 � 20 J.

W2 � (0.20 J/s2)(10 s)2 � 20 J.W1 � (0.20 J/s2)(0 s)2 � 0 J

t � 10 st � 0

�P�

t2 � 10 st1 � 0�P�

W(t) � (0.20 J/s2)t2

Thus, during the 10 s interval, the cable does work at the averagerate of 2.0 joules per second.

(b) What is the instantaneous power P due to the cable’s force at, and is P then increasing or decreasing?

S O L U T I O N � The Ke y I d e a here is that the instantaneouspower P at is the time derivative of the work dW/dt evalu-ated at (Eq. 9-48). Taking the derivative of W(t) gives us

This result tells us that as time t increases, so does P. Evaluating Pfor , we find

(Answer)

Thus, at , the cable is doing work at the rate of 1.20 joulesper second, and that rate is increasing.

t � 3.0 s

P � (0.40 J/s2)(3.0 s) � 1.20 W.

t � 3.0 s

P �dWdt

�ddt

[(0.20 J/s2)t2] � (0.40 J/s2)t.

t � 3.0 st � 3.0 s

t � 3.0 s

Problems

FIGURE 9-22 � Problem 2. A largemeteorite skips across the atmo-sphere in the sky above the Ottawaregion.

the initial kinetic energy is transferred to thermal energy, sound,vibrations, and so on. Find the mass of the caboose. Hint: Transla-tional momentum is conserved.

10. Two Chunks An 8.0 kg body is traveling at 2.0 m/s with no ex-ternal force acting on it. At a certain instant an internal explosionoccurs, splitting the body into two chunks of 4.0 kg mass each. Theexplosion gives the chunks an additional 16 J of kinetic energy.Neither chunk leaves the line of original motion. Determine thespeed and direction of motion of each of the chunks after the ex-plosion. Hint: Translational momentum is conserved.

11. Kinetic Energy and Impulse A ball having a mass of 150 gstrikes a wall with a speed of 5.2 m/s and rebounds with only 50%of its initial kinetic energy. (a) What is the speed of the ball immedi-ately after rebounding? (b) What is the magnitude of the impulseon the wall from the ball? (c) If the ball was in contact with the wallfor 7.6 ms, what was the magnitude of the average force on the ballfrom the wall during this time interval?

12. Unmanned Space Probe A 2500 kg unmanned space probe ismoving in a straight line at a constant speed of 300 m/s. Control rock-ets on the space probe execute a burn in which a thrust of 3000 Nacts for 65.0 s. (a) What is the change in the magnitude of the probe’stranslational momentum if the thrust is backward, forward, or di-rectly sideways? (b) What is the change in kinetic energy under thesame three conditions? Assume that the mass of the ejected burnproducts is negligible compared to the mass of the space probe.

SEC. 9-6 � WORK FOR A ONE-DIMENSIONAL

VARIABLE FORCE

13. Graph of Acceleration Figure9-23 gives the acceleration of a 2.00 kg particle as it moves fromrest along an x axis while a con-stant applied force acts on it x � 0 to x � 9 m. How much workhas the force done on the particlewhen the particle reaches (a) x � 4m, (b) x � 7 m, and (c) x � 9 m?What is the particle’s speed anddirection of travel when it reaches(d) x � 4 m, (e) x � 7 m, and (f) x � 9 m?

14. Can of Nuts and Bolts A canof nuts and bolts is pushed 2.00 malong an x axis by a broom alongthe greasy (frictionless) floor of acar repair shop in a version of shuf-fleboard. Figure 9-24 gives thework W done on the can by theconstant horizontal force from thebroom, versus the can’s position x.(a) What is the magnitude of thatforce? (b) If the can had an initialkinetic energy of 3.00 J, moving inthe positive direction of the x axis,what is its kinetic energy at the endof the 2.00 m displacement?

15. Single Force A single forceacts on a body that moves along an

F: app

x axis. Figure 9-25 shows the velocity component vx versus time t forthe body. For each of the intervals AB, BC, CD, and DE, give thesign (plus or minus) of the work done by the force on the body orstate that the work is zero.

