POTENTIAL ENERGY AND ENERGY CONSERVATION Books... · 2019. 9. 27. · Conservation of Mechanical...

7LEARNING GOALSBy studying this chapter, you will learn:

• How to use the concept of gravita-tional potential energy in problemsthat involve vertical motion.

• How to use the concept of elasticpotential energy in problems thatinvolve a moving body attached toa stretched or compressed spring.

• The distinction between conserva-tive and nonconservative forces,and how to solve problems inwhich both kinds of forces act on amoving body.

• How to calculate the properties ofa conservative force if you knowthe corresponding potential-energyfunction.

• How to use energy diagrams tounderstand the motion of an objectmoving in a straight line under theinfluence of a conservative force.

213

POTENTIAL ENERGYAND ENERGY

CONSERVATION

?As this diver entersthe water, is the forceof gravity doing positiveor negative work onhim? Is the water doingpositive or negativework on him?

When a diver jumps off a high board into a swimming pool, he hits thewater moving pretty fast, with a lot of kinetic energy. Where does thatenergy come from? The answer we learned in Chapter 6 was that the

gravitational force (his weight) does work on the diver as he falls. The diver’skinetic energy—energy associated with his motion—increases by an amountequal to the work done.

However, there is a very useful alternative way to think about work andkinetic energy. This new approach is based on the concept of potential energy,which is energy associated with the position of a system rather than its motion. Inthis approach, there is gravitational potential energy even while the diver isstanding on the high board. Energy is not added to the earth–diver system as thediver falls, but rather a storehouse of energy is transformed from one form(potential energy) to another (kinetic energy) as he falls. In this chapter we’ll seehow the work–energy theorem explains this transformation.

If the diver bounces on the end of the board before he jumps, the bent boardstores a second kind of potential energy called elastic potential energy. We’ll dis-cuss elastic potential energy of simple systems such as a stretched or compressedspring. (An important third kind of potential energy is associated with the posi-tions of electrically charged particles relative to each other. We’ll encounter thispotential energy in Chapter 23.)

We will prove that in some cases the sum of a system’s kinetic and potentialenergy, called the total mechanical energy of the system, is constant during themotion of the system. This will lead us to the general statement of the law of con-servation of energy, one of the most fundamental and far-reaching principles inall of science.

7.1 Gravitational Potential Energy 215214 C HAPTE R 7 Potential Energy and Energy Conservation

7.1 As a basketball descends, gravitationalpotential energy is converted to kineticenergy and the basketball’s speedincreases.

y1

y2 2 y1 , 0,so w doespositive work andgravitationalpotential energydecreases:

(a) A body moves downward

y2 2 y1

y2

O

y2

y2 2 y1

y1

FotherS

w 5 mgS S

S

S

(b) A body moves upward

O

FotherS

w 5 mgS S

Motion

Motion

DUgrav , 0.

y2 2 y1 . 0,so w doesnegative workand gravitationalpotential energyincreases:DUgrav . 0.

7.2 When a body moves vertically froman initial height to a final height thegravitational force does work and thegravitational potential energy changes.

wSy2 ,y1

7.1 Gravitational Potential EnergyWe learned in Chapter 6 that a particle gains or loses kinetic energy because itinteracts with other objects that exert forces on it. During any interaction, thechange in a particle’s kinetic energy is equal to the total work done on the particleby the forces that act on it.

In many situations it seems as though energy has been stored in a system, tobe recovered later. For example, you must do work to lift a heavy stone over yourhead. It seems reasonable that in hoisting the stone into the air you are storingenergy in the system, energy that is later converted into kinetic energy when youlet the stone fall.

This example points to the idea of an energy associated with the position ofbodies in a system. This kind of energy is a measure of the potential or possibilityfor work to be done; when a stone is raised into the air, there is a potential forwork to be done on it by the gravitational force, but only if the stone is allowed tofall to the ground. For this reason, energy associated with position is calledpotential energy. Our discussion suggests that there is potential energy associ-ated with a body’s weight and its height above the ground. We call this gravi-tational potential energy (Fig. 7.1).

We now have two ways to describe what happens when a body falls withoutair resistance. One way is to say that gravitational potential energy decreases andthe falling body’s kinetic energy increases. The other way, which we learned inChapter 6, is that a falling body’s kinetic energy increases because the force ofthe earth’s gravity (the body’s weight) does work on the body. Later in this sec-tion we’ll use the work–energy theorem to show that these two descriptions areequivalent.

To begin with, however, let’s derive the expression for gravitational potentialenergy. Suppose a body with mass m moves along the (vertical) y-axis, as inFig. 7.2. The forces acting on it are its weight, with magnitude and pos-sibly some other forces; we call the vector sum (resultant) of all the other forces

We’ll assume that the body stays close enough to the earth’s surface thatthe weight is constant. (We’ll find in Chapter 12 that weight decreases with alti-tude.) We want to find the work done by the weight when the body moves down-ward from a height above the origin to a lower height (Fig. 7.2a). Theweight and displacement are in the same direction, so the work done on thebody by its weight is positive;

(7.1)

This expression also gives the correct work when the body moves upward and is greater than (Fig. 7.2b). In that case the quantity is negative, and

is negative because the weight and displacement are opposite in direction.Equation (7.1) shows that we can express in terms of the values of the

quantity mgy at the beginning and end of the displacement. This quantity, theproduct of the weight mg and the height y above the origin of coordinates, iscalled the gravitational potential energy,

(gravitational potential energy) (7.2)

Its initial value is and its final value is The changein is the final value minus the initial value, or Wecan express the work done by the gravitational force during the displacementfrom to as

(7.3)Wgrav 5 Ugrav, 1 2 Ugrav, 2 5 2 1Ugrav, 2 2 Ugrav, 1 2 5 2DUgrav

y2y1

Wgrav

DUgrav 5 Ugrav, 2 2 Ugrav, 1 .Ugrav

Ugrav, 2 5 mgy2 .Ugrav, 1 5 mgy1

Ugrav 5 mgy

Ugrav :

Wgrav

Wgrav

1 y1 2 y2 2y1

y2

Wgrav 5 Fs 5 w 1 y1 2 y2 2 5 mgy1 2 mgy2

Wgrav

y2y1

FS

other .

w 5 mg,

The negative sign in front of is essential. When the body moves up, yincreases, the work done by the gravitational force is negative, and the gravita-tional potential energy increases When the body moves down, ydecreases, the gravitational force does positive work, and the gravitational poten-tial energy decreases It’s like drawing money out of the bank(decreasing and spending it (doing positive work). As Eq. (7.3) shows, theunit of potential energy is the joule (J), the same unit as is used for work.

CAUTION To what body does gravitational potential energy “belong”? It is notcorrect to call the “gravitational potential energy of the body.” The reason isthat gravitational potential energy is a shared property of the body and the earth. Thevalue of increases if the earth stays fixed and the body moves upward, away from theearth; it also increases if the body stays fixed and the earth is moved away from it. Noticethat the formula involves characteristics of both the body (its mass m) and theearth (the value of g). ❚

Conservation of Mechanical Energy (Gravitational Forces Only)To see what gravitational potential energy is good for, suppose the body’s weightis the only force acting on it, so The body is then falling freely with noair resistance, and can be moving either up or down. Let its speed at point be and let its speed at be The work–energy theorem, Eq. (6.6), says that thetotal work done on the body equals the change in the body’s kinetic energy:

If gravity is the only force that acts, then from Eq. (7.3),Putting these together, we get

which we can rewrite as

(if only gravity does work) (7.4)

or

(if only gravity does work) (7.5)

The sum of kinetic and potential energy is called E, the total mechanicalenergy of the system. By “system” we mean the body of mass m and the earthconsidered together, because gravitational potential energy U is a shared propertyof both bodies. Then is the total mechanical energy at and

is the total mechanical energy at Equation (7.4) says thatwhen the body’s weight is the only force doing work on it, That is, E isconstant; it has the same value at and But since the positions and arearbitrary points in the motion of the body, the total mechanical energy E has thesame value at all points during the motion:

(if only gravity does work)

A quantity that always has the same value is called a conserved quantity. Whenonly the force of gravity does work, the total mechanical energy is constant—thatis, is conserved (Fig. 7.3). This is our first example of the conservation ofmechanical energy.

When we throw a ball into the air, its speed decreases on the way up askinetic energy is converted to potential energy; and On theway back down, potential energy is converted back to kinetic energy and theball’s speed increases; and But the total mechanical energy(kinetic plus potential) is the same at every point in the motion, provided that noforce other than gravity does work on the ball (that is, air resistance must be

DUgrav , 0.DK . 0

DUgrav . 0.DK , 0

E 5 K 1 Ugrav 5 constant

y2y1y2 .y1

E1 5 E2 .y2 .E2 5 K2 1 Ugrav, 2

y1E1 5 K1 1 Ugrav, 1

K 1 Ugrav

1

2 mv1

2 1 mgy1 51

2 mv2

2 1 mgy2

K1 1 Ugrav, 1 5 K2 1 Ugrav, 2

DK 5 2DUgrav or K2 2 K1 5 Ugrav, 1 2 Ugrav, 2

Wtot 5 Wgrav 5 2DUgrav 5 Ugrav, 1 2 Ugrav, 2 .Wtot 5 DK 5 K2 2 K1 .

v2 .y2

v1y1

FS

other 5 0.

Ugrav 5 mgy

Ugrav

Ugrav

Ugrav 5 mgy

Ugrav)1DUgrav , 0 2 .

1DUgrav . 0 2 .DUgrav


w 5 mgr r

Moving down:• K increases.• Ugrav decreases.• E 5 K 1 Ugrav stays the same.

Moving up:• K decreases.• Ugrav increases.• E 5 K 1 Ugrav stays the same.

7.3 While this athlete is in midair, only gravity does work on him (if we neglect theminor effects of air resistance). Mechanical energy E—the sum of kinetic and gravita-tional potential energy—is conserved.

negligible). It’s still true that the gravitational force does work on the body as itmoves up or down, but we no longer have to calculate work directly; keepingtrack of changes in the value of takes care of this completely.

CAUTION Choose “zero height” to be wherever you like When working withgravitational potential energy, we may choose any height to be If we shift the ori-gin for y, the values of and change, as do the values of and But this shifthas no effect on the difference in height or on the difference in gravitational poten-tial energy As the following example shows, the physi-cally significant quantity is not the value of at a particular point, but only thedifference in between two points. So we can define to be zero at whatever pointwe choose without affecting the physics. ❚

UgravUgrav

Ugrav

Ugrav, 2 2 Ugrav, 1 5 mg 1 y2 2 y1 2 .y2 2 y1

Ugrav, 2 .Ugrav, 1y2y1

y 5 0.

Ugrav

When Forces Other Than Gravity Do WorkIf other forces act on the body in addition to its weight, then in Fig. 7.2 isnot zero. For the pile driver described in Example 6.4 (Section 6.2), the forceapplied by the hoisting cable and the friction with the vertical guide rails areexamples of forces that might be included in The gravitational work is still given by Eq. (7.3), but the total work is then the sum of and thework done by We will call this additional work so the total workdone by all forces is Equating this to the change in kineticenergy, we have

(7.6)

Also, from Eq. (7.3), so

which we can rearrange in the form

(if forces other thangravity do work)

(7.7)

Finally, using the appropriate expressions for the various energy terms, we obtain

(if forces other thangravity do work)

(7.8)

The meaning of Eqs. (7.7) and (7.8) is this: The work done by all forces otherthan the gravitational force equals the change in the total mechanical energy

of the system, where is the gravitational potential energy.When is positive, E increases, and is greater than When is negative, E decreases (Fig. 7.5). In the special case in which noforces other than the body’s weight do work, The total mechanicalenergy is then constant, and we are back to Eq. (7.4) or (7.5).

Wother 5 0.Wother

K1 1 Ugrav, 1 .K2 1 Ugrav, 2Wother

UgravE 5 K 1 Ugrav

1

2 mv1

2 1 mgy1 1 Wother 51

2 mv2

2 1 mgy2

K1 1 Ugrav, 1 1 Wother 5 K2 1 Ugrav, 2

Wother 1 Ugrav, 1 2 Ugrav, 2 5 K2 2 K1

Wgrav 5 Ugrav, 1 2 Ugrav, 2 ,

Wother 1 Wgrav 5 K2 2 K1

Wtot 5 Wgrav 1 Wother .Wother ,F

S

other .WgravWtot

WgravFS

other .

FS

other

Example 7.1 Height of a baseball from energy conservation

You throw a 0.145-kg baseball straight up in the air, giving it aninitial upward velocity of magnitude Find how high itgoes, ignoring air resistance.

SOLUTION

IDENTIFY: After the ball leaves your hand, the only force doingwork on the ball is gravity. Hence we can use conservation ofmechanical energy.

SET UP: We’ll use Eqs. (7.4) and (7.5), taking point 1 to be wherethe ball leaves your hand and point 2 to be where it reaches itsmaximum height. As in Fig. 7.2, we take the positive y-direction tobe upward. The ball’s speed at point 1 is at its max-imum height the ball is instantaneously at rest, so

We want to know how far the ball moves vertically between thetwo points, so our target variable is the displacement If wetake the origin to be where the ball leaves your hand (point 1), then

(Fig. 7.4) and the target variable is just

EXECUTE: Since the potential energy at point 1 isFurthermore, since the ball is at rest at point 2,

the kinetic energy at that point is HenceEq. (7.4), which says that becomes

As the energy bar graphs in Fig. 7.4 show, the kinetic energy of theball at point 1 is completely converted to gravitational potentialenergy at point 2. At point 1 the kinetic energy is

K1 51

2 mv1

2 51

2 10.145 kg 2 1 20.0 m/s 2 2 5 29.0 J

K1 5 Ugrav, 2

K1 1 Ugrav, 1 5 K2 1 Ugrav, 2 ,K2 5 1

2 mv2

2 5 0.Ugrav, 1 5 mgy1 5 0.

y1 5 0,

y2 .y1 5 0

y2 2 y1 .

v2 5 0.v1 5 20.0 m/s;

20.0 m/s.

y1 5 0

y2

v1 5 20.0 m/sEnergy at y1

Energy at y2

m 5 0.145 kg

v2 5 0E UgravK5 1

E UgravK5 1

zero

zero

After the ball leaves your hand, the only forceacting on it is gravity ...

... so the mechanical energyE 5 K 1 U stays constant.

7.4 After a baseball leaves your hand, mechanical energyis conserved.E 5 K 1 U

EVALUATE: The mass divides out, as we should expect; welearned in Chapter 2 that the motion of a body in free fall doesn’tdepend on its mass. Indeed, we could have derived the result

using Eq. (2.13).In our calculation we chose the origin to be at point 1, so

and What happens if we make a differentchoice? As an example, suppose we choose the origin to be 5.0 mbelow point 1, so Then the total mechanical energy aty1 5 5.0 m.

Ugrav, 1 5 0.y1 5 0

y2 5 v1

2/2g

point 1 is part kinetic and part potential, while at point 2 it’s purelypotential energy. If you work through the calculation again withthis choice of origin, you’ll find this is 20.4 m abovepoint 1, just as with the first choice of origin. In problems like this,the choice of height at which is up to you; don’t agonizeover the choice, though, because the physics of the answer doesn’tdepend on your choice.

Ugrav 5 0

y2 5 25.4 m;

This equals the gravitational potential energy atpoint 2, so

We can also solve the equation algebraically for

y2 5v1

2

2g5

120.0 m/s 2 2

2 19.80 m/s2 2 5 20.4 m

1

2 mv1

2 5 mgy2

y2 :K1 5 Ugrav, 2

y2 5Ugrav, 2

mg5

29.0 J10.145 kg 2 1 9.80 m/s2 2 5 20.4 m

Ugrav, 2 5 mgy2

7.5 As this skydiver moves downward,the upward force of air resistance doesnegative work on him. Hence thetotal mechanical energy decreases: The skydiver’s speed andkinetic energy stay the same, while thegravitational potential energy goesdown.

UK

E 5 K 1 UWother

Problem-Solving Strategy 7.1 Problems Using Mechanical Energy I

IDENFITY the relevant concepts: Decide whether the problemshould be solved by energy methods, by using directly,or by a combination of these. The energy approach is best whenthe problem involves varying forces, motion along a curved path(discussed later in this section), or both. If the problem involveselapsed time, the energy approach is usually not the best choice,because it doesn’t involve time directly.

SET UP the problem using the following steps:1. When using the energy approach, first decide what the initial

and final states (the positions and velocities) of the system are.Use the subscript 1 for the initial state and the subscript 2 for

gFS

5 maSthe final state. It helps to draw sketches showing the initial andfinal states.

2. Define your coordinate system, particularly the level at whichYou will use it to compute gravitational potential ener-

gies. We suggest that you always choose the positive y-directionto be upward because this is what Eq. (7.2) assumes.

3. Identify all forces that do work that can’t be described in termsof potential energy. (So far this means any forces other thangravity. But later in this chapter we’ll see that the work done byan ideal spring can also be expressed as a change in potentialenergy.) A free-body diagram is always helpful.

y 5 0.

