THE WORK–ENERGY THEOREM
5.7 WORK DONE BY A VARYING FORCE
Suppose an object is displaced along the x-axis under the action of a force Fx that acts in the x-direction and varies with position, as shown in Figure 5.28. The object
5.7 Work Done by a Varying Force 147
vCM = 0 CM
h
Extension Free flight
CM
CM vCM vCM
= 0 H
Liftoff
FIGURE 5.27 Extension and free fl ight in
the vertical jump. TABLE 5.1
Maximum Power Output from Humans over Various Periods
Power Time 2 hp, or 1 500 W 6 s 1 hp, or 750 W 60 s 0.35 hp, or 260 W 35 min 0.2 hp, or 150 W 5 h 0.1 hp, or 75 W 8 h (safe daily level )
APPLICATION
Diet Versus Exercise in Weight-loss Programs
is displaced in the direction of increasing x from x xi to x xf. In such a situa- tion, we can’t use Equation 5.1 to calculate the work done by the force because this relationship applies only when FS is constant in magnitude and direction. However, if we imagine that the object undergoes the small displacement x shown in Figure 5.28a, then the x-component Fx of the force is nearly constant over this interval and we can approximate the work done by the force for this small displacement as
W1 Fxx [5.26]
This quantity is just the area of the shaded rectangle in Figure 5.28a. If we imagine that the curve of Fx versus x is divided into a large number of such intervals, then the total work done for the displacement from xi to xf is approximately equal to the sum of the areas of a large number of small rectangles:
W F1x1 F2x2 F3x3 [5.27]
Now imagine going through the same process with twice as many intervals, each half the size of the original x. The rectangles then have smaller widths and will better approximate the area under the curve. Continuing the process of increas- ing the number of intervals while allowing their size to approach zero, the number of terms in the sum increases without limit, but the value of the sum approaches a defi nite value equal to the area under the curve bounded by Fx and the x-axis in Figure 5.28b. In other words, the work done by a variable force acting on an object that undergoes a displacement is equal to the area under the graph of Fx versus x.
A common physical system in which force varies with position consists of a block on a horizontal, frictionless surface connected to a spring, as discussed in Section 5.4. When the spring is stretched or compressed a small distance x from its equilib- rium position x 0, it exerts a force on the block given by Fxkx, where k is the force constant of the spring.
Now let’s determine the work done by an external agent on the block as the spring is stretched very slowly from xi 0 to xf xmax, as in Active Figure 5.29a.
This work can be easily calculated by noting that at any value of the displacement, Newton’s third law tells us that the applied force FSapp is equal in magnitude to the spring force F
S
s and acts in the opposite direction, so that Fapp (kx) kx. A plot of Fapp versus x is a straight line, as shown in Active Figure 5.29b. Therefore, the work done by this applied force in stretching the spring from x 0 to x xmax is the area under the straight line in that fi gure, which in this case is the area of the shaded triangle:
WFapp512kxmax2
During this same time the spring has done exactly the same amount of work, but that work is negative, because the spring force points in the direction opposite the motion. The potential energy of the system is exactly equal to the work done by the applied force and is the same sign, which is why potential energy is thought of as stored work.
ACTIVE FIGURE 5.29
(a) A block being pulled from xi 0 to xf xmax on a frictionless sur- face by a force F
S
app. If the process is carried out very slowly, the applied force is equal in magnitude and oppo- site in direction to the spring force at all times. (b) A graph of Fapp versus x.
xi = 0 xf = xmax
s app
(a)
O Fapp
xmax x (b)
F F
FIGURE 5.28 (a) The work done by the force component Fx for the small displacement x is Fxx, which equals the area of the shaded rectangle. The total work done for the displacement from xi to xf is approximately equal to the sum of the areas of all the rectangles. (b) The work done by the component Fx of the varying force as the particle moves from xi to xf is exactly equal to the area under the curve shown.
(a) Fx
Area = ΔA = Fx Δx
Fx
xf x xi
Δx
(b) Fx
xf x xi
Work
5.7 Work Done by a Varying Force 149
EXAMPLE 5.15 Work Required to Stretch a Spring Goal Apply the graphical method of fi nding work.
Problem One end of a horizontal spring (k 80.0 N/m) is held fi xed while an external force is applied to the free end, stretching it slowly from x 0 to x 4.00 cm. (a) Find the work done by the applied force on the spring. (b) Find the additional work done in stretching the spring from x 4.00 cm to x 7.00 cm.
Strategy For part (a), simply fi nd the area of the smaller triangle in Figure 5.30, using A512bh, one-half the base times the height. For part (b), the easiest way to fi nd the additional work done from x 4.00 cm to x 7.00 cm is to fi nd the area of the new, larger triangle and subtract the area of the smaller triangle.
