7
I
n this chapter we introduce the idea that force times distance is an important new physical quantity, which we refer to as work. Closely related to work is the energy of motion, or kineticenergy. Together, work and kinetic energy set the stage for a deeper understanding of the physical world.
Work and Kinetic Energy
▲ The work done by this cyclist can increase his energy of motion. The relationship between work (force times distance) and energy of motion (kinetic energy) is developed in detail in this chapter.
Big
Ideas
1
Work is force times distance.2
Kinetic energy is one-half mass times velocity squared.3
Power is the rate atwhich work is done.
7-1
Work Done by a Constant Force
In this section we define work—in the physics sense of the word—and apply our defi-nition to a variety of physical situations. We start with the simplest case—namely, the work done when force and displacement are in the same direction.
Force in the Direction of Displacement
When we push a shopping cart in a store or pull a suitcase through an airport, we do work. The greater the force, the greater the work; the greater the distance, the greater the work. These simple ideas form the basis for our definition of work.
To be specific, suppose we push a box with a constant force F >, as shown in FIGURE 7-1. If we move the box in the direction of F > through a displacement d>, the work W we have done is Fd:
Definition of Work, W, When a Constant Force Is in the Direction of Displacement
W = Fd 7-1
SI unit: newton-meter (N
#
m) = joule, JWork is the product of two magnitudes, and hence it is a scalar. In addition, notice that a small force acting over a large distance gives the same work as a large force acting over a small distance. For example, W = 11 N21400 m2 = 1400 N211 m2.
F> F>
d> A constant force of
magnitude F, c cacting in the direction of adisplacement of magnitude d, c cdoes workW = Fd on
the object.
▲ FIGURE 7-1 Work: constant force in the direction of motion A constant force F > pushes a box through a displacement d >. in this special case, where the force and displacement are in the same direction, the work done on the box by the force is W = Fd .
The dimensions of work are newtons (force) times meters (distance), or N·m. This combination of dimensions is called the joule (rhymes with “school,” as commonly pronounced) in honor of James Prescott Joule (1818–1889), a dedicated physicist who is said to have conducted physics experiments even while on his honeymoon. We define a joule as follows:
Definition of the joule, J
1 joule = 1 J = 1 N
#
m = 11kg#
m>s22#
m = 1 kg#
m2>s2 7-2To get a better feeling for work and the associated units, suppose you exert a force of 82.0 N on the box in Figure 7-1 and move it in the direction of the force through a distance of 3.00 m. The work you have done is
W = Fd = 182.0 N213.00 m2 = 246 N
#
m = 246 JSimilarly, if you do 5.00 J of work to lift a book through a vertical distance of 0.750 m, the force you have exerted on the book is
7-1 WoRK donE by A ConsTAnT FoRCE 197
EXERCISE 7-1
WORK DONE
One species of Darwin’s finch, Geospiza magnirostris, can exert a force of 217 N with its beak as it cracks open a Tribulus seed case. If its beak moves through a distance of 0.42 cm during this operation, how much work does the finch do to get the seed?
REASONING AND SOLUTION
The work the finch does is force times distance. In this case, the force is 217 N and the dis-tance is 0.42 cm = 0.0042 m. Thus,
W = Fd = 1217 N210.0042 m2 = 0.91 J
This is a relatively small amount of work for a human, but significant for a small bird.
Just how much work is a joule, anyway? Well, you do one joule of work when you lift a gallon of milk through a height of about an inch, or lift an apple a meter. One joule of work lights a 100-watt lightbulb for 0.01 second or heats a glass of water 0.00125 degree Celsius. Clearly, a joule is a modest amount of work in everyday terms. Additional examples of work are listed in Table 7-1.
TABLE 7-1 Typical Values of Work
Activity Equivalent work (J) Annual U.s. energy use 8 * 1019
Mt. st. Helens eruption 1018
burning 1 gallon of gas 108
Human food intake/day 107
Melting an ice cube 104
Lighting a 100-W bulb for 1 minute
6000
Heartbeat 0.5
Turning a page of a book 10-3
Hop of a flea 10-7
breaking a bond in dnA 10-20
EXAMPLE 7-2
HEADING FOR THE ER
An intern pushes an 87-kg patient on an 18-kg gurney, producing an acceleration of 0.55 m>s2. (a) How much work
does the intern do in pushing the patient and gurney through a distance of 1.9 m? Assume the gurney moves without friction. (b) How far must the intern push the gurney to do 140 J of work?
PICTURE THE PROBLEM
Our sketch shows the physical situation for this problem. Notice that the force exerted by the intern is in the same direction as the displacement of the gurney; therefore, we know that the work is
W = Fd.
REASONING AND STRATEGY
We are not given the magnitude of the force, F, so we cannot apply Equation 7-1 1W = Fd2 directly. However, we are given the mass and acceleration of the patient and gurney, and from them we can calculate the force with F = ma. The work done by the intern is then W = Fd, where d = 1.9 m.
Known Mass of patient, 87 kg; mass of gurney, 18 kg; accelera-tion, a = 0.55 m>s2; (a) pushing distance, d = 1.9 m;
(b) work, W = 140 J.
Unknown (a) Work done, W = ? (b) Pushing distance, d = ?
SOLUTION
Part (a)
1. First, find the force F exerted by the intern: F = ma = 187 kg + 18 kg210.55 m>s22 = 58 N
2. The work done by the intern, W, is the force times the distance: W = Fd = 158 N211.9 m2 = 110 J
Part (b)
3. Use W = Fd to solve for the distance d: d =WF = 140 J58 N = 2.4 m
INSIGHT
You might wonder whether the work done by the intern depends on the speed of the gurney. The answer is no. The work done on an object, W = Fd, doesn’t depend on whether the object moves through the distance d quickly or slowly. What does depend on the speed of the gurney is the rate at which work is done, which we discuss in detail in Section 7-4.
PRACTICE PROBLEM — PREDICT/CALCULATE
(a) If the total mass of the gurney plus patient is halved and the acceleration is doubled, does the work done by the intern increase, decrease, or remain the same? Explain. (b) Determine the work in this case. [Answer: (a) The work remains the same because the two changes offset one another; that is, F = ma = 1m>2212a2. (b) The work is 110 J, as before.]
Some related homework problems: Problem 3, Problem 4
F>
d = 1.9 m
Zero Distance Implies Zero Work
Before moving on, let’s note an interesting point about our definition of work. It’s clear from Equation 7-1 that the work W is zero if the distance d is zero—and this is true regardless of how great the force might be, as illus-trated in FIGURE 7-2. For example, if you push against a solid wall, you do no work on it, even though you may become tired from your efforts. Similarly, if you stand in one place holding a 50-pound suitcase in your hand, you do no work on the suitcase, even though you soon feel worn out. The fact that we become tired when we push against a wall or hold a heavy object is due to the repeated contraction and expansion of indi-vidual cells within our muscles. Thus, even when we are “at rest,” our muscles are doing mechanical work on the microscopic level.Force at an Angle to the Displacement
In FIGURE 7-3 we see a person pulling a suitcase on a level surface with a strap that makes an angle u with the horizontal—in this case the force is at an angle to the direction of motion. How do we calculate the work now? Well, instead of force times distance, we say that work is the component of force in the direction of displacement times the mag-nitude of the displacement. In Figure 7-3, the component of force in the direction of the displacement is F cos u and the magnitude of the displacement is d. Therefore, the work is F cos u times d:
Definition of Work When the Angle Between a Constant Force and the Displacement Is U
W = 1F cos u2d = Fd cos u 7-3
SI unit: joule, J
Of course, in the case where the force is in the direction of motion, the angle u is zero; then W = Fd cos 0 = Fd
#
1 = Fd, in agreement with Equation 7-1.▲ FIGURE 7-2Visualizing Concepts
Force, Work, and Displacement The person doing push ups does positive work as he moves upward, but does no work when he holds himself in place. The weightlifter does positive work as she raises the weights, but no work is done on the weights as she holds them motionless above her head.
