Cutnell_9th_problems_ch_1_thru_10_converted

34.
“Rocket Man” has a propulsion unit strapped to his back. He starts from rest on the ground, fires the
unit, and accelerates straight upward. At a height of 16 m, his speed is
. His mass, including the propulsion unit, has
the approximately constant value of 136 kg. Find the work done by the force generated by the
propulsion unit.
35.
A 55.0-kg skateboarder starts out with a speed of
. He does
of work on himself by
pushing with his feet against the ground. In addition, friction does
of work on him. In both cases, the
forces doing the work are nonconservative. The final speed of the skateboarder is
.
(a) Calculate the change
in the gravitational potential energy.
Answer:
REASONING AND SOLUTION
(a) The work done by non-conservative forces is given by Equation 6.7b as
Now
and
(b)

so
This answer is negative, so that the skater's vertical position has changed by
and the
skater is
.
(b) How much has the vertical height of the skater changed, and is the skater above or below the
starting point?
Answer:
The skater is 2.01 m below the starting point.
REASONING AND SOLUTION
(a) The work done by non-conservative forces is given by Equation 6.7b as
Now
and
(b)
so
This answer is negative, so that the skater's vertical position has changed by
and the
skater is
.
REASONING AND SOLUTION
(a) The work done by non-conservative forces is given by Equation 6.7b as

Now
and
(b)
so
This answer is negative, so that the skater's vertical position has changed by
and the skater is
.
Section 6.5
T he
Conservatio
n of
Mechanical

Energy

36. A 35-kg girl is bouncing on a trampoline. During a certain interval after she leaves the surface of
the trampoline, her kinetic energy decreases to 210 J from 440 J. How high does she rise during this
interval? Neglect air resistance.

37.

A gymnast is swinging on a high bar. The distance between his waist and the bar is 1.1 m, as the
drawing shows. At the top of the swing his speed is momentarily zero. Ignoring friction and treating
the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.

Problem 37

Answer:

REASONING The only two forces that act on the gymnast are his weight and the force exerted on his
hands by the high bar. The latter is the (non-conservative) reaction force to the force exerted on the
bar by the gymnast, as predicted by Newton's third law. This force, however, does no work because it
points perpendicular to the circular path of motion. Thus,

, and we can apply the principle of conservation of mechanical energy.

SOLUTION The conservation principle gives

Since the gymnast's speed is momentarily zero at the top of the swing,

If we take

at the bottom

of the swing, then

,where is the radius of the circular path followed by the gymnast's waist. Making these

substitutions in the above expression and solving for , we obtain

38.

The skateboarder in the draw-ing starts down the left side of the ramp with an initial speed of

. Neglect

nonconservative forces, such as friction and air resistance, and find the height h of the highest point
reached by the skateboarder on the right side of the ramp.

39.

A slingshot fires a pebble from the top of a building at a speed of

. The building is 31.0 m tall.

Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is
fired (a) horizontally,

Answer:

(b) vertically straight up, and

Answer:

(c) vertically straight down.

Answer:

40.

The drawing shows two boxes resting on frictionless ramps. One box is relatively light and sits on a
steep ramp.

The other box is heavier and rests on a ramp that is less steep. The boxes are released from rest at A
and allowed to slide down the ramps. The two boxes have masses of 11 and 44 kg. If A and B are 4.5
and 1.5 m, respectively, above the ground, determine the speed of

(a) the lighter box and
(b) the heavier box when each reaches B.
(c) What is the ratio of the kinetic energy of the heavier box to that of the lighter box at B?
41. A 47.0-g golf ball is driven from the tee with an initial speed of
and rises to a height of 24.6 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
Answer:
52.2 J
(b) What is its speed when it is 8.0 m below its highest point?
Answer:

42.
Consult Conceptual Example 9 in preparation for this problem. The drawing shows a person who,
starting from rest at the top of a cliff, swings down at the end of a rope, releases it, and falls into the

water below. There are two paths by which the person can enter the water. Suppose he enters the water
at a speed of

via path 1. How

fast is he moving on path 2 when he releases the rope at a height of 5.20 m above the water? Ignore
the effects of air resistance.

*43.