FIGURE 5.30 (Example 5.15) A graph of the external force required to stretch a spring that obeys Hooke’s law versus the elongation of the spring.
Fapp
Fapp = (80.0 N/m)(x)
x (cm)
O 4.00 7.00
Solution
(a) Find the work from x 0 cm to x 4.00 cm.
Compute the area of the smaller triangle: W512kx2512180.0 N/m2 10.040 m225 0.064 0 J (b) Find the work from x 4.00 cm to x 7.00 cm.
Compute the area of the large triangle and subtract the area of the smaller triangle:
W512kx2212kx2
W512180.0 N/m2 10.070 0 m2220.064 0 J 0.196 J 0.064 0 J
0.132 J
Remarks Only simple geometries—rectangles and triangles—can be solved exactly with this method. More com- plex shapes require calculus or the square-counting technique in the next worked example.
QUESTION 5.15
True or False: When stretching springs, half the displacement requires half as much work.
EXERCISE 5.15
How much work is required to stretch this same spring from xi 5.00 cm to xf 9.00 cm?
Answer 0.224 J
EXAMPLE 5.16 Estimating Work by Counting Boxes Goal Use the graphical method and counting
boxes to estimate the work done by a force.
Problem Suppose the force applied to stretch a thick piece of elastic changes with position as indi- cated in Figure 5.31a. Estimate the work done by the applied force.
Strategy To fi nd the work, simply count the number of boxes underneath the curve and mul- tiply that number by the area of each box. The curve will pass through the middle of some boxes, in which case only an estimated fractional part should be counted.
10.0
6.0 8.0
2.0 4.0
0.0
(a)
0.2 0.4 0.6 0.8 1.0 Fapp (N)
x (m) 100
60 80
20 40
0
(b)
0.1 0.3 0.5 0.7
Fapp (N)
x (m)
FIGURE 5.31 (a) (Example 5.16) (b) (Exercise 5.16)
Solution
There are 62 complete or nearly complete boxes under the curve, 6 boxes that are about half under the curve, and a triangular area from x 0 m to x 0.10 m that is equivalent to 1 box, for a total of about 66 boxes. Because the area of each box is 0.10 J, the total work done is approximately 66 0.10 J 6.6 J.
Remarks Mathematically, there are a number of other methods for creating such estimates, all involving adding up regions approximating the area. To get a better estimate, make smaller boxes.
QUESTION 5.16
In developing such an estimate, is it necessary for all boxes to have the same length and width?
EXERCISE 5.16
Suppose the applied force necessary to pull the drawstring on a bow is given by Figure 5.31b. Find the approximate work done by counting boxes.
Answer About 50 J. (Individual answers may vary.)
SUMMARY
5.1 Work
The work done on an object by a constant force is
W (F cos u)x [5.2]
where F is the magnitude of the force, x is the object’s displacement, and u is the angle between the direction of the force F
S
and the displacement DxS. Solving simple prob- lems requires substituting values into this equation. More complex problems, such as those involv ing friction, often require using Newton’s second law, mSa 5 SF
S
, to deter- mine forces.
5.2 Kinetic Energy and the Work–Energy Theorem The kinetic energy of a body with mass m and speed v is given by
KE;12mv2 [5.5]
The work–energy theorem states that the net work done on an object of mass m is equal to the change in its kinetic energy, or
Wnet5KEf2KEi5 DKE [5.6]
Work and energy of any kind carry units of joules. Solving problems involves fi nding the work done by each force act- ing on the object and summing them up, which is Wnet, fol- lowed by substituting known quantities into Equation 5.6, solving for the unknown quantity.
Conservative forces are special: Work done against them can be recovered—it’s conserved. An example is grav- ity: The work done in lifting an object through a height is effectively stored in the gravity fi eld and can be recovered in the kinetic energy of the object simply by letting it fall.
Nonconservative forces, such as surface friction and drag, dissipate energy in a form that can’t be readily recovered.
To account for such forces, the work–energy theorem can be rewritten as
Wnc1Wc5 DKE [5.7]
where Wnc is the work done by nonconservative forces and Wc is the work done by conservative forces.
5.3 Gravitational Potential Energy
The gravitational force is a conservative fi eld. Gravitational potential energy is another way of accounting for gravita- tional work Wg:
Wg5 21PEf2PEi2 5 21mg yf2mg yi2 [5.11]
To fi nd the change in gravitational potential energy as an object of mass m moves between two points in a gravitational fi eld, substitute the values of the object’s y-coordinates.