F>
d> F>
F cos u F cos u The component of force in the direction of displacement is F cos u. This is the only component of the force that does work.
The work in this case is W = (F cos u)d.
u u
▲ FIGURE 7-3 Work: force at an angle to the direction of motion A person pulls a suitcase with a strap at an angle u to the direction of motion. The component of force in the direction of motion is F cos u, and the work done by the person is W = (F cos u)d .
Equally interesting is a situation in which the force and the displacement are at right angles to one another. In this case u = 90° and the work done by the force F is zero; W = Fd cos 90° = 0.
This result leads naturally to an alternative way to think about the expression W = Fd cos u. In FIGURE 7-4 we show the displacement and the force for the suitcase in Figure 7-3. Notice that the displacement is equivalent to a displacement in the direction of the force of magnitude 1d cos u2 plus a displacement at right angles to the force of magnitude 1d sin u2. Since the displacement at right angles to the force cor-responds to zero work and the displacement in the direction of the force corcor-responds to a work W = F1d cos u2, it follows that the work done in this case is Fd cos u, as given in Equation 7-3. Thus, the work done by a force can be thought of in the following two equivalent ways:
F>
d>
d sin u
u d cos u
▲ FIGURE 7-4 Force at an angle to direc-tion of modirec-tion: another look The dis-placement of the suitcase in Figure 7-3 is equivalent to a displacement of magnitude
d cos u in the direction of the force F>, plus a displacement of magnitude d sin u perpendicular to the force. only the dis-placement parallel to the force results in nonzero work; hence the total work done is F1d cos u2 as expected.
7-1 WoRK donE by A ConsTAnT FoRCE 199
(i) Work is the component of force in the direction of the displacement times the magnitude of
the displacement.
(ii) Work is the component of displacement in the direction of the force times the magnitude of
the force.
In both interpretations, the mathematical expression for work is exactly the same, W = Fd cos u, where u is the angle between the force vector and the displacement vec-tor when they are placed tail-to-tail. This definition of u is illustrated in Figure 7-4.
Finally, we can also express work as the dot product of the vectors F> and d>; that is, W = F >
#
d >= Fd cos u. Notice that the dot product, which is always a scalar, is simply the magnitude of one vector times the magnitude of the second vector times the cosine of the angle between them. We discuss the dot product in greater detail in Appendix A.PHYSICS
IN CONTEXT
Looking Back
Even though work is a scalar quantity, no-tice that vector components (Chapter 3) are used in the definition of work.
EXAMPLE 7-3
GRAVITY ESCAPE SYSTEM
RWP* In a gravity escape system (GES), an enclosed lifeboat on a large ship is deployed by letting it slide down a ramp and then continue in free fall to the water below. Suppose a 4970-kg lifeboat slides a distance of 5.00 m on a ramp, drop-ping through a vertical height of 2.50 m. How much work does gravity do on the boat?
PICTURE THE PROBLEM
From our sketch, we see that the force of gravity mg > and the displacement d > are at an angle u relative to one another when placed tail-to-tail, and that u is also the angle the ramp makes with the vertical. In addition, we note that the vertical height of the ramp is h = 2.50 m and the length of the ramp is d = 5.00 m.
REASONING AND STRATEGY
By definition, the work done on the lifeboat by gravity is W = Fd cos u, where
F = mg, d = 5.00 m, and u is the angle between mg > and d >. We are not given u in the problem statement, but from the right triangle that forms the ramp we see that cos u = h>d. Once u is determined from the geometry of our sketch, it is straightfor-ward to calculate W.
Known Mass of lifeboat, m = 4970 kg; sliding distance, d = 5.00 m; vertical height, h = 2.50 m. Unknown Work done by gravity, W = ?
SOLUTION
1. First, find the component of F > = mg > in the direction of motion:
2. Multiply by distance to find the work: W = 1F cos u2d = 124,400 N215.00 m2 = 122,000 J
3. Alternatively, cancel d algebraically before substituting numerical values:
INSIGHT
The work is simply W = mgh, exactly the same as if the lifeboat had fallen straight down through the height h. Working the problem symbolically, as in Step 3, results in two distinct advantages. First, it makes for a simpler ex-pression for the work. Second, and more important, it shows that the distance d cancels; hence the work depends on the height h but not on the distance. Such a result is not apparent when we work solely with numbers, as in Steps 1 and 2.
PRACTICE PROBLEM
Suppose the lifeboat slides halfway to the water, gets stuck for a moment, and then starts up again and continues to the end of the ramp. What is the work done by gravity in this case? [Answer: The work done by gravity is exactly the same, W = mgh, independent of how the boat moves down the ramp.]
Some related homework problems: Problem 10, Problem 11
d> d> u u u mg> mg> h = 2.50 m F cos u = 1mg2ahdb = 14970 kg219.81 m>s22a2.50 m 5.00 mb =24,400 N W = Fd cos u = 1mg21d2ahdb = mgh = 14970 kg219.81 m>s2212.50 m2 = 122,000 J
Next, we present a Conceptual Example that compares the work required to move an object along two different paths.
Big Idea
1
Work is force times distance when force and displace-ment are in the same direction. More generally, work is the component of force in the direction of displacement times the distance.Negative Work and Total Work
Work depends on the angle between the force, F >, and the displacement (or direction of motion), d >. This dependence gives rise to three distinct possibilities, as shown in FIGURE 7-5:
(i) Work is positive if the force has a component in the direction of motion 1-90° 6 u 6 90°2. (ii) Work is zero if the force has no component in the direction of motion 1u = {90°2. (iii) Work is negative if the force has a component opposite to the direction of motion
(90° 6 u 6 270°).
Thus, whenever we calculate work, we must be careful about its sign and not just assume it is positive.
CONCEPTUAL EXAMPLE 7-4
PATH DEPENDENCE OF WORK
You want to load a box into the back of a truck, as shown in the sketch. One way is to lift it straight up with constant speed through a height h, as shown, doing a work W1. Alternatively, you can slide the box up a loading ramp with constant speed
a distance L, doing a work W2. Assuming the box slides on the ramp without friction, which of the following statements is correct: (a) W1 6 W2, (b) W1= W2, (c) W1 7 W2? f f f F1 > F2 > h> L> L W1 W2 mg sin f mg cos f h mg> mg>
REASONING AND DISCUSSION
You might think that W2 is less than W1, because the force needed to slide the box up the ramp, F2, is less than the force needed to lift it straight up. On the other hand, the distance up the ramp, L, is greater than the vertical distance, h, so per-haps W2 should be greater than W1. In fact, these two effects cancel exactly, giving W1= W2.