The drawing shows a skateboarder moving at

along a horizontal section of a track that is slanted

upward by

above the horizontal at its end, which is 0.40 m above the ground. When she leaves the track, she
follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the
maximum height H to which she rises above the end of the track.

Answer:

0.60 m

REASONING To find the maximum height above the end of the track we will analyze the projectile
motion of the skateboarder after she leaves the track. For this analysis we will use the principle of
conservation of mechanical energy, which applies because friction and air resistance are being
ignored. In applying this principle to the projectile motion, however, we will need to know the speed
of the skateboarder when she leaves the track. Therefore, we will begin by determining this speed,
also using the conservation principle in the process. Our approach, then, uses the conservation
principle twice.

SOLUTION Applying the conservation of mechanical energy in the form of Equation 6.9b, we have
We designate the flat portion of the track as having a height

and note from the drawing that its end is at a

height of
above the ground. Solving for the final speed at the end of the track gives

This speed now becomes the initial speed

for the next application of the conservation principle. At the

maximum height of her trajectory she is traveling horizontally with a speed that equals the horizontal
component of her launch velocity. Thus, for the next application of the conservation principle

. Applying

the conservation of mechanical energy again, we have

Recognizing that

and

and solving for give

*44.

A small lead ball, attached to a 1.5-m rope, is being whirled in a circle that lies in the vertical plane.
The ball is whirled at a constant rate of three revolutions per second and is released on the upward
part of the circular motion when it is 0.75 m above the ground. The ball travels straight upward. In the
absence of air resistance, to what maximum height above the ground does the ball rise?

*45.

Two pole-vaulters just clear the bar at the same height. The first lands at a speed of

, and the second

lands at a speed of

. The first vaulter clears the bar at a speed of

. Ignore air resistance and

friction and determine the speed at which the second vaulter clears the bar.

Answer:

REASONING Gravity is the only force acting on the vaulters, since friction and air resistance are
being ignored. Therefore, the net work done by the nonconservative forces is zero, and the principle
of conservation of mechanical energy holds.

SOLUTION Let

represent the total mechanical energy of the second vaulter at the ground, and

represent the

total mechanical energy of the second vaulter at the bar. Then, the principle of mechanical energy is
written for the second vaulter as

Since the mass of the vaulter appears in every term of the equation, can be eliminated algebraically.
The quantity is

, where is the height of the bar. Furthermore, when the vaulter is at ground level,

. Solving

for

we have

(1)

In order to use Equation 1, we must first determine the height of the bar. The height can be determined
by applying the principle of conservation of mechanical energy to the first vaulter on the ground and
at the bar. Using notation similar to that above, we have

Where

corresponds to the total mechanical energy of the first vaulter at the bar. The height of the bar is,
therefore,

The speed at which the second vaulter clears the bar is, from Equation 1,

*46. A pendulum consists of a small object hanging from the ceiling at the end of a string of
negligible mass. The string has a length of 0.75 m. With the string hanging vertically, the object is
given an initial velocity of parallel to

the ground and swings upward on a circular arc. Eventually, the object comes to a momentary halt at a
point where the string makes an angle with its initial vertical orientation and then swings back
downward. Find the angle .

*47.

A semitrailer is coasting downhill along a mountain highway when its brakes fail. The driver pulls
onto a runaway-truck ramp that is inclined at an angle of

above the horizontal. The semitrailer coasts to a stop after

traveling 154 m along the ramp. What was the truck's initial speed? Neglect air resistance and friction.

Answer:

*48.

The drawing shows two frictionless inclines that begin at ground level

and slope upward at the same

angle . One track is longer than the other, however. Identical blocks are projected up each track with
the same initial speed

. On the longer track the block slides upward until it reaches a maximum height H above the ground.
On the shorter track the block slides upward, flies off the end of the track at a height

above the ground, and then follows

the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block
is a height above the end of the track. The initial total mechanical energy of each block is the same
and is all kinetic energy. The initial speed of each block is

, and each incline slopes upward at an angle of

. The block

on the shorter track leaves the track at a height of

above the ground. Find

(a) the height H for the block on the longer track and

(b) the total height

for the block on the shorter track.

**49. A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill,
as the drawing illustrates. The crest of the second hill is circular, with a radius of

. Neglect friction and air resistance. What

must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the
second hill?