Continued

5.2 Upward-Moving Elevator Stops

5.3 Stopping a Downward-Moving Elevator

5.6 Skier Speed

O N L I N E


Gravitational Potential Energy for Motion Along a Curved PathIn our first two examples the body moved along a straight vertical line. Whathappens when the path is slanted or curved (Fig. 7.7a)? The body is acted on bythe gravitational force and possibly by other forces whose resultant wecall To find the work done by the gravitational force during this displace-ment, we divide the path into small segments Fig. 7.7b shows a typical seg-ment. The work done by the gravitational force over this segment is the scalarproduct of the force and the displacement. In terms of unit vectors, the force is

and the displacement is so the work doneby the gravitational force is

The work done by gravity is the same as though the body had been displaced ver-tically a distance with no horizontal displacement. This is true for every seg-ment, so the total work done by the gravitational force is multiplied by thetotal vertical displacement

This is the same as Eq. (7.1) or (7.3), in which we assumed a purely vertical path.So even if the path a body follows between two points is curved, the total workdone by the gravitational force depends only on the difference in height betweenthe two points of the path. This work is unaffected by any horizontal motion thatmay occur. So we can use the same expression for gravitational potential energywhether the body’s path is curved or straight.

Wgrav 5 2mg 1 y2 2 y1 2 5 mgy1 2 mgy2 5 Ugrav, 1 2 Ugrav, 2

1 y2 2 y1 2 : 2mgDy,

wS # D sS 5 2mge # 1Dxd 1 Dye 2 5 2mgDy

D sS 5 Dxd 1 Dye,wS 5 mgS 5 2mge

D sS;FS

other .wS 5 mgS

The work done by the gravitational force depends only on the vertical component of displacement Dy.

In this caseDy is negative.

y1

y2O

(a)

FotherS

w 5 mgS S

(b)

Dx

Dy

w 5 mgS S

DsS

7.7 Calculating the change in gravita-tional potential energy for a displacementalong a curved path.

4. List the unknown and known quantities, including the coordi-nates and velocities at each point. Decide which unknowns areyour target variables.

EXECUTE the solution: Write expressions for the initial and finalkinetic and potential energies—that is, and Then relate the kinetic and potential energies and the work done byother forces, using Eq. (7.7). (You will have to calculate

in terms of these forces.) If no other forces do work, thisexpression becomes Eq. (7.4). It’s helpful to draw bar graphsWother

Wother ,

Ugrav, 2 .Ugrav, 1 ,K2 ,K1 ,

showing the initial and final values of K, and Then solve to find whatever unknown quantity is required.

EVALUATE your answer: Check whether your answer makesphysical sense. Keep in mind, here and in later sections, that thework done by each force must be represented either in

or as but never in both places. The gravi-tational work is included in so make sure you did notinclude it again in Wother .

DUgrav ,Wother ,Ugrav, 2 5 2DUgrav

Ugrav, 1 2

E 5 K 1 Ugrav .Ugrav ,

Example 7.2 Work and energy in throwing a baseball

In Example 7.1, suppose your hand moves up 0.50 m while you arethrowing the ball, which leaves your hand with an upward velocityof Again ignore air resistance. (a) Assuming that yourhand exerts a constant upward force on the ball, find the magnitudeof that force. (b) Find the speed of the ball at a point 15.0 m abovethe point where it leaves your hand.

SOLUTION

IDENTIFY: In Example 7.1 we used conservation of mechanicalenergy because only gravity did work. In this example, however,we must also include the nongravitational work done by your hand.

SET UP: Figure 7.6 shows a diagram of the situation, including afree-body diagram for the ball while it is being thrown. We letpoint 1 be where your hand first starts to move, point 2 be wherethe ball leaves your hand, and point 3 be where the ball is 15.0 mabove point 2. The nongravitational force of your hand acts onlybetween points 1 and 2. Using the same coordinate system as inExample 7.1, we have and The ball starts at rest at point 1, so and we are given thatv1 5 0,

y3 5 15.0 m.y2 5 0,y1 5 20.50 m,

FS

20.0 m/s.

the ball’s speed as it leaves your hand is Our targetvariables are (a) the magnitude F of the force of your hand and(b) the speed at point 3.

EXECUTE: (a) To determine the magnitude of we’ll first useEq. (7.7) to calculate the work done by this force. We have

The initial potential energy is negative because the ball wasinitially below the origin. (Don’t worry about having a potentialenergy that’s less than zero. Remember, all that matters is thedifference in potential energy from one point to another.) Accord-ing to Eq. (7.7), so

The kinetic energy of the ball increases by andthe potential energy increases by thesum is the change in total mechanical energy, which isequal to

Assuming the upward force that your hand applies is con-stant, the work done by this force is equal to the magnitude Fof the force multiplied by the upward displacement overwhich it acts:

This is about 40 times greater than the weight of the ball.(b) To find the speed at point 3, note that between points 2 and

3, total mechanical energy is conserved; the force of your hand nolonger acts, so We can then find the kinetic energy atpoint 3 using Eq. (7.4):

K3 5 1K2 1 Ugrav, 2 2 2 Ugrav, 3

Ugrav, 3 5 mgy3 5 10.145 kg 2 19.80 m/s2 2 115.0 m 2 5 21.3 J


Wother 5 0.

F 5Wother

y2 2 y15

29.7 J

0.50 m5 59 N

Wother 5 F 1 y2 2 y1 2y2 2 y1

Wother

FS

Wother .E2 2 E1 ,

Ugrav, 2 2 Ugrav, 1 5 0.71 J;K2 2 K1 5 29.0 J,

5 129.0 J 2 0 2 1 10 2 120.71 J 2 2 5 29.7 J

Wother 5 1K2 2 K1 2 1 1Ugrav, 2 2 Ugrav, 1 2K1 1 Ugrav, 1 1 Wother 5 K2 1 Ugrav, 2 ,

Ugrav, 1

Ugrav, 2 5 mgy2 5 10.145 kg 2 19.80 m/s2 2 1 0 2 5 0

K2 51

2 mv2

2 51

2 10.145 kg 2 1 20.0 m/s 2 2 5 29.0 J

Ugrav, 1 5 mgy1 5 10.145 kg 2 19.80 m/s2 2 120.50 m 2 5 20.71 J

K1 5 0

Wother

FS

,

v3

v2 5 20.0 m/s.

(a)

E UgravK5 1

E UgravK5

5

1

1

zero

After the ball leaves your hand, the only forceacting on it is gravity ...

As you throw the ball,you do positive workWother on it ...

... so the totalmechanical energyE increases.

... so the total mechan-ical energy E 5 K 1 Ustays constant.

zero

E UgravK

(b)

w

F

x

y

0.50 m

y2 � 0

y1 5 20.50 m

v1 5 0

y3 5 15.0 mv3

v2 5 20.0 m/s

7.6 (a) Applying energy ideas to a ball thrown vertically upward.(b) Free-body diagram for the ball as you throw it.

Since where is the y-component of the ball’svelocity at point 3, we have

The significance of the plus-or-minus sign is that the ball passespoint 3 twice, once on the way up and again on the way down.The total mechanical energy E is constant and equal to 29.0 Jwhile the ball is in free fall, and the potential energy at point 3 is

whether the ball is moving up or down. So at point3, the ball’s kinetic energy and speed don’t depend on the direc-K3

Ugrav, 3 5 21.3 J

v3y 5 6 Å2K3

m5 6 Å

2 17.7 J 20.145 kg

5 610 m/s

v3yK3 5 12 mv3y

2,

5 129.0 J 1 0 J 2 2 21.3 J 5 7.7 J tion the ball is moving. The velocity is positive when the ball is moving up and negative when it ismoving down; the speed is in either case.

EVALUATE: As a check on our result, recall from Example 7.1 thatthe ball reaches a maximum height At that point all ofthe kinetic energy that the ball had when it left your hand at has been converted to gravitational potential energy. At

the ball is about three-fourths of the way to its maxi-mum height, so about three-fourths of its mechanical energyshould be in the form of potential energy. (This is shown in theenergy bar graphs in Fig. 7.6a.) Can you show that this is true fromour results for and Ugrav, 3 ?K3

y 5 15.0 m,

y 5 0y 5 20.4 m.

10 m/sv3

1210 m/s 2 1110 m/s 2v3y

Conceptual Example 7.3 Energy in projectile motion

A batter hits two identical baseballs with the same initial speed andheight but different initial angles. Prove that at a given height h,both balls have the same speed if air resistance can be neglected.

SOLUTION

If there is no air resistance, the only force acting on each ballafter it is hit is its weight. Hence the total mechanical energy foreach ball is constant. Figure 7.8 shows the trajectories of two ballsbatted at the same height with the same initial speed, and thus thesame total mechanical energy, but with different initial angles. Atall points at the same height the potential energy is the same. Thusthe kinetic energy at this height must be the same for both balls,and the speeds are the same.

At y 5 h

At y 5 0

y

h

xO

zero

E K5 1

E Ugrav

Ugrav

K5 1

7.8 For the same initial speed and initial height, the speed of aprojectile at a given elevation h is always the same, neglecting airresistance.


At each point, the normal forceacts perpendicular to the directionof Throcky’s displacement, so onlythe force of gravity (w) does the workon him.

R 5 3.00 m

Ov1 5 0

v2

(a)

Reference level

Point 1

Point 2

zero

zero

E K5 1

E Ugrav

Ugrav

K5 1

w

w

w

ww

R

n 5 0

n

nn n

(b)

Point 1

Point 2

At point 2

At point 1

7.9 (a) Throcky skateboarding down a frictionless circular ramp. The total mechanical energy is constant. (b) Free-body diagrams forThrocky and his skateboard at various points on the ramp.

Example 7.4 Calculating speed along a vertical circle

Your cousin Throckmorton skateboards down a curved playgroundramp. If we treat Throcky and his skateboard as a particle, hemoves through a quarter-circle with radius (Fig. 7.9).The total mass of Throcky and his skateboard is 25.0 kg. He startsfrom rest and there is no friction. (a) Find his speed at the bottomof the ramp. (b) Find the normal force that acts on him at the bot-tom of the curve.

SOLUTION

IDENTIFY: We can’t use the constant-acceleration equationsbecause Throcky’s acceleration isn’t constant; the slope decreasesas he descends. Instead, we’ll use the energy approach. SinceThrocky moves along a circular arc, we’ll also use what we learnedabout circular motion in Section 5.4.

SET UP: Since there is no friction, the only force other thanThrocky’s weight is the normal force exerted by the ramp(Fig. 7.9b). Although this force acts all along the path, it does zerowork because is perpendicular to Throcky’s displacement atevery point. Hence and mechanical energy is conserved.

We take point 1 at the starting point and point 2 at the bottom ofthe curved ramp, and we let be at the bottom of the ramp(Fig. 7.9a). Then and (We are treating Throcky as ifhis entire mass were concentrated at his center.) Throcky starts atrest at the top, so Our target variable in part (a) is his speedat the bottom, In part (b) we want to find the magnitude n of thenormal force at point 2. Because this force does no work, it doesn’tappear in the energy equation, so we’ll use Newton’s second lawinstead.

EXECUTE: (a) The various energy quantities are

From conservation of mechanical energy,

5 "2 19.80 m/s2 2 1 3.00 m 2 5 7.67 m/s v2 5 "2gR

0 1 mgR 51

2 mv2

2 1 0


K1 5 0

K2 51

2 mv2

2

Ugrav, 1 5 mgR

Ugrav, 2 5 0

v2 .v1 5 0.

y2 5 0.y1 5 Ry 5 0

Wother 5 0nS

nS

R 5 3.00 m

Notice that this answer doesn’t depend on the ramp being circular;no matter what the shape of the ramp, Throcky will have the samespeed at the bottom. This would be true even if thewheels of his skateboard lost contact with the ramp during the ride,because only the gravitational force would still do work. In fact,the speed is the same as if Throcky had fallen vertically through aheight R. The answer is also independent of his mass.

(b) To find n at point 2 using Newton’s second law, we need thefree-body diagram at that point (Fig. 7.9b). At point 2, Throcky ismoving at speed in a circle of radius R; his accelera-tion is toward the center of the circle and has magnitude

If we take the positive y-direction to be upward, the y-componentof Newton’s second law is

At point 2 the normal force is three times Throcky’s weight. Thisresult is independent of the radius of the circular ramp. We learnedin Example 5.9 (Section 5.2) and Example 5.24 (Section 5.4) thatthe magnitude of n is the apparent weight, so Throcky feels asthough he weighs three times his true weight mg. But as soon as hereaches the horizontal part of the ramp to the right of point 2, thenormal force decreases to and Throcky feels normalagain. Can you see why?

EVALUATE: This example shows a general rule about the role offorces in problems in which we use energy techniques: What mat-ters is not simply whether a force acts, but whether that force doeswork. If the force does no work, like the normal force in thisexample, then it does not appear at all in Eq. (7.7),

Notice we had to use both the energy approach and Newton’ssecond law to solve this problem; energy conservation gave us thespeed and gave us the normal force. For each part ofthe problem we used the technique that most easily led us to theanswer.

gFS

5 maS

Wother 5 K2 1 Ugrav, 2 .K1 1 Ugrav, 1 1

nS

w 5 mg

5 3 125.0 kg 2 19.80 m/s2 2 5 735 N

n 5 w 1 2mg 5 3mg

aFy 5 n 1 12w 2 5 marad 5 2mg

arad 5v2

2

R5

2gR

R5 2g

v2 5 !2gR

v2 5 "2gR

Example 7.5 A vertical circle with friction

In Example 7.4, suppose that the ramp is not frictionless and thatThrocky’s speed at the bottom is only What work wasdone by the friction force acting on him?

SOLUTION

IDENTIFY: Figure 7.10 shows that again the normal force does nowork, but now there is a friction force that does do work. Hencethe nongravitational work done on Throcky between points 1 and2, is not zero.

SET UP: We use the same coordinate system and the same initialand final points as in Example 7.4 (see Fig. 7.10). Our target vari-able is the work done by friction, since friction is the onlyforce other than gravity that does work, this is just equal to We’ll find using Eq. (7.7).

EXECUTE: The energy quantities are

From Eq. (7.7),

The work done by the friction force is and the totalmechanical energy decreases by 285 J. Do you see why has tobe negative?

Wf

2285 J,

5 450 J 1 0 2 0 2 735 J 5 2285 J

Wf 5 K2 1 Ugrav, 2 2 K1 2 Ugrav, 1

Ugrav, 2 5 0

K2 51

2 mv2

2 51

2 125.0 kg 2 1 6.00 m/s 2 2 5 450 J

Ugrav, 1 5 mgR 5 125.0 kg 2 19.80 m/s2 2 1 3.00 m 2 5 735 J

K1 5 0

Wf

Wother .Wf ;

Wother ,

fS

6.00 m/s.

EVALUATE: Throcky’s motion is determined by Newton’s secondlaw, But it would be very difficult to apply the secondlaw directly to this problem because the normal and friction forcesand the acceleration are continuously changing in both magnitudeand direction as Throcky moves. The energy approach, by contrast,relates the motions at the top and bottom of the ramp withoutinvolving the details of what happens in between. Many problemsare easy if energy considerations are used but very complex if wetry to use Newton’s laws directly.

gFS

5 maS.

Point 1

Point 2f

f

f

f

w

w

w

ww

R 5 3.00 m

f 5 0

n 5 0

n

nn n

zero

zero

E K5 1E Ugrav UgravK5 1

The friction force( f ) does negative work onThrocky as he descends,so the total mechanicalenergy decreases.

At point 2At point 1

7.10 Free-body diagram and energy bar graphs for Throckyskateboarding down a ramp with friction.

Example 7.6 An inclined plane with friction

We want to load a 12-kg crate into a truck by sliding it up a ramp2.5 m long, inclined at A worker, giving no thought to friction,calculates that he can get the crate up the ramp by giving it an ini-tial speed of at the bottom and letting it go. But friction isnot negligible; the crate slides 1.6 m up the ramp, stops, and slidesback down (Fig. 7.11). (a) Assuming that the friction force actingon the crate is constant, find its magnitude. (b) How fast is thecrate moving when it reaches the bottom of the ramp?

SOLUTION

IDENTIFY: The friction force does work on the crate as it slides.As in Example 7.2, we’ll use the energy approach in part (a) to findthe magnitude of the nongravitational force that does work (in thiscase, friction). In part (b) we’ll calculate how much nongravita-tional work this force does as the crate slides back down and thenuse the energy approach to find the crate’s speed at the bottom ofthe ramp.

SET UP: The first part of the motion is from point 1, at the bottomof the ramp, to point 2, where the crate stops instantaneously. Inthe second part of the motion, the crate returns to the bottom of theramp, which we’ll also call point 3 (Fig. 7.11a). We take (and hence to be at ground level, so y1 5 0,Ugrav 5 0)

y 5 0

5.0 m/s

30°.

The force of friction does negative work onthe crate as it moves, so the total mechanicalenergy E 5 K 1 Ugrav decreases.

The crate is movingat speed v3 when itreturns to point 3.

The crate slides up from point1 to point 2, then back downto its starting position(point 3).

(a)

(b)

2.5 m

1.6 m

v1 5 5.0 m/s

v2 5 0

0.80 m30°

Point 1 , 3

Point2

zero

zero

E K5 1E Ugrav Ugrav

zero

E K5 1UgravK5 1

At point 2 At point 3At point 1

7.11 (a) A crate slides partway up the ramp, stops, and slidesback down. (b) Energy bar graphs for points 1, 2, and 3.

Continued

7.2 Elastic Potential Energy 223222 C HAPTE R 7 Potential Energy and Energy Conservation

We proceed just as we did for gravitational potential energy. We begin withthe work done by the elastic (spring) force and then combine this with thework–energy theorem. The difference is that gravitational potential energy is ashared property of a body and the earth, but elastic potential energy is stored justin the spring (or other deformable body).

Figure 7.13 shows the ideal spring from Fig. 6.18, with its left end held sta-tionary and its right end attached to a block with mass m that can move along thex-axis. In Fig. 7.13a the body is at when the spring is neither stretched norcompressed. We move the block to one side, thereby stretching or compressingthe spring, and then let it go. As the block moves from one position to anotherposition how much work does the elastic (spring) force do on the block?