The work–energy theorem can be generalized to include gravitational potential energy:
Wnc5 1KEf2KEi2 1 1PEf2PEi2 [5.12]
Gravitational work and gravitational potential energy should not both appear in the work–energy theorem at the same time, only one or the other, because they’re equiva- lent. Setting the work due to nonconservative forces to zero and substituting the expressions for KE and PE, a form of the conservation of mechanical energy with gravitation can be obtained:
1
2mvi21mgyi512mvf21mgyf [5.14]
To solve problems with this equation, identify two points in the system—one where information is known and the other where information is desired. Substitute and solve for the unknown quantity.
The work done by other forces, as when frictional forces are present, isn’t always zero. In that case, identify two points as before, calculate the work due to all other forces, and solve for the unknown in Equation 5.12.
5.4 Spring Potential Energy
The spring force is conservative, and its potential energy is given by
PEs;12kx2 [5.16]
Spring potential energy can be put into the work–energy theorem, which then reads
Wnc5 1KEf2KEi2 1 1PEgf2PEgi2 1 1PEsf2PEsi2 [5.17]
When nonconservative forces are absent, Wnc 0 and mechanical energy is conserved.
5.5 Systems and Energy Conservation
The principle of the conservation of energy states that energy can’t be created or destroyed. It can be transformed, but the total energy content of any isolated system is always constant. The same is true for the universe at large. The work done by all nonconservative forces acting on a system equals the change in the total mechanical energy of the system:
Wnc5 1KEf1PEf2 2 1KEi 1PEi2 5Ef2Ei [5.20–21]
where PE represents all potential energies present.
5.6 Power
Average power is the amount of energy transferred divided by the time taken for the transfer:
5 W
Dt [5.22]
This expression can also be written
5F v# [5.23]
where v# is the object’s average speed. The unit of power is the watt (W J/s). To solve simple problems, substitute given quantities into one of these equations. More diffi cult problems usually require fi nding the work done on the object using the work–energy theorem or the defi nition of work.
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MULTIPLE-CHOICE QUESTIONS
1. A worker pushes a wheelbarrow 5.0 m along a level sur- face, exerting a constant horizontal force of 50.0 N. If a frictional force of 43 N acts on the wheelbarrow in a direction opposite to that of the worker, what net work is done on the wheelbarrow? (a) 250 J (b) 215 J (c) 35 J (d) 15 J (e) 45 J
2. A skier leaves a ski jump at 15.0 m/s at some angle u. At what speed is he traveling at his maximum height of 4.50 m above the level of the end of the ski jump?
(Neglect air friction.) (a) 11.7 m/s (b) 16.3 m/s (c) 12.2 m/s (d) 8.55 m/s (e) 17.4 m/s
3. A 40.0-N crate starting at rest slides down a rough 6.00-m-long ramp, inclined at 30.0 with the horizon- tal. The magnitude of the force of friction between the crate and the ramp is 6.0 N. What is the speed of the crate at the bottom of the incline? (a) 1.60 m/s (b) 3.32 m/s (c) 4.5 m/s (d) 6.42 m/s (e) 7.75 m/s 4. What average mechanical power must be delivered by
the muscles of a 70.0-kg mountain climber who climbs a summit of height 325 m in 95.0 min? Note: Due to inef- fi ciencies in converting chemical energy to mechanical energy, the amount calculated here is only a fraction of the power that must be produced by the climber’s body. See Chapter 12. (a) 39.1 W (b) 54.6 W (c) 25.5 W (d) 67.0 W (e) 88.4 W
5. The work required to accelerate an object on a friction- less surface from a speed v to a speed 2v is (a) equal to the work required to accelerate the object from v 0 to v, (b) twice the work required to accelerate the object from v 0 to v, (c) three times the work required to accelerate the object from v 0 to v, (d) four times the work required to accelerate the object from 2v to 3v, or (e) not known without knowledge of the acceleration.
6. Alex and John are loading identical cabinets onto a truck. Alex lifts his cabinet straight up from the ground to the bed of the truck, whereas John slides his cabinet
up a rough ramp to the truck. Which statement is cor- rect? (a) Alex and John do the same amount of work.
(b) Alex does more work than John. (c) John does more work than Alex. (d) None of these statements is neces- sarily true because the force of friction is unknown.
(e) None of these statements is necessarily true because the angle of the incline is unknown.
7. Mark and David are loading identical cement blocks onto David’s pickup truck. Mark lifts his block straight up from the ground to the truck, whereas David slides his block up a ramp on massless, frictionless rollers.
Which statement is true? (a) Mark does more work than David. (b) Mark and David do the same amount of work. (c) David does more work than Mark. (d) None of these statements is necessarily true because the angle of the incline is unknown. (e) None of these statements is necessarily true because the mass of one block is not given.