To see this, we first calculate W1. The force needed to lift the box with constant speed is F1= mg, and the height is h;
therefore W1 = mgh.
Next, the work to slide the box up the ramp with constant speed is W2= F2L, where F2 is the force required to push
against the tangential component of gravity. In the sketch we see that F2= mg sin f. The sketch also shows that sin f = h>L; thus W2 = 1mg sin f2L = 1mg21h>L2L = mgh = W1: The two works are identical!
Clearly, the ramp is a useful device—it reduces the force required to move the box upward from F1= mg to F2= mg1h>L2. Even so, it doesn’t decrease the amount of work we need to do. As we have seen, the reduced force on the ramp is offset by the increased distance.
ANSWER (b) W1= W2 u u F> F> d> d> -908 6 u 6 08 u = {908 (a) Positive work
(b) Zero work u u u F> F> F> d> d> d> -908 6 u 6 08 908 6 u 6 2708 u = {908 (a) Positive work
(b) Zero work (c) Negative work u u u F> F> F> d> d> d> -908 6 u 6 08 908 6 u 6 2708 u = {908 (a) Positive work
(b) Zero work
(c) Negative work ▲ FIGURE 7-5 Positive, negative, and zero work Work is positive when the force is in the same general direction as the displacement and is negative if the force is generally opposite to the displacement. Zero work is done if the force is at right angles to the displacement.
7-1 WoRK donE by A ConsTAnT FoRCE 201
When more than one force acts on an object, the total work is the sum of the work done by each force separately. Thus, if force F >1 does work W1, force F
>
2 does work W2,
and so on, the total work is
Wtotal = W1 + W2 + W3 + g = a W 7-4
Equivalently, the total work can be calculated by first performing a vector sum of all the forces acting on an object to obtain F >total and then using our basic definition of work:
Wtotal = 1Ftotal cos u2d = Ftotal d cos u 7-5
In this expression, u is the angle between F >total and the displacement d
>
. In the next two Examples we calculate the total work in each of these ways.
P R O B L E M - S O L V I N G N O T E Be Careful About the Angle U
In calculating W = Fd cos u, be sure that the angle you use in the cosine is the angle between the force and the displacement vec-tors when they are placed tail-to-tail. Some-times u may be used to label a different angle in a given problem. For example, u is often used to label the angle of a slope, in which case it may have nothing to do with the angle between the force and the displace-ment. To summarize: Just because an angle is labeled u doesn’t mean it’s automatically the correct angle to use in the work formula.
EXAMPLE 7-5
A COASTING CAR I
A car of mass m coasts down a hill inclined at an angle f below the horizontal. The car is acted on by three forces: (i) the normal force N > exerted by the road, (ii) a force due to air resistance, F >air, and (iii) the force of gravity, mg>. Find the total
work done on the car as it travels a distance d along the road.
PICTURE THE PROBLEM
Because f is the angle the slope makes with the horizontal, it is also the angle between mg> and the downward normal direction, as was shown in Figure 5-22. It follows that the angle between mg> and the displacement d > is u = 90° - f. Our sketch also shows that the angle between N > and d > is u = 90°, and the angle between F >air and d
>
is u = 180°.
REASONING AND STRATEGY
For each force we calculate the work using W = Fd cos u, where u is the angle between that particular force and the displacement d >. The total work is the sum of the work done by each of the three forces.
Known Mass of car, m; angle of incline, f; distance, d; forces acting on car, N >, F >air, and mg>.
Unknown Work done on car, W = ?
SOLUTION
1. We start with the work done by the normal force, N >. WN = Nd cos u = Nd cos 90° = Nd102 = 0
From the figure we see that u = 90° for this force:
2. For the force of air resistance, u = 180°: Wair = Fair d cos 180° = Fair d1-12 = -Fair d
3. For gravity the angle u is u = 90° - f, as indicated in the sketch. Wmg = mgd cos 190° - f2 = mgd sin f
Recall that cos190° - f2 = sin f (see Appendix A ):
4. The total work is the sum of the individual works: Wtotal = WN + Wair + Wmg= 0 - Faird + mgd sin f
INSIGHT
The normal force is perpendicular to the motion of the car, and thus does no work. Air resistance points in a direction that op-poses the motion, so it does negative work. On the other hand, gravity has a component in the direction of motion; therefore, its work is positive. The physical significance of positive, negative, and zero work will be discussed in detail in the next section.
PRACTICE PROBLEM
Calculate the total work done on a 1550-kg car as it coasts 20.4 m down a hill with f = 5.00°. Let the force due to air resistance be 15.0 N.
[Answer: Wtotal = WN+ Wair+ Wmg= 0 - Faird + mgd sin f = 0 - 306 J + 2.70 * 104 J = 2.67 * 104 J]
Some related homework problems: Problem 15, Problem 63
f f f f Fair > Fair > N> N> d> d> d> d> mg> mg> u = 908 u = 1808 u = 908 - f f f f f Fair > Fair > N> N> d> d> d> d> mg> mg> u = 908 u = 1808 u = 908 - f
In the next Example, we first sum the forces acting on the car to find Ftotal. Once the
total force is determined, we calculate the total work using Wtotal= Ftotald cos u.
EXAMPLE 7-6
A COASTING CAR II
Consider the car described in Example 7-5. Calculate the total work done on the car using Wtotal = Ftotald cos u.
PICTURE THE PROBLEM
First, we choose the x axis to point down the slope, and the y axis to be at right angles to the slope. With this choice, there is no ac-celeration in the y direction, which means that the total force in that direction must be zero. As a result, the total force acting on the car is in the x direction. The magnitude of the total force is
mg sin f - Fair, as can be seen in our sketch.
REASONING AND STRATEGY
We begin by finding the x component of each force vector, and then we sum them to find the total force acting on the car. As can be seen from the sketch, the total force points in the positive
x direction—that is, in the same direction as the displacement.
Therefore, the angle u in W = Ftotald cos u is zero.
Known Mass of car, m; angle of incline, f; distance, d; forces acting on car, N >, F >air, and mg >. Unknown Work done on car, W = ?
SOLUTION
1. Referring to the sketch, we see that the magnitude of the total Ftotal = mg sin f - Fair
force is mg sin f minus Fair:
2. The direction of F >total is the same as the direction of d >; thus u = 0°. We can now calculate Wtotal:
INSIGHT
Notice that we were careful to calculate both the magnitude and the direction of the total force. The magnitude (which is always positive) gives Ftotal and the direction gives u = 0°, allowing us to use Wtotal = Ftotald cos u.
PRACTICE PROBLEM
Suppose the total work done on a 1620-kg car as it coasts 25.0 m down a hill with f = 6.00° is Wtotal = 3.75 * 104 J.