We found in Section 6.3 that the work we must do on the spring to move oneend from an elongation to a different elongation is

(work done on a spring)

where k is the force constant of the spring. If we stretch the spring farther, we dopositive work on the spring; if we let the spring relax while holding one end, wedo negative work on it. We also saw that this expression for work is still correct ifthe spring is compressed, not stretched, so that or or both are negative. Nowwe need to find the work done by the spring. From Newton’s third law the twoquantities of work are just negatives of each other. Changing the signs in thisequation, we find that in a displacement from to the spring does an amountof work given by

(work done by a spring)

The subscript “el” stands for elastic. When and are both positive and(Fig. 7.13b), the spring does negative work on the block, which moves in

the while the spring pulls on it in the The springstretches farther, and the block slows down. When and are both positive and

(Fig. 7.13c), the spring does positive work as it relaxes and the blockspeeds up. If the spring can be compressed as well as stretched, or or bothmay be negative, but the expression for is still valid. In Fig. 7.13d, both and are negative, but is less negative than the compressed spring doespositive work as it relaxes, speeding the block up.

Just as for gravitational work, we can express the work done by the spring interms of a given quantity at the beginning and end of the displacement. Thisquantity is and we define it to be the elastic potential energy:

(elastic potential energy) (7.9)

Figure 7.14 is a graph of Eq. (7.9). The unit of is the joule (J), the unit usedfor all energy and work quantities; to see this from Eq. (7.9), recall that the unitsof k are and that

We can use Eq. (7.9) to express the work done on the block by the elasticforce in terms of the change in elastic potential energy:

(7.10)

When a stretched spring is stretched farther, as in Fig. 7.13b, is negative andincreases; a greater amount of elastic potential energy is stored in the spring.

When a stretched spring relaxes, as in Fig. 7.13c, x decreases, is positive, anddecreases; the spring loses elastic potential energy. Negative values of x referUel

Wel

Uel

Wel

Wel 51

2 kx1

2 21

2 kx2

2 5 Uel, 1 2 Uel, 2 5 2DUel

Wel

1 N # m 5 1 J.N/m

Uel

Uel 51

2 kx2

12 kx2

,

x1 ;x2x2

x1Wel

x2x1

x2 , x1

x2x1

2x-direction.1x-directionx2 . x1

x2x1

Wel 51

2 kx1

2 21

2 kx2

2

Wel

x2x1

x2x1

W 51

2 kx2

2 21

2 kx1

2

x2x1

x2 ,x1

x 5 0

7.2 Elastic Potential EnergyThere are many situations in which we encounter potential energy that is notgravitational in nature. One example is a rubber-band slingshot. Work is done onthe rubber band by the force that stretches it, and that work is stored in the rubberband until you let it go. Then the rubber band gives kinetic energy to theprojectile.

This is the same pattern we saw with the pile driver in Section 7.1: Do workon the system to store energy, which can later be converted to kinetic energy.We’ll describe the process of storing energy in a deformable body such as aspring or rubber band in terms of elastic potential energy (Fig. 7.12). A body iscalled elastic if it returns to its original shape and size after being deformed.

To be specific, we’ll consider storing energy in an ideal spring, like the oneswe discussed in Section 6.3. To keep such an ideal spring stretched by a distancex, we must exert a force where k is the force constant of the spring. Theideal spring is a useful idealization because many elastic bodies show this samedirect proportionality between force and displacement x, provided that x is suf-ficiently small.

FS

F 5 kx,

and We are given thatand (the crate is instantaneously at rest at

point 2). Our target variable in part (a) is the magnitude of thefriction force. In part (b) our target variable is the speed at thebottom of the ramp.

EXECUTE: (a) The energy quantities are

Here is the unknown magnitude of the friction force andUsing Eq. (7.7), we find

The friction force of 35 N, acting over 1.6 m, causes the mechani-cal energy of the crate to decrease from 150 J to 94 J (Fig. 7.11b).

5 2

10 1 94 J 2 2 1150 J 1 0 21.6 m

5 35 N

f 5 2

1K2 1 Ugrav, 2 2 2 1K1 1 Ugrav, 1 2s

Wother 5 2fs 5 1K2 1 Ugrav, 2 2 2 1K1 1 Ugrav, 1 2 K1 1 Ugrav, 1 1 Wother 5 K2 1 Ugrav, 2

s 5 1.6 m.f

Wother 5 2fs

Ugrav, 2 5 112 kg 2 19.8 m/s2 2 10.80 m 2 5 94 J

K2 5 0

Ugrav, 1 5 0

K1 51

2 112 kg 2 15.0 m/s 2 2 5 150 J

v3 ,f,

v2 5 0v1 5 5.0 m/sy3 5 0.y2 5 11.6 m 2 sin 30° 5 0.80 m, (b) On the way down from point 2 to point 3 at the bottom of

the ramp, the friction force and the displacement both reversedirection but have the same magnitudes, so the frictional work hasthe same negative value as from point 1 to point 2. The total workdone by friction between points 1 and 3 is

From part (a), and Equation (7.7) then gives

The crate returns to the bottom of the ramp with only 38 J of theoriginal 150 J of mechanical energy (Fig. 7.11b). Usingwe get

EVALUATE: The crate’s speed when it returns to the bottom of theramp, is less than the speed at which itleft that point. That’s good—energy was lost due to friction.

In part (b) we applied Eq. (7.7) to points 1 and 3, consideringthe entire round trip as a whole. Alternatively, we could have con-sidered the second part of the motion by itself and appliedEq. (7.7) to points 2 and 3. Try it and see whether you get the sameresult for v3 .

v1 5 5.0 m/sv3 5 2.5 m/s,

v3 5 Å2K3

m5 Å

2 138 J 212 kg

5 2.5 m/s

K3 5 12 mv3

2 ,

5 150 J 1 0 2 0 1 12112 J 2 5 38 J

K3 5 K1 1 Ugrav, 1 2 Ugrav, 3 1 Wother

K1 1 Ugrav, 1 1 Wother 5 K3 1 Ugrav, 3

Ugrav, 1 5 0.K1 5 150 J

Wother 5 Wfric 5 22fs 5 22 135 N 2 11.6 m 2 5 2112 J

Block I Block II mm

y1 y1

y2 y2

Test Your Understanding of Section 7.1 The figure shows two differentfrictionless ramps. The heights and are the same for both ramps. If a block ofmass m is released from rest at the left-hand end of each ramp, which block arrivesat the right-hand end with the greater speed? (i) block I; (ii) block II; (iii) the speed is thesame for both blocks.

y2y1

❚

7.12 The Achilles tendon, which runsalong the back of the ankle to the heelbone, acts like a natural spring. When itstretches and then relaxes, this tendonstores and then releases elastic potentialenergy. This spring action reduces theamount of work your leg muscles must doas you run.

Here the spring isneither stretchednor compressed.

As the spring stretches, it does negativework on the block.

As the spring relaxes, it does positivework on the block.

A compressed springalso does positivework on the block asit relaxes.

(a)

(b)

(c)

(d)

x 5 0

xO

m

xO

x1

x2

Fspringr

sS

m

x1

xO

x2

Fspringr

sS

m

x1

xO

x2

FspringS

sS

m

7.13 Calculating the work done by aspring attached to a block on a horizontalsurface. The quantity x is the extension orcompression of the spring.

Uel

Spring iscompressed:x , 0.

x

O Spring isstretched:x . 0.

7.14 The graph of elastic potential energyfor an ideal spring is a parabola:

where x is the extension orcompression of the spring. Elastic potentialenergy is never negative.Uel

Uel 5 12 kx2,


7.15 The fall of a bungee jumper involvesan interplay among kinetic energy, gravita-tional potential energy, and elastic poten-tial energy. Due to air resistance andfrictional forces within the bungee cord,mechanical energy is not conserved. (Ifmechanical energy were conserved, thebungee jumper would keep bouncing upand down forever!)

Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II

Problem-Solving Strategy 7.1 (Section 7.1) is equally useful insolving problems that involve elastic forces as well as gravitationalforces. The only new wrinkle is that the potential energy U nowincludes the elastic potential energy where x is the dis-Uel 5 1

2 kx2,

placement of the spring from its unstretched length. The work doneby the gravitational and elastic forces is accounted for by theirpotential energies; the work of the other forces, has to beincluded separately.

Wother ,

to a compressed spring. But, as Fig. 7.14 shows, is positive for both positiveand negative x, and Eqs. (7.9) and (7.10) are valid for both cases. The more aspring is compressed or stretched, the greater its elastic potential energy.

CAUTION Gravitational potential energy vs. elastic potential energy An impor-tant difference between gravitational potential energy and elastic potentialenergy is that we do not have the freedom to choose to be wherever wewish. To be consistent with Eq. (7.9), must be the position at which the spring isneither stretched nor compressed. At that position, its elastic potential energy and the forcethat it exerts are both zero. ❚

The work–energy theorem says that no matter what kind offorces are acting on a body. If the elastic force is the only force that does work onthe body, then

The work–energy theorem then gives us

(if only the elastic force does work) (7.11)

Here is given by Eq. (7.9), so

(7.12)

In this case the total mechanical energy —the sum of kinetic andelastic potential energy—is conserved. An example of this is the motion of theblock in Fig. 7.13, provided the horizontal surface is frictionless so that no forcedoes work other than that exerted by the spring.

For Eq. (7.12) to be strictly correct, the ideal spring that we’ve been dis-cussing must also be massless. If the spring has a mass, it also has kinetic energyas the coils of the spring move back and forth. We can neglect the kinetic energyof the spring if its mass is much less than the mass m of the body attached to thespring. For instance, a typical automobile has a mass of 1200 kg or more. Thesprings in its suspension have masses of only a few kilograms, so their mass canbe neglected if we want to study how a car bounces on its suspension.

Situations with Both Gravitational and Elastic Potential EnergyEquations (7.11) and (7.12) are valid when the only potential energy in the sys-tem is elastic potential energy. What happens when we have both gravitationaland elastic forces, such as a block attached to the lower end of a vertically hang-ing spring? And what if work is also done by other forces that cannot bedescribed in terms of potential energy, such as the force of air resistance on amoving block? Then the total work is the sum of the work done by the gravita-tional force the work done by the elastic force and the work doneby other forces Then the work–energy the-orem gives

The work done by the gravitational force is and the workdone by the spring is Hence we can rewrite the work–energytheorem for this most general case as

(7.13)(valid in general)

K1 1 Ugrav, 1 1 Uel, 1 1 Wother 5 K2 1 Ugrav, 2 1 Uel, 2

Wel 5 Uel, 1 2 Uel, 2 .Wgrav 5 Ugrav, 1 2 Ugrav, 2

Wgrav 1 Wel 1 Wother 5 K2 2 K1

Wtot 5 Wgrav 1 Wel 1 Wother .1Wother 2 : 1Wel 2 ,1Wgrav 2 ,

E 5 K 1 Uel

(if only the elasticforce does work)

1

2 mv1

2 11

2 kx1

2 51

2 mv2

2 11

2 kx2

2

Uel

K1 1 Uel, 1 5 K2 1 Uel, 2

Wtot 5 K2 2 K1

Wtot 5 Wel 5 Uel, 1 2 Uel, 2

Wtot 5 K2 2 K1 ,

x 5 0x 5 0Uel 5 1

2 kx2Ugrav 5 mgy

Uel or, equivalently,

(valid in general) (7.14)

where is the sum of gravitational potentialenergy and elastic potential energy. For short, we call U simply “the potentialenergy.”

Equation (7.14) is the most general statement of the relationship amongkinetic energy, potential energy, and work done by other forces. It says:

The work done by all forces other than the gravitational force or elastic forceequals the change in the total mechanical energy of the system,where is the sum of the gravitational potential energy and theelastic potential energy.

The “system” is made up of the body of mass m, the earth with which it interactsthrough the gravitational force, and the spring of force constant k.

If is positive, increases; if is negative, E decreases. Ifthe gravitational and elastic forces are the only forces that do work on the body,then and the total mechanical energy (including both gravitational andelastic potential energy) is conserved. (You should compare Eq. (7.14) toEqs. (7.7) and (7.8), which describe situations in which there is gravitationalpotential energy but no elastic potential energy.)

Bungee jumping (Fig. 7.15) is an example of transformations among kineticenergy, elastic potential energy, and gravitational potential energy. As the jumperfalls, gravitational potential energy decreases and is converted into the kineticenergy of the jumper and the elastic potential energy of the bungee cord. Beyonda certain point in the fall, the jumper’s speed decreases so that both gravitationalpotential energy and kinetic energy are converted into elastic potential energy.

Wother 5 0

WotherE 5 K 1 UWother

U 5 Ugrav 1 Uel

E 5 K 1 U

U 5 Ugrav 1 Uel 5 mgy 1 12 kx2

K1 1 U1 1 Wother 5 K2 1 U2

Example 7.7 Motion with elastic potential energy

SET UP: Figure 7.16 shows our sketches. The spring force is theonly force doing work on the glider, so and we may useWother 5 0

7.16 Our sketches and energy bar graphs for this problem.

Continued

A glider with mass sits on a frictionless horizontalair track, connected to a spring with force constant You pull on the glider, stretching the spring 0.100 m, and thenrelease it with no initial velocity. The glider begins to move backtoward its equilibrium position What is its x-velocitywhen

SOLUTION

IDENTIFY: Because the spring force varies with position, thisproblem can’t be solved with the equations for motion with con-stant acceleration. Instead, we’ll use the idea that as the gliderstarts to move, elastic potential energy is converted into kineticenergy. (The glider remains at the same height throughout themotion, so gravitational potential energy is not a factor. HenceU 5 Uel 5 1

2 kx2.)

x 5 0.080 m?1 x 5 0 2 .

k 5 5.00 N/m.m 5 0.200 kg

5.4 Inverse Bungee Jumper

5.5 Spring-Launched Bowler

O N L I N E


Example 7.8 Motion with elastic potential energy and work done by other forces

For the system of Example 7.7, suppose the glider is initially atrest at with the spring unstretched. You then apply a con-stant force in the with magnitude 0.610 N to theglider. What is the glider’s velocity when it has moved to

SOLUTION

IDENTIFY: Although the force you apply is constant, the springforce isn’t, so the acceleration of the glider won’t be constant.Total mechanical energy is not conserved because of the workdone by the force so we must use the generalized energy rela-tionship given by Eq. (7.13). (As in Example 7.7, we ignore grav-itational potential energy because the glider’s height doesn’tchange. Hence we have only elastic potential energy, and so

SET UP: Let point 1 be at where the velocity is and let point 2 be at (These points are different fromthe ones labeled in Fig. 7.16.) Our target variable is the veloc-ity at point 2.


(To calculate we multiplied the magnitude of the force bythe displacement, since both are in the Initially, thetotal mechanical energy is zero; the work done by the forceincreases the total mechanical energy to 0.0610 J, of which

FS

1x-direction.)Wother

Wother 5 10.610 N 2 1 0.100 m 2 5 0.0610 J

U2 51

2 kx2

2 51

2 15.00 N/m 2 10.100 m 2 2 5 0.0250 J

K2 51

2 mv2x

2

U1 51

2 kx1

2 5 0

K1 5 0

v2x ,x 5 0.100 m.

v1x 5 0,x 5 0,

U 5 Uel 5 12 kx2.)

FS

,

FS

x 5 0.100 m ?

1x-directionFS

x 5 0,0.0250 J is elastic potential energy. The remainder is kineticenergy. From Eq. (7.13),

We choose the positive square root because the glider is moving inthe

EVALUATE: To test our answer, think what would be different ifwe disconnected the glider from the spring. Then would be theonly force doing work, there would be zero potential energy at alltimes, and Eq. (7.13) would give us

We found a lower velocity than this value because the spring doesnegative work on the glider as it stretches (see Fig. 7.13b).

If you stop pushing on the glider when it reaches the pointbeyond that point the only force that does work on

the glider is the spring force. Hence for the totalmechanical energy is conserved and maintains thesame value of 0.0610 J. The glider will slow down as the springcontinues to stretch, so the kinetic energy K will decrease as thepotential energy increases. The glider will come to rest at a point

at this point the kinetic energy is zero and the potentialenergy is equal to the total mechanical energy0.0610 J. You should be able to show that the glider comes to restat which means that it moves an additional 0.056 mafter the force is removed at (Since there’s nofriction, the glider will not remain at rest but will start movingback toward due to the force of the stretched spring.)x 5 0

x2 5 0.100 m.FS

x3 5 0.156 m,

U 5 Uel 5 12 kx3

2x 5 x3 ;

E 5 K 1 Ux . 0.100 m,

x 5 0.100 m,

v2x 5 Å2K2

m5 Å

2 10.0610 J 20.200 kg

5 0.78 m/s

K2 5 K1 1 Wother 5 0 1 0.0610 J

FS

1x-direction.

v2x 5 Å2K2

m5 Å

2 10.0360 J 20.200 kg

5 0.60 m/s

5 0 1 0 1 0.0610 J 2 0.0250 J 5 0.0360 J

K2 5 K1 1 U1 1 Wother 2 U2

K1 1 U1 1 Wother 5 K2 1 U2

Eq. (7.11). We designate the point where the glider is released aspoint 1 and as point 2. We know the velocity at point1 our target variable is the x-velocity at point 2,


Then from Eq. (7.11),

v2x 5 6 Å2K2

m5 6 Å

2 10.0090 J 20.200 kg

5 60.30 m/s

K2 5 K1 1 U1 2 U2 5 0 1 0.0250 J 2 0.0160 J 5 0.0090 J

U2 51

2 kx2

2 51

2 15.00 N/m 2 1 0.080 m 2 2 5 0.0160 J

K2 51

2 mv2x

2

U1 51

2 kx1

2 51

2 15.00 n/m 2 10.100 m 2 2 5 0.0250 J

K1 51

2 mv1x

2 51

2 10.200 kg 2 10 2 2 5 0

v2x .1v1x 5 0 2 ;x 5 0.080 mWe choose the negative root because the glider is moving in the

the answer we want is

EVALUATE: What is the meaning of the second solution,Eventually the spring will compress and push

the glider back to the right in the positive x-direction (seeFig. 7.13d). The second solution tells us that when the gliderpasses through while moving to the right, its speedwill be —the same speed as when it passed through thispoint while moving to the left.

