CHAPTER 6 WORK AND ENERGY
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
3.
REASONING AND SOLUTION The work done by the retarding force is given by
Equation 6.1: W = (F cosθ )s. Since the force is a retarding force, it must point opposite to
the direction of the displacement, so that θ =180°. Thus, we have
3
6
W = (F cosθ )s = (3.0 ×10 N)(cos 180°)(850 m) = –2.6 ×10 J
The work done by this force is negative , because the retarding force is directed opposite
to the direction of the displacement of the truck.
7.
SSM REASONING AND SOLUTION
a. In both cases, the lift force L is perpendicular to the displacement of the plane, and,
therefore, does no work. As shown in the drawing in the text, when the plane is in the dive,
there is a component of the weight W that points in the direction of the displacement of the
plane. When the plane is climbing, there is a component of the weight that points opposite
to the displacement of the plane. Thus, since the thrust T is the same for both cases, the net
force in the direction of the displacement is greater for the case where the plane is diving.
Since the displacement s is the same in both cases, more net work is done during the dive .
b. The work done during the dive is Wdive = (T + W cos 75°) s, while the work done during
the climb is Wclimb = (T + W cos 115°) s. Therefore, the difference between the net work
done during the dive and the climb is
Wdive – Wclimb = (T + W cos 75°) s – (T + W cos 115°) s = Ws (cos 75° – cos 115°)
4
3
7
= (5.9 ×10 N)(1.7× 10 m)(cos 75° – cos 115°) = 6.8 ×10 J
13. REASONING The work done by the catapult Wcatapult is one contribution to the work done
by the net external force that changes the kinetic energy of the plane. The other contribution
is the work done by the thrust force of the plane’s engines Wthrust. According to the workenergy theorem (Equation 6.3), the work done by the net external force Wcatapult + Wthrust is
equal to the change in the kinetic energy. The change in the kinetic energy is the given
kinetic energy of 4.5 × 107 J at lift-off minus the initial kinetic energy, which is zero since
the plane starts at rest. The work done by the thrust force can be determined from Equation
6.1 [W = (F cos θ) s], since the magnitude F of the thrust is 2.3 × 105 N and the magnitude s
of the displacement is 87 m. We note that the angle θ between the thrust and the
displacement is 0º, because they have the same direction. In summary, we will calculate
Wcatapult from Wcatapult + Wthrust = KEf − KE0.
SOLUTION According to the work-energy theorem, we have
Wcatapult + Wthrust = KEf − KE0
Using Equation 6.1 and noting that KE0 = 0 J, we can write the work energy theorem as
follows:
Wcatapult + ( F cosθ ) s = KE f
1 4 2 43
Work done by thrust
Solving for Wcatapult gives
Wcatapult = KEf − ( F cosθ ) s
1 4 2 43
Work done by thrust
= 4.5 ×107 J − ( 2.3 ×105 N ) cos 0° (87 m ) = 2.5 ×107 J
15. SSM REASONING
The car’s kinetic energy depends upon its mass and speed via
KE = 12 mv 2 (Equation 6.2). The total amount of work done on the car is equal to the
difference between its final and initial kinetic energies: W = KE f − KE 0 (Equation 6.3). We
will use these two relationships to determine the car’s mass.
SOLUTION Combining KE = 12 mv 2 (Equation 6.2) and W = KE f − KE 0 (Equation 6.3),
we obtain
1 m v2 − v2 = W
or
W = 12 mvf2 − 12 mv02
f
0
2
(
)
Solving this expression for the car’s mass m, and noting that 185 kJ = 185 000 J, we find
that
2 (185 000 J )
2W
m= 2 2 =
= 1450 kg
vf − v0 ( 28.0 m/s )2 − ( 23.0 m/s )2
31. REASONING The work done by the weight of the basketball is given by
Equation 6.1 as W = ( F cos θ ) s , where F = mg is the magnitude of the
weight, θ is the angle between the weight and the displacement, and s is the
magnitude of the displacement. The drawing shows that the weight and
displacement are parallel, so that θ = 0°. The potential energy of the mg
basketball is given by Equation 6.5 as PE = mgh, where h is the height of the
ball above the ground.
s
SOLUTION
a. The work done by the weight of the basketball is
2
W = ( F cos θ ) s = mg (cos 0°)(h0 − hf) = (0.60 kg)(9.80 m/s )(6.1 m − 1.5 m) = 27 J
b. The potential energy of the ball, relative to the ground, when it is released is
PE0 = mgh0 = (0.60 kg)(9.80 m/s2)(6.1 m) = 36 J
(6.5)
c. The potential energy of the ball, relative to the ground, when it is caught is
PEf = mghf = (0.60 kg)(9.80 m/s2)(1.5 m) = 8.8 J
(6.5)
d. The change in the ball’s gravitational potential energy is
ΔPE = PEf − PE0 = 8.8 J – 36 J = −27 J
We see that the change in the gravitational potential energy is equal to –27 J =
−W ,
where W is the work done by the weight of the ball (see part a).