16. Block Attached to a Spring The block in Fig. 9-7 lies on a hor-izontal frictionless surface and is attached to the free end of thespring, with a spring constant of 50 N/m. Initially, the spring is atits relaxed length and the block is stationary at position x � 0.Then an applied force with a constant magnitude of 3.0 N pullsthe block in the positive direction of the x axis, stretching thespring until the block stops. When that stopping point is reached,what are (a) the position of the block, (b) the work that has beendone on the block by the applied force, and (c) the work that hasbeen done on the block by the spring force? During the block’sdisplacement, what are (d) the block’s position when its kineticenergy is maximum and (e) the value of that maximum kineticenergy?

17. Luge Rider A luge and its rider, with a total mass of 85 kg,emerges from a downhill track onto a horizontal straight track withan initial speed of 37 m/s. If they slow at a constant rate of 2.0 m/s2,(a) what magnitude F is required for the slowing force, (b) whatdistance d do they travel while slowing, and (c) what work W isdone on them by the slowing force? What are (d) F, (e) d, and (f) Wif the luge and the rider slow at a rate of 4.0 m/s2?

18. Work from Graph A 5.0kg block moves in a straightline on a horizontal frictionlesssurface under the influence of aforce that varies with positionas shown in Fig. 9-26. Howmuch work is done by the forceas the block moves from theorigin to x � 8.0 m?

19. A 10 kg brick moves alongan x axis. Its acceleration as afunction of its position isshown in Fig. 9-27. What is thenet work done on the brick bythe force causing the accelera-tion as the brick moves from x� 0 to x � 8.0 m?

20. Velodrome (a) In 1975 theroof of Montreal’s Velodrome,with a weight of 360 kN, waslifted by 10 cm so that it couldbe centered. How much workwas done on the roof by theforces making the lift? (b) In1960, Mrs. Maxwell Rogers ofTampa, Florida, reportedly raised one end of a car that had fallenonto her son when a jack failed. If her panic lift effectively raised4000 N (about of the car’s weight) by 5.0 cm, how much work didher force do on the car?

21. Two Pulleys and a Canister In Fig. 9-28, a cord runs around twomassless, frictionless pulleys; a canister with mass m � 20 kg hangsfrom one pulley; and you exert a force on the free end of thecord. (a) What must be the magnitude of if you are to lift thecanister at a constant speed? (b) To lift the canister by 2.0 cm, howfar must you pull the free end of the cord? During that lift, what is

F:

F:

14


a (m

/s2 )

6

0

–6

2 4 6 8x (m)

FIGURE 9-23 � Problem 13.

W (

J)

6

4

2

0 1x (m)

2

t

B C

vx

A

+

–

D

E



Forc

e (N

) 10

5

0

–5

–104

Position (m)6 820

a (

m/s

2 )

20

15

10

5

0

x (m)4 5 6 7 83210



the work done on the canister by (c)your force (via the cord) and (d) thegravitational force on the canister?(Hint: When a cord loops around apulley as shown, it pulls on the pul-ley with a net force that is twice thetension in the cord.)

22. Spring at MIT During springsemester at MIT, residents of theparallel buildings of the East Cam-pus dorms battle one another withlarge catapults that are made withsurgical hose mounted on a windowframe. A balloon filled with dyedwater is placed in a pouch attachedto the hose, which is then stretchedthrough the width of the room. Assume that the stretching of thehose obeys Hooke’s law with a spring constant of 100 N/m. If thehose is stretched by 5.00 m and then released, how much work doesthe force from the hose do on the balloon in the pouch by the timethe hose reaches its relaxed length?

23. Plot F(x) The force on a particle is directed along an x axis andgiven by � F0(x/x0 � 1). Find the work done by the force in mov-ing the particle from x � 0 to x � 2x0 by (a) plotting (x) and mea-suring the work from the graph and (b) integrating (x).

24. Block Dropped on a Spring A 250 gblock is dropped onto a relaxed verticalspring that has a spring constant of k �2.5 N/cm (Fig. 9-29). The block becomesattached to the spring and compressesthe spring 12 cm before momentarilystopping. While the spring is being com-pressed, what work is done on the blockby (a) the gravitational force on it and(b) the spring force? (c) What is thespeed of the block just before it hits thespring? (Assume that friction is negligi-ble.) (d) If the speed at impact is dou-bled, what is the maximum compressionof the spring?

25. Bird Cage A spring with a springconstant of 15 N/cm has a cage attachedto one end (Fig. 9-30). (a) How much work does the spring force doon the cage when the spring is stretched from its relaxed length by7.6 mm? (b) How much additional work is done by the spring forcewhen the spring is stretched by an additional 7.6 mm?