When the glider passes through the point the spring isrelaxed and all of the mechanical energy is in the form of kineticenergy. Can you show that the speed of the glider at this point is0.50 m/s?

x 5 0,

0.30 m/sx 5 0.080 m

v2x 5 10.30 m/s?

v2x 5 20.30 m/s.2x-direction; Example 7.9 Motion with gravitational, elastic, and friction forces

In a “worst-case” design scenario, a 2000-kg elevator with brokencables is falling at when it first contacts a cushioningspring at the bottom of the shaft. The spring is supposed to stop theelevator, compressing 2.00 m as it does so (Fig. 7.17). During themotion a safety clamp applies a constant 17,000-N frictional forceto the elevator. As a design consultant, you are asked to determinewhat the force constant of the spring should be.

SOLUTION

IDENTIFY: We’ll use the energy approach to determine the forceconstant, which appears in the expression for elastic potentialenergy. Note that this problem involves both gravitational andelastic potential energy. Furthermore, total mechanical energy isnot conserved because the friction force does negative work on the elevator.

SET UP: Since mechanical energy isn’t conserved and more thanone kind of potential energy is involved, we’ll use the most generalform of the energy relationship, Eq. (7.13). We take point 1 as theposition of the bottom of the elevator when it initially contacts thespring, and take point 2 as its position when it is at rest. We choosethe origin to be at point 1, so and With thischoice the coordinate of the upper end of the spring is the same asthe coordinate of the elevator, so the elastic potential energy at anypoint between point 1 and point 2 is (The gravitationalpotential energy is as usual.) We know the initial andfinal speeds of the elevator and the magnitude of the friction force,so the only unknown is the force constant k (our target variable).

EXECUTE: The elevator’s initial speed is so theinitial kinetic energy is

The elevator stops at point 2, so The potential energy atpoint 1, is zero; is zero because and because the spring is not yet compressed. At point 2 there is bothgravitational and elastic potential energy, so

The gravitational potential energy at point 2 is

The other force is the 17,000-N friction force, acting opposite tothe 2.00-m displacement, so

Putting these terms into we have

so the force constant of the spring is

5 1.06 3 104 N/m 5

2 316,000 J 1 1234,000 J 2 2 1239,200 J 2 4122.00 m 2 2

k 52 1K1 1 Wother 2 mgy2 2

y2

2

K1 1 0 1 Wother 5 0 1 1mgy2 11

2 ky2

2 2K1 1 U1 1 Wother 5 K2 1 U2 ,

Wother 5 2 117,000 N 2 12.00 m 2 5 234,000 J

mgy2 5 12000 kg 2 19.80 m/s2 2 122.00 m 2 5 239,200 J

U2 5 mgy2 11

2 ky2

2

Uel 5 0y1 5 0,UgravU1 ,K2 5 0.

K1 51

2 mv1

2 51

2 12000 kg 2 1 4.00 m/s 2 2 5 16,000 J

v1 5 4.00 m/s,

Ugrav 5 mgyUel 5 1

2 ky2.

y2 5 22.00 m.y1 5 0

Wother

4.00 m/s

This is about one-tenth the force constant of a spring in an automo-bile suspension.

EVALUATE: Let’s note what might seem to be a paradox in thisproblem. The elastic potential energy in the spring at point 2 is

This is more than the total mechanical energy at point 1:

But the friction force caused the mechanical energy of the systemto decrease by 34,000 J between point 1 and point 2. Does thismean that energy appeared from nowhere? Don’t panic; there is noparadox. At point 2 there is also negative gravitational potentialenergy, because point 2 is below the origin.The total mechanical energy at point 2 is

This is just the initial mechanical energy of 16,000 J, minus34,000 J lost to friction.

Will the elevator stay at the bottom of the shaft? At point 2the compressed spring exerts an upward force of magni-tude while thedownward force of gravity on the elevator is only

So if there were no friction,there would be a net upward force of

and the elevator would bounce back upward. However,there is friction in the safety clamp, which can exert a force of asmuch as 17,000 N; hence the clamp can keep the elevator fromrebounding.

1600 N21,200 N 2 19,600 N 5

12000 kg 2 19.80 m/s2 2 5 19,600 N.w 5 mg 5

Fspring 5 11.06 3 104 N/m 2 1 2.00 m 2 5 21,200 N,

5 0 1 21,200 J 1 1239,200 J 2 5 218,000 J

E2 5 K2 1 U2 5 0 11

2 ky2

2 1 mgy2

mgy2 5 239,200 J,

E1 5 K1 1 U1 5 16,000 J 1 0 5 16,000 J

1

2 ky2

2 51

2 11.06 3 104 N/m 2 122.00 m 2 2 5 21,200 J

2.00 m

w 5 mg

v1 5 4.00 m/s

m 5 2000 kg

f 5 17,000 N

v2 5 0

Point 1

Point 2

7.17 The fall of an elevator is stopped by a spring and by a con-stant friction force.

7.3 Conservative and Nonconservative Forces 229228 C HAPTE R 7 Potential Energy and Energy Conservation

Test Your Understanding of Section 7.2 Consider the situation in Exam-ple 7.9 at the instant when the elevator is still moving downward and the spring iscompressed by 1.00 m. Which of the energy bar graphs in the figure most accu-rately shows the kinetic energy K, gravitational potential energy and elastic poten-tial energy at this instant?Uel

Ugrav ,

( i) ( ii) (iii) (iv)

Ugrav Uel UelK

UgravK

Uel

UgravK

UelUgravK

❚

Finalposition

Initialposition

The work done by thegravitational force is thesame for all three paths,because this force isconservative.

7.18 The work done by a conservative force such as gravity depends only on the endpoints of a path, not on the specific path taken between those points.

stays close to the surface of the earth, the gravitational force is independent ofheight, and the work done by this force depends only on the change in height. Ifthe body moves around a closed path, ending at the same point where it started,the total work done by the gravitational force is always zero.

The work done by a conservative force always has four properties:

1. It can be expressed as the difference between the initial and final values ofa potential-energy function.

2. It is reversible.3. It is independent of the path of the body and depends only on the starting

and ending points.4. When the starting and ending points are the same, the total work is zero.

When the only forces that do work are conservative forces, the total mechanicalenergy is constant.

Nonconservative ForcesNot all forces are conservative. Consider the friction force acting on the cratesliding on a ramp in Example 7.6 (Section 7.1). When the body slides up and thenback down to the starting point, the total work done on it by the friction force isnot zero. When the direction of motion reverses, so does the friction force, andfriction does negative work in both directions. When a car with its brakes lockedskids across the pavement with decreasing speed (and decreasing kinetic energy),the lost kinetic energy cannot be recovered by reversing the motion or in anyother way, and mechanical energy is not conserved. There is no potential-energyfunction for the friction force.

In the same way, the force of fluid resistance (see Section 5.3) is not conserva-tive. If you throw a ball up in the air, air resistance does negative work on the ballwhile it’s rising and while it’s descending. The ball returns to your hand with lessspeed and less kinetic energy than when it left, and there is no way to get back thelost mechanical energy.

A force that is not conservative is called a nonconservative force. The workdone by a nonconservative force cannot be represented by a potential-energy func-tion. Some nonconservative forces, like kinetic friction or fluid resistance, causemechanical energy to be lost or dissipated; a force of this kind is called a dissipativeforce. There are also nonconservative forces that increase mechanical energy. Thefragments of an exploding firecracker fly off with very large kinetic energy,thanks to a chemical reaction of gunpowder with oxygen. The forces unleashed bythis reaction are nonconservative because the process is not reversible. (The frag-ments never spontaneously reassemble themselves into a complete firecracker!)

E 5 K 1 U

mgS

7.3 Conservative and Nonconservative ForcesIn our discussions of potential energy we have talked about “storing” kineticenergy by converting it to potential energy. We always have in mind that later wemay retrieve it again as kinetic energy. For example, when you throw a ball up inthe air, it slows down as kinetic energy is converted into potential energy. But onthe way down, the conversion is reversed, and the ball speeds up as potentialenergy is converted back to kinetic energy. If there is no air resistance, the ball ismoving just as fast when you catch it as when you threw it.

Another example is a glider moving on a frictionless horizontal air track thatruns into a spring bumper at the end of the track. The glider stops as it com-presses the spring and then bounces back. If there is no friction, the glider endsup with the same speed and kinetic energy it had before the collision. Again,there is a two-way conversion from kinetic to potential energy and back. In bothcases we can define a potential-energy function so that the total mechanicalenergy, kinetic plus potential, is constant or conserved during the motion.

Conservative ForcesA force that offers this opportunity of two-way conversion between kinetic andpotential energies is called a conservative force. We have seen two examples ofconservative forces: the gravitational force and the spring force. (Later in thisbook we will study another conservative force, the electric force betweencharged objects.) An essential feature of conservative forces is that their work isalways reversible. Anything that we deposit in the energy “bank” can later bewithdrawn without loss. Another important aspect of conservative forces is that abody may move from point 1 to point 2 by various paths, but the work done by aconservative force is the same for all of these paths (Fig. 7.18). Thus, if a body

Example 7.10 Frictional work depends on the path

You are rearranging your furniture and wish to move a 40.0-kgfuton 2.50 m across the room. However, the straight-line path isblocked by a heavy coffee table that you don’t want to move.Instead, you slide the futon in a dogleg path over the floor; thedoglegs are 2.00 m and 1.50 m long. Compared to the straight-linepath, how much more work must you do to push the futon in thedogleg path? The coefficient of kinetic friction is 0.200.

SOLUTION

IDENTIFY: Here work is done both by you and by the force offriction, so we must use the energy relationship that includesforces other than elastic or gravitational forces. We’ll use this rela-tionship to find a connection between the work that you do and thework done by friction.

SET UP: Figure 7.19 shows our sketch. The futon is at rest at bothpoint 1 and point 2, so There is no elastic potentialK1 5 K2 5 0.

7.19 Our sketch for this problem.

Continued

7.3 Conservative and Nonconservative Forces 231230 C HAPTE R 7 Potential Energy and Energy Conservation

7.21 When 1 liter of gasoline is burned inan automotive engine, it releases

of internal energy. Hencewhere the minus

sign means that the amount of energystored in the gasoline has decreased. Thisenergy can be converted into kineticenergy (making the car go faster) or intopotential energy (enabling the car to climbuphill).

23.3 3 107 J,DUint 53.3 3 107 JExample 7.11 Conservative or nonconservative?

In a certain region of space the force on an electron is where C is a positive constant. The electron moves in a counter-clockwise direction around a square loop in the xy-plane(Fig. 7.20). The corners of the square are at

and Calculate the work done on the elec-tron by the force during one complete trip around the square. Isthis force conservative or nonconservative?

SOLUTION

IDENTIFY: In Example 7.10 the force of friction was constant inmagnitude and always opposite to the displacement, so it waseasy to calculate the work done. Here, however, the force is notconstant and in general is not in the same direction as thedisplacement.

SET UP: To calculate the work done by the force we’ll use themore general expression for work, Eq. (6.14):

where is an infinitesimal displacement. Let’s calculate the workdone on each leg of the square and then add the results to find thework done on the round trip.

EXECUTE: On the first leg, from to the force variesbut is everywhere perpendicular to the displacement. So

and the work done on the first leg is The force has the same value everywhere on the second leg from

to The displacement on this leg is in theso and

The work done on the second leg is then

On the third leg, from to is again perpendicular tothe displacement so The force is zero on the final leg,W3 5 0.

FS10, L 2 ,1L, L 2

W2 5 31L, L21L, 02 F

S # d lS

5 3y5L

y50

CL dy 5 CL3L

0

dy 5 CL2

FS # d l

S

5 CL e # dy e 5 CL dy

d lS

5 dye1y-direction,1L, L 2 .1L, 0 2 F

S

5 CL e

W1 5 0.FS # d l

S

5 0,

1L, 0 2 ,10, 0 2

d lS

W 5 3P2

P1

FS # d l

S

FS

,

FS

FS

10, L 2 .1L, L 2 ,1L, 0 2 , 1 x, y 2 5 10, 0 2 ,FS

5 Cx e,

from to so no work is done and The workdone by the force on the round trip is

The starting and ending points are the same, but the total workdone by is not zero. This is a nonconservative force; it cannot berepresented by a potential-energy function.

EVALUATE: Because W is positive, the mechanical energyincreases as the electron goes around the loop. This is not a mathe-matical curiosity; it’s a description of what happens in an electricalgenerating plant. A loop of wire is moved through a magnetic field,which gives rise to a nonconservative force similar to the one in thisexample. Electrons in the wire gain energy as they move around theloop, and this energy is carried via transmission lines to the con-sumer. (We’ll discuss how this works in detail in Chapter 29.)

If the electron went around the loop clockwise instead of coun-terclockwise, the force would be unaffected but the direction ofeach infinitesimal displacement would reverse. Thus the sign ofwork would also reverse, and the work for a clockwise round tripwould be This is a different behavior than the non-conservative friction force. When a body slides over a stationarysurface with friction, the work done by friction is always negative,no matter what the direction of motion (see Example 7.6 inSection 7.1).

W 5 2CL2.

d lS

FS

FS

W 5 W1 1 W2 1 W3 1 W4 5 0 1 CL2 1 0 1 0 5 CL2

FS

W4 5 0.10, 0 2 ,10, L 2Leg 1

Leg 3

y

x

Leg 2

Leg 4

(L, 0)

(L, L)

(0, 0)

(0, L)

FS

FS

F 5 0S

dlS

dlS

dlS

dlS

F 5 CL jS

^

7.20 An electron moving around a square loop while being actedon by the force F

S

5 Cx e.

The Law of Conservation of EnergyNonconservative forces cannot be represented in terms of potential energy. Butwe can describe the effects of these forces in terms of kinds of energy other thankinetic and potential energy. When a car with locked brakes skids to a stop, thetires and the road surface both become hotter. The energy associated with thischange in the state of the materials is called internal energy. Raising the temper-ature of a body increases its internal energy; lowering the body’s temperaturedecreases its internal energy.

To see the significance of internal energy, let’s consider a block sliding on arough surface. Friction does negative work on the block as it slides, and thechange in internal energy of the block and surface (both of which get hotter) ispositive. Careful experiments show that the increase in the internal energy isexactly equal to the absolute value of the work done by friction. In other words,

where is the change in internal energy. If we substitute this into Eq. (7.7) or(7.14), we find

Writing and we can finally express this as

(law of conservation of energy) (7.15)

This remarkable statement is the general form of the law of conservation ofenergy. In a given process, the kinetic energy, potential energy, and internalenergy of a system may all change. But the sum of those changes is always zero.If there is a decrease in one form of energy, it is made up for by an increase in theother forms (Fig. 7.21). When we expand our definition of energy to includeinternal energy, Eq. (7.15) says: Energy is never created or destroyed; it onlychanges form. No exception to this rule has ever been found.

The concept of work has been banished from Eq. (7.15); instead, it sug-gests that we think purely in terms of the conversion of energy from oneform to another. For example, when you throw a baseball straight up, you con-vert a portion of the internal energy of your molecules into kinetic energy of thebaseball. This is converted into gravitational potential energy as the ball climbsand back to kinetic energy as the ball falls. If there is air resistance, part of theenergy is used to heat up the air and the ball and increase their internal energy.Energy is converted back into the kinetic form as the ball falls. If you catch theball in your hand, whatever energy was not lost to the air once again becomesinternal energy; the ball and your hand are now warmer than they were at thebeginning.

In Chapters 19 and 20, we will study the relationship of internal energy totemperature changes, heat, and work. This is the heart of the area of physicscalled thermodynamics.

DK 1 DU 1 DUint 5 0

DU 5 U2 2 U1 ,DK 5 K2 2 K1

K1 1 U1 2 DUint 5 K2 1 U2

DUint

DUint 5 2Wother

energy (there are no springs), and the gravitational potentialenergy does not change because the futon moves only horizontally,so From Eq. (7.14) it follows that The otherwork done on the futon is the sum of the positive work you do,

and the negative work done by the kinetic friction force.Since the sum of these is zero, we have

Thus to determine we’ll calculate the work done by friction.

EXECUTE: Because the floor is horizontal, the normal force on thefuton equals its weight mg, and the magnitude of the friction forceis The work you must do over each path is then

(straight-line path) 5 196 J

5 10.200 2 140.0 kg 2 1 9.80 m/s2 2 1 2.50 m 2 Wyou 5 2Wfric 5 2 12fk s 2 5 1mk mgs

fk 5 mk n 5 mk mg.

Wyou ,

Wyou 5 2Wfric

WfricWyou ,

Wother 5 0.U1 5 U2 .(dogleg path)

The extra work you must do is

EVALUATE: The work done by friction is on the straight-line path and on the dogleg path. The workdone by friction depends on the path taken, which illustrates thatfriction is a nonconservative force.

2274 JWfric 5 2Wyou 5 2196 J

274 J 2 196 J 5 78 J.

5 274 J

5 10.200 2 140.0 kg 2 1 9.80 m/s2 2 1 2.00 m 1 1.50 m 2 Wyou 5 2Wfric

?