Find the magnitude of the force due to air resistance. [Answer: Faird = -Wtotal + mgd sin f = 4030 J; thus
Fair= 14030 J2>d = 161 N]
Some related homework problems: Problem 15, Problem 63
Wtotal = Ftotald cos u = 1mg sin f - Fair2d cos 0°
= mgd sin f - Faird mg> N mg cos f f mg sin f Ftotal = mg sin f - Fair x y Fair mg sin f x force components (Enlarged) Fair >
Enhance Your Understanding
(Answers given at the end of the chapter) 1. A block slides a distance d to the righton a horizontal surface, as indicated in
FIGURE 7-6. Rank the four forces that act on the block 1F, N, mg, ƒk2 in order of
the amount of work they do, from most negative to most positive. Indicate ties where appropriate.
Section Review
• The work done by a force is W = 1F cos u2d = Fd cos u, where F is the force, d is the distance, and u is the angle between the force and displacement.
CONTINUED F d N fk mg ▲ FIGURE 7-6 P R O B L E M - S O L V I N G N O T E
Work Can Be Positive, Negative, or Zero When you calculate work, be sure to keep track of whether it is positive or negative. The distinction is important, since positive work increases speed, whereas negative work decreases speed. Zero work, of course, has no effect on speed.
7-2 KinETiC EnERgy And THE WoRK–EnERgy THEoREM 203
• Work is positive when the force and displacement are in the same direction, zero if the force is perpendicular to the displacement, and negative if the force and displace-ment are in opposite directions.
7-2
Kinetic Energy and the Work–Energy Theorem
Suppose you drop an apple. As it falls, gravity does positive work on it, as indi-cated in FIGURE 7-7, and its speed increases. If you toss the apple upward, gravity does negative work on it, and the apple slows down. In general, whenever the total work done on an object is positive, its speed increases; whenever the total work is negative, its speed decreases. In this section we derive an important result, the
work–energy theorem, which makes this connection between work and change
in speed precise.
Work and Kinetic Energy
To begin, consider an apple of mass m falling through the air, and suppose that two forces act on the apple—gravity, mg >, and the average force of air resistance, F >air. The total force acting on the apple, F>
total, gives the apple a constant
downward acceleration of magnitude
a = Ftotalm
Since the total force is downward and the motion is downward, the work done on the apple is positive.
Now, suppose that the initial speed of the apple is vi, and that after falling a
dis-tance d its speed increases to vf. The apple falls with constant acceleration a; hence
constant-acceleration kinematics (Equation 2–12) gives vf2 = vi2 + 2ad
With a slight rearrangement we find
2ad = vf2 - vi2
Next, we substitute a = Ftotal >m into this equation:
2aFtotalm bd = vf2 - vi2
Multiplying both sides by m and dividing by 2 yields Ftotald = 12mvf2 -12mvi2
In this expression, Ftotald is simply the total work done on the apple. Thus we find
Wtotal = 12mvf2 -12mvi2
This shows that the total work is directly related to the change in speed. Notice that Wtotal 7 0 means vf 7 vi, Wtotal 6 0 means vf 6 vi, and Wtotal= 0 means vf = vi.
The quantity 12mv2 in the equation for W
total has a special significance in physics.
We call it the kinetic energy, K:
Definition of Kinetic Energy, K
K = 12 mv2 7-6
SI unit: kg
#
m2>s2= joule, JIn general, the kinetic energy of an object is the energy due to its motion. We measure kinetic energy in joules, the same units as work, and both kinetic energy and work are scalars. Unlike work, however, kinetic energy is never negative. Instead, K is always greater than or equal to zero, independent of the direction of motion or the direction of any forces. d> mg > d> mg> Force is in the direction of displacement c cso positive work is done on the apple. This causes the apple to speed up.
Negative work done on the apple c ccauses it to slow down. Apple falling: W + 0, speed increases
Apple tossed upward:
W * 0, speed decreases
(a)
(b)
▲ FIGURE 7-7 Gravitational work (a) The work done by gravity on an apple that moves downward is positive. if the apple is in free fall, this positive work will result in an increase in speed. (b) in contrast, the work done by gravity on an apple that moves upward is negative. if the apple is in free fall, the negative work done by gravity will result in a decrease of speed.
To get a feeling for typical values of kinetic energy, consider your kinetic energy when jogging. If we assume a mass of about 62 kg and a speed of 2.5 m>s, your kinetic energy is K =12162 kg212.5 m>s22 = 190 J. Additional examples of kinetic energy are
given in Table 7-2.
EXERCISE 7-7
KINETIC ENERGY TAKES OFF
A small airplane moving along the runway during takeoff has a mass of 690 kg and a kinetic energy of 25,000 J. (a) What is the speed of the plane? (b) By what multiplicative factor does the kinetic energy of the plane change if its speed is tripled?
REASONING AND SOLUTION
(a) The kinetic energy of an object is given by K =12mv2. We know that K = 25,000 J
and m = 690 kg; therefore the speed is
v = B 2K m = B 2125,000 J2 690 kg = 8.5 m>s
(b) Kinetic energy depends on the speed squared, and hence tripling the speed increases the kinetic energy by a factor of nine.
The Work–Energy Theorem
In the preceding discussion we used a calculation of work to derive an expression for the kinetic energy of an object. The precise connec-tion we derived between these quantities is known as the work–energy theorem. This theorem can be stated as follows:Work–Energy Theorem
The total work done on an object is equal to the change in its kinetic energy:
Wtotal = ∆K = 21mvf2- 12mvi2 7-7
Thus, the work–energy theorem says that when a force acts on an object over a distance—doing work on it—the result is a change in the speed of the object, and hence a change in its energy of motion. Equation 7-7 is the quantitative expression of this connection.
We have derived the work–energy theorem for a force that is constant in direc-tion and magnitude, but it is valid for any force. In fact, the work–energy theorem is completely general, making it one of the most important and fundamental results in physics. It is also a very handy tool for problem solving, as we shall see many times throughout this text.
EXERCISE 7-8
WORK TO ACCELERATE
A 220-kg motorcycle is cruising at 14 m>s. What is the total work that must be done on the motorcycle to increase its speed to 19 m>s?
REASONING AND SOLUTION
From the work–energy theorem, we know that the total work required to change an object’s kinetic energy is Wtotal = ∆K =12mvf2- 12mvi2. Given that m = 220 kg, vi= 14 m>s, and
vf= 19 m>s, we find
Wtotal = 12mvf2- 12mvi2
= 121220 kg2119 m>s22 - 1
21220 kg2114 m>s22 = 18,000 J This much energy could lift a 90-kg person through a vertical distance of about 20 m.
We now present Examples showing how the work–energy theorem is used in prac-tical situations.
PHYSICS
IN CONTEXT
Looking Back
The kinematic equations of motion for constant acceleration (Chapters 2 and 4) are used in the derivation of kinetic energy.
PHYSICS
IN CONTEXT
Looking Ahead
In Chapter 8 we introduce the concept of potential energy. The combination of ki-netic and potential energy is referred to as mechanical energy, which will play a cen-tral role in our discussion of the conserva-tion of energy.
TABLE 7-2 Typical Kinetic Energies
Source Approximate kinetic energy (J) Jet aircraft at 500 mi>h 109 Car at 60 mi>h 106 Home-run baseball 102
Person at walking speed 50 Housefly in flight 10-4
PHYSICS
IN CONTEXT
Looking Ahead
The kinetic energy before and after a col-lision is an important characterizing fea-ture, as we shall see in Chapter 9. In ad-dition, look for kinetic energy to reappear when we discuss rotational motion in Chapter 10, and again when we study ideal gases in Chapter 17.
