Cutnell_9th_problems_ch_1_thru_10_converted

Answer:

332 m

REASONING AND SOLUTION Since the magnitude of the centripetal acceleration is given by
Equation 5.2,

, we can solve for and find that

*10.

A computer is reading data from a rotating CD-ROM. At a point that is 0.030 m from the center of the
disc, the centripetal acceleration is

. What is the centripetal acceleration at a point that is 0.050 m from the center of

the disc?

*11.

A centrifuge is a device in which a small container of material is rotated at a high speed on a circular
path.

Such a device is used in medical laboratories, for instance, to cause the more dense red blood cells to
settle through the less dense blood serum and collect at the bottom of the container. Suppose the
centripetal acceleration of the sample is

times as large as the acceleration due to gravity. How many revolutions per minute is the sample

making, if it is located at a radius of 5.00 cm from the axis of rotation?

Answer:

*12. The earth rotates once per day about an axis passing through the north and south poles, an axis
that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of

, determine the speed and

centripetal acceleration of a person situated

(a) at the equator and

(b) at a latitude of

north of the equator.

Section 5.3 Centripetal Force

13.

Review Example 3, which deals with the bobsled in Figure 5.5. Also review Conceptual Example 4.

The mass of the sled and its two riders in Figure 5.5 is 350 kg. Find the magnitude of the centripetal
force that acts on the sled during the turn with a radius of

(a) 33 m and

Answer:

REASONING In Example 3, it was shown that the magnitudes of the centripetal acceleration for the
two cases are

[Radius = 33 m]

[Radius = 24 m]

According to Newton's second law, the centripetal force is

(see Equation 5.3).

SOLUTION
(a) Therefore, when the sled undergoes the turn of radius 33 m,
(b) Similarly, when the radius of the turn is 24 m,
(b) 24 m.
Answer:
REASONING In Example 3, it was shown that the magnitudes of the centripetal acceleration for the
two cases are
[Radius = 33 m]
[Radius = 24 m]
According to Newton's second law, the centripetal force is
(see Equation 5.3).
SOLUTION
(a) Therefore, when the sled undergoes the turn of radius 33 m,
(b) Similarly, when the radius of the turn is 24 m,
REASONING In Example 3, it was shown that the magnitudes of the centripetal acceleration for the
two cases are
[Radius = 33 m]
[Radius = 24 m]
According to Newton's second law, the centripetal force is
(see Equation 5.3).
SOLUTION

(a) Therefore, when the sled undergoes the turn of radius 33 m,

(b) Similarly, when the radius of the turn is 24 m,

14. At an amusement park there is a ride in which cylindrically shaped chambers spin around a central
axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall
moves at a speed of

,

and an 83-kg person feels a 560-N force pressing against his back. What is the radius of the chamber?

15.

Multiple-Concept Example 7 reviews the concepts that play a role in this problem. Car A uses tires for
which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at
which the car can negotiate this curve is

. Car B uses tires for which the coefficient of static friction is 0.85 on the same curve.

What is the maximum speed at which car B can negotiate the curve?

Answer:

16. A speed skater goes around a turn that has a radius of 31 m. The skater has a speed of and
experiences a

centripetal force of 460 N. What is the mass of the skater?

17. For background pertinent to this problem, review Conceptual Example 6. In Figure 5.7 the man
hanging upside down

is holding a partner who weighs 475 N. Assume that the partner moves on a circle that has a radius of
6.50 m. At a swinging speed of

, what force must the man apply to his partner in the straight-down position?

Answer:

594 N

18.

A car is safely negotiating an unbanked circular turn at a speed of

. The road is dry, and the maximum

static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum
static frictional force to one-third of its dry-road value. If the car is to continue safely around the
curve, to what speed must the driver slow the car?

19. See Conceptual Example 6 to review the concepts involved in this problem. A 9.5-kg monkey is
hanging by one arm from a branch and is swinging on a vertical circle. As an approximation, assume
a radial distance of 85 cm between the branch and the point where the monkey's mass is located. As
the monkey swings through the lowest point on the circle, it has a speed of

. Find

(a) the magnitude of the centripetal force acting on the monkey and

Answer:

88 N

(b) the magnitude of the tension in the monkey's arm.
Answer:

181 N

20.

Multiple-Concept Example 7 deals with the concepts that are important in this problem. A penny is
placed at the outer edge of a disk

that rotates about an axis perpendicular to the plane of the disk at its center.