Example 7.12 Work done by friction

Let’s look again at Example 7.5 (Section 7.1), in which yourcousin Throcky skateboards down a curved ramp. He starts withzero kinetic energy and 735 J of potential energy, and at the bottomhe has 450 J of kinetic energy and zero potential energy. So

and The work done bythe nonconservative friction forces is so the change ininternal energy is The wheels, theDUint 5 2Wother 5 1285 J.

2285 J,Wother 5 WfricDU 5 2735 J.DK 5 1450 J

bearings, and the ramp all get a little warmer as Throcky rollsdown. In accordance with Eq. (7.15), the sum of the energychanges equals zero:

The total energy of the system (including nonmechanical forms ofenergy) is conserved.

DK 1 DU 1 DUint 5 1450 J 1 12735 J 2 1 285 J 5 0

5.7 Modified Atwood Machine

O N L I N E

7.4 Force and Potential Energy 233232 C HAPTE R 7 Potential Energy and Energy Conservation

Test Your Understanding of Section 7.3 In a hydroelectric generating sta-tion, falling water is used to drive turbines (“water wheels”), which in turn runelectric generators. Compared to the amount of gravitational potential energyreleased by the falling water, how much electrical energy is produced? (i) the same;(ii) more; (iii) less.

❚

U 5 kx212 dU

dx

U

xO

Fx

Fx 5 2 5 2kx

xO

(a) Spring potential energy and force as functions of x

U

y

U 5 mgy

O

dUdy

Fy 5 2 5 2mg

(b) Gravitational potential energy and force as function of y

Fy

yO

Potential energy isa minimum at x 5 0.

Potential energydecreases as y decreases.

For x . 0, Fx , 0;force pushes bodytoward x 5 0.

For all y, Fy , 0; force pushesbody toward decreasing y.

For x , 0, Fx . 0;force pushes bodytoward x 5 0.

7.22 A conservative force is the negative derivative of the corresponding potential energy.

Example 7.13 An electric force and its potential energy

An electrically charged particle is held at rest at the point while a second particle with equal charge is free to move along thepositive x-axis. The potential energy of the system is

where C is a positive constant that depends on the magnitude of thecharges. Derive an expression for the x-component of force actingon the movable charged particle, as a function of its position.

SOLUTION

IDENTIFY: We are given the potential-energy function andwe want to find the force function

SET UP: We’ll use Eq. (7.16), Fx 1 x 2 5 2dU 1 x 2 /dx.

Fx 1 x 2 . U 1 x 2 ,

U 1 x 2 5C

x

x 5 0, EXECUTE: The derivative with respect to x of the function isSo the force on the movable charged particle for is

EVALUATE: The x-component of force is positive, correspondingto a repulsion between like electric charges. The potential energy isvery large when the particles are close together (small x) andapproaches zero as the particles move farther apart (large x); theforce pushes the movable particle toward large positive values ofx, for which the potential energy is less. The force gets weaker as the particles move farther apart (x increases). We’llstudy electric forces in greater detail in Chapter 21.

Fx 1 x 2 5 C/x2

Fx 1 x 2 5 2

dU 1 x 2dx

5 2C 12

1

x2 2 5C

x2

x . 021/x2.1/x

7.4 Force and Potential EnergyFor the two kinds of conservative forces (gravitational and elastic) we have stud-ied, we started with a description of the behavior of the force and derived fromthat an expression for the potential energy. For example, for a body with mass min a uniform gravitational field, the gravitational force is We foundthat the corresponding potential energy is To stretch an idealspring by a distance x, we exert a force equal to By Newton’s third law theforce that an ideal spring exerts on a body is opposite this, or The cor-responding potential energy function is

In studying physics, however, you’ll encounter situations in which you aregiven an expression for the potential energy as a function of position and have tofind the corresponding force. We’ll see several examples of this kind when westudy electric forces later in this book: it’s often far easier to calculate the elec-tric potential energy first and then determine the corresponding electric forceafterward.

Here’s how we find the force that corresponds to a given potential-energyexpression. First let’s consider motion along a straight line, with coordinate x. Wedenote the x-component of force, a function of x, by and the potentialenergy as This notation reminds us that both and U are functions of x.Now we recall that in any displacement, the work W done by a conservative forceequals the negative of the change in potential energy:

Let’s apply this to a small displacement The work done by the force during this displacement is approximately equal to We have to say“approximately” because may vary a little over the interval But it is atleast approximately true that

You can probably see what’s coming. We take the limit as in this limit,the variation of becomes negligible, and we have the exact relationship

(force from potential energy, one dimension) (7.16)

This result makes sense; in regions where changes most rapidly with x(that is, where is large), the greatest amount of work is done during agiven displacement, and this corresponds to a large force magnitude. Also, when

is in the positive x-direction, decreases with increasing x. So and should indeed have opposite signs. The physical meaning ofEq. (7.16) is that a conservative force always acts to push the system towardlower potential energy.

As a check, let’s consider the function for elastic potential energy, Substituting this into Eq. (7.16) yields

Fx 1 x 2 5 2

d

dx 112 kx2 2 5 2kx

12 kx2.

U 1 x 2 5dU 1 x 2 /dx

Fx 1 x 2U 1 x 2Fx 1 x 2dU 1 x 2 /dx

U 1 x 2Fx 1 x 2 5 2

dU 1 x 2dx

Fx

Dx S 0;

Fx 1 x 2 Dx 5 2DU and Fx 1 x 2 5 2

DU

Dx

Dx.Fx 1 x 2 Fx 1 x 2 Dx.Fx 1 x 2Dx.

W 5 2DU

DU

FxU 1 x 2 . Fx 1 x 2 ,

U 1 x 2 5 12 kx2.

Fx 5 2kx.1kx.

U 1 y 2 5 mgy.Fy 5 2mg.

which is the correct expression for the force exerted by an ideal spring (Fig. 7.22a). Similarly, for gravitational potential energy we have taking care to change x to y for the choice of axis, we get

which is the correct expression for gravitational force(Fig. 7.22b).2d 1mgy 2 /dy 5 2mg,

Fy 5 2dU/dy 5U 1 y 2 5 mgy;

Force and Potential Energy in Three DimensionsWe can extend this analysis to three dimensions, where the particle may move inthe x-, y-, or z-direction, or all at once, under the action of a conservative forcethat has components and Each component of force may be a function ofthe coordinates x, y, and z. The potential-energy function U is also a functionof all three space coordinates. We can now use Eq. (7.16) to find each componentof force. The potential-energy change when the particle moves a small dis-tance in the x-direction is again given by it doesn’t depend on and

which represent force components that are perpendicular to the displacementand do no work. So we again have the approximate relationship

The y- and z-components of force are determined in exactly the same way:

To make these relationships exact, we take the limits andso that these ratios become derivatives. Because U may be a function ofDz S 0

Dy S 0,Dx S 0,

Fy 5 2

DU

Dy Fz 5 2

DU

Dz

Fx 5 2

DU

Dx

Fz ,Fy2Fx Dx;Dx

DU

Fz .Fy ,Fx ,

7.5 Energy Diagrams 235234 C HAPTE R 7 Potential Energy and Energy Conservation

(a)

2A O A

x

U

K

U

2A O Ax

E 5 K 1 U

U 5 kx212

On the graph, the limits of motion are the pointswhere the U curve intersects the horizontal linerepresenting total mechanical energy E.

(b)

The limits of the glider’s motionare at x 5 A and x 5 2A.

7.23 (a) A glider on an air track. Thespring exerts a force (b) Thepotential-energy function.

Fx 5 2kx.

Example 7.14 Force and potential energy in two dimensions

A puck slides on a level, frictionless air-hockey table. The coordi-nates of the puck are x and y. It is acted on by a conservative forcedescribed by the potential-energy function

Derive an expression for the force acting on the puck, and find anexpression for the magnitude of the force as a function of position.

SOLUTION

IDENTIFY: Starting with the function we need to find thevector components and magnitude of the corresponding conserva-tive force

SET UP: We’ll find the components of the force from using Eq. (7.18). This function doesn’t depend on z, so the partialderivative of U with respect to z is and the force has noz-component. We’ll then determine the magnitude of the forceusing the formula for the magnitude of a vector:

EXECUTE: The x- and y-components of the force are

From Eq. (7.18) this corresponds to the vector expression

FS

5 2k 1 x d 1 ye 2

Fx 5 2

'U

'x5 2kx Fy 5 2

'U

'y5 2ky

F 5 "Fx

2 1 Fy

2 .

'U/'z 5 0

U 1 x, y 2FS

.

U 1 x, y 2 ,

U 1 x, y 2 51

2 k 1 x2 1 y2 2

Now is just the position vector of the particle, so we canrewrite this expression as This represents a force that ateach point is opposite in direction to the position vector of thepoint—that is, a force that at each point is directed toward the ori-gin. The potential energy is minimum at the origin, so again theforce pushes in the direction of decreasing potential energy.

The magnitude of the force at any point is

where r is the particle’s distance from the origin. This is the forcethat would be exerted on the puck if it were attached to one end ofa spring that obeys Hooke’s law and has a negligibly small length(compared to the other distances in the problem) when it is notstretched. (The other end is attached to the air-hockey table at theorigin.)

EVALUATE: To check our result, note that the potential-energyfunction can also be expressed as Written this way, U isa function of a single coordinate r, so we can find the force usingEq. (7.16) with x replaced by r:

Just as we calculated above, the force has magnitude the minussign indicates that the force is radially inward (toward the origin).

kr;

Fr 5 2

dU

dr5 2

d

dr 112 kr 2 2 5 2kr

U 5 12 kr 2.

F 5 "12kx 2 2 1 12ky 2 2 5 k "x2 1 y2 5 kr

FS

5 2krS.rSx d 1 ye

Test Your Understanding of Section 7.4 A particle moving along the x-axis is acted on by a conservative force At a certain point, the force is zero.(a) Which of the following statements about the value of the potential-energyfunction at that point is correct? (i) (ii) (iii) (iv) not enough information is given to decide. (b) Which of the following statementsabout the value of the derivative of at that point is correct? (i) (ii) (iii) (iv) not enough information is given to decide.

❚dU 1 x 2 /dx , 0;dU 1 x 2 /dx . 0;

dU 1 x 2 /dx 5 0;U 1 x 2U 1 x 2 , 0;U 1 x 2 . 0;U 1 x 2 5 0;U 1 x 2

Fx .

all three coordinates, we need to remember that when we calculate each of thesederivatives, only one coordinate changes at a time. We compute the derivative ofU with respect to x by assuming that y and z are constant and only x varies, and soon. Such a derivative is called a partial derivative. The usual notation for a partialderivative is and so on; the symbol is a modified d. So we write

(7.17)

We can use unit vectors to write a single compact vector expression for the force

(force from potential energy) (7.18)

The expression inside the parentheses represents a particular operation on thefunction U, in which we take the partial derivative of U with respect to each coor-dinate, multiply by the corresponding unit vector, and then take the vector sum.This operation is called the gradient of U and is often abbreviated as Thusthe force is the negative of the gradient of the potential-energy function:

(7.19)

As a check, let’s substitute into Eq. (7.19) the function for gravitationalpotential energy:

This is just the familiar expression for the gravitational force.

FS

5 2=S 1mgy 2 5 2 1' 1mgy 2

'x d 1

' 1mgy 2'y

e 1' 1mgy 2

'z k 2 5 12mg 2e

U 5 mgy

FS

5 2=S

U

=S

U.

FS

5 2 1'U

'x d 1

'U

'y e 1

'U

'z k 2

FS

:

(force from potential energy)

Fx 5 2

'U

'x Fy 5 2

'U

'y Fz 5 2

'U

'z

''U/'x

7.5 Energy DiagramsWhen a particle moves along a straight line under the action of a conservativeforce, we can get a lot of insight into its possible motions by looking at the graphof the potential-energy function Figure 7.23a shows a glider with mass mthat moves along the x-axis on an air track. The spring exerts on the glider a forcewith x-component Figure 7.23b is a graph of the corresponding poten-tial-energy function If the elastic force of the spring is the onlyhorizontal force acting on the glider, the total mechanical energy isconstant, independent of x. A graph of E as a function of x is thus a straight hori-zontal line. We use the term energy diagram for a graph like this, which showsboth the potential-energy function and the energy of the particle subjectedto the force that corresponds to

The vertical distance between the U and E graphs at each point represents thedifference equal to the kinetic energy K at that point. We see that K isgreatest at It is zero at the values of x where the two graphs cross, labeledA and in the diagram. Thus the speed is greatest at and it is zero at

the points of maximum possible displacement from for a givenvalue of the total energy E. The potential energy U can never be greater than thetotal energy E; if it were, K would be negative, and that’s impossible. The motionis a back-and-forth oscillation between the points and

At each point, the force on the glider is equal to the negative of the slope ofthe curve: (see Fig. 7.22a). When the particle is at theslope and the force are zero, so this is an equilibrium position. When x is positive,the slope of the curve is positive and the force is negative, directedtoward the origin. When x is negative, the slope is negative and is positive,again toward the origin. Such a force is called a restoring force; when the glideris displaced to either side of the force tends to “restore” it back to An analogous situation is a marble rolling around in a round-bottomed bowl. Wesay that is a point of stable equilibrium. More generally, any minimum ina potential-energy curve is a stable equilibrium position.

Figure 7.24a shows a hypothetical but more general potential-energy functionFigure 7.24b shows the corresponding force Points and

are stable equilibrium points. At each of these points, is zero because theslope of the curve is zero. When the particle is displaced to either side, theforce pushes back toward the equilibrium point. The slope of the curve isalso zero at points and and these are also equilibrium points. But when theparticle is displaced a little to the right of either point, the slope of the curve becomes negative, corresponding to a positive that tends to push the par-ticle still farther from the point. When the particle is displaced a little to the left,

is negative, again pushing away from equilibrium. This is analogous to a mar-ble rolling on the top of a bowling ball. Points and are called unstable equi-librium points; any maximum in a potential-energy curve is an unstableequilibrium position.

CAUTION Potential energy and the direction of a conservative force The direc-tion of the force on a body is not determined by the sign of the potential energy U. Rather,it’s the sign of that matters. As we discussed in Section 7.1, the physicallysignificant quantity is the difference in the value of U between two points, which is just

Fx 5 2dU/dx

x4x2

Fx

Fx

U 1 x 2x4 ,x2

U 1 x 2U 1 x 2 Fxx3

x1Fx 5 2dU/dx.U 1 x 2 .x 5 0

x 5 0.x 5 0,

Fx

FxU 1 x 2x 5 0,Fx 5 2dU/dxU 1 x 2 Fx

x 5 2A.x 5 A

x 5 0x 5 6A,x 5 0,v2A

x 5 0.E 2 U,

U 1 x 2 .U 1 x 2E 5 K 1 U

U 1 x 2 5 12 kx2.

Fx 5 2kx.

U 1 x 2 .

237

236 C HAPTE R 7 Potential Energy and Energy Conservation

E3

xc xa x1 xdx2 x3 x4xb

U

E2

E1

E0

Ox

(a) A hypothetical potential-energy function U(x)

Stable equilibrium points, are mimimain the potential-energy curve.

Minimum possible energy is E0; the particle is at rest at x1.

If E = E1, the particle is trapped between xa and xb.

If E = E2, the particle is trapped between xc and xd.

If the total energy E . E3, the particle can “escape” to x . x4.

Unstable equilibrium points are maximain the potential-energy curve.

dU/dx . 0Fx � 0

(b) The corresponding x-component of force Fx(x) 5 dU(x)/dx

Fx

O x1 x2 x3 x4x

dU/dx . 0Fx , 0

dU/dx , 0Fx . 0

dU/dx , 0Fx . 0

dU/dx , 0Fx . 0

7.24 The maxima and minima of a potential-energy function correspond to points where Fx 5 0.U 1 x 2

Test Your Understanding of Section 7.5 The curve in Fig. 7.24b has amaximum at a point between and Which statement correctly describes whathappens to the particle when it is at this point? (i) The particle’s acceleration iszero. (ii) The particle accelerates in the positive x-direction; the magnitude of the acceler-ation is less than at any other point between and (iii) The particle accelerates in thepositive x-direction; the magnitude of the acceleration is greater than at any other pointbetween and (iv) The particle accelerates in the negative x-direction; the magnitudeof the acceleration is less than at any other point between and (v) The particleaccelerates in the negative x-direction; the magnitude of the acceleration is greater than atany other point between and

❚x3 .x2

x3 .x2

x3 .x2

x3 .x2

x3 .x2

what the derivative measures. This means that you can always add a con-stant to the potential-energy function without changing the physics of the situation. ❚

If the total energy is and the particle is initially near it can move only inthe region between and determined by the intersection of the and Ugraphs (Fig. 7.24a). Again, U cannot be greater than because K can’t be nega-tive. We speak of the particle as moving in a potential well, and and are theturning points of the particle’s motion (since at these points, the particle stopsand reverses direction). If we increase the total energy to the level the particlecan move over a wider range, from to If the total energy is greater than the particle can “escape” and move to indefinitely large values of x. At the otherextreme, represents the least possible total energy the system can have.E0

E3 ,xd .xc

E2 ,

xbxa

E1

E1xbxa

x1 ,E1

Fx 5 2dU/dx

CHAPTER 7 SUMMARY

Gravitational potential energy and elastic potentialenergy: The work done on a particle by a constantgravitational force can be represented as a change in thegravitational potential energy This energyis a shared property of the particle and the earth. Apotential energy is also associated with the elastic force

exerted by an ideal spring, where x is theamount of stretch or compression. The work done bythis force can be represented as a change in the elasticpotential energy of the spring, Uel 5 1

2 kx2.