Big Idea
2
Kinetic energy is one-half mass times velocity squared. The total work done on an object is equal to the change in its kinetic energy.7-2 KinETiC EnERgy And THE WoRK–EnERgy THEoREM 205
In the previous Example the initial speed was zero. This is not always the case, of course. The next Example considers a case with a nonzero initial speed.
A 4.10-kg box of books is lifted vertically from rest a distance of 1.60 m with a constant, upward applied force of 52.7 N. Find (a) the work done by the applied force, (b) the work done by gravity, and (c) the final speed of the box.
PICTURE THE PROBLEM
Our sketch shows that the direction of motion of the box is upward. In addi-tion, we see that the applied force, F >app, is upward and the force of gravity, mg >,
is downward. Finally, the box is lifted from rest (vi= 0) through a distance
∆y = 1.60 m.
REASONING AND STRATEGY
The applied force is in the direction of motion, so the work it does, Wapp, is
positive. Gravity is opposite in direction to the motion; thus its work, Wmg, is negative. The total work is the sum of Wapp and Wmg and the final speed of
the box is found by applying the work–energy theorem, Wtotal = ∆K. Known Mass of box, m = 4.10 kg; vertical distance, ∆y = 1.60 m; applied
force, Fapp= 52.7 N.
Unknown (a) Work done by applied force, Wapp= ? (b) work done by gravity,
Wmg = ? (c) Final speed of box, vf = ?
SOLUTION
Part (a)
1. First we find the work done by the applied force. Wapp = Fapp cos 0° ∆y = 152.7 N211211.60 m2 = 84.3 J
In this case, u = 0° and the distance is ∆y = 1.60 m: Part (b)
2. Next, we calculate the work done by gravity. The distance is ∆y = 1.60 m, as before, but now u = 180°: Part (c)
3. The total work done on the box, Wtotal, is the Wtotal = Wapp+ Wmg= 84.3 J - 64.4 J = 19.9 J
sum of Wapp and Wmg:
4. To find the final speed, vf, we apply the work–energy
theorem. Recall that the box started at rest; thus vi= 0:
INSIGHT
As a check on our result, we can find vf in a completely different way. First, calculate the acceleration of the box with the
result a = 1Fapp - mg2>m = 3.04 m>s2. Next, use this result in the kinematic equation v2= v02+ 2a∆y. With v0 = 0
and ∆y = 1.60 m, we find v = 3.12 m>s, in agreement with the results using the work–energy theorem.
PRACTICE PROBLEM — PREDICT/CALCULATE
(a) If the box is lifted only a quarter of the distance, ∆y = 11.60 m2>4 = 0.400 m, is the final speed 1>8, 1>4, or 1>2 of the value found in this Example? Explain. (b) Determine vf in this case. [Answer: (a) Work depends linearly on ∆y,
and vf depends on the square root of the work. As a result, it follows that the final speed is 21>4 = 12 the value found in
Step 4. (b) We find vf= 1213.12 m>s2 = 1.56 m>s.]
Some related homework problems: Problem 28, Problem 29, Problem 65
Wmg= mg cos 180° ∆y = 14.10 kg219.81 m>s221-1211.60 m2 = -64.4 J Wtotal = 12 mvf2- 21 mvi2 = 12 mvf2 vf= B 2Wtotal m = B 2119.9 J2 4.10 kg = 3.12 m>s
EXAMPLE 7-9
HIT THE BOOKS
Fapp > mg> vi = 0 vf = ? ∆y = 1.60 m y
EXAMPLE 7-10
GOING FOR A STROLL
When a father takes his daughter out for a spin in a stroller, he exerts a force F > with a magnitude of 44.0 N at an angle of 32.0° below the horizontal, as shown in the sketch. The stroller and child together have a mass m = 22.7 kg. (a) How much work does the father do on the stroller when he pushes it through a horizontal distance d = 1.13 m? (b) If the initial speed of the stroller is vi= 1.37 m>s, what is its final speed? (Assume the stroller rolls with negligible friction.)
The final speeds in the previous Examples could have been found using Newton’s laws and the constant-acceleration kinematics of Chapter 2, as indicated in the Insight in Example 7-9. The work–energy theorem provides an alternative method of calcula-tion that is often much easier to apply than Newton’s laws. We return to this point in Chapter 8.
PICTURE THE PROBLEM
Our sketch shows the direction of motion and the directions of each of the forces. The normal forces and the force due to gravity are ver-tical, whereas the displacement is horizontal. The force exerted by the father has a vertical component, -F sin 32.0°, and a horizontal component, F cos 32.0°, where F = 44.0 N.
REASONING AND STRATEGY
a. The normal forces 1 N >1 and N
>
22 and the weight (mg >) do no
work because they are at right angles to the horizontal displace-ment. The force exerted by the father, however, has a horizontal component that does positive work on the stroller. Therefore, the total work is simply the work done by the father.
b. After calculating the work in part (a), we can find the final speed vf by applying the work–energy theorem with vi= 1.37 m>s.
Known Force, F = 44.0 N; angle below horizontal, u = 32.0°; mass, m = 22.7 kg; distance, d = 1.13 m; initial speed, vi = 1.37 m>s.
Unknown (a) Work done by the father, Wfather = ? (b) Final speed, vf= ?
SOLUTION
Part (a)
1. The work done by the father is the horizontal component of his force times the distance, 1F cos u2d. This is also the total work done on the stroller:
Part (b)
2. Use the work–energy theorem to solve for the
final speed:
3. Substitute numerical values to get the desired vf =
B 2142.2 J2 22.7 kg + 11.37 m>s22 = 2.37 m>s result: INSIGHT
Notice that the speed of the stroller increased by 1.00 m>s, from 1.37 m>s to 2.37 m>s. If the stroller had started from rest, would its speed increase from zero to 1.00 m>s? No. The work–energy theorem depends on the square of the speeds, and hence speeds don’t simply add or subtract in an intuitive “linear” way. The final speed in this case is greater than 1.00 m>s, as we show in the following Practice Problem.
PRACTICE PROBLEM
Suppose the stroller starts at rest. What is its final speed in this case? [Answer: vf = 22Wtotal>m = 1.93 m>s, almost
twice what simple “linear reasoning” would suggest]
Some related homework problems: Problem 29, Problem 62
Wfather = 1F cos u2d
= 144.0 N21cos 32.0°211.13 m2 = 42.2 J = Wtotal Wtotal = ∆K = 12 mvf2- 12 mvi2 1 2 mvf2 = Wtotal + 12 mvi2 vf = B 2Wtotal m + vi2 N2 > N1 > mg> 32.08 32.08 F> F> F cos 32.08 -F sin 32.08 Components of F (expanded view) > d> P R O B L E M - S O L V I N G N O T E Be Careful About Linear Reasoning Though some relationships are linear—if you double the mass, you double the kinetic energy—others are not. For example, if you
double the speed, you quadruple the kinetic
energy. Be careful not to jump to conclu-sions based on linear reasoning.