Answer:



18 m

**50. A person starts from rest at the top of a large frictionless spherical surface, and slides into the
water below (see the drawing). At what angle does the person leave the surface?
(Hint: When the person leaves the surface, the normal force is zero.)
Section 6.6
Nonconservative
Forces and the
Work–Energy
T heo rem
51.
A projectile of mass 0.750 kg is shot straight up with an initial speed of
.
(a) How high would it go if there were no air resistance?
Answer:
16.5 m
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
andslows 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
)

The mass can be eliminated algebraically from this equation since it appears as a factor in every term.
Solving for the final height
gives
Setting
and
, the final height, in the absence of air resistance, is
(b The work-energy theorem is
)
(6.6)
where
is the nonconservative work done by air resistance. According to Equation 6.1, the work

can be written as

, where

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

and

. The magnitude of the displacement is the difference between

the final and initial heights,

. With these substitutions, the work-energy

theorem becomes

Solving for

gives

(b) If the projectile rises to a maximum height of only 11.8 m, determine the magnitude of the
average force due to air resistance.

Answer:

2.9 N

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
andslows 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
)
The mass can be eliminated algebraically from this equation since it appears as a factor in every term.
Solving for the final height
gives
Setting
and
, the final height, in the absence of air resistance, is
(b The work-energy theorem is
)
(6.6)
where
is the nonconservative work done by air resistance. According to Equation 6.1, the work
can be written as
, where
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
and
. The magnitude of the displacement is the difference between
the final and initial heights,
. With these substitutions, the work-energy

theorem becomes

Solving for

gives

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
andslows 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

The mass can be eliminated algebraically from this equation since it appears as a factor in every term.
Solving for the final height

gives

Setting

and

, the final height, in the absence of air resistance, is

(b) The work-energy theorem is

(6.6)

where

is the nonconservative work done by air resistance. According to Equation 6.1, the work can be

written as

, where

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 and

. The magnitude of the displacement is the difference between the final and initial heights,

. With these substitutions, the work-energy theorem becomes

Solving for
gives
52. A basketball player makes a jump shot. The 0.600-kg ball is released at a height of 2.00 m above
the floor with a speed of
. The ball goes through the net 3.10 m above the floor at a speed of

. What is the work done

on the ball by air resistance, a nonconservative force?

53. Starting from rest, a 93-kg firefighter slides down a fire pole. The average frictional force
exerted on him by the pole has a magnitude of 810 N, and his speed at the bottom of the pole is

. How far did he slide down the pole?

Answer:

5.3 m

54.

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a
nonconservative force) acts on her. The student has a mass of 83.0 kg, and the height of the water
slide is 11.8 m. If the kinetic frictional force does

of work, how fast is the student going at the bottom of the slide?

55.

The (nonconservative) force propelling a

car up a mountain road does

of work on the

car. The car starts from rest at sea level and has a speed of

at an altitude of

above sea level.

Obtain the work done on the car by the combined forces of friction and air resistance, both of which
are nonconservative forces.

Answer:

REASONING The work-energy theorem can be used to determine the net work done on the car by the
nonconservative forces of friction and air resistance. The work-energy theorem is, according to
Equation 6.8, The nonconservative forces are the force of friction, the force of air resistance, and the
force provided by the engine.

Thus, we can rewrite Equation 6.8 as

This expression can be solved for

.

SOLUTION We will measure the heights from sea level, where
. Since the car starts from rest,
.
Thus, we have
56.
In the sport of skeleton a participant jumps onto a sled (known as a skeleton) and proceeds to slide
down an icy track, belly down and head first. In the 2010 Winter Olympics, the track had sixteen turns
and dropped 126 m in elevation from top to bottom.

(© Frank Gunn/AP/Wide World Photos)
(a) In the absence of nonconservative forces, such as friction and air resistance,
what would be the speed of a rider at the bottom of the track? Assume that
the speed at the beginning of the run is relatively small and can be ignored.
(b) In reality, the gold-medal winner (Canadian Jon Montgomery) reached the bottom in one heat with
a speed of
(about 91 mi/h). How much work was done on him and his sled (assuming a total mass of 118

kg) by nonconservative forces during this heat?

*57.