Fx 5 2kx

Ugrav 5 mgy.

(7.1), (7.3)

(7.10)

5 Uel, 1 2 Uel, 2 5 2DUel

Wel 51

2 kx1

2 21

2 kx2

2

5 2DUgrav

5 Ugrav,1 2 Ugrav,2

Wgrav 5 mgy1 2 mgy2

Ugrav,2 5 mgy2

Uel 5 kx2

x 5 0 xx

O

y

Ugrav,1 5 mgy1 12

At y 5 h

At y 5 0

y

h

xO

zero

E 5 K 1 Ugrav

E 5 K 1 Ugrav

Point 2

Point 1

f

f

w

w

w

R

f 5 0n 5 0

nn

zero

E K5 1Ugrav

At point 2

zero

E UgravK5 1

At point 1

When total mechanical energy is conserved: The totalpotential energy U is the sum of the gravitational andelastic potential energy: If no forcesother than the gravitational and elastic forces do work ona particle, the sum of kinetic and potential energy is con-served. This sum is called the total mechan-ical energy. (See Examples 7.1, 7.3, 7.4, and 7.7.)

E 5 K 1 U

U 5 Ugrav 1 Uel .

(7.4), (7.11)K1 1 U1 5 K2 1 U2

When total mechanical energy is not conserved:When forces other than the gravitational and elasticforces do work on a particle, the work done bythese other forces equals the change in total mechanicalenergy (kinetic energy plus total potential energy). (SeeExamples 7.2, 7.5, 7.6, 7.8, and 7.9.)

Wother

(7.14)K1 1 U1 1 Wother 5 K2 1 U2

Conservative forces, nonconservative forces, and thelaw of conservation of energy: All forces are eitherconservative or nonconservative. A conservative force isone for which the work–kinetic energy relationship iscompletely reversible. The work of a conservative forcecan always be represented by a potential-energy func-tion, but the work of a nonconservative force cannot.The work done by nonconservative forces manifestsitself as changes in the internal energy of bodies. Thesum of kinetic, potential, and internal energy is alwaysconserved. (See Examples 7.10–7.12.)

(7.15)DK 1 DU 1 DUint 5 0 v

v 5 0As friction slows block,mechanical energy is convertedto internal energy of block and ramp.

zero

zero

zero

E K5 1UgravE UgravK5 1

Determining force from potential energy: For motionalong a straight line, a conservative force is thenegative derivative of its associated potential-energyfunction U. In three dimensions, the components of aconservative force are negative partial derivatives of U.(See Examples 7.13 and 7.14.)

Fx 1 x 2 Unstable equilibria

Stable equilibria

U

Ox

(7.16)

(7.17)

(7.18)FS

5 2 1'U

'x d 1

'U

'y e 1

'U

'z k 2

Fz 5 2

'U

'y

Fx 5 2

'U

'x Fy 5 2

'U

'y

Fx 1 x 2 5 2

dU 1 x 2dx

Exercises 239238 C HAPTE R 7 Potential Energy and Energy Conservation

Key Termspotential energy, 214gravitational potential energy, 214total mechanical energy, 215conservation of mechanical energy, 215elastic potential energy, 223

conservative force, 228nonconservative force, 229dissipative force, 229internal energy, 231law of conservation of energy, 231

gradient, 234energy diagram, 235stable equilibrium, 235unstable equilibrium, 236

Answer to Chapter Opening Question ?Gravity is doing positive work on the diver, since this force is inthe same downward direction as his displacement. This corre-sponds to a decrease in gravitational potential energy. The water isdoing negative work on the diver; it exerts an upward force of fluidresistance as he moves downward. This corresponds to an increasein internal energy of the diver and the water (see Section 7.3).

Answers to Test Your Understanding Questions7.1 Answer: (iii) The initial kinetic energy the initialpotential energy and the final potential energy

are the same for both blocks. Mechanical energy isconserved in both cases, so the final kinetic energy isalso the same for both blocks. Hence the speed at the right-handend is the same in both cases!

K 5 12 mv2

2U2 5 mgy2

U1 5 mgy1 ,K1 5 0,

7.2 Answer: (iii) The elevator is still moving downward, so thekinetic energy K is positive (remember that K can never be nega-tive); the elevator is below point 1, so and and thespring is compressed, so 7.3 Answer: (iii) Because of friction in the turbines and betweenthe water and turbines, some of the potential energy goes into rais-ing the temperatures of the water and the mechanism.7.4 Answers: (a) (iv), (b) (i) If at a point, then the deriva-tive of must be zero at that point because However, this tells us absolutely nothing about the value of at that point.7.5 Answers: (iii) Figure 7.24b shows the x-component of force,

Where this is maximum (most positive), the x-component offorce and the x-acceleration have more positive values than atadjacent values of x.

Fx .

U 1 x 2Fx 5 2dU 1 x 2 /dx.U 1 x 2 Fx 5 0

Uel . 0.Ugrav , 0;y , 0

PROBLEMS For instructor-assigned homework, go to www.masteringphysics.com

Discussion Questions

Q7.1. A baseball is thrown straight up with initial speed If airresistance cannot be ignored, when the ball returns to its initialheight its speed is less than Explain why, using energy concepts.Q7.2. A projectile has the same initial kinetic energy no matterwhat the angle of projection. Why doesn’t it rise to the same maxi-mum height in each case?Q7.3. Does an object’s speed at the bottom of a frictionless rampdepend on the shape of the ramp or just on its height? Explain.What if the ramp is not frictionless?Q7.4. An egg is released from rest from the roof of a building andfalls to the ground. Its fall is observed by a student on the roof of thebuilding, who uses coordinates with origin at the roof, and by a stu-dent on the ground, who uses coordinates with origin at the ground.Do the two students assign the same or different values to the initialgravitational potential energy, the final gravitational potentialenergy, the change in gravitational potential energy, and the kineticenergy of the egg just before it strikes the ground? Explain.Q7.5. A physics teacher had a bowling ball suspended from a verylong rope attached to the high ceiling of a large lecture hall. Toillustrate his faith in conservation of energy, he would back up toone side of the stage, pull the ball far to one side until the taut ropebrought it just to the end of his nose, and then release it. The mas-sive ball would swing in a mighty arc across the stage and thenreturn to stop momentarily just in front of the nose of the station-ary, unflinching teacher. However, one day after the demonstrationhe looked up just in time to see a student at the other side of thestage push the ball away from his nose as he tried to duplicate thedemonstration. Tell the rest of the story and explain the reason forthe potentially tragic outcome.

v0 .

v0 .

Q7.6. Lost Energy? The principle of the conservation of energytells us that energy is never lost, but only changes from one form toanother. Yet in many ordinary situations, energy may appear to belost. In each case, explain what happens to the “lost” energy. (a) Abox sliding on the floor comes to a halt due to friction. How didfriction take away its kinetic energy, and what happened to thatenergy? (b) A car stops when you apply the brakes. What happenedto its kinetic energy? (c) Air resistance uses up some of the originalgravitational potential energy of a falling object. What type ofenergy did the “lost” potential energy become? (d) When a return-ing space shuttle touches down on the runway, it has lost almost allits kinetic energy and gravitational potential energy. Where did allthat energy go?Q7.7. Is it possible for a frictional force to increase the mechanicalenergy of a system? If so, give examples.Q7.8. A woman bounces on a trampoline, going a little higher witheach bounce. Explain how she increases the total mechanical energy.Q7.9. Fractured Physics. People often call their electric bill apower bill, yet the quantity on which the bill is based is expressedin kilowatt-hours. What are people really being billed for?Q7.10. A rock of mass m and a rock of mass 2m are both releasedfrom rest at the same height and feel no air resistance as they fall.Which statements about these rocks are true? (There may be morethan one correct choice.) (a) Both have the same initial gravitationalpotential energy. (b) Both have the same kinetic energy when theyreach the ground. (c) Both reach the ground with the same speed.(d) When it reaches the ground, the heavier rock has twice thekinetic energy of the lighter one. (e) When it reaches the ground, theheavier rock has four times the kinetic energy of the lighter one.Q7.11. On a friction-free ice pond, a hockey puck is pressed against(but not attached to) a fixed ideal spring, compressing the spring

by a distance The maximum energy stored in the spring is the maximum speed the puck gains after being released is andits maximum kinetic energy is Now the puck is pressed so itcompresses the spring twice as far as before. In this case, (a) whatis the maximum potential energy stored in the spring (in terms of

and (b) what are the puck’s maximum kinetic energy andspeed (in terms of and Q7.12. When people are cold, they often rub their hands together towarm them up. How does doing this produce heat? Where did theheat come from?Q7.13. You often hear it said that most of our energy ultimatelycomes from the sun. Trace each of the following energies back tothe sun. (a) the kinetic energy of a jet plane; (b) the potential energygained by a mountain climber; (c) the electrical energy used to runa computer; (d) the electrical energy from a hydroelectric plant.Q7.14. A box slides down a ramp and work is done on the box bythe forces of gravity and friction. Can the work of each of theseforces be expressed in terms of the change in a potential-energyfunction? For each force explain why or why not.Q7.15. In physical terms, explain why friction is a nonconservativeforce. Does it store energy for future use?Q7.16. A compressed spring is clamped in its compressed positionand then is dissolved in acid. What becomes of its potential energy?Q7.17. Since only changes in potential energy are important in anyproblem, a student decides to let the elastic potential energy of aspring be zero when the spring is stretched a distance The studentdecides, therefore, to let Is this correct? Explain.Q7.18. Figure 7.22a shows the potential-energy function for theforce Sketch the potential-energy function for the force

For this force, is a point of equilibrium? Is thisequilibrium stable or unstable? Explain.Q7.19. Figure 7.22b shows the potential-energy function associ-ated with the gravitational force between an object and the earth.Use this graph to explain why objects always fall toward the earthwhen they are released.Q7.20. For a system of two particles we often let the potentialenergy for the force between the particles approach zero as theseparation of the particles approaches infinity. If this choice ismade, explain why the potential energy at noninfinite separation ispositive if the particles repel one another and negative if theyattract.Q7.21. Explain why the points and in Fig. 7.23bare called turning points. How are the values of E and U related ata turning point?Q7.22. A particle is in neutral equilibrium if the net force on it iszero and remains zero if the particle is displaced slightly in anydirection. Sketch the potential-energy function near a point of neu-tral equilibrium, for the case of one-dimensional motion. Give anexample of an object in neutral equilibrium.Q7.23. The net force on a particle of mass m has the potential-energy function graphed in Fig. 7.24a. If the total energy is graph the speed of the particle versus its position x. At whatvalue of x is the speed greatest? Sketch versus x if the totalenergy is Q7.24. The potential-energy function for a force is where is a positive constant. What is the direction of

ExercisesSection 7.1 Gravitational Potential Energy7.1. In one day, a 75-kg mountain climber ascends from the1500-m level on a vertical cliff to the top at 2400 m. The next day,

FS

?aU 5 ax3,F

S

E2 .v

vE1 ,

x 5 2Ax 5 A

x 5 0Fx 5 1kx.Fx 5 2kx.

U 5 12 k 1 x 2 x1 2 2.

x1 .

x0)?K0

U0),

K0 .v0 ,

U0 ,x0 . she descends from the top to the base of the cliff, which is at anelevation of 1350 m. What is her change in gravitational potentialenergy (a) on the first day and (b) on the second day?7.2. A 5.00-kg sack of flour is lifted vertically at a constant speedof through a height of 15.0 m. (a) How great a force isrequired? (b) How much work is done on the sack by the liftingforce? What becomes of this work?7.3. A 120-kg mail bag hangs by a vertical rope 3.5 m long. Apostal worker then displaces the bag to a position 2.0 m sidewaysfrom its original position, always keeping the rope taut. (a) Whathorizontal force is necessary to hold the bag in the new position?(b) As the bag is moved to this position, how much work is done(i) by the rope and (ii) by the worker?7.4. A 72.0-kg swimmer jumps into the old swimming hole from adiving board 3.25 m above the water. Use energy conservation tofind his speed just he hits the water (a) if he just holds his nose anddrops in, (b) if he bravely jumps straight up (but just beyond theboard!) at and (c) if he manages to jump downward at

7.5. A baseball is thrown from the roof of a 22.0-m-tall buildingwith an initial velocity of magnitude and directed at anangle of above the horizontal. (a) What is the speed of theball just before it strikes the ground? Use energy methods andignore air resistance. (b) What is the answer for part (a) if the ini-tial velocity is at an angle of below the horizontal? (c) If theeffects of air resistance are included, will part (a) or (b) give thehigher speed?7.6. A crate of mass M starts from rest at the top of a frictionlessramp inclined at an angle above the horizontal. Find its speed atthe bottom of the ramp, a distance d from where it started. Do this intwo ways: (a) Take the level at which the potential energy is zeroto be at the bottom of the ramp with y positive upward. (b) Takethe zero level for potential energy to be at the top of the ramp withy positive upward. (c) Why did the normal force not enter into yoursolution?7.7. Answer part (b) of Example 7.6 (Section 7.1) by applyingEq. (7.7) to points 2 and 3, rather than to points 1 and 3 as wasdone in the example.7.8. An empty crate is given an initial push down a ramp, startingit with a speed and reaches the bottom with speed and kineticenergy K. Some books are now placed in the crate, so that the totalmass is quadrupled. The coefficient of kinetic friction is constantand air resistance is negligible. Starting again with at the top ofthe ramp, what are the speed and kinetic energy at the bottom?Explain the reasoning behind your answers.7.9. A small rock with mass0.20 kg is released from rest atpoint A, which is at the topedge of a large, hemisphericalbowl with radius(Fig. 7.25). Assume that the sizeof the rock is small compared toR, so that the rock can betreated as a particle, and assume that the rock slides rather thanrolls. The work done by friction on the rock when it moves frompoint A to point B at the bottom of the bowl has magnitude 0.22 J.(a) Between points A and B, how much work is done on the rockby (i) the normal force and (ii) gravity? (b) What is the speed ofthe rock as it reaches point B? (c) Of the three forces acting onthe rock as it slides down the bowl, which (if any) are constantand which are not? Explain. (d) Just as the rock reaches point B,what is the normal force on it due to the bottom of the bowl?

R 5 0.50 m

v0

vv0 ,

a

53.1°

53.1°12.0 m/s

2.50 m/s.2.50 m/s,

3.50 m/s

RA

B

v

Figure 7.25 Exercise 7.9.

Problems 241240 C HAPTE R 7 Potential Energy and Energy Conservation

7.10. A stone of mass m is thrown upward at an angle above thehorizontal and feels no appreciable air resistance. Use conserva-tion of energy to show that at its highest point, it is a distance

above the point where it was launched. (Hint:

7.11. You are testing a new amusement park roller coaster with anempty car with mass 120 kg. One part of the track is a vertical loopwith radius 12.0 m. At the bottom of the loop (point A) the car hasspeed and at the top of the loop (point B) it has speed

As the car rolls from point A to point B, how much workis done by friction?7.12. Tarzan and Jane. Tarzan, in one tree, sights Jane inanother tree. He grabs the end of a vine with length 20 m thatmakes an angle of with the vertical, steps off his tree limb, andswings down and then up to Jane’s open arms. When he arrives,his vine makes an angle of with the vertical. Determinewhether he gives her a tender embrace or knocks her off her limbby calculating Tarzan’s speed just before he reaches Jane. You canignore air resistance and the mass of the vine.7.13. A 10.0-kg microwave oven is pushed 8.00 m up the slopingsurface of a loading ramp inclined at an angle of above thehorizontal, by a constant force with a magnitude 110 N and actingparallel to the ramp. The coefficient of kinetic friction between theoven and the ramp is 0.250. (a) What is the work done on the ovenby the force (b) What is the work done on the oven by the fric-tion force? (c) Compute the increase in potential energy for theoven. (d) Use your answers to parts (a), (b), and (c) to calculate theincrease in the oven’s kinetic energy. (e) Use to calcu-late the acceleration of the oven. Assuming that the oven is initiallyat rest, use the acceleration to calculate the oven’s speed after trav-eling 8.00 m. From this, compute the increase in the oven’s kineticenergy, and compare it to the answer you got in part (d).7.14. Pendulum. A small rock with mass 0.12 kg is fastened to amassless string with length 0.80 m to form a pendulum. The pendu-lum is swinging so as to make a maximum angle of with the ver-tical. Air resistance is negligible. (a) What is the speed of the rockwhen the string passes through the vertical position? (b) What is thetension in the string when it makes an angle of with the vertical?(c) What is the tension in the string as it passes through the vertical?

Section 7.2 Elastic Potential Energy7.15. A force of 800 N stretches a certain spring a distance of0.200 m. (a) What is the potential energy of the spring when it isstretched 0.200 m? (b) What is its potential energy when it is com-pressed 5.00 cm?7.16. An ideal spring of negligible mass is 12.00 cm long whennothing is attached to it. When you hang a 3.15-kg weight from it,you measure its length to be 13.40 cm. If you wanted to store10.0 J of potential energy in this spring, what would be its totallength? Assume that it continues to obey Hooke’s law.7.17. A spring stores potential energy when it is compressed adistance from its uncompressed length. (a) In terms of howmuch energy does it store when it is compressed (i) twice as muchand (ii) half as much? (b) In terms of how much must it becompressed from its uncompressed length to store (i) twice asmuch energy and (ii) half as much energy?7.18. A slingshot will shoot a 10-g pebble 22.0 m straight up.(a) How much potential energy is stored in the slingshot’s rubberband? (b) With the same potential energy stored in the rubberband, how high can the slingshot shoot a 25-g pebble? (c) Whatphysical effects did you ignore in solving this problem?

x0 ,

U0 ,x0

U0

45°

45°

gFS

5 maS

FS

?