_____________________________________________________________________________
32. REASONING Gravitational potential energy is PE = mgh (Equation 6.5), where m is the
mass, g is the magnitude of the acceleration due to gravity, and h is the height. The change
in the potential energy is the final minus the initial potential energy. The weight is mg, so it
can be obtained from the change in the potential energy.
SOLUTION The change in the pole-vaulter's potential energy is
PEf – PE0 = mghf – mgh0
With hf = 0 m and h0 = 5.80 m we find that
0 J – 3.70 ×103 J
= 6.38 ×102 N
hf − h0
0 m – 5.80 m
______________________________________________________________________________
mg =
PEf − PE0
=
34. REASONING The gravitational force on Rocket Man is conservative, but the force
generated by the propulsion unit is nonconservative. Therefore, the work WP done by the
propulsive force is equal to the net work done by all external nonconservative forces:
WP = Wnc. We will use Wnc = Ef − E0 (Equation 6.8) to calculate Wnc in terms of Rocket
Man’s initial mechanical energy and final mechanical energy. Because Rocket Man’s speed
and height increase after he leaves the ground, both his kinetic energy KE = 12 mv 2
(Equation 6.2) and potential energy PE = mgh (Equation 6.5) increase. Therefore, his final
mechanical energy Ef = KEf + PEf is greater than his initial mechanical energy
E0 = KE0 + PE0, and the work WP = Wnc = Ef − E0 done by the propulsive force is positive.
SOLUTION We make the simplifying assumption that Rocket Man’s initial height is
h0 = 0 m. Thus, his initial potential energy PE 0 = mgh0 is zero, and his initial kinetic energy
KE0 = 12 mv02 is also zero (because he starts from rest). This means he has no initial
mechanical energy, and we see that the work done by the propulsive force is equal to his
final mechanical energy: WP = Ef − 0 = Ef . Expressing Rocket Man’s final mechanical
energy Ef in terms of his final kinetic energy and his final potential energy, we obtain the
work done by the propulsive force:
WP = 12 mvf2 + mghf
2
= 12 (136 kg )(5.0 m/s ) + (136 kg ) 9.80 m/s 2 (16 m ) = +2.3 ×10 4 J
(
)
38. REASONING The distance h in the drawing in the text is the difference between the
skateboarder’s final and initial heights (measured, for example, with respect to the ground),
or h = hf − h0. The difference in the heights can be determined by using the conservation of
mechanical energy. This conservation law is applicable because nonconservative forces are
negligible, so the work done by them is zero (Wnc = 0 J). Thus, the skateboarder’s final total
mechanical energy Ef is equal to his initial total mechanical energy E0:
1 mv 2 + mgh = 1 mv 2 + mgh
12 4 4f 2 4 43f
12 4 402 4 430
Ef
(6.9b)
E0
Solving Equation 6.9b for hf − h0, we find that
1 v2
2 0
− 12 vf2
h −h =
14f 2 430
g
h
SOLUTION Using the fact that v0 = 5.4 m/s and vf = 0 m/s (since the skateboarder comes
to a momentary rest), the distance h is
h=
1 v2
2 0
− 12 vf2
1 ( 5.4
2
2
m/s ) − 12 ( 0 m/s )
2
=
= 1.5 m
g
9.80 m/s 2
______________________________________________________________________________
51.
SSM REASONING
a. Since there is no air friction, the only force that acts on the projectile is the conservative
gravitational force (its weight). The initial and final speeds of the ball are known, so the
conservation of mechanical energy can be used to find the maximum height that the
projectile attains.
b. When air resistance, a nonconservative force, is present, it does negative work on the
projectile and slows it down. Consequently, the projectile does not rise as high as when
there is no air resistance. The work-energy theorem, in the form of Equation 6.6, may be
used to find the work done by air friction. Then, using the definition of work, Equation 6.1,
the average force due to air resistance can be found.