In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of

. However,

this speed is inadequate to compensate for the kinetic friction between the puck and the ice. As a
result, the puck travels only one-half the distance between the players before sliding to a halt. What
minimum initial speed should the puck have been given so that it reached the teammate, assuming that
the same force of kinetic friction acted on the puck everywhere between the two players?

Answer:

*58. A pitcher throws a 0.140-kg baseball, and it approaches the bat at a speed of

. The bat does

of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball
leaves the bat and is 25.0 m above the point of impact.

*59.

A 67.0-kg person jumps from rest off a 3.00-m-high tower straight down into the water. Neglect air
resistance.

She comes to rest 1.10 m under the surface of the water. Determine the magnitude of the average
force that the water exerts on the diver. This force is nonconservative.

Answer:

2450 N

*60. At a carnival, you can try to ring a bell by striking a target with a 9.00-kg hammer. In response, a
0.400-kg metal piece is sent upward toward the bell, which is 5.00 m above. Suppose that 25.0% of the
hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the
hammer be moving when it strikes the target so that the bell just barely rings?

**61.

A truck is traveling at

down a hill when the brakes on all four wheels lock. The hill makes an

angle of

with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is
0.750. How far does the truck skid before coming to a stop?

Answer:

13.5 m

REASONING After the wheels lock, the only nonconservative force acting on the truck is
friction.The work done by this conservative force can be determined from the work-energy theorem.
According to Equation 6.8,

(1)
where
is the work done by friction. According to Equation 6.1,
; since the force of kinetic
friction points opposite to the displacement,
. According to Equation 4.8, the kinetic frictional force has
a magnitude of
, where
is the coefficient of kinetic friction and
is the magnitude of the normal
force. Thus,
(2)
Since the truck is sliding down an incline, we refer to the free-body diagram in order to determine the
magnitude of the normal force
. The free-body diagram at the right shows the three forces that act on the truck. They are
the normal force

, the force of kinetic friction , and the weight

. The weight has been resolved into its

components, and these vectors are shown as dashed arrows. From the free body diagram and the fact
that the truck does not accelerate perpendicular to the incline, we can see that the magnitude of the
normal force is given by Therefore, Equation 2 becomes

(3)

Combining Equations 1, 2, and 3, we have

(4)

This expression may be solved for .

SOLUTION When the car comes to a stop,

. If we take

at ground level, then, from the

drawing at the right, we see that

. Equation 4 then gives

Sectio

n 6.7

Power

62. A person is making homemade ice cream. She exerts a force of magnitude 22 N on the free end of
the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.28 m. The
force is always applied parallel to the motion of the handle. If the handle is turned once every 1.3 s,
what is the average power being expended?

63.

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the
average power per kilogram generated by seven-time-winner Lance Armstrong

is 6.50 W per kilogram of his
body mass.
(a) How much work does he do during a 135-km race in which his average speed is
?
Answer:
REASONING The work done is equal to the average power multiplied by the time , or (6.10a)
The average power is the average power generated per kilogram of body mass multiplied by
Armstrong's mass. The time of the race is the distance traveled divided by the average speed , or
(see Equation
2.1).
SOLUTION
(a) Substituting
into Equation 6.10a gives
(b) Since 1
nutritional calories, the work done is
(b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work
done in part (a) in terms of nutritional Calories, noting that
nutritional Calories.
Answer:
nutritional calories
REASONING The work done is equal to the average power multiplied by the time , or (6.10a)
The average power is the average power generated per kilogram of body mass multiplied by
Armstrong's mass. The time of the race is the distance traveled divided by the average speed , or
(see Equation
2.1).
SOLUTION
(a) Substituting

into Equation 6.10a gives
(b) Since 1
nutritional calories, the work done is

REASONING The work done is equal to the average power multiplied by the time , or (6.10a)
The average power is the average power generated per kilogram of body mass multiplied by
Armstrong's mass. The time of the race is the distance traveled divided by the average speed , or
(see Equation 2.1).

SOLUTION
(a) Substituting
into Equation 6.10a gives
(b) Since 1
nutritional calories, the work done is

64.
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the
oars) toward you, it moves a distance of 1.2 m in a time of 1.5 s. The readout on the display indicates
that the average power you are producing is 82 W. What is the magnitude of the force that you exert
on the handle?