FS

36.9°

30°

45°

8.0 m/s.25.0 m/s,

v0

2 5 v0x

2 1 v0y

2.)v0

2 1 sin2 u 2 /2g

u 7.19. A spring of negligible mass has force constant (a) How far must the spring be compressed for

of potential energy to be stored in it? (b) You place the spring ver-tically with one end on the floor. You then drop a 1.20-kg bookonto it from a height of 0.80 m above the top of the spring. Findthe maximum distance the spring will be compressed.7.20. A 1.20-kg piece of cheese is placed on a vertical spring ofnegligible mass and force constant that is com-pressed 15.0 cm. When the spring is released, how high does thecheese rise from this initial position? (The cheese and the springare not attached.)7.21. Consider the glider of Example 7.7 (Section 7.2) andFig. 7.16. As in the example, the glider is released from rest withthe spring stretched 0.100 m. What is the displacement x of theglider from its equilibrium position when its speed is (You should get more than one answer. Explain why.)7.22. Consider the glider of Example 7.7 (Section 7.2) andFig. 7.16. (a) As in the example, the glider is released from restwith the spring stretched 0.100 m. What is the speed of the gliderwhen it returns to (b) What must the initial displacement ofthe glider be if its maximum speed in the subsequent motion is tobe 7.23. A 2.50-kg mass is pushed against a horizontal spring of forceconstant on a frictionless air table. The spring isattached to the tabletop, and the mass is not attached to the springin any way. When the spring has been compressed enough to store11.5 J of potential energy in it, the mass is suddenly released fromrest. (a) Find the greatest speed the mass reaches. When does thisoccur? (b) What is the greatest acceleration of the mass, and whendoes it occur?7.24. (a) For the elevator of Example 7.9 (Section 7.2), what is thespeed of the elevator after it has moved downward 1.00 m frompoint 1 in Fig. 7.17? (b) When the elevator is 1.00 m below point 1in Fig. 7.17, what is its acceleration?7.25. You are asked to design a spring that will give a 1160-kgsatellite a speed of relative to an orbiting space shuttle.Your spring is to give the satellite a maximum acceleration of5.00g. The spring’s mass, the recoil kinetic energy of the shuttle,and changes in gravitational potential energy will all be negligible.(a) What must the force constant of the spring be? (b) What dis-tance must the spring be compressed?

Section 7.3 Conservative and Nonconservative Forces7.26. A 75-kg roofer climbs a vertical 7.0-m ladder to the flat roofof a house. He then walks 12 m on the roof, climbs down anothervertical 7.0-m ladder, and finally walks on the ground back to hisstarting point. How much work is done on him by gravity (a) as heclimbs up; (b) as he climbs down; (c) as he walks on the roof andon the ground? (d) What is the total work done on him by gravityduring this round trip? (e) On the basis of your answer to part (d),would you say that gravity is a conservative or nonconservativeforce? Explain.7.27. A 10.0-kg box is pulled by a horizontal wire in a circle on arough horizontal surface for which the coefficient of kinetic fric-tion is 0.250. Calculate the work done by friction during one com-plete circular trip if the radius is (a) 2.00 m and (b) 4.00 m. (c) Onthe basis of the results you just obtained, would you say that fric-tion is a conservative or nonconservative force? Explain.7.28. In an experiment, one of the forces exerted on a proton is

where (a) How much work does dowhen the proton moves along the straight-line path from the point

FS

a 5 12 N/m2.FS

5 2ax2 d,

2.50 m/s

25.0 N/cm

2.50 m/s?

x 5 0?

0.20 m/s?

k 5 1800 N/m

3.20 J1600 N/m.k 5 to the point (b) Along the straight-

line path from the point to the point (c) Along the straight-line path from the point to thepoint (d) Is the force conservative? Explain. If isconservative, what is the potential-energy function for it? Let

when 7.29. A 0.60-kg book slides on a horizontal table. The kinetic fric-tion force on the book has magnitude 1.2 N. (a) How much work isdone on the book by friction during a displacement of 3.0 m to theleft? (b) The book now slides 3.0 m to the right, returning to itsstarting point. During this second 3.0-m displacement, how muchwork is done on the book by friction? (c) What is the total workdone on the book by friction during the complete round trip?(d) On the basis of your answer to part (c), would you say that thefriction force is conservative or nonconservative? Explain.7.30. You and three friends standat the corners of a square whosesides are 8.0 m long in the mid-dle of the gym floor, as shown inFig. 7.26. You take your physicsbook and push it from one personto the other. The book has a massof 1.5 kg, and the coefficient ofkinetic friction between the bookand the floor is (a) Thebook slides from you to Beth and then from Beth to Carlos, alongthe lines connecting these people. What is the work done by fric-tion during this displacement? (b) You slide the book from you toCarlos along the diagonal of the square. What is the work done byfriction during this displacement? (c) You slide the book to Kimwho then slides it back to you. What is the total work done by fric-tion during this motion of the book? (d) Is the friction force on thebook conservative or nonconservative? Explain.7.31. A block with mass m is attached to an ideal spring that hasforce constant k. (a) The block moves from to where

How much work does the spring force do during thisdisplacement? (b) The block moves from to and then from

to How much work does the spring force do during the dis-placement from to What is the total work done by thespring during the entire displacement? Explain whyyou got the answer you did. (c) The block moves from towhere How much work does the spring force do duringthis displacement? The block then moves from to Howmuch work does the spring force do during this displacement?What is the total work done by the spring force during the

displacement? Compare your answer to the answerin part (a), where the starting and ending points are the same butthe path is different.

Section 7.4 Force and Potential Energy7.32. The potential energy of a pair of hydrogen atoms separatedby a large distance x is given by where is apositive constant. What is the force that one atom exerts on theother? Is this force attractive or repulsive?7.33. A force parallel to the x-axis acts on a particle moving alongthe x-axis. This force produces potential energy given by

where What is the force (magnitudeand direction) when the particle is at 7.34. Gravity in One Dimension. Two point masses, and

lie on the x-axis, with held in place at the origin and atposition x and free to move. The gravitational potential energy of

m2m1m2 ,m1

x 5 20.800 m ?a 5 1.20 J/m4.U 1 x 2 5 ax4,

U 1 x 2

C6U 1 x 2 5 2C6/x6,

x1 S x3 S x2

x2 .x3

x3 . x2 .x3 ,x1

x1 S x2 S x1

x1 ?x2

x1 .x2

x2x1

x2 . x1 .x2 ,x1

mk 5 0.25.

x 5 0.U 5 0

FS

FS10.10 m, 0 2 ? 10.30 m, 0 21 0.30 m, 0 2 ?10.10 m, 0 21 0.10 m, 0.40 m 2 ?10.10 m, 0 2 these masses is found to be where G is a

constant (called the gravitational constant). You’ll learn more aboutgravitation in Chapter 12. Find the x-component of the force actingon due to Is this force attractive or repulsive? How do youknow?7.35. Gravity in Two Dimen-sions. Two point masses,and lie in the xy-plane, with

held in place at the origin andfree to move a distance r away

at a point P having coordinates xand y (Fig. 7.27). The gravita-tional potential energy of thesemasses is found to be

where G isthe gravitational constant. (a) Show that the components of theforce on due to are

(Hint: First write r in terms of x and y.) (b) Show that the magni-tude of the force on is (c) Does attract orrepel How do you know?7.36. An object moving in the xy-plane is acted on by a conserva-tive force described by the potential-energy function

where is a positive constant. Derive an expres-sion for the force expressed in terms of the unit vectors and

Section 7.5 Energy Diagrams7.37. The potential energy of two atoms in a diatomic molecule isapproximated by where r is the spacingbetween atoms and a and b are positive constants. (a) Find theforce on one atom as a function of r. Make two graphs, oneof versus r and one of versus r. (b) Find the equilib-rium distance between the two atoms. Is this equilibrium stable?(c) Suppose the distance between the two atoms is equal to theequilibrium distance found in part (b). What minimum energymust be added to the molecule to dissociate it—that is, to separatethe two atoms to an infinite distance apart? This is called the diss-ociation energy of the molecule. (d) For the molecule CO, theequilibrium distance between the carbon and oxygen atoms is

and the dissociation energy is permolecule. Find the values of the constants a and b.7.38. A marble moves along thex-axis. The potential-energyfunction is shown in Fig. 7.28.(a) At which of the labeledx-coordinates is the force on themarble zero? (b) Which of thelabeled x-coordinates is a posi-tion of stable equilibrium?(c) Which of the labeled x-coor-dinates is a position of unstable equilibrium?

Problems7.39. At a construction site, a 65.0-kg bucket of concrete hangsfrom a light (but strong) cable that passes over a light friction-freepulley and is connected to an 80.0-kg box on a horizontal roof(Fig. 7.29). The cable pulls horizontally on the box, and a 50.0-kg

1.54 3 10218 J1.13 3 10210 m

F 1 r 2U 1 r 2F 1 r 2U 1 r 2 5 a/r 12 2 b/r 6,

e.d

aa 11/x2 1 1/y2 2 , U 1 x, y 2 5m2 ?

m1F 5 Gm1 m2/r 2.m2

Fx 5 2

Gm1 m2 x

1 x2 1 y2 2 3/2 and Fy 5 2

Gm1 m2 y

1 x2 1 y2 2 3/2

m1m2

U 1 r 2 5 2Gm1 m2/r,

m2

m1

m2 ,m1

m1 .m2

U 1 x 2 5 2Gm1 m2/x,

8.0 mBeth Carlos

8.0 m

You Kim


y

xO

m2 P(x, y)

m1

r


Ox

a b c d

U



m 5 2.00 kgk 5 400 N/m

37.0°

0.220 m

Figure 7.30 Problem 7.42.

k 5 100 N/m m 5 0.50 kg

1.00 m

0.20 m

Figure 7.31 Problem 7.43.bag of gravel rests on top of the box. The coefficients of frictionbetween the box and roof are shown. (a) Find the friction forceon the bag of gravel and on the box. (b) Suddenly a worker picksup the bag of gravel. Use energy conservation to find the speed ofthe bucket after it has descended 2.00 m, from rest. (You cancheck your answer by solving this problem using Newton’s laws.)

entering the rough horizontal region, the stone travels 100 m andthen runs into a very long, light spring with force constant

The coefficients of kinetic and static friction betweenthe stone and the horizontal ground are 0.20 and 0.80, respec-tively. (a) What is the speed of the stone when it reaches point B?(b) How far will the stone compress the spring? (c) Will the stonemove again after it has been stopped by the spring?7.50. A 2.8-kg block slidesover the smooth, icy hill shownin Fig. 7.35. The top of the hillis horizontal and 70 m higherthan its base. What minimumspeed must the block have atthe base of the hill so that itwill not fall into the pit on thefar side of the hill?7.51. Bungee Jump. A bungee cord is 30.0 m long and, whenstretched a distance x, it exerts a restoring force of magnitude kx.Your father-in-law (mass 95.0 kg) stands on a platform 45.0 mabove the ground, and one end of the cord is tied securely to hisankle and the other end to the platform. You have promised himthat when he steps off the platform he will fall a maximum dis-tance of only 41.0 m before the cord stops him. You had severalbungee cords to select from, and you tested them by stretchingthem out, tying one end to a tree, and pulling on the other end witha force of 380.0 N. When you do this, what distance will thebungee cord that you should select have stretched?7.52. Ski Jump Ramp. You are designing a ski jump ramp forthe next Winter Olympics. You need to calculate the vertical heighth from the starting gate to the bottom of the ramp. The skiers pushoff hard with their ski poles at the start, just above the starting gate,so they typically have a speed of as they reach the gate.For safety, the skiers should have a speed of no more than when they reach the bottom of the ramp. You determine that for a85.0-kg skier with good form, friction and air resistance will dototal work of magnitude 4000 J on him during his run down theslope. What is the maximum height h for which the maximum safespeed will not be exceeded?7.53. The Great Sandini is a 60-kg circus performer who is shotfrom a cannon (actually a spring gun). You don’t find many men ofhis caliber, so you help him design a new gun. This new gun has avery large spring with a very small mass and a force constant of

that he will compress with a force of 4400 N. The insideof the gun barrel is coated with Teflon, so the average friction forcewill be only 40 N during the 4.0 m he moves in the barrel. At whatspeed will he emerge from the end of the barrel, 2.5 m above hisinitial rest position?7.54. You are designing a delivery ramp for crates containing exer-cise equipment. The 1470-N crates will move at at the topof a ramp that slopes downward at The ramp exerts a 550-Nkinetic friction force on each crate, and the maximum static fric-tion force also has this value. Each crate will compress a spring atthe bottom of the ramp and will come to rest after traveling a totaldistance of 8.0 m along the ramp. Once stopped, a crate must notrebound back up the ramp. Calculate the force constant of thespring that will be needed in order to meet the design criteria.7.55. A system of two paint buckets connected by a lightweightrope is released from rest with the 12.0-kg bucket 2.00 m abovethe floor (Fig. 7.36). Use the principle of conservation of energy tofind the speed with which this bucket strikes the floor. You canignore friction and the mass of the pulley.

22.0°.1.8 m/s

1100 N/m

30.0 m/s2.0 m/s

2.00 N/m.

12.0 kg

4.0 kg

2.00 m


Rocket islaunchedupward.

Rocket starts here.

53°


ms 5 0.700

Box

Gravel

Concrete

mk 5 0.400


7.40. Two blocks with different mass are attached to either end of alight rope that passes over a light, frictionless pulley that is sus-pended from the ceiling. The masses are released from rest, and themore massive one starts to descend. After this block has descended1.20 m, its speed is If the total mass of the two blocks is15.0 kg, what is the mass of each block?7.41. Legal Physics. In an auto accident, a car hit a pedestrianand the driver then slammed on the brakes to stop the car. Duringthe subsequent trial, the driver’s lawyer claimed that he was obey-ing the posted speed limit, but that the legal speed was toohigh to allow him to see and react to the pedestrian in time. Youhave been called in as the state’s expert witness. Your investigationof the accident found that the skid marks made while the brakeswere applied were 280 ft long, and the tread on the tires produced acoefficient of kinetic friction of 0.30 with the road. (a) In your testi-mony in court, will you say that the driver was obeying the postedspeed? You must be able to back up your conclusion with clear rea-soning because one of the lawyers will surely cross-examine you.(b) If the driver’s speeding ticket were $10 for each mile per hourhe was driving above the posted speed limit, would he have to paya fine? If so, how much would it be?7.42. A 2.00-kg block is pushed against a spring with negligiblemass and force constant compressing it 0.220 m.When the block is released, it moves along a frictionless, horizontalsurface and then up a frictionless incline with slope(Fig. 7.30). (a) What is the speed of the block as it slides along thehorizontal surface after having left the spring? (b) How far doesthe block travel up the incline before starting to slide back down?

37.0°

k 5 400 N/m,

35 mi/h

3.00 m/s.

7.44. On a horizontal surface, a crate with mass 50.0 kg is placedagainst a spring that stores 360 J of energy. The spring is released,and the crate slides 5.60 m before coming to rest. What is thespeed of the crate when it is 2.00 m from its initial position?7.45. Bouncing Ball. A 650-gram rubber ball is dropped from aninitial height of 2.50 m, and on each bounce it returns to 75% of itsprevious height. (a) What is the initial mechanical energy of theball, just after it is released from its initial height? (b) How muchmechanical energy does the ball lose during its first bounce? Whathappens to this energy? (c) How much mechanical energy is lostduring the second bounce?7.46. Riding a Loop-the-Loop.A car in an amusement park riderolls without friction around thetrack shown in Fig. 7.32. Itstarts from rest at point A at aheight h above the bottom of theloop. Treat the car as a particle.(a) What is the minimum value of h (in terms of R) such that the carmoves around the loop without falling off at the top (point B)? (b) If

and compute the speed, radial acceleration,and tangential acceleration of the passengers when the car is at pointC, which is at the end of a horizontal diameter. Show these accelera-tion components in a diagram, approximately to scale.7.47. A 2.0-kg piece of woodslides on the surface shown inFig. 7.33. The curved sides areperfectly smooth, but the roughhorizontal bottom is 30 m longand has a kinetic friction coefficient of 0.20 with the wood. Thepiece of wood starts from rest 4.0 m above the rough bottom.(a) Where will this wood eventually come to rest? (b) For themotion from the initial release until the piece of wood comes torest, what is the total amount of work done by friction?7.48. Up and Down the Hill. A 28-kg rock approaches the footof a hill with a speed of This hill slopes upward at a con-stant angle of above the horizontal. The coefficients of staticand kinetic friction between the hill and the rock are 0.75 and 0.20,respectively. (a) Use energy conservation to find the maximumheight above the foot of the hill reached by the rock. (b) Will therock remain at rest at its highest point, or will it slide back downthe hill? (c) If the rock does slide back down, find its speed when itreturns to the bottom of the hill.7.49. A 15.0-kg stone slides downa snow-covered hill (Fig. 7.34),leaving point A with a speed of

There is no friction onthe hill between points A and B,but there is friction on the levelground at the bottom of the hill,between B and the wall. After

10.0 m/s.

40.0°15 m/s.

R 5 20.0 m,h 5 3.50R

A

CB

Rh


Rough bottom

Wood


Rough15 m

20 m

B

A


120 m

70 m

40 m

50 m


7.56. A 1500-kg rocket is to be launched with an initial upwardspeed of In order to assist its engines, the engineers willstart it from rest on a ramp that rises 53° above the horizontal(Fig. 7.37). At the bottom, the ramp turns upward and launches therocket vertically. The engines provide a constant forward thrust of2000 N, and friction with the ramp surface is a constant 500 N.How far from the base of the ramp should the rocket start, asmeasured along the surface of the ramp?