SOLUTION
a. The conservation of mechanical energy, as expressed by Equation 6.9b, states that
1 mv 2 + mgh = 1 mv 2 + mgh
12 44f2 4 43f 12 4402 4 430
Ef
E0
The mass m can be eliminated algebraically from this equation since it appears as a factor in
every term. Solving for the final height hf gives
1
2
hf =
(v02 − vf2 ) + h
0
g
Setting h0 = 0 m and vf = 0 m/s, the final height, in the absence of air resistance, is
hf =
vo2 − vf2
2g
=
(18.0 m / s )2 − (0 m/s )2
2 9.80 m / s 2
(
)
= 16.5 m
b. The work-energy theorem is
Wnc =
(
1 mv 2
f
2
2
− 12 mv0
) + (mgh
f
− mgh0 )
(6.6)
where Wnc is the nonconservative work done by air resistance. According to Equation 6.1,
the work can be written as Wnc = ( FR cos180° ) s , where FR is the average force of air
resistance. As the projectile moves upward, the force of air resistance is directed downward,
so the angle between the two vectors is θ = 180° and cos θ = –1. The magnitude s of the
displacement is the difference between the final and initial heights, s = hf – h0 = 11.8 m.
With these substitutions, the work-energy theorem becomes
− FR s = 12 m vf2 − vo2 + mg ( hf − h0 )
(
Solving for FR gives
)
FR =
=
1m
2
(vf2 − vo2 ) + mg ( hf − h0 )
−s
1
2
( 0.750 kg ) ⎡⎣( 0 m/s )2 − (18.0 m/s )2 ⎤⎦ + (0.750 kg ) (9.80 m/s 2 ) (11.8 m )
=
− (11.8 m )
2.9 N
52. REASONING The ball’s initial speed and vertical position are given, as are the ball’s final
speed and vertical position. Therefore, we can use the work-energy theorem as given in
Equation 6.8 to determine the work done by the nonconservative force of air resistance.
SOLUTION According to Equation 6.8, the
work Wnc done by the nonconservative force of
air resistance is
Wnc =
3.10 m
2.00 m
(1 4mv4 2+ 4mgh43 ) − (1 4mv4 2+ 4mgh43 )
1
2
2
f
f
E
1
2
2
0
0
E
f
0
Thus, we find that
2
1
Wnc = ⎡ ( 0.600 kg )( 4.20 m / s ) + ( 0.600 kg ) 9.80 m/s 2 (3.10 m )⎤
⎣ 2
⎦
(
)
2
1
− ⎡ ( 0.600 kg )( 7.20 m / s ) + ( 0.600 kg ) 9.80 m/s 2 ( 2.00 m )⎤ = −3.8 J
⎣ 2
⎦
______________________________________________________________________________
(
)
in energy
65. REASONING The average power is given by Equation 6.10b ( P = ChangeTime
). The time
is given. Since the road is level, there is no change in the gravitational potential energy, and
the change in energy refers only to the kinetic energy. According to Equation 6.2, the
kinetic energy is 12 mv 2 . The speed v is given, but the mass m is not. However, we can obtain
the mass from the given weights, since the weight is mg.
SOLUTION
a. Using Equations 6.10b and 6.2, we find that the average power is
2
2
Change in energy KE f − KE 0 12 mvf − 12 mv0
P=
=
=
Time
Time
Time
Since the car starts from rest, v0 = 0 m/s, and since the weight is W = mg, the mass is
m = W/g. Therefore, the average power is
P=
1
2
mvf2 − 12 mv02
Time
=
Wvf2
2 g ( Time )
(1)
Using the given values for the weight, final speed, and the time, we find that
2
9.0 ×103 N ) ( 20.0 m/s )
Wvf2
(
P=
=
= 3.3 ×104 W (44 hp)
2
2 g ( Time )
2 (9.80 m/s ) (5.6 s )
b. From Equation (1) it follows that
2
1.4 ×104 N ) ( 20.0 m/s )
Wvf2
(
P=
=
= 5.1×104 W (68 hp)
2
2 g ( Time )
2 (9.80 m/s ) (5.6 s )
66. REASONING The average power is defined as the work divided by the time,
Equation 6.10a, so both the work and time must be known. The time is given. The work
can be obtained with the aid of the work-energy theorem as formulated in
(
)
Wnc = 12 mvf2 − 12 mv02 + ( mghf − mgh0 ) (Equation 6.6). Wnc is the work done by the lifting
force acting on the helicopter. In using this equation, we note that two types of energy are
changing: the kinetic energy
( mv ) and the gravitational potential energy (mgh).