8. An athlete jumping vertically on a trampoline leaves the surface with a velocity of 8.5 m/s upward. What maximum height does she reach? (a) 13 m (b) 2.3 m (c) 3.7 m (d) 0.27 m (e) The answer can’t be determined because the mass of the athlete isn’t given.
9. A certain truck has twice the mass of a car. Both are moving at the same speed. If the kinetic energy of the truck is K, what is the kinetic energy of the car? (a) K/4 (b) K/2 (c) 0.71K (d) K (e) 2K
10. If the speed of a particle is doubled, what happens to its kinetic energy? (a) It becomes four times larger. (b) It becomes two times larger. (c) It becomes !2 times larger. (d) It is unchanged. (e) It becomes half as large.
11. If the net work done on a particle is zero, which of the following statements must be true? (a) The velocity is zero. (b) The velocity is decreased. (c) The velocity is unchanged. (d) The speed is unchanged. (e) More information is needed.
Multiple-Choice Questions 151
12. A block of mass m is dropped from the fourth fl oor of an offi ce building, subsequently hitting the sidewalk at speed v. From what fl oor should the mass be dropped to double that impact speed? (a) the sixth fl oor (b) the eighth fl oor (c) the tenth fl oor (d) the twelfth fl oor (e) the sixteenth fl oor
13. A car accelerates uniformly from rest. When does the car require the greatest power? (a) when the car fi rst accelerates from rest (b) just as the car reaches its maxi- mum speed (c) when the car reaches half its maximum speed (d) The question is misleading because the power required is constant. (e) More information is needed.
CONCEPTUAL QUESTIONS
1. Consider a tug-of-war as in Figure Q5.1, in which two teams pulling on a rope are evenly matched so that no motion takes place. Is work done on the rope? On the pullers? On the ground? Is work done on anything?
demonstrator be safe if the ball were given a push from its starting position at her nose?
FIGURE Q5.1
Arthur Tilley/FPG/Getty Images
2. Discuss whether any work is being done by each of the following agents and, if so, whether the work is positive or negative: (a) a chicken scratching the ground, (b) a person studying, (c) a crane lifting a bucket of concrete, (d) the force of gravity on the bucket in part (c), (e) the leg muscles of a person in the act of sitting down.
3. If the height of a playground slide is kept constant, will the length of the slide or whether it has bumps make any difference in the fi nal speed of children playing on it? Assume that the slide is slick enough to be consid- ered frictionless. Repeat this question, assuming that the slide is not frictionless.
4. (a) Can the kinetic energy of a system be negative?
(b) Can the gravitational potential energy of a system be negative? Explain.
5. Roads going up mountains are formed into switchbacks, with the road weaving back and forth along the face of the slope such that there is only a gentle rise on any portion of the roadway. Does this confi guration require any less work to be done by an automobile climbing the mountain, compared with one traveling on a road- way that is straight up the slope? Why are switchbacks used?
6. A bowling ball is suspended from the ceiling of a lec- ture hall by a strong cord. The ball is drawn away from its equilibrium position and released from rest at the tip of the demonstrator’s nose, as shown in Figure Q5.6. If the demonstrator remains stationary, explain why the ball does not strike her on its return swing. Would this
FIGURE Q5.6
7. As a simple pendulum swings back and forth, the forces acting on the suspended object are the force of grav- ity, the tension in the supporting cord, and air resis- tance. (a) Which of these forces, if any, does no work on the pendulum? (b) Which of these forces does nega- tive work at all times during the pendulum’s motion?
(c) Describe the work done by the force of gravity while the pendulum is swinging.
8. During a stress test of the cardiovascular system, a patient walks and runs on a treadmill. (a) Is the energy expended by the patient equivalent to the energy of walking and running on the ground? Explain. (b) What effect, if any, does tilting the treadmill upward have?
Discuss.
9. When a punter kicks a football, is he doing any work on the ball while the toe of his foot is in contact with it? Is he doing any work on the ball after it loses contact with his toe? Are any forces doing work on the ball while it is in fl ight?
10. The driver of a car slams on her brakes to avoid collid- ing with a deer crossing the highway. What happens to the car’s kinetic energy as it comes to rest?
11. A weight is connected to a spring that is suspended ver- tically from the ceiling. If the weight is displaced down- ward from its equilibrium position and released, it will oscillate up and down. If air resistance is neglected, will the total mechanical energy of the system (weight plus Earth plus spring) be conserved? How many forms of potential energy are there for this situation?
12. The feet of a standing person of mass m exert a force equal to mg on the fl oor, and the fl oor exerts an equal and opposite force upwards on the feet, which we call the normal force. During the extension phase of a vertical jump (see page 147), the feet exert a force on the fl oor that is greater than mg, so the normal force is greater than mg. As you learned in Chapter 4, we can