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CHAPTER 6 WORK AND ENERGY
CONCEPTUAL QUESTIONS
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2.
REASONING AND SOLUTION
The force P acts along the displacement;
therefore, it does positive work. Both the normal force FN and the weight mg are
perpendicular to the displacement; therefore, they do zero work. The kinetic
frictional force fk acts opposite to the direction of the displacement; therefore, it does
negative work.
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4.
REASONING AND SOLUTION The sailboat moves at constant velocity, and,
therefore, has zero acceleration. From Newton's second law, we know that the net
external force on the sailboat must be zero.
a. There is no work done on the sailboat by a zero net external force.
b. Work is done by the individual forces that act on the boat; namely the wind that
propels the boat forward and the water that resists the motion of the boat. Since the
wind propels the boat forward, it does positive work on the boat. Since the force of
the water is a resistive force, it acts opposite to the displacement of the boat, and,
therefore, it does negative work. Since the total work done on the boat is zero, each
force must do an equal amount of work with one quantity being positive, and the
other being negative.
Note: The answer to part (a) could have been deduced from the work-energy
theorem as well. Since the velocity of the boat is constant, the kinetic energy of the
boat does not change and the total work done on the boat is zero.
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6.
REASONING AND SOLUTION The kinetic energy of an object of mass m
2
moving with speed v is given by KE  12 mv . The kinetic energy depends on both
the mass and the speed of the object. The mass of an automobile is significantly
greater than the mass of a motorcycle. Therefore, even if an automobile is moving
2
slowly, it is possible that the product 12 mv is greater for the car than it is for the
faster-moving motorcycle.
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7.
REASONING AND SOLUTION A net external force acts on a particle. This net
force is not zero. From Newton's second law, we can conclude that the net external
force causes the particle to accelerate. Since the particle experiences an acceleration,
its velocity must change. The change in velocity, however, may occur as a change
in magnitude only, a change in direction only, or a change in both magnitude and
direction.
a. This is sufficient information to conclude that the velocity of the particle changes;
however, there is not sufficient information to determine exactly how the velocity
changes.
b. There is not sufficient information to determine if the kinetic energy of the
particle changes. In terms of the work-energy theorem, the kinetic energy will
change if the net external force does work on the particle. But without knowing the
direction of the net external force with respect to the particle's displacement, we
cannot know if work is done.
c. There is not sufficient information to determine if the speed of the particle
2
changes. Kinetic energy is 12 mv , so the speed v will change if the kinetic energy
changes. But, as explained in part (b), there is insufficient information to determine
whether the kinetic energy changes.
12. REASONING AND SOLUTION If the total mechanical energy of an object is
conserved, then the sum of the kinetic energy and the potential energy must be
constant.
a. If the kinetic energy decreases, the gravitational potential energy must increase by
the same amount that the kinetic energy decreases.
b. If the potential energy decreases, the kinetic energy must increase by the same
amount that the potential energy decreases.
c. If the kinetic energy does not change, the potential energy cannot change either.
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PROBLEMS
3.
REASONING AND SOLUTION According to Equation 6.1, W = Fs cos the
work is

a.
W = (94.0 N)(35.0 m) cos 25.0° = 2980 J
b.
W = (94.0 N)(35.0 m) cos 0° = 3290 J
9.
SSM REASONING AND SOLUTION According to Equation 6.1, the work
done by the husband and wife are, respectively,
Husband
WH  ( FH cos  H )s
 Wife
WW  ( FW cos  W )s
Since both the husband and the wife do the same amount of work,
( FH cos  H )s  ( FW cos  W )s
Since the displacement has the same magnitude s in both cases, the magnitude of the
force exerted by the wife is
FW  FH
cos  H
cos 58
 (67 N)
 45 N
cos  W
cos 38
15.
SSM WWW REASONING AND SOLUTION The work required to bring
each car up to speed is, from the work-energy theorem,
1
1
2
2
W  KEf  KE 0  mvf  mv0 . Therefore,
2
2

2

2

1
2

1
2

WB  m vf  v0  (1.20  10 kg)  (40.0 m/s)   0 m/s 
1
2
2
3
2


2


WB  m vf  v0  (2.00  10 kg)  (40.0 m/s)   0 m/s 
1
2
2
3
2

5
9.60  10 J
2
6
  1.60  10

J
The additional work required to bring car B up to speed is, therefore,
6
5
5
WB  WA  (1.6 10 J) – (9.60 10 J) = 6.4 10 J
20.
REASONING Since the person has an upward acceleration, there must be a net
force acting in the upward direction. The net force Fy is related to the acceleration ay by
Newton’s second law, Fy  ma y , where m is the mass of the person. This relation will
allow us to determine the tension in the cable. The work done by the tension and the
person’s weight can be found directly from the definition of work, Equation 6.1.
SOLUTION
a. The free-body diagram at the right shows the two forces that act on
the person. Applying Newton’s second law, we have
T
s
mg
+y
T  mg  ma y
Fy
Solving for the magnitude of the tension in the cable yields
T = m(ay + g) = (79 kg)(0.70 m/s2 + 9.80 m/s2) = 8.3  102 N
b. The work done by the tension in the cable is
WT  T cos   s  (8.3  102 N) (cos 0°) (11 m) = 9.1  103 J
(6.1)
c. The work done by the person’s weight is
WW   mg cos   s  (79 kg)  9.8 m/s2  (cos 180°) (11 m) = 8.5  103 J
(6.1)
d. The work-energy theorem relates the work done by the two forces to the
change in the kinetic energy of the person. The work done by the two forces is W = WT +
WW:
WT + WW  12 mvf2  12 mv02
(6.3)
W
Solving this equation for the final speed of the person gives
vf  v02 
2
WT  WW 
m
2
 9.1  103 J  8.5  103 J   4 m / s
79 kg
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
 0 m/s 
2