1
2
2
The
kinetic energy is increasing, because the speed of the helicopter is increasing. The
gravitational potential energy is increasing, because the height of the helicopter is
increasing.
SOLUTION The average power is
P=
Wnc
t
(6.10a)
where Wnc is the work done by the nonconservative lifting force and t is the time. The work is
related to the helicopter’s kinetic and potential energies by Equation 6.6:
Wnc =
(
1
2
)
mvf2 − 12 mv02 + ( mghf − mgh0 )
Thus, the average power is
W
P = nc =
t
P=
1
2
(
1 mv 2
f
2
2
− 12 mv0
) + ( mgh
f
t
− mgh0 )
=
1m
2
2
f
(v
2
− v0
) + mg ( h
f
t
(810 kg ) ⎡⎣(7.0 m /s )2 − (0 m /s )2 ⎤⎦ + (810 kg ) (9.80 m /s 2 )[8.2 m − 0 m ]
3.5 s
− h0 )
4
= 2.4 × 10 W
74. REASONING The work done during each interval is equal to the area under the force vs.
displacement curve over that interval. In the interval from 2.0 to 4.0 m, the area under the
curve has two distinct regions, one triangular and the other rectangular. The total work done
from 2.0 to 4.0 m is the sum of the areas of these two regions.
SOLUTION
a. Since the area under the curve from 0 to 1.0 m is triangular, the area under the curve over
this interval is (1/2) × height × base:
1
2
W01 = (+6.0 N)(1.0 m − 0 m) = 3.0 J
b. Since there is no area under the curve from 1.0 to 2.0 m, the work done is zero . This is
reasonable since there is no net force acting on the object during this interval.
c. In the interval from 2.0 to 4.0 m, the area under the curve has two distinct regions. From
2.0 to 3.0 m, the area is triangular in shape, while from 3.0 to 4.0 m the region is
rectangular. The total work done from 2.0 to 4.0 m is the sum of the areas of these two
regions.
W23 =
1
2
( −6.0 N )(3.0 m − 2.0 m) = −3.0 J
and
W34 = ( −6.0 N )( 4.0 m − 3.0 m ) = −6.0 J
Thus, the total work done in the interval from 2.0 to 4.0 m is
W24 = W23 + W34 = ( −3.0 J ) + ( −6.0 J ) = −9.0 J
Notice in the interval from 2.0 to 4.0 m, the area under the curve (and hence the work done)
is negative, because the force is negative in this interval.
______________________________________________________________________________
75. REASONING The final speed vf of the object is related to its final kinetic energy by
KEf = 12 mvf2 (Equation 6.2). The object’s mass m is given, and we will use the work-energy
theorem W = KE f − KE 0 (Equation 6.3) to determine its final kinetic energy. We are helped
by the fact that the object starts from rest, and therefore has no initial kinetic energy KE0.
Thus, the net work done on the object equals its final kinetic energy: W = KEf = 12 mvf2 . In a
graph of F cos θ versus s, such as that given, the net work W is the total area under the
graph. For the purpose of calculation, it is convenient to divide this area into two pieces, a
triangle (from s = 0 m to s = 10.0 m) and a rectangle (from s = 10.0 m to s = 20.0 m). Both
pieces have the same width (10.0 m) and height (10.0 N), so the triangle has half the area of
the rectangle.
2W
.
m
done on the object from s = 0 m to s = 10.0 m
SOLUTION Solving W = 12 mvf2 for the final speed of the object, we obtain vf =
The net work W is the sum of the work W0,10
(the triangular area) and the work W10,20 done on the object from s = 10.0 m to s = 20.0 m
(the rectangular area). We calculate the work W10,20 by multiplying the width (10.0 m) and
height (10.0 N) of the rectangle: W10,20 = ( Width )( Height ) . The triangle’s area, which is
the work W0,10, is half this amount: W0,10 = 12 W10,20 . Therefore, the net work
W = W0,10 + W10,20 done during the entire interval from s = 0 m to s = 20.0 m is
W = 12 W10,20 + W10,20 = 23 W10,20 = 23 ( Width )( Height )
We can now calculate the final speed of the object from vf =
vf =
2
2W
:
m
( 32 ) ( Width )( Height ) = 3 (10.0 m )(10.0 N ) = 7.07 m/s
m
6.00 kg