7-3 WoRK donE by A VARiAbLE FoRCE 207
CONCEPTUAL EXAMPLE 7-11
COMPARE THE WORK
PREDICT/EXPLAIN
(a) To accelerate a certain car from rest to the speed v requires the work W1, as shown in the sketch. The work needed to
accelerate the car from v to 2v is W2. Which of the following statements is correct: W2= W1, W2= 2W1, W2 = 3W1,
W2= 4W1? (b) Which of the following is the best explanation for your prediction?
I. The increase in speed is the same, so the work is also the same. II. To double the speed requires double the work.
III. Kinetic energy depends on v2, and hence it takes four times as
much work to increase the speed to 2v.
IV. Four times as much work is required to go from 0 to 2v as to go from 0 to v. Therefore, the work required to increase the speed from v to 2v is three times the original work.
REASONING AND DISCUSSION
A common misconception is to reason that because we increase the speed by the same amount in each case, the work re-quired is the same. It is not, and the reason is that work depends on the speed squared rather than on the speed itself.
To see how this works, first calculate W1, the work needed to go from rest to a speed v. From the work–energy theorem, with vi= 0 and vf= v, we find W1= 12 mvf2- 12 mvi2= 12 mv2. Similarly, the work W2 needed to go from vi = v to vf = 2v
is W2= 12m(2v)2- 1
2mv2= 3
A
12mv2B
= 3W1.ANSWER
(a) The required work is W2 = 3W1. (b) The best explanation is IV.
y 2y
y = 0
W1 W2
Enhance Your Understanding
(Answers given at the end of the chapter) 2. An object has an initial kinetic energy of 100 J. A few minutes later, after a singleexternal force has acted on the object, its kinetic energy is 200 J. Is the work done by the force positive, negative, or zero? Explain.
Section Review
• The kinetic energy of an object is one-half its mass times its velocity squared. • The total work done on an object is equal to the change in its kinetic energy.
7-3
Work Done by a Variable Force
Thus far we have calculated work only for constant forces, yet most forces in nature vary with position. For example, the force exerted by a spring depends on how far the spring is stretched, and the force of gravity between planets depends on their separation. In this section we show how to calculate the work for a force that varies with position.
A Graphical Interpretation of Work
First, let’s review briefly the case of a constant force, and develop a graphical interpretation of work. FIGURE 7-8 shows a constant force plotted versus position, x. If the force acts in the positive x direction and moves an object a distance d, from x1 to x2, the work it does is W = Fd = F1x2 - x12. Referring to the figure,we see that the work is equal to the shaded area1 between the force line and the x axis.
Next, consider a force that has the value F1 from x = 0 to x = x1 and a different
value F2 from x = x1 to x = x2, as in FIGURE 7-9 (a). The work in this case is the sum of
the works done by F1 and F2. Therefore, W = F1x1 + F21x2 - x12. This, again, is the area
between the force lines and the x axis. Clearly, this type of calculation can be extended to a force with any number of different values, as indicated in FIGURE 7-9 (b).
If a force varies continuously with position, we can approximate it with a series of con-stant values that follow the shape of the curve, as shown in FIGURE 7-10 (a). It follows that the work done by the continuous force is approximately equal to the area of the corresponding
1Usually, area has the dimensions of (length) * (length), or length2. In this case, however, the vertical axis is force and the horizontal axis is distance. As a result, the dimensions of area are (force) * (distance), which in SI units is N
#
m = J. For ce Position F d x1 x2 O Area = Fd = W▲ FIGURE 7-8 Graphical representation of the work done by a constant force A constant force F acting through a dis-tance d does the work W = Fd. note that
Fd is also equal to the shaded area
208 CHAPTER 7 WoRK And KinETiC EnERgy
rectangles, as FIGURE 7-10 (b) shows. The approximation can be made better by using more rectangles, as illustrated in FIGURE 7-10 (c). In the limit of an infinite number of vanishingly small rectangles, the area of the rectangles becomes identical to the area under the force curve. Hence this area is the work done by the continuous force. To summarize:
The work done by a force in moving an object from one position to another is equal to the corresponding area between the force curve and the x axis.
Work for the Spring Force
A case of particular interest is a spring. If we recall that the force exerted by a spring is given by Fx = -kx (Section 6-2), it follows that the force wemust exert to hold the spring at the position x is +kx. This is illustrated in FIGURE 7-11, where we also show that the corresponding force curve is a straight line extending from the origin. Therefore, the work we do in stretching a spring from x = 0 (equilibrium) to the general position x is the shaded triangular area shown in FIGURE 7-12. This area is equal to 121base21height2, where in this case the base is x and the height is kx. As a result, the work is 121x21kx2 =12kx2. Similar reasoning shows that the work needed to compress
a spring a distance x is also 12kx2. Therefore:
Work to Stretch or Compress a Spring a Distance x from Equilibrium
W = 12kx2 7-8 SI unit: joule, J Position x1 x2 (c) A better approximation For ce For ce Position x1 x2
(b) Approximating the work done by a continuous force
For
ce
Position
x1 x2
(a) Approximating a continuous force O
O
O
▼ FIGURE 7-10 Work done by a continu-ously varying force (a) A continuously varying force can be approximated by a series of constant values that follow the shape of the curve. (b) The work done by the continuous force is approximately equal to the area of the small rectangles corresponding to the constant values of force shown in part (a). (c) in the limit of an infinite number of vanishingly small rectangles, we see that the work done by the force is equal to the area between the force curve and the x axis.
Force of spring Applied force
Equilibrium position of spring x -kx +kx F = kx x = 0 x Applied f or ce Position O
▲ FIGURE 7-11 Stretching a spring The force we must exert on a spring to stretch it a distance x is +kx. Thus, ap-plied force versus position for a spring is a straight line of slope k.
For ce Position Area = W kx O x
▲ FIGURE 7-12 Work needed to stretch a spring a distance x The work done is equal to the shaded area, which is a right triangle. The area of the triangle is 1 21x21kx2 = 12kx2. For ce Position F2 F1 x1 x2 F2(x2 - x1) F1x1 (a) O For ce Position (b) O For ce Position F1 x1 x2 F2(x2 - x1) F1x1 (a) O For ce Position (b) O
▲ FIGURE 7-9 Work done by a nonconstant force (a) A force with a value F1 from 0 to x1 and a value F2 from x1 to x2 does the work W = F1x1 + F2(x2 - x1). This is simply the combined
area of the two shaded rectangles. (b) if a force takes on a number of different values, the work it does is still the total area between the force lines and the x axis, just as in part (a).
7-3 WoRK donE by A VARiAbLE FoRCE 209
We can get a feeling for the amount of work required to compress a typical spring in the following Exercise.
EXERCISE 7-12
SPRING WORK
(a) A toy snake has a spring with a force constant of 230 N>m. How much work is required to stretch this spring 2.5 cm? (b) The coiled suspension spring in a car has a force constant of 21,000 N>m. If 3.5 J of work is done to compress this spring, what is the compression distance?
REASONING AND SOLUTION
Work is related to the force constant and compression (or stretch) by W = 12kx2. We can
apply this relationship to both questions. (a) Substitute k = 230 N>m and x = 0.025 m into W = 12kx2:
W = 12kx2 = 1
21230 N>m210.025 m22= 0.072 J (b) Solve W =12kx2 for x, then substitute W = 3.5 J and k = 21,000 N>m:
x = B 2W k = B 213.5 J2 21,000 N>m= 0.018 m
The work done in compressing or expanding a spring varies with the second power of x, the displacement from equilibrium. The consequences of this dependence are explored throughout the rest of this section.