65.
A car accelerates uniformly from rest to
in 5.6 s along a level stretch of road. Ignoring friction,
determine the average power required to accelerate the car if
(a) the weight of the car is
and

Answer:

(b) the weight of the car is
.
Answer:
66.
A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force
does work in raising the helicopter. An 810-kg helicopter rises from rest to a speed of
in a time of 3.5 s. During this time it
climbs to a height of 8.2 m. What is the average power generated by the lifting force?
*67.
Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the
fastest-accelerating animals, because it can go from rest to
in 4.0 s. If its mass is 110 kg,
determine the average power developed by the cheetah during the acceleration phase of its motion.
Express your answer in
(a) watts and
Answer:
(b) horsepower.
Answer:

13 hp

*68.
In 2.0 minutes, a ski lift raises four skiers at constant speed to a height of 140 m. The average mass of
each skier is 65 kg. What is the average power provided by the tension in the cable pulling the lift?
**69.
The motor of a ski boat generates an average power of
when the boat is moving at a constant
speed of
. When the boat is pulling a skier at the same speed, the engine must generate an average power of
. What is the tension in the tow rope that is pulling the skier?
Answer:

REASONING AND SOLUTION In the following drawings, the positive direction is to the right. When
the boat is not pulling a skier, the engine must provide a force
to overcome the resistive force of the water
.
Since the boat travels with constant speed, these forces must be equal in magnitude and opposite in
direction.
or
(1)
When the boat is pulling a skier, the engine must provide a force
to balance the resistive force of the water
and
the tension in the tow rope .
or
(2)
The average power is given by
, according to Equation 6.11 in the text. Since the boat moves with the same
speed in both cases,
, and we have
Using Equations 1 and 2, this becomes
Solving for gives

The force
can be determined from
, thereby giving
**70. A 1900-kg car experiences a combined force of air resistance and friction that has the same
magnitude whether the car goes up or down a hill at
. Going up a hill, the car's engine produces 47 hp more power to sustain
the constant velocity than it does going down the same hill. At what angle is the hill inclined above the
horizontal?
Section
6.9
Work
Done
by a
Variable
Force

71.
The drawing shows the force–displacement graph for two different bows. These graphs give the
force that an archer must apply to draw the bowstring.
Problem 71
(a) For which bow is more work required to draw the bow fully
from
to
? Give your reasoning.
Answer:
Bow 1 requires more work.
REASONING The area under the force-versus-
displacement graph over any displacement interval is equal to
the work done over that displacement interval.
SOLUTION
(a) Since the area under the force-versus-displacement graph
from
is greater for bow 1, it follows that
.
(b) The work required to draw bow 1 is equal to the area of
the triangular region under the force-versus-displacement
curve for bow 1. Since the area of a triangle is equal one-
half times the base of the triangle times the height of the
triangle, we have
For bow 2, we note that each small square under the
force-versus-displacement graph has an area of

We estimate that there are approximately 31.3 squares
under the force-versus-displacement graph for bow 2;
therefore, the total work done is
Therefore, the additional work required to stretch bow 1
as compared to bow 2 is
(b) Estimate the additional work required for the bow identified in part (a) compared to the other bow.
Answer:

25 J

REASONING The area under the force-versus-displacement graph over any displacement interval is
equal to the work done over that displacement interval.

SOLUTION
(a) Since the area under the force-versus-displacement graph from
is greater for bow 1, it
follows that
.
(b) The work required to draw bow 1 is equal to the area of the triangular region under the force-

versus-displacement curve for bow 1. Since the area of a triangle is equal one-half times the base of
the triangle times the height of the triangle, we have

For bow 2, we note that each small square under the force-versus-displacement graph has an area of
We estimate that there are approximately 31.3 squares under the force-versus-displacement graph for
bow 2; therefore, the total work done is

Therefore, the additional work required to stretch bow 1 as compared to bow 2 is

REASONING The area under the force-versus-displacement graph over any displacement interval is
equal to the work done over that displacement interval.

SOLUTION

(a) Since the area under the force-versus-displacement graph from

is greater for bow 1, it follows

that

.