SEC. 9-7 � FORCE AND DISPLACEMENT IN MORE THAN

ONE DIMENSION

26. Constant Force A constant force of magnitude 10 N makes anangle of 150° (measured counterclockwise) with the positive x di-

F:

F:

F:

Problems 253

rection as it acts on a 2.0 kg object moving in the xy plane. Howmuch work is done on the object by the force as the object movesfrom the origin to the point with position vector (2.0 m) � (4.0 m) ?

27. Force on a Particle A force � (4.0 N) � (c N) acts on aparticle as the particle goes through displacement � (3.0 m) �(2.0 m) . (Other forces also act on the particle.) What is the value of cif the work done on the particle by force is (a) zero, (b) 17 J, and(c) �18 J?

28. Crate on an Incline To push a 25.0 kg crate up a frictionless in-cline, angled at 25.0° to the horizontal, a worker exerts a force of 209N, parallel to the incline. As the crate slides 1.50 m, how much workis done on the crate by (a) the worker’s applied force, (b) the gravi-tational force on the crate, and (c) the normal force exerted by theincline on the crate? (d) What is the total work done on the crate?

29. Cargo Canister Figure 9-31shows an overhead view of threehorizontal forces acting on a cargocanister that was initially stationarybut that now moves across a fric-tionless floor. The force magnitudesare FA � 3.00 N, FB � 4.00 N, andFC � 10.0 N. What is the net workdone on the canister by the threeforces during the first 4.00 m of dis-placement?

30. A Particle Moves A particle moves along a straight paththrough displacement � (8 m) � (cm) while force �(2 N) � (4 N) acts on it. (Other forces also act on the particle.)What is the value of c if the work done by on the particle is (a) zero, (b) positive, and (c) negative?

31. Worker Pulling Crate To pull a 50 kg crate across a horizontalfrictionless floor, a worker applies a force of 210 N, directed 20°above the horizontal. As the crate moves 3.0 m, what work is doneon the crate by (a) the worker’s force, (b) the gravitational force onthe crate, and (c) the normal force on the crate from the floor? (d)What is the total work done on the crate?

32. Floating Ice Blocks A floating ice block is pushed through adisplacement � (15 m) � (12 m) along a straight embankmentby rushing water, which exerts a force � (210 N) � (150 N) onthe block. How much work does the force do on the block duringthe displacement?

33. Coin on a Frictionless Plane A coin slides over a frictionlessplane and across an xy coordinate system from the origin to a pointwith xy coordinates (3.0 m, 4.0 m) while a constant force acts on it.The force has magnitude 2.0 N and is directed at a counterclockwiseangle of l00° from the positive direction of the x axis. How muchwork is done by the force on the coin during the displacement?

34. Work Done by 2-D Force What work is done by a force � ((2 N/m)x) � (3 N) , with x in meters, that moves a particle from aposition � (2 m) � (3 m) to a position � � (4 m) � (3 m) ?

SEC. 9-9 � NET WORK AND TRANSLATIONAL

KINETIC ENERGY

35. Cold Hot Dogs Figure 9-32 shows a cold package of hot dogssliding rightward across a frictionless floor through a distance d �20.0 cm while three forces are applied to it. Two of the forces arehorizontal and have the magnitudes FA � 5.00 N and FB � 1.00 N;

jir:2jir:1

jiF:

jiF:

jid:

F:

jiF:

jid:

F:

jid

:jiF

:

ji

mF


FIGURE 9-29 �

Problem 24.

7.6 mm 7.6 mm

FA

FB

FC

y

x

50°

35°


the third force is angled down by � � 60.0° and has the magnitudeFC � 4.00 N. (a) For the 20.0 cm displacement, what is the net workdone on the package by the three applied forces, the gravitationalforce on the package, and the normal force on the package? (b) Ifthe package has a mass of 2.0 kg and an initial kinetic energy of 0,what is its speed at the end of the displacement?


36. Air Track A 1.0 kg standard body is at rest on a frictionlesshorizontal air track when a constant horizontal force acting inthe positive direction of an x axis along the track is applied to thebody. A stroboscopic graph of the position of the body as it slides tothe right is shown in Fig. 9-33. The force is applied to the body att1 � 0, and the graph records the position of the body at 0.50 s in-tervals. How much work is done on the body by the applied force between t1 � 0 and t2 � 2.0 s?