“Springs” in Other Contexts
Our results for a spring apply to more than just the clas-sic case of a helical coil of wire. In fact, any flexible structure satisfies the relationshipsFx = -kx and W = 12kx2, given the appropriate value of the force constant, k, and small
enough displacements, x. Several examples were mentioned in Section 6-2.
Here we consider an example from the field of nanotechnology—namely, the can-tilevers used in atomic-force microscopy (AFM). As we show in Example 7-13, a typical atomic-force cantilever is basically a thin silicon bar about 250 mm in length, supported at one end like a diving board, with a sharp, hanging point at the other end. When the point is pulled across the surface of a material—like an old-fashioned pho-nograph needle in the groove of a record—individual atoms on the surface cause the point to move up and down, deflecting the cantilever. These deflections, which can be measured by reflecting a laser beam from the top of the cantilever, are then converted into an atomic-level picture of the surface, as shown in FIGURE 7-13.
A typical force constant for an AFM cantilever is on the order of 1 N>m, much smaller than the 100–500-N>m force constant of a common lab spring. The implica-tions of this are discussed in the following Example.
▲ FIGURE 7-13 Human chromosomes, as imaged by an atomic-force microscope.
EXAMPLE 7-13
FLEXING AN AFM CANTILEVER
The work required to deflect a typical AFM cantilever by 0.10 nm is 1.2 * 10-20 J. (a) What is the force constant of the cantilever,
treat-ing it as an ideal sprtreat-ing? (b) How much work is required to increase the deflection of the cantilever from 0.10 nm to 0.20 nm?
PICTURE THE PROBLEM
The upper sketch shows the cantilever and its sharp point being dragged across the surface of a material. In the lower sketch, we show an exaggerated view of the cantilever’s deflection, and indicate that it is equivalent to the stretch of an “effective” ideal spring with a force constant k.
REASONING AND STRATEGY
a. Given that W = 1.2 * 10-20 J for a deflection of x = 0.10 nm, we
can find the effective force constant k using W =12 kx2.
b. To find the work required to deflect from x = 0.10 nm to
x = 0.20 nm, W1S2, we calculate the work to deflect from x = 0 to
x = 0.20 nm, W0S2, and then subtract the work needed to deflect
from x = 0 to x = 0.10 nm, W0S1. (Notice that we cannot simply
assume that the work to go from x = 0.10 nm to x = 0.20 nm is the same as the work to go from x = 0 to x = 0.10 nm.)
Silicon rod
x = 0.10 nm
k Effectiveideal spring
BIO Another example of the work required to stretch a flexible structure is the
work done inside your eye. The human eye can accommodate, or focus on objects at dif-ferent distances, by using ciliary muscles to alter the shape of the flexible lens behind the pupil of the eye. (The physiology and optical properties of the eye are explored in more detail in Chapter 27.) The shorter the distance you attempt to focus your eye, the harder your ciliary muscles must work to change the shape of the lens. This is why long periods of viewing close-up objects, such as a book or a computer screen, can lead to asthenopia, or eye strain.
Using Average Force to Calculate Work
An equivalent way to calculate the work for a variable force is to multiply the average force, Fav, by the distance, d:W = Favd 7-9
For a spring that is stretched a distance x from equilibrium, the force varies linearly from 0 to kx. Thus, the average force is Fav =12 kx, as indicated in FIGURE 7-14. Therefore,
the work is
W = 12 kx(x) = 12 kx2
As expected, our result agrees with Equation 7-8.
Finally, when you stretch or compress a spring from its equilibrium position, the work you do is always positive. The work done by a spring, however, may be positive or nega-tive, depending on the situation. For example, consider a block sliding to the right with an initial speed v0 on a smooth, horizontal surface, as shown in FIGURE 7-15 (a). When
the block begins to compress the spring, as in FIGURE 7-15 (b), the spring exerts a force Known Deflection distance, x = 0.10 nm = 0.10 * 10-9 m; work to deflect, W = 1.2 * 10-20 J.
Unknown (a) Force constant, k = ? (b) Work to increase deflection, W1S2 = ?
SOLUTION
Part (a)
1. Solve W = 12 kx2 for the force constant k: k = 2W
x2 =
211.2 * 10-20 J2
10.10 * 10-9 m22= 2.4 N>m
Part (b)
2. First, calculate the work needed to deflect the cantilever from x = 0 to x = 0.20 nm:
3. Subtract from the above result the work to deflect from x = 0 to x = 0.10 nm, which the problem statement gives as 1.2 * 10-20 J:
INSIGHT
Our results show that more energy is needed to deflect the cantilever the second 0.10 nm than to deflect it the first 0.10 nm. Why? The reason is that the force of the cantilever increases with distance; thus, the average force over the second 0.10 nm is greater than the average force over the first 0.10 nm. In fact, we can see from the graph that the average force between 0.10 nm and 0.20 nm 10.36 nN2 is three times the average force between 0 and
0.10 nm 10.12 nN2. It follows that the work required for the second 0.10 nm is three times the work required for the first 0.10 nm.
PRACTICE PROBLEM — PREDICT/CALCULATE
A second cantilever has half the force constant of the cantilever in this Example. (a) Is the work required to deflect the second cantilever by 0.20 nm greater than, less than, or equal to the work required to deflect the cantilever in this Example by 0.10 nm? Explain. (b) Determine the work required to deflect the second cantilever 0.20 nm. [Answer: (a) Halving the force constant halves the work, but doubling the deflection quadruples the work. The net effect is that the work increases by a factor of two. (b) The work required is 2.4 * 10-20 J.]
Some related homework problems: Problem 31, Problem 36
W0S2 = 12 kx2 =1212.4 N>m210.2 * 10-9 m22= 4.8 * 10-20 J W1S2 = W0S2 - W0S1 = 4.8 * 10-20 J - 1.2 * 10-20 J = 3.6 * 10-20 J F 0.24 nN 0.12 nN 0.20 nm O x 0.36 nN 0.48 nN 0.10 nm k = 2.4 N/m 1 2 For ce Position x kx Fav = kx O
▲ FIGURE 7-14 Work done in stretching a spring: average force The average force to stretch a spring from x = 0 to
x is Fav = 12kx , and the work done is
7-3 WoRK donE by A VARiAbLE FoRCE 211
on the block to the left—that is, opposite to the block’s direction of motion. As a result, the spring does negative work on the block, which causes the block’s speed to decrease. Eventually the negative work done by the spring, W = -12 kx2, is equal in magnitude to
the initial kinetic energy of the block. At this point, FIGURE 7-15 (c), the block comes to rest momentarily, and W = ∆K = Kf - Ki = 0 - Ki = -Ki= -21mv02 = -12 kx2.
After the block comes to rest, the spring expands back to its equilibrium position, as we see in FIGURE 7-15 (d)–(f). During this expansion the force exerted by the spring is in the same direction as the block’s motion, and hence it does positive work in the amount W = 12 kx2. As a result, the block leaves the spring with the same speed it had
initially.