The period of the rotation is 1.80 s. Find the minimum coefficient of friction necessary to allow the
penny to rotate along with the disk.

*21.

The hammer throw is a track-and-field event in which a 7.3-kg ball (the “hammer”) is whirled around
in a circle several times and released. It then moves upward on the familiar curving path of projectile
motion and eventually returns to earth some distance away. The world record for this distance is
86.75 m, achieved in 1986 by Yuriy Sedykh.

Ignore air resistance and the fact that the ball is released above the ground rather than at ground level.
Furthermore, assume that the ball is whirled on a circle that has a radius of 1.8 m and that its velocity
at the instant of release is directed

above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the
moment of release.

Answer:

3500 N

REASONING Let
be the initial speed of the ball as it begins its projectile motion. Then, the centripetal
force is given by Equation 5.3:
. We are given the values for and ; however, we must determine the
value of
from the details of the projectile motion after the ball is released.
In the absence of air resistance, the component of the projectile motion has zero acceleration, while
the component of the motion is subject to the acceleration due to gravity. The horizontal distance
traveled by the ball is given by Equation 3.5a (with
):
with equal to the flight time of the ball while it exhibits projectile motion. The time can be found by
considering the vertical motion. From Equation 3.3b,
After a time ,
. Assuming that up and to the right are the positive directions, we have
and
Using the fact that
, we have

(1)

Equation 1 (with upward and to the right chosen as the positive directions) can be used to determine
the speed with

which the ball begins its projectile motion. Then Equation 5.3 can be used to find the centripetal force.

SOLUTION Solving equation 1 for

, we have

Then, from Equation 5.3,

*22.

An 830-kg race car can drive around an unbanked turn at a maximum speed of

without slipping. The

turn has a radius of curvature of 160 m. Air flowing over the car's wing exerts a downward-pointing
force (called the downforce) of 11 000 N on the car.

(a) What is the coefficient of static friction between the track and the car's tires?

(b) What would be the maximum speed if no downforce acted on the car?

*23.

A “swing” ride at a carnival consists of chairs that are swung in a circle by 15.0-m cables attached to a
vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant is 179
kg.

(a) Determine the tension in the cable attached to the chair.

Answer:

3510 N

(b) Find the speed of the chair.

Answer:

Section

5.4

Banked

Curves
24. On a banked race track, the smallest circular path on which cars can move has a radius of 112 m,
while the largest has a radius of 165 m, as the drawing illustrates. The height of the outer wall is 18 m.
Find
(a) the smallest and
(b) the largest speed at which cars can move on this track without relying on friction.

25.
Before attempting this problem, review Examples 7 and 8. Two curves on a highway have the same
radii.
However, one is unbanked and the other is banked at an angle A car can safely travel along the
unbanked curve at a maximum speed
under conditions when the coefficient of static friction between the tires and the road is
. The banked curve is frictionless, and the car can negotiate it at the same maximum speed
. Find the

angle of the banked curve.
Answer:
26.
A woman is riding a Jet Ski at a speed of
and notices a seawall straight ahead. The farthest she can lean
the craft in order to make a turn is
. This situation is like that of a car on a curve that is banked at an angle of
. If she tries to make the turn without slowing down, what is the minimum distance from the seawall
that she can begin making her turn and still avoid a crash?
27. Two banked curves have the same radius. Curve A is banked at an angle of

, and curve B is banked at an angle
of
. A car can travel around curve A without relying on friction at a speed of
. At what speed can this
car travel around curve B without relying on friction?
Answer:
*28.
A racetrack has the shape of an inverted cone, as the drawing shows. On this surface the cars race in
circles that are parallel to the ground. For a speed of
, at what value of the distance d should a driver locate his car if he
wishes to stay on a circular path without depending on friction?
*29.
A jet flying at
banks to make a horizontal circular turn. The radius of the turn is 3810 m, and the mass
of the jet is
. Calculate the magnitude of the necessary lifting force.
Answer:
**30. The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way
down the slope and is going around at a constant speed on a circle
as the carousel turns. The coefficient of static
friction between the suitcase and the carousel is 0.760, and the angle in the drawing is
. How much
time is required for your suitcase to go around once?
Section
5.5
Satellites

in

Circular

Orbits

Section 5.6 Apparent Weightlessness and Artificial Gravity

31.

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 360 km
above the earth's surface, while that for satellite B is at a height of 720 km. Find the orbital speed for
each satellite.