37. Three Forces Figure 9-34shows three forces applied to atrunk that moves leftward by 3.00 mover a frictionless floor. The forcemagnitudes are FA � 5.00 N, FB �9.00 N, and FC � 3.00 N. During thedisplacement, (a) what is the network done on the trunk by thethree forces and (b) does the ki-netic energy of the trunk increaseor decrease?

38. Block of Ice In Fig. 9-35, ablock of ice slides down a friction-less ramp at angle � � 50°, while anice worker pulls up the ramp (via arope) with a force of magnitude Fr

� 50 N. As the block slides throughdistance d � 0.50 m along the ramp,its kinetic energy increases by 80 J.How much greater would its kineticenergy have been if the rope hadnot been attached to the block?

39. Helicopter A helicopter hoists a 72 kg astronaut 15 m verti-cally from the ocean by means of a cable. The acceleration of theastronaut is g/10. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force onher? What are the (c) kinetic energy and (d) speed of the astronautjust before she reaches the helicopter?

40. Given x(t) A force acts on a 3.0 kg particle-like object in such away that the position of the object as a function of time is given by

F:

F:

F:

x � (3 m/s)t � (4 m/s2)t2 � (1 m/s3)t3 with x in meters and t in sec-onds. Find the work done on the object by the force from t1 � 0 to t2 � 4.0 s. (Hint: What are the speeds at those times?)

41. Lowering a Block A cord is used to vertically lower an initiallystationary block of mass M at a constant downward acceleration ofg/4. When the block has fallen a distance d, find (a) the work doneby the cord’s force on the block, (b) the work done by the gravita-tional force on the block, (c) the kinetic energy of the block, and(d) the speed of the block.

42. Force Applied Downward In Fig. 9-36a, 2.0 N force is appliedto a 4.0 kg block at a downward angle � as the block moves right-ward through 1.0 m across a frictionless floor. Find an expressionfor the speed v2 of the block at the end of that distance if theblock’s initial velocity is (a) 0 and (b) 1.0 m/s to the right. (c) Thesituation in Fig. 9-36b is similar in that the block is initially movingat 1.0 m/s to the right, but now the 2.0 N force is directed downwardto the left. Find an expression for the speed v2 of the block at theend of the 1.0 m distance. (d) Graph all three expressions for v2 ver-sus downward angle �, for � � 0° to � � 90°. Interpret the graphs.


43. Canister and One Force The only force acting on a 2.0 kg can-ister that is moving in an xy plane has a magnitude of 5.0 N. Thecanister initially has a velocity of 4.0 m/s in the positive x direction,and some time later has a velocity of 6.0 m/s in the positive y direc-tion. How much work is done on the canister by the 5.0 N forceduring this time?

44. Block of Ice Slides A 45 kg block of ice slides down a friction-less incline 1.5 m long and 0.91 m high. A worker pushes up againstthe ice, parallel to the incline, so that the block slides down at con-stant speed. (a) Find the magnitude of the worker’s force. Howmuch work is done on the block by (b) the worker’s force, (c) thegravitational force on the block, (d) the normal force on the blockfrom the surface of the incline, and (e) the net force on the block?

45. Cave Rescue A cave rescue team lifts an injured spelunker di-rectly upward and out of a sinkhole by means of a motor-driven ca-ble. The lift is performed in three stages, each requiring a verticaldistance of 10.0 m: (a) the initially stationary spelunker is acceler-ated to a speed of 5.00 m/s; (b) he is then lifted at the constantspeed of 5.00 m/s; (c) finally he is slowed to zero speed. How muchwork is done on the 80.0 kg rescueeby the force lifting him during eachstage?

46. Work-Kinetic Energy The onlyforce acting on a 2.0 kg body as thebody moves along the x axis variesas shown in Fig. 9-37. The velocity ofthe body at x � 0 is 4.0 m/s. (a)What is the kinetic energy of thebody at x � 3.0 m? (b) At whatvalue of x will the body have a ki-


FB FA

d

FC

θ

0 0.2 0.4 0.6 0.8

t = 0 0.5 s 1.0 s 1.5 s 2.0 s

x (m)

60°FA

FC

FB


θ

d

Fr


θ

F

θ

F

(a) (b)

4

Fx (N)

0

–4

x (m)54321


netic energy of 8.0 J? (c) What is the maximum kinetic energy at-tained by the body between x � 0 and x � 5.0 m?