QUICK EXAMPLE 7-14
A BLOCK COMPRESSES A SPRING
A block with a mass of 1.5 kg and an initialspeed of vi= 2.2 m>s slides on a frictionless,
horizontal surface. The block comes into con-tact with a spring that is in its equilibrium posi-tion, and compresses it until the block comes to rest momentarily. Find the maximum compres-sion of the spring, assuming its force constant is 475 N>m.
REASONING AND SOLUTION
Our sketch shows the block just as it comes into contact with the spring with an initial speed of vi= 2.2 m>s. We also show the spring
at maximum compression, x, when the block is momentarily at rest. As the block comes to
rest, the force of the spring is opposite to the CONTINUED
Spring is doing negative work on the block – force and displacement are in opposite directions. Equilibrium position of spring Initial speed of block is v0. Equilibrium position of spring Final speed of block is again v0. F2 F2 F1 F1 v = 0 (e) (d) (c) (b) v v (a) v0 (f) v0 v = 0
Spring is doing positive work on the block – force and displacement are in the same direction.
▲ FIGURE 7-15 The work done by a spring can be positive or negative (a) A block slides to the right on a frictionless surface with a speed v0 until it encounters a spring. (b) The spring
now exerts a force to the left—opposite to the block’s motion—and hence it does negative work on the block. This causes the block’s speed to decrease. (c) The negative work done by the spring eventually is equal in magnitude to the block’s initial kinetic energy, at which point the block comes to rest momentarily. As the spring expands, (d) and (e), it does positive work on the block and increases its speed. (f) When the block leaves the spring, its speed is again equal to v0. Equilibrium position of spring Maximum compression F x v = 0 vi
block’s displacement, and hence the work done by the spring is negative. The (negative) work done by the spring, W = -12 kx2, is equal to the (negative) change in kinetic energy,
ΔK. Thus, we can set -12 kx2 = ∆K and solve for x to find the maximum compression.
1. Calculate the initial and final kinetic Ki= 12mvi2
= 1211.5 kg212.2 m>s22= 3.6 J
Kf= 0
energies of the block:
2. Calculate the change in kinetic ∆K = Kf- Ki= -3.6 J
energy of the block:
3. Set the negative work done by the -12 kx2 = ∆K = -3.6 J
spring equal to the negative change in kinetic energy of the block:
4. Solve for the compression, x, and x = B
-2∆K
k
substitute numerical values:
= B
-21-3.6 J2
475 N>m = 0.12 m When the spring brings the block to rest, the kinetic energy it had initially is not lost—it is stored in the spring itself and can be released later. We discuss situations like this, and their connection with energy conservation, in Chapter 8.
Enhance Your Understanding
(Answers given at the end of the chapter) 3. As an object moves along the positive x axis, the force shown in FIGURE 7-16 acts on it.Is the work done by the force from x = 0 to x = 4 m greater than, less than, or equal to the work done by the force from x = 5 m to x = 9 m? Explain.
1 2 3 4 5 6 7 8 9 10 4 O 1 2 3 5 6 7 8 9 10 For ce, F ( N ) Position x, (m) ▲ FIGURE 7-16
Section Review
• Work is equal to the area between a force curve and the x axis. • Work is also equal to the average force times displacement.
7-4
Power
Power is a measure of how quickly work is done. To be precise, suppose the work W is
per-formed in the time t. The average power delivered during this time is defined as follows:
Definition of Average Power, P
P = Wt 7-10
7-4 PoWER 213
For simplicity of notation we drop the usual subscript av for an average quantity and simply understand that the power P refers to an average power unless stated otherwise.
The Dimensions of Power
The dimensions of power are joules (work) per second (time). We define one joule per second to be a watt (W), after James Watt (1736–1819), the Scottish engineer and inventor who played a key role in the development of practi-cal steam engines:1 watt = 1 W = 1 J>s 7-11
Of course, the watt is the unit of power used to rate the output of lightbulbs. Another common unit of power is the horsepower (hp), which is used to rate the output of car engines. It is defined as follows:
1 horsepower = 1 hp = 746 W 7-12
Though it sounds like a horse should be able to produce one horsepower, in fact, a horse can generate only about 2>3 hp for sustained periods. The reason for the discrep-ancy is that when James Watt defined the horsepower—as a way to characterize the output of his steam engines—he purposely chose a unit that was overly generous to the horse, so that potential investors couldn’t complain he was overstating the capability of his engines.
Human Power
To get a feel for the magnitude of the watt and the horsepower, consider the power you might generate when walking up a flight of stairs. Suppose, for example, that an 80.0-kg person walks up a flight of stairs in 20.0 s, and that the altitude gain is 12.0 ft (3.66 m). Referring to Example 7-3 and Conceptual Example 7-4, we find that the work done by the person is W = mgh = 180.0 kg219.81 m>s2213.66 m2 = 2870 J. Tofind the power, we simply divide the work by the time: P = W>t = (2870 J)>(20.0 s) = 144 W = 0.193 hp. Thus, a leisurely stroll up the stairs requires about 1>5 hp or 150 W. Similarly, the power produced by a sprinter bolting out of the starting blocks is about 1 hp, and the greatest power most people can produce for sustained periods of time is roughly 1>3 to 1>2 hp. Further examples of power are given in Table 7-3.
Human-powered flight is a feat just barely within our capabilities, since the most efficient human-powered airplanes require a steady power output of about 1>3 hp. In 1979 the Gossamer Albatross became the first (and so far the only) human-powered air-craft to fly across the English Channel. This 22.25-mile flight—from Folkestone, Eng-land, to Cap Gris-Nez, France—took 2 hours 49 minutes and required a total energy output roughly equivalent to climbing to the top of the Empire State Building 10 times. The Gossamer Albatross is shown in midflight in FIGURE 7-17.
RWP Power output is also an important factor in the performance of a car. For example, suppose it takes a certain amount of work, W, to accelerate a car from 0 to 60 mi>h. If the average power provided by the engine is P, then according to Equation 7-10, the amount of time required to reach 60 mi>h is t = W>P. Clearly, the greater the power P, the less the time required to accelerate. Thus, in a loose way of speaking, we can say that of “how fast it can go fast.”
Big Idea
3
Power is the rate at which work is done. The more work that is done in a shorter time, the greater the power.TABLE 7-3 Typical Values of Power
Source Approximate power (W)
Hoover dam 1.34 * 109
Car moving at 40 mi>h 7 * 104
Home stove 1.2 * 104
sunlight falling on one square meter
1380
Refrigerator 615
Television 200
Person walking up stairs 150
Human brain 20
▲ FIGURE 7-17 The Gossamer Albatross Twice the aircraft touched the surface of the water, but the pilot was able to main-tain control.
EXAMPLE 7-15
PASSING FANCY
To pass a slow-moving truck, you want your fancy 1.30 * 103@kg car to accelerate from 13.4 m>s 130.0 mi>h2 to 17.9 m>s
140.0 mi>h2 in 3.00 s. What is the minimum power required for this pass?
PICTURE THE PROBLEM
Our sketch shows the car accelerating from an initial speed of vi= 13.4 m>s to a final speed of vf= 17.9 m>s. We assume
the road is level, so that no work is done against gravity, and that friction and air resistance may be ignored.