(b) The work required to draw bow 1 is equal to the area of the triangular region under the force-
versus-displacement curve for bow 1. Since the area of a triangle is equal one-half times the base of
the triangle times the height of the triangle, we have

For bow 2, we note that each small square under the force-versus-displacement graph has an area of
We estimate that there are approximately 31.3 squares under the force-versus-displacement graph for
bow 2; therefore, the total work done is

Therefore, the additional work required to stretch bow 1 as compared to bow 2 is

72. The graph shows how the force component

along the displacement varies with the magnitude s of the

displacement. Find the work done by the force.

(Hint: Recall how the area of a triangle is related to the triangle's base and height.)

73. Review Example 14, in which the work done in drawing the bowstring in Figure 6.21 from to
is
determined. In part b of the figure, the force component
reaches a maximum value at
. Find the
percentage of the total work that is done when the bowstring is moved
(a) from
to 0.306 m and
Answer:
54%
(b) from
to 0.500 m.
Answer:
46%
74. The force component along the displacement varies with the magnitude of the displacement, as
shown in the graph.

Find the work done by the force in the interval from

(a) 0 to 1.0 m,

(b) 1.0 to 2.0 m, and

(c) 2.0 to 4.0 m. (Note: In the last interval the force component is negative, so the work is negative.)
75.

A net external force is applied to a 6.00-kg object that is initially at rest. The net force component
along the displacement of the object varies with the magnitude of the displacement as shown in the
drawing. What is the speed of the object at

?

Answer:

Copyright © 2012 John Wiley & Sons, Inc. Al rights reserved.

Problems

Sect io n 7.1 T he Impulse–Mo ment um T heo rem

1.

A 46-kg skater is standing still in front of a wall. By pushing against the wall she propels herself
backward with a velocity of

. Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find

the magnitude and direction of the average force she exerts on the wall (which has the same
magnitude as, but opposite direction to, the force that the wall applies to her).

Answer:

69 N, directed opposite to the skater's velocity

REASONING The impulse that the wall exerts on the skater can be found from the impulse-
momentum theorem, Equation 7.4. The average force exerted on the skater by the wall is the only
force exerted on her in the horizontal direction, so it is the net force;

.

SOLUTION From Equation 7.4, the average force exerted on the skater by the wall is From Newton's
third law, the average force exerted on the wall by the skater is equal in magnitude and opposite in
direction to this force. Therefore,

The plus sign indicates that this force points

of the skater.

2. A model rocket is constructed with a motor that can provide a total impulse of

. The mass of the rocket is

0.175 kg. What is the speed that this rocket achieves when launched from rest? Neglect the effects of
gravity and air resistance.

3. Before starting this problem, review Conceptual Example 3. Suppose that the hail described there
bounces off the roof of the car with a velocity of

. Ignoring the weight of the hailstones, calculate the force exerted by the hail

on the roof. Compare your answer to that obtained in Example 2 for the rain, and verify that your
answer is consistent with the conclusion reached in Conceptual Example 3.

Answer:

1.8 N, directed straight downward

4.

In a performance test, each of two cars takes 9.0 s to accelerate from rest to

. Car A has a mass of 1400

kg, and car B has a mass of 1900 kg. Find the net average force that acts on each car during the test.

5.

A volleyball is spiked so that its incoming velocity of

is changed to an outgoing velocity of

. The mass of the volleyball is 0.35 kg. What impulse does the player apply to the ball?

Answer:

REASONING The impulse that the volleyball player applies to the ball can be found from the
impulse-momentum theorem, Equation 7.4. Two forces act on the volleyball while it's being spiked:
an average force exerted by the player, and the weight of the ball. As in Example 1, we will assume
that is much greater than the weight of the ball, so the weight can be neglected. Thus, the net average
force

is equal to .

SOLUTION From Equation 7.4, the impulse that the player applies to the volleyball is The minus sign
indicates that the direction of the impulse is the same as that of the final velocity of the ball.

6. Two arrows are fired horizontally with the same speed of

. Each arrow has a mass of 0.100 kg. One is fired

due east and the other due south. Find the magnitude and direction of the total momentum of this two-
arrow system.

Specify the direction with respect to due east.

7. Refer to Conceptual Example 3 as an aid in understanding this problem. A hockey goalie is
standing on ice. Another player fires a puck

at the goalie with a velocity of

.

(a) If the goalie catches the puck with his glove in a time of

s, what is the average force (magnitude

and direction) exerted on the goalie by the puck?