47. Block at Rest A 1.5 kg block is initially at rest on a horizontalfrictionless surface when a horizontal force in the positive directionof an x axis is applied to the block. The force is given by (x) �(2.5 N � x2 N/m2) , where x is in meters and the initial position ofthe block is x � 0. (a) What is the kinetic energy of the block as itpasses through x � 2.0 m? (b) What is the maximum kinetic energyof the block between x � 0 and x � 2.0 m?

SEC. 9-10 � POWER

48. Average Rate of Work The loaded cab of an elevator has amass of 3.0 � 103 kg and moves 210 m up the shaft in 23 s at con-stant speed. At what average rate does the force from the cable dowork on the cab?

49. Block Pulled at Constant Speed A 100 kg block is pulled at aconstant speed of 5.0 m/s across a horizontal floor by an appliedforce of 122 N directed 37° above the horizontal. What is the rate atwhich the force does work on the block?

50. Resistance to Motion Resistance to the motion of an automo-bile consists of road friction, which is almost independent of speed,and air drag, which is proportional to speed-squared. For a certaincar with a weight of 12,000 N, the total resistant force is given by

� 300 N � (1.8 N � s2/m2)v2, where is in newtons and v is inmeters per second. Calculate the power (in horsepower) requiredto accelerate the car at 0.92 m/s2 when the speed is 80 km/h.

51. A Force Acts on a Body A force of 5.0 N acts on a 15 kg bodyinitially at rest. Compute the work done by the force in (a) the first,(b) the second, and (c) the third seconds and (d) the instantaneouspower due to the force at the end of the third second.

52. Rope Tow A skier is pulled by a tow rope up a frictionless skislope that makes an angle of 12° with the horizontal. The ropemoves parallel to the slope with a constant speed of 1.0 m/s. Theforce of the rope does 900 J of work on the skier as the skiermoves a distance of 8.0 m up the incline. (a) If the rope movedwith a constant speed of 2.0 m/s, how much work would the forceof the rope do on the skier as the skier moved a distance of 8.0 mup the incline? At what rate is the force of the rope doing work onthe skier when the rope moves with a speed of (b) 1.0 m/s and (c)2.0 m/s?

53. Freight Elevator A fully loaded, slow-moving freight elevatorhas a cab with a total mass of 1200 kg, which is required to travelupward 54 m in 3.0 min, starting and ending at rest. The elevator’scounterweight has a mass of only 950 kg, so the elevator motormust help pull the cab upward. What average power is required ofthe force the motor exerts on the cab via the cable?

54. Ladle Attached to Spring A 0.30 kg ladle sliding on a horizon-tal frictionless surface is attached to one end of a horizontal spring(with k � 500 N/m) whose other end is fixed. The ladle has a kineticenergy of 10 J as it passes through its equilibrium position (thepoint at which the spring force is zero). (a) At what rate is the

F:

F:

F:

iF:

Additional Problems 255

spring doing work on the ladle as the ladle passes through its equi-librium position? (b) At what rate is the spring doing work on theladle when the spring is compressed 0.10 m and the ladle is movingaway from the equilibrium position?

55. Towing a Boat The force (but not the power) required to tow aboat at constant velocity is proportional to the speed. If a speed of4.0 km/h requires 7.5 kW, how much power does a speed of 12 km/hrequire?

56. Transporting Boxes Boxes are transported from one locationto another in a warehouse by means of a conveyor belt that moveswith a constant speed of 0.50 m/s. At a certain location the con-veyor belt moves for 2.0 m up an incline that makes an angle of 10°with the horizontal, then for 2.0 m horizontally, and finally for 2.0 mdown an incline that makes an angle of 10° with the horizontal. As-sume that a 2.0 kg box rides on the belt without slipping. At whatrate is the force of the conveyor belt doing work on the box (a) asthe box moves up the 10° incline, (b) as the box moves horizontally,and (c) as the box moves down the 10° incline?

57. Horse Pulls Cart A horse pulls a cart with a force of 40 lb atan angle of 30° above the horizontal and moves along at a speed of6.0 mi/h. (a) How much work does the force do in 10 min? (b) Whatis the average power (in horsepower) of the force?

58. Object Accelerates Horizontally An initially stationary 2.0 kgobject accelerates horizontally and uniformly to a speed of 10 m/sin 3.0 s. (a) In that 3.0 s interval, how much work is done on the ob-ject by the force accelerating it? What is the instantaneous powerdue to that force (b) at the end of the interval and (c) at the end ofthe first half of the interval?