26.
REASONING The work done by the weight of the basketball is given by
Equation 6.1 as W   F cos   s , where F = mg is the magnitude of the weight,
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  = 0. The potential energy of the basketball is given by Equation 6.5 as
mg
PE = mgh, where h is the height of the ball above the ground.
SOLUTION
a.
The work done by the weight of the basketball is
s
2
W   F cos   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
c.
(6.5)
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
d.
(6.5)
The change in the ball’s gravitational potential energy is
PE = 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).
37. REASONING AND SOLUTION
Mechanical energy is conserved
if no friction acts. Ef = E0 or
1 mv 2
f
2
 mgh 
h = 5.0 m
1 mv 2
0
2
Taking h0 = 0 m and rearranging
yields
2
v f =?
vf  v0  2 gh 
v o = 11.0 m/s
11.0 m/s 2  2 9.80 m/s2  5.0 m  
4.8 m/s
41.
SSM REASONING Friction and air resistance are being ignored. The normal
force from the slide is perpendicular to the motion, so it does no work. Thus, no net work
is done by nonconservative forces, and the principle of conservation of mechanical
energy applies.
SOLUTION Applying the principle of conservation of mechanical energy to the
swimmer at the top and the bottom of the slide, we have
1
2
mvf  mghf
2
Ef
1
2
2
 mv0  mgh0
E0
If we let h be the height of the bottom of the slide above the water, hf  h , and
h0  H . Since the swimmer starts from rest, v0  0 m/s, and the above expression
becomes
1 2
v
2 f
 gh  gH
Solving for H, we obtain
v2
H  h f
2g
Before we can calculate H, we must find vf and h. Since the velocity in the
horizontal direction is constant,
x 5.00 m
vf 

 10.0 m/s
t 0.500 s
The vertical displacement of the swimmer after leaving the slide is, from Equation
3.5b (with down being negative),
y  2 ay t 2  2 (–9.80 m/s 2 )(0.500 s)2  1.23 m
1
1
Therefore, h = 1.23 m. Using these values of vf and h in the above expression for
H, we find
vf2
(10.0 m/s)2
H  h
 1.23 m 
2  6.33 m
2g
2(9.80 m/s )
44.
REASONING AND SOLUTION If air resistance is ignored, the only
nonconservative force that acts on the person is the normal force exerted on the person by
the surface. Since this force is always perpendicular to the direction of the displacement,
the work done by the normal force is zero. We can conclude, therefore, that mechanical
energy is conserved.
1
2
mv 0 
2
(1)
1
2
2
mgh 0  mv f  mgh f
where the final state pertains to the position where
the person leaves the surface. Since the person starts from
rest v0 = 0 m/s. Since the radius of the surface is r,
h0 = r, and hf = r cos f where f is the angle at which the
person leaves the surface.
Equation (1) becomes
1
2
mgr  mv f  mg(r cos f )
2
(2)
In general, as the person slides down the surface, the
two forces that act on him are the normal force FN and the
weight mg. The centripetal force required to keep the person
moving in the circular path is the resultant of FN and the radial
component of the weight, mg cos .
When the person leaves the surface, the normal force is
zero, and the radial component of the weight provides the
centripetal force.
mg cos f 
mv 2f
r
v2f  gr cos f (3)

2
Substituting this expression for v f into Equation (2) gives
mgr  12 mg (r cos f )  mg (r cos f )
Solving for  f gives
2
 
 f  cos 1   = 48
3
47.
REASONING AND SOLUTION
a. The work done by non-conservative forces is given by Equation 6.7b as
Wnc = KE + PE
so
PE = Wnc – KE
Now
KE =
and
1
2
mvf2 
1
2
mv02 =
1
2
(55.0 kg)[(6.00 m/s) 2  (1.80 m/s) 2 ] = 901 J
PE = 80.0 J – 265 J – 901 J = 1086 J
b.
PE = mg (h – h0) so
h – h0 
 PE
–1086 J

 –2.01 m
mg 55.0 kg  9.80 m/s 2


Thus, the skater’s vertical position has changed by 2.01 m , and the skater is
below the starting point .
53.
SSM REASONING AND SOLUTION According to the work-energy
theorem as given in Equation 6.8, we have
Wnc 

1
2
mvf
2
 mghf

1
2
mv0
2
 mgh0

The metal piece starts at rest and is at rest just as it barely strikes the bell, so that


2
vf  v0  0 m/s. In addition, hf  h and h0  0 m, while Wnc  0.25 12 M v , where M
and v are the mass and speed of the hammer. Thus, the work-energy theorem becomes
0.25

1 M v2
2
  mgh
Solving for the speed of the hammer, we find
v 
2mgh

0.25M
2(0.400 kg)(9.80 m/s 2 )(5.00 m)
 4.17 m/s
0.25 (9.00 kg)
60. REASONING AND SOLUTION
a. The power developed by the engine is
P = Fv = (2.00  102 N)(20.0 m/s) = 4.00  103 W
b. The force required of the engine in order to maintain a constant speed up the
slope is
F = Fa + mg sin 37.0°
The power developed by the engine is then
P = Fv = (Fa + mg sin 37.0°)v
P = [2.00  102 N + (2.50  102 kg)(9.80 m/s2)sin 37.0°](20.0 m/s) =
3.35 10 4 W
(6.11)
77.
REASONING AND SOLUTION The conservation of energy applied between
point A and the top of the trajectory gives
KEA + mghA = mgh
where h = 4.00 m. Rearranging, we find
KEA = mg(h – hA)
or

vA  2 g  h  hA   2 9.80 m/s
2
  4.00 m  3.00 m   4.43 m/s
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