Answer:

(b) Instead of catching the puck, the goalie slaps it with his stick and returns the puck straight back to
the player with a velocity of

. The puck and stick are in contact for a time of

. Now what is the

average force exerted on the goalie by the puck? Verify that your answers to parts (a) and (b) are
consistent

with the conclusion of Conceptual Example 3.

Answer:

8.

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid
injury is to bend your knees upon landing to reduce the force of the impact. A 75-kg man just before
contact with the ground has a speed of

.

(a) In a stiff-legged landing he comes to a halt in 2.0 ms. Find the average net force that acts on him
during this time.

(b) When he bends his knees, he comes to a halt in 0.10 s. Find the average net force now.

(c) During the landing, the force of the ground on the man points up-ward, while the force due to
gravity points downward. The average net force acting on the man includes both of these forces.
Taking into account the directions of the forces, find the force of the ground on the man in parts (a)
and (b).

9. A space probe is traveling in outer space with a momentum that has a magnitude of

. A

retrorocket is fired to slow down the probe. It applies a force to the probe that has a magnitude of and
a

direction opposite to the probe's motion. It fires for a period of 12 s. Determine the momentum of the
probe after the retrorocket ceases to fire.

Answer:

*10.

A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The
incident water stream has a velocity of

, while the exiting water stream has a velocity of

. The mass of

water per second that strikes the blade is

. Find the magnitude of the average force exerted on the water by

the blade.

*11.

A student

falls freely from rest and strikes the ground. During the collision with the ground, he

comes to rest in a time of 0.040 s. The average force exerted on him by the ground is +18 000 N,
where the upward direction is taken to be the positive direction. From what height did the student fall?
Assume that the only force acting on him during the collision is that due to the ground.

Answer:

6.7 m

*12.

A golf ball strikes a hard, smooth floor at an angle of

and, as the drawing shows, rebounds at the same
angle. The mass of the ball is 0.047 kg, and its speed is
just before and after striking the floor. What is the
magnitude of the impulse applied to the golf ball by the floor?

(Hint: Note that only the vertical component of the ball's momentum changes during impact with the
floor, and ignore the weight of the ball.)
*13.
An 85-kg jogger is heading due east at a speed of
. A 55-kg jogger is heading
north of east at a
speed of
. Find the magnitude and direction of the sum of the momenta of the two joggers.
Answer:
north of east
*14.
A basketball
is dropped from rest. Just before striking the floor, the ball has a momentum whose
magnitude is
. At what height was the basketball dropped?
**15.
A dump truck is being filled with sand. The sand falls straight downward from rest from a height of
2.00
m above the truck bed, and the mass of sand that hits the truck per second is

. The truck is parked

on the platform of a weight scale. By how much does the scale reading exceed the weight of the truck
and sand?

Answer:

344 N

REASONING AND SOLUTION The excess weight of the truck is due to the force exerted on the
truck by the sand. Newton's third law requires that this force be equal in magnitude to the force
exerted on the sand by the truck. In time , a mass of sand falls into the truck bed and comes to rest. The
impulse is
The sand gains a speed
in falling a height so
The velocity of the sand just before it hits the truck is
, where the downward direction is
taken to be the negative direction. The final velocity of the sand is
. Thus, the average force
exerted on the sand is
Section 7.2
T he Principle
of
Conservation
of Linear
Momentum
16.
In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer space. From stationary
positions they push against each other. Bonzo flies off with a velocity of
, while Ender recoils with a velocity of
.
(a) Without doing any calculations, decide which person has the greater mass. Give your reasoning.
(b) Determine the ratio
of the masses of these two enemies.
17. A 2.3-kg cart is rolling across a frictionless, horizontal track toward a 1.5-kg cart that is held

initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus,
the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is

, and the second cart's velocity is

.

(a) What is the total momentum of the system of the two carts at this instant?

Answer:

(b) What was the velocity of the first cart when the second cart was still at rest?

Answer:

18.

As the drawing illustrates, two disks with masses

and

are moving horizontally to the right at a speed of .

They are on an air-hockey table, which supports them with an essentially frictionless cushion of air.
They move as a unit, with a compressed spring between them, which has a negligible mass. Then the
spring is released and allowed to