59. A Sprinter A sprinter who weighs 670 N runs the first 7.0 m ofa race in 1.6 s, starting from rest and accelerating uniformly. Whatare the sprinter’s (a) speed and (b) kinetic energy at the end of the1.6 s? (c) What average power does the sprinter generate during the1.6 s interval?

60. The Queen Elizabeth 2 The luxury liner Queen Elizabeth 2 hasa diesel-electric powerplant with a maximum power of 92 MW at acruising speed of 32.5 knots. What forward force is exerted on theship at this speed? (1 knot � 1.852 km/h.)

61. Swimmer A swimmer moves through the water at an averagespeed of 0.22 m/s. The average drag force opposing this motion is110 N. What average power is required of the swimmer?

62. Auto Starts from Rest A 1500 kg automobile starts from reston a horizontal road and gains a speed of 72 km/h in 30 s. (a) Whatis the kinetic energy of the auto at the end of the 30 s? (b) What isthe average power required of the car during the 30 s interval? (c)What is the instantaneous power at the end of the 30 s interval, as-suming that the acceleration is constant?

63. A Locomotive A locomotive with a power capability of 1.5 MWcan accelerate a train from a speed of 10 m/s to 25 m/s in 6.0 min. (a)Calculate the mass of the train. Find (b) the speed of the train and (c)the force accelerating the train as functions of time (in seconds) dur-ing the 6.0 min interval. (d) Find the distance moved by the train dur-ing the interval.

64. Estimate, Then Integrate (a) Estimate the work done by theforce represented by the graph of Fig. 9-38 in displacing a particlefrom x � 1 m to x � 3 m. (b) The curve is given by Fx � a/x2, witha � 9 N� m2. Calculate the work using integration.


65. Explosion at Ground Level An explosion at ground levelleaves a crater with a diameter that is proportional to the energy ofthe explosion raised to the power; an explosion of 1 megaton ofTNT leaves a crater with a 1 km diameter. Below Lake Huron inMichigan there appears to be an ancient impact crater with a 50 kmdiameter. What was the kinetic energy associated with that impact,in terms of (a) megatons of TNT (1 megaton yields 4.2 � 1015 J)and (b) Hiroshima bomb equivalents (13 kilotons of TNT each)?(Ancient meteorite or comet impacts may have significantly alteredEarth’s climate and contributed to the extinction of the dinosaursand other life-forms.)

66. Pushing a Block A hand pushes a 3 kg block along a tablefrom point A to point C as shown in Fig. 9-39. The table has beenprepared so that the left half of the table (from A to B) is friction-less. The right half (from B to C) has a nonzero coefficient of fric-tion equal to �kin. The hand pushes the block from A to C using aconstant force of 5 N. The block starts off at rest at point A andcomes to a stop when it reaches point C. The distance from A to Bis meter and the distance from B to C is also meter.


(a) Describe in words the motion of the block as it moves from Ato C.(b) Draw a free-body diagram for the block when it is at point P.(c) What is the direction of the acceleration of the block at pointP? If it is 0, state that explicitly. Explain your reasoning.(d) Does the magnitude of the acceleration increase, decrease, orremain the same as the block moves from B to C? Explain yourreasoning.(e) What is the net work done on the object as it moves from A toB? From B to C? (f) Calculate the coefficient of friction �kin.

12

12

13

67. Continental Drift According to some recent highly accuratemeasurements made from satellites, the continent of North Amer-ica is drifting at a rate of about 1 cm per year. Assuming a continentis about 50 km thick, estimate the kinetic energy the continentalUnited States has as a result of this motion.68. Fan Carts P&E Twofan carts labeled A and Bare placed on oppositesides of a table with theirfans pointed in the samedirection as shown in Fig.9-40. Cart A is weightedwith iron bars so it istwice as massive as cartB. When each fan isturned on, it provides thesame constant force on the cart independent of its mass. Assumethat friction is small enough to be neglected. The fans are set with atimer so that after they are switched on, they stay on for a fixedlength of time, �t, and then turn off.(a) Just after the fans turn off, which of the following statements aretrue about the magnitude of the momenta of the two carts?

(i) pA pB

(ii) pA � pB

(iii) pA � pB

(b) Just after the fans turn off, which of the following statements aretrue about the kinetic energies of the two carts?

(i) KA KB

(ii) KA � KB

(iii) KA � KB

(c) Which of the following statements are true? You may choose asmany as you like, or none. If you choose none, write N.