50.0 m/s.

7.43. A block with mass 0.50 kg is forced against a horizontalspring of negligible mass, compressing the spring a distance of0.20 m (Fig. 7.31). When released, the block moves on a horizon-tal tabletop for 1.00 m before coming to rest. The spring constant kis What is the coefficient of kinetic friction betweenthe block and the tabletop?

mk100 N/m.

7.57. A machine part of mass m is attached to a horizontal idealspring of force constant k that is attached to the edge of a friction-free horizontal surface. The part is pushed against the spring,compressing it a distance and then released from rest. Findthe maximum (a) speed and (b) acceleration of the machinepart. (c) Where in the motion do the maxima in parts (a) and(b) occur? (d) What will be the maximum extension of thespring? (e) Describe the subsequent motion of this machine part.Will it ever stop permanently?7.58. A wooden rod of negligible mass and length 80.0 cm is piv-oted about a horizontal axis through its center. A white rat withmass 0.500 kg clings to one end of the stick, and a mouse withmass 0.200 kg clings to the other end. The system is releasedfrom rest with the rod horizontal. If the animals can manage tohold on, what are their speeds as the rod swings through a verticalposition?7.59. A 0.100-kg potato is tied to a string with length 2.50 m, andthe other end of the string is tied to a rigid support. The potato isheld straight out horizontally from the point of support, with thestring pulled taut, and is then released. (a) What is the speed of thepotato at the lowest point of its motion? (b) What is the tension inthe string at this point?

x0 ,


m 5 0.200 kg

B C

A

3.00 m

R 5 1.60 m


Distance 5 ?LIcy road

Ski’sVan Lines

Truck ramp

v0

ba


a


F 2

1

Unstretchedlength

a

u

S


7.60. These data are from a computer simulation for a batted base-ball with mass 0.145 kg, including air resistance:

t x y

0 0 03.05 s 70.2 m 53.6 m 06.59 s 124.4 m 0

(a) How much work was done by the air on the baseball as itmoved from its initial position to its maximum height? (b) Howmuch work was done by the air on the baseball as it moved fromits maximum height back to the starting elevation? (c) Explainwhy the magnitude of the answer in part (b) is smaller than themagnitude of the answer in part (a).7.61. Down the Pole. A fireman of mass m slides a distance ddown a pole. He starts from rest. He moves as fast at the bottom asif he had stepped off a platform a distance above the groundand descended with negligible air resistance. (a) What averagefriction force did the fireman exert on the pole? Does your answermake sense in the special cases of and (b) Find anumerical value for the average friction force a 75-kg firemanexerts, for and (c) In terms of g, h, and d,what is the speed of the fireman when he is a distance y above thebottom of the pole?7.62. A 60.0-kg skier starts from rest at the top of a ski slope65.0 m high. (a) If frictional forces do of work on heras she descends, how fast is she going at the bottom of the slope?(b) Now moving horizontally, the skier crosses a patch of softsnow, where If the patch is 82.0 m wide and the aver-age force of air resistance on the skier is 160 N, how fast is shegoing after crossing the patch? (c) The skier hits a snowdrift andpenetrates 2.5 m into it before coming to a stop. What is the aver-age force exerted on her by the snowdrift as it stops her?7.63. A skier starts at the top of a very large, frictionless snowball,with a very small initial speed, and skis straight down the side(Fig. 7.38). At what point does she lose contact with the snowballand fly off at a tangent? That is, at the instant she loses contactwith the snowball, what angle does a radial line from the centerof the snowball to the skier make with the vertical?

a

mk 5 0.20.

210.5 kJ

h 5 1.0 m.d 5 2.5 m

h 5 0 ?h 5 d

h # d

228.7 m/s11.9 m/s18.6 m/s

40.0 m/s30.0 m/svyvx

7.65. In a truck-loading station at a post office, a small 0.200-kgpackage is released from rest at point A on a track that is one-quarterof a circle with radius 1.60 m (Fig. 7.39). The size of the package ismuch less than 1.60 m, so the package can be treated as a particle. Itslides down the track and reaches point B with a speed ofFrom point B, it slides on a level surface a distance of 3.00 m topoint C, where it comes to rest. (a) What is the coefficient of kineticfriction on the horizontal surface? (b) How much work is done onthe package by friction as it slides down the circular arc from A to B?

4.80 m/s.

block with weight w is moved, and the spring to which it isattached is stretched from position 1 to position 2. The spring hasnegligible mass and force constant k. The end of the spring movesin an arc of radius a. Calculate the work done by the force 7.69. A 0.150-kg block of ice is placed against a horizontal, com-pressed spring mounted on a horizontal tabletop that is 1.20 mabove the floor. The spring has force constant and is ini-tially compressed 0.045 m. The mass of the spring is negligible.The spring is released, and the block slides along the table, goesoff the edge, and travels to the floor. If there is negligible frictionbetween the block of ice and the tabletop, what is the speed of theblock of ice when it reaches the floor?7.70. A 3.00-kg block is con-nected to two ideal horizontalsprings having force constants

and (Fig. 7.42). The sys-

tem is initially in equilibrium on a horizontal, frictionless surface.The block is now pushed 15.0 cm to the right and released fromrest. (a) What is the maximum speed of the block? Where in themotion does the maximum speed occur? (b) What is the maximumcompression of spring 1? 7.71. An experimental apparatus with mass m is placed on a verti-cal spring of negligible mass and pushed down until the spring iscompressed a distance x. The apparatus is then released andreaches its maximum height at a distance h above the point whereit is released. The apparatus is not attached to the spring, and at itsmaximum height it is no longer in contact with the spring. Themaximum magnitude of acceleration the apparatus can have with-out being damaged is a, where (a) What should the forceconstant of the spring be? (b) What distance x must the spring becompressed initially?7.72. If a fish is attached to a vertical spring and slowly loweredto its equilibrium position, it is found to stretch the spring by anamount d. If the same fish is attached to the end of theunstretched spring and then allowed to fall from rest, throughwhat maximum distance does it stretch the spring? (Hint: Calcu-late the force constant of the spring in terms of the distance d andthe mass m of the fish.)7.73. A wooden block with mass 1.50 kg is placed against a com-pressed spring at the bottom of an incline of slope (point A).When the spring is released, it projects the block up the incline. Atpoint B, a distance of 6.00 m up the incline from A, the block is mov-ing up the incline at and is no longer in contact with thespring. The coefficient of kinetic friction between the block and theincline is The mass of the spring is negligible. Calculatethe amount of potential energy that was initially stored in the spring.7.74. A 2.00-kg package isreleased on a incline,4.00 m from a long springwith force constant

that is attachedat the bottom of the incline(Fig. 7.43). The coeffi-cients of friction betweenthe package and theincline are and

The mass ofthe spring is negligible.(a) What is the speed of the package just before it reaches thespring? (b) What is the maximum compression of the spring?(c) The package rebounds back up the incline. How close does itget to its initial position?

mk 5 0.20.ms 5 0.40

120 N/m

53.1°

mk 5 0.50.

7.00 m/s

30.0°

a . g.

20.0 N/cmk2 5k1 5 25.0 N/cm

1900 N/m

FS

.

7.75. A 0.500-kg block, attached to a spring with length 0.60 mand force constant is at rest with the back of the block atpoint A on a frictionless, horizontal air table (Fig. 7.44). The massof the spring is negligible. You move the block to the right alongthe surface by pulling with a constant 20.0-N horizontal force.(a) What is the block’s speed when the back of the block reachespoint B, which is 0.25 m to the right of point A? (b) When the backof the block reaches point B, you let go of the block. In the subse-quent motion, how close does the block get to the wall where theleft end of the spring is attached?

40.0 N/m,

m 5 2.00 kg

53.1°

4.00 m


k 5 40.0 N/m m 5 0.500 kg

F 5 20.0 N

0.60 m 0.25 m

A B


7.64. A rock is tied to a cord and the other end of the cord is heldfixed. The rock is given an initial tangential velocity that causes itto rotate in a vertical circle. Prove that the tension in the cord at thelowest point exceeds the tension at the highest point by six timesthe weight of the rock.

7.66. A truck with mass m has a brake failure while going down anicy mountain road of constant downward slope angle (Fig. 7.40).Initially the truck is moving downhill at speed After careeningdownhill a distance L with negligible friction, the truck driversteers the runaway vehicle onto a runaway truck ramp of constantupward slope angle The truck ramp has a soft sand surface forwhich the coefficient of rolling friction is What is the distancethat the truck moves up the ramp before coming to a halt? Solveusing energy methods.

mr .b.

v0 .a

7.67. A certain spring is found not to obey Hooke’s law; it exerts arestoring force if it is stretched or com-pressed, where and The mass ofthe spring is negligible. (a) Calculate the potential-energy function

for this spring. Let when (b) An object withmass 0.900 kg on a frictionless, horizontal surface is attached tothis spring, pulled a distance 1.00 m to the right (the to stretch the spring, and released. What is the speed of the objectwhen it is 0.50 m to the right of the equilibrium position?7.68. A variable force is maintained tangent to a frictionless,semicircular surface (Fig. 7.41). By slow variations in the force, a

FS

x 5 0

1x-direction)

x 5 0.U 5 0U 1 x 2b 5 18.0 N/m2.a 5 60.0 N/m

Fx 1 x 2 5 2ax 2 bx2

k1 k2


7.76. Fraternity Physics. The brothers of Iota Eta Pi fraternitybuild a platform, supported at all four corners by vertical springs,in the basement of their frat house. A brave fraternity brother wear-ing a football helmet stands in the middle of the platform; hisweight compresses the springs by 0.18 m. Then four of his frater-nity brothers, pushing down at the corners of the platform, com-press the springs another 0.53 m until the top of the brave brother’shelmet is 0.90 m below the basement ceiling. They then simultane-ously release the platform. You can ignore the masses of thesprings and platform. (a) When the dust clears, the fraternity asksyou to calculate their fraternity brother’s speed just before his hel-met hit the flimsy ceiling. (b) Without the ceiling, how high wouldhe have gone? (c) In discussing their probation, the dean of stu-dents suggests that the next time they try this, they do it outdoorson another planet. Would the answer to part (b) be the same if thisstunt were performed on a planet with a different value of g?Assume that the fraternity brothers push the platform down 0.53 mas before. Explain your reasoning.7.77. A particle with mass m is acted on by a conservative forceand moves along a path given by and where and are constants. (a) Find the components of theforce that acts on the particle. (b) Find the potential energy of theparticle as a function of x and y. Take when and

(c) Find the total energy of the particle when (i) and (ii)

7.78. When it is burned, 1 gallon of gasoline produces of energy. A 1500-kg car accelerates from rest to in 10 s.The engine of this car is only 15% efficient (which is typical),meaning that only 15% of the energy from the combustion of thegasoline is used to accelerate the car. The rest goes into things likethe internal kinetic energy of the engine parts as well as heating ofthe exhaust air and engine. (a) How many gallons of gasoline doesthis car use during the acceleration? (b) How many such accelera-tions will it take to burn up 1 gallon of gas?7.79. A hydroelectric dam holds back a lake of surface area

that has vertical sides below the water level. Thewater level in the lake is 150 m above the base of the dam. Whenthe water passes through turbines at the base of the dam, itsmechanical energy is converted into electrical energy with 90%efficiency. (a) If gravitational potential energy is taken to be zero at

3.0 3 106 m2

37 m/s1.3 3 108 J

y 5 y0 .x 5 0,y 5 0x 5 x0 ,y 5 0.

x 5 0U 5 0

v0y0 ,x0 ,y 5 y0 sin v0 t,x 5 x0 cos v0 t

246 C HAPTE R 7 Potential Energy and Energy Conservation

the base of the dam, how much energy is stored in the top meter ofthe water in the lake? The density of water is (b) Whatvolume of water must pass through the dam to produce 1000 kilo-watt-hours of electrical energy? What distance does the level ofwater in the lake fall when this much water passes through the dam?7.80. How much total energy is stored in the lake in Problem 7.79?As in that problem, take the gravitational potential energy to bezero at the base of the dam. Express your answer in joules and inkilowatt-hours. (Hint: Break the lake up into infinitesimal horizon-tal layers of thickness dy, and integrate to find the total potentialenergy.)7.81. Gravity in Three Dimensions. A point mass is heldin place at the origin, and another point mass is free to movea distance r away at a point P having coordinates x, y, and z.The gravitational potential energy of these masses is found to be

where G is the gravitational constant (seeExercises 7.34 and 7.35). (a) Show that the components of theforce on due to are

(Hint: First write r in terms of x, y, and z.) (b) Show that the mag-nitude of the force on is (c) Does attract orrepel How do you know?7.82. (a) Is the force where C is a negative constantwith units of conservative or nonconservative? Justify youranswer. (b) Is the force where C is a negative constantwith units of conservative or nonconservative? Justify youranswer.7.83. A cutting tool under microprocessor control has severalforces acting on it. One force is a force in the nega-tive y-direction whose magnitude depends on the position of thetool. The constant is Consider the displacement ofthe tool from the origin to the point (a) Calculate the work done on the tool by if this displacement isalong the straight line that connects these two points.(b) Calculate the work done on the tool by if the tool is firstmoved out along the x-axis to the point andthen moved parallel to the y-axis to the point

(c) Compare the work done by along these twopaths. Is conservative or nonconservative? Explain.7.84. An object has several forces acting on it. One force is

a force in the x-direction whose magnitude depends onthe position of the object. (See Problem 6.96.) The constant is

The object moves along the following path: (1) Itstarts at the origin and moves along the y-axis to the point

(2) it moves parallel to the x-axis to the point(3) it moves parallel to the y-axis to they 5 1.50 m;x 5 1.50 m,

y 5 1.50 m;x 5 0,

a 5 2.00 N/m2.

FS

5 axyd,

FS

FS

y 5 3.00 m.x 5 3.00 m,

x 5 3.00 m, y 5 0FS

y 5 xFS

y 5 3.00 m.x 5 3.00 m,a 5 2.50 N/m3.

FS

5 2axy2 e,

N/m2,FS

5 Cy2 d,

N/m2,FS

5 Cy2 e,

m2 ?m1F 5 Gm1 m2/r 2.m2

Fz 5 2Gm1 m2z

1 x2 1 y2 1 z2 2 3/2

Fy 5 2

Gm1 m2 y

1 x2 1 y2 1 z2 2 3/2 Fx 5 2

Gm1 m2x

1 x2 1 y2 1 z2 2 3/2

m1m2

U 1 r 2 5 2Gm1 m2/r,

m2

m1

1000 kg/m3.point (4) it moves parallel to the x-axis back tothe origin. (a) Sketch this path in the xy-plane. (b) Calculate thework done on the object by for each leg of the path and for thecomplete round trip. (c) Is conservative or nonconservative?Explain.7.85. A Hooke’s law force and a constant conservative forceF in the act on an atomic ion. (a) Show that a possiblepotential-energy function for this combination of forces is

Is this the only possible function?Explain. (b) Find the stable equilibrium position. (c) Graph (in units of versus x (in units of for values of x between

and (d) Are there any unstable equilibrium positions?(e) If the total energy is what are the maximum andminimum values of x that the ion reaches in its motion? If the ionhas mass m, find its maximum speed if the total energy is

For what value of x is the speed maximum?7.86. A particle moves along thex-axis while acted on by a singleconservative force parallel to thex-axis. The force corresponds tothe potential-energy functiongraphed in Fig. 7.45. The particleis released from rest at point A.(a) What is the direction of theforce on the particle when it is at point A? (b) At point B? (c) Atwhat value of x is the kinetic energy of the particle a maximum?(d) What is the force on the particle when it is at point C? (e) Whatis the largest value of x reached by the particle during its motion?(f) What value or values of x correspond to points of stable equilib-rium? (g) Of unstable equilibrium?

Challenge Problem7.87. A proton with mass m moves in one dimension. The poten-tial-energy function is where and arepositive constants. The proton is released from rest at (a) Show that can be written as

Graph Calculate and thereby locate the point onthe graph. (b) Calculate the speed of the proton as a functionof position. Graph and give a qualitative description of themotion. (c) For what value of x is the speed of the proton a maxi-mum? What is the value of that maximum speed? (d) What is theforce on the proton at the point in part (c)? (e) Let the proton bereleased instead at Locate the point on the graph of

Calculate and give a qualitative description of themotion. (f ) For each release point what arethe maximum and minimum values of x reached during themotion?

1 x 5 x0 and x 5 x1 2 ,v 1 x 2U 1 x 2 . x1x1 5 3a/b.

v 1 x 2v 1 x 2 ,x0U 1 x0 2U 1 x 2 .

U 1 x 2 5a

x0

2 S 1 x0

x 2 2

2x0

x TU 1 x 2 x0 5 a/b.

baU 1 x 2 5 a/x2 2 b/x,

E 5 F 2/k.

E 5 F 2/k,5F/k.25F/k

F/k)F 2/k)U 1 x 2U 1 x 2 5 1

2 kx2 2 Fx 2 F 2/2k.

1x-direction2kx

FS

FS

y 5 0;x 5 1.50 m,

U(J)

4.0

2.0

0

22.0

A

B

C

x (m)0.5 1.5 2.52.0


POTENTIAL ENERGY AND ENERGY CONSERVATION Books... · 2019. 9. 27. · Conservation of Mechanical...

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Transcript of POTENTIAL ENERGY AND ENERGY CONSERVATION Books... · 2019. 9. 27. · Conservation of Mechanical...