(i) After the fans are is turned on, each cart moves at a con-stant velocity, but the two velocities are different fromeach other.

(ii) The kinetic energy of each cart is conserved.(iii) The momentum of each cart is conserved.

69. Sticky Carts Two identical carts labeled A and B are initiallyresting on an air track. The coordinate system is shown in Fig. 9-41a.The cart on the right, cart B, is given a push to the left and is re-leased. The clock is then started. At t � 0, cart B moves in the direc-tion shown with a speed v1. The carts hit and stick to each other.The graphs in Fig. 9-41b describe some of the variables associatedwith the motion as a function of time, but without labels on the ver-tical axis. For the experiment described and for each item in the listbelow, identify which graph (or graphs) is a possible display of thatvariable as a function of time, assuming a proper scale and units.“The system” refers to carts A and B together. Friction is so smallthat it can be ignored. If none apply, write N.


Additional ProblemsF x

(N

)

12

10

8

6

4

2

0

x (m)2 3 410

No friction FrictionP

A B C

Initial positionof block

Final positionof block

1/2 m1/2 m

reduce art to 95%

B

A


initially at rest

+x v1

Cart A Cart B

FIGURE 9-41a � Problem 69.

(a) The x-component of momentum of cart B(b) The x-component of force on cart A(c) The x-component of total momentum of the system(d) The kinetic energy of cart B(e) The total kinetic energy of the system

70. Graphs and Carts Two identical carts are riding on an air track.Cart A is given a quick push in the positive x direction toward cartB. When the carts hit, they stick to each other. The graphs shown inFig. 9-42 describe some of the variables associated with the motionas a function of time beginning just after the push is completed. Forthe experiment described and for each item in the list below, iden-tify which graph (or graphs) is a possible display of that variable asa function of time.

Additional Problems 257

(a) The momentum of cart A(b) The total momentum of the two carts(c) The kinetic energy of cart A(d) The force on cart A(e) The force on cart B

71. Rebound to the Left A 5.0 kg block travels to the right on arough, horizontal surface and collides with a spring. The speed ofthe block just before the collision is 3.0 m/s. The block continues tomove to the right, compressing the spring to some maximum extent.The spring then forces the block to begin moving to the left. As theblock rebounds to the left, it leaves the now uncompressed springat 2.2 m/s. If the coefficient of kinetic friction between the blockand surface is 0.30, determine (a) the work done by friction whilethe block is in contact with the spring and (b) the maximum dis-tance the spring is compressed.

72. Rescue A helicopter lifts a stretcher with a 74 kg accident vic-tim in it out of a canyon by applying a vertical force on the stretcher.The stretcher is attached to a guide rope, which is 50 meterslong and makes an angle of 37° with respect to the horizontal. SeeFig. 9-43. What is the work done by the helicopter on the injuredperson and stretcher?


73. A Spring Idealized data for a spring’sdisplacement � from its equilibrium po-sition as a function of an external force,

, are shown in Fig. 9-44.

(a) Draw a properly scaled and carefullylabeled graph of vs. � for thesedata.(b) Does this spring obey Hooke’s law?Why or why not?(c) What is the value of its spring constant k?(d) Shade the area on your graph that represents the amount ofwork done in stretching the spring from a displacement or exten-sion of 0 cm to one of 5 cm. Also shade the area on the graph thatrepresents the amount of work done in stretching the spring from adisplacement or extension of 15 cm to one of 20 cm. Are the shadedareas approximately the same size? What does the size of theshaded area indicate about the work done in these two cases?(e) Explain why the amount of work done in the second case is dif-ferent from the amount done in the first case, even though thechange in length of the spring is the same in both cases.

74. Variable Force The center of mass of a cart having a mass of0.62 kg starts with a velocity of �2.5 m/s along an x axis. The cartstarts from a position of 5.0 m and moves without any noticeablefriction acting on it to a position of 0.0 meters. During this motion, afan assembly exerts a force on the cart in a positive x direction.

x:F: ext

F: ext

x:

9-41bW-0363b

00 0.5 1 1.5 2 2.5

–0.5–1

–1.5–2

Time #1

00 0.5 1 1.5 2 2.5

–0.5–1

–1.5–2

Time #3

2

0 0.5 1 1.5 2 2.5

1.51

0.50

Time #2

1

0 0.5 1 1.5 2 2.5

0.80.60.40.2

0

Time #4

2

0 0.5 1 1.5 2 2.5

1.50

0.50

Time #6

2

0 0.5 1 1.5 2 2.5

1.51

0.50

Time #5

0 0.5 1 1.5 2 2.5Time #7

#9

0 0.5 1 1.5 2 2.5

Time #8

None of thegraphs shown

FIGURE 9-41b � Problem 69.

A

0

–

+

t

B

0

–

+

t

C

0

–

+

t

D

0

–

+

t

E

0

–

+

t

F

0

–

+

t

G H

0

–

+

t 0

–

+

t

I

None of thegraphs shown


Accidentvictim

Guide rope

F

37°

F ext [N]x [cm]∆05

101520

0.01.02.03.04.0

FIGURE 9-44 �

Problem 73.

However, instead of being powered by batteries,the fan is driven by a voltage source that is pro-grammed to change with its distance from a mo-tion detector. This program leads to a variableforce as shown in Fig. 9-45.

(a) What is the work done on the cart by thefan as the cart moves from 5.0 m to 0.0 m? (b)What is the change in kinetic energy of the cartbetween 5.0 m and 0.0 m? (c) What is the finalvelocity of the cart when it is at 0.0 m?

75. Given a Shove An ice skater of mass m isgiven a shove on a frozen pond. After the shoveshe has a speed of v1 � 2 m/s. Assume that theonly horizontal force that acts on her is a slightfrictional force between the blades of the skates and the ice.

(a) Draw a free-body diagram showing the horizontal force andthe two vertical forces that act on the skater. Identify these forces.(b) Use the work-energy theorem to find the distance the skatermoves before coming to rest. Assume that the coefficient of kineticfriction between the blades of the skates and the ice is �kin � 0.12.

76. Karate Board Tester The karate board tester shown in Fig. 9-46a is a destructive testing device that allows one to deter-mine the deformation of the center of a pine karate board as afunction of the forces applied to it.The displacement component, � , of the center of a pine board fromits equilibrium position increases as a function of the x-componentof an external force, , applied to it as shown in the data table ofFig. 9-46b.

(a) Draw a properly scaled and carefully la-beled graph of vs. � for these data.(b) Does this pine board obey Hooke’s law? Why or why not?(c) What is the value of the effective spring constant k for theboard?

x:F: ext

F extx

x


(d) Shade the area on the graph that represents the amount ofwork done in stretching the center of the board from a displace-ment or extension of 0.000 cm to one of 0.156 cm. Also shade thearea on the graph that represents the amount of work done instretching the board from a displacement or extension of 0.446 cmto one of 0.602 cm. Are the shaded areas approximately the samesize? What does the size of the shaded area indicate about the workdone in these two cases?(e) Explain why the amount of work done in the second case is dif-ferent from the amount done in the first case, even though thechange in displacement of the board is the same in both cases.

77. Karate Chop Movie In the movie DSON012 (available inVideoPoint or from your instructor) a physics student breaks astack of eight pine boards. The thickness of the stack of boards withspacers is 0.34 m. In answering the following questions, treat anywork done by gravitational forces on the student’s hand as negligi-ble.

(a) Use video analysis software to analyze the motion of the stu-dent’s hand in the vertical or y direction. By using data from frames3–5, find the velocity of the student’s hand just before he hits theboards. By using data from frames 7–9, find the velocity of the stu-dent’s hand just after he breaks all the boards.(b) Assume that the effective mass of the student’s hand is 1.0 kg.Use the work-energy theorem to find the work done on the stu-dent’s hand by the boards.

0.6

0.3

0.2

0.0 1.0 2.0x [m]

3.0 4.0 5.0

F x [

N]

Karate BoardThis board will be deformed and broken bythe weightplaced on the mass platform.

Mass PlatformBricks are placed on thisplatform to provide the downward force needed tobend and ultimately break the karate board.

The BreakerThis metal bar acts like a knifeand bends the karate board asmore weights are placed on the platform. A gauge allowsstudents to determine the amount of board deformation.The bar will ultimately break the boardas the critical weight limit is exceeded.


Deformation force

Board

Deformation, ∆x0 0.000

167 0.156

326 0.312

461 0.446

567 0.602

�x[cm]Fxext

[N]

FIGURE 9-46b � Problem 76.

FIGURE 9-46a � Problem 76.

Kinetic Energy and Work - John Wiley & Sons

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