Answer:

,
32. A rocket is used to place a synchronous satellite in orbit about the earth. What is the speed of the
satellite in orbit?
33.
A satellite is in a circular orbit around an unknown planet. The satellite has a speed of
, and the
radius of the orbit is
. A second satellite also has a circular orbit around this same planet. The orbit of
this second satellite has a radius of
. What is the orbital speed of the second satellite?
Answer:
REASONING Equation 5.5 gives the orbital speed for a satellite in a circular orbit around the earth. It
can be modified to determine the orbital speed around any planet P by replacing the mass of the earth
by the mass of the
planet
:
.
SOLUTION The ratio of the orbital speeds is, therefore,

Solving for
gives
34. Multiple-Concept Example 13 offers a helpful perspective for this problem. Suppose the surface
of the
space station in Figure 5.18 is rotating at
. What must be the value of r for the astronauts to weigh one-half
of their earth-weight?
35.
A satellite is in a circular orbit about the earth
. The period of the satellite is
. What is the speed at which the satellite travels?
Answer:
36. A satellite circles the earth in an orbit whose radius is twice the earth's radius. The earth's mass is
,
and its radius is
. What is the period of the satellite?
*37.
A satellite moves on a circular earth orbit that has a radius of
. A model airplane is flying on a 15-m
guideline in a horizontal circle. The guideline is parallel to the ground. Find the speed of the plane
such that the plane and the satellite have the same centripetal acceleration.
Answer:
REASONING Equation 5.2 for the centripetal acceleration applies to both the plane and the satellite,
and the centripetal acceleration is the same for each. Thus, we have
The speed of the satellite can be obtained directly from Equation 5.5.
SOLUTION Using Equation 5.5, we can express the speed of the satellite as
Substituting this expression into the expression obtained in the reasoning for the speed of the plane

gives

*38.
A satellite has a mass of 5850 kg and is in a circular orbit
above the surface of a planet. The period
of the orbit is 2.00 hours. The radius of the planet is
. What would be the true weight of the satellite if it
were at rest on the planet's surface?
*39.

Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The
orbital speeds of the planets are determined to be 43.3 km/s and 58.6 km/s. The slower planet's orbital
period is 7.60 years.
(a) What is the mass of the star?
Answer:
(b) What is the orbital period of the faster planet, in years?
Answer:
3.07 yr
**40. Multiple-Concept Example 14 deals with the issues on which this problem focuses. To create
artificial gravity, the space station shown in the drawing is rotating at a rate of 1.00 rpm. The radii of
the cylindrically shaped chambers have the ratio
. Each chamber A simulates an acceleration due to gravity of
. Find values
for (a)
, (b)
, and (c) the acceleration due to gravity that is simulated in chamber B.
Problem 40
Section 5.7 Vertical Circular Motion
41.
A motorcycle has a constant speed of
as it passes over the top of a hill whose radius of curvature is
126 m. The mass of the motorcycle and driver is 342 kg. Find the magnitudes of
(a) the centripetal force and
Answer:
REASONING As the motorcycle passes over the top of the hill, it will experience a centripetal force,
the magnitude of which is given by Equation 5.3:
. The centripetal force is provided by
the net force on the cycle + driver system. At that instant, the net force on the system is composed of

the normal force, which points upward, and the weight, which points downward. Taking the direction
toward the center of the circle (downward) as the positive direction, we have

. This expression can

be solved for

, the normal force.

SOLUTION

(a) The magnitude of the centripetal force is

(b) The magnitude of the normal force is

(b) the normal force that acts on the cycle.

Answer:

REASONING As the motorcycle passes over the top of the hill, it will experience a centripetal force,
the magnitude of which is given by Equation 5.3:

. The centripetal force is provided by

the net force on the cycle + driver system. At that instant, the net force on the system is composed of
the normal force, which points upward, and the weight, which points downward. Taking the direction
toward the center of the circle (downward) as the positive direction, we have

. This expression can

be solved for

, the normal force.

SOLUTION

(a) The magnitude of the centripetal force is

(b) The magnitude of the normal force is

REASONING As the motorcycle passes over the top of the hill, it will experience a centripetal force,
the magnitude of which is given by Equation 5.3:

. The centripetal force is provided by the net force on the

cycle + driver system. At that instant, the net force on the system is composed of the normal force,
which points upward, and the weight, which points downward. Taking the direction toward the center
of the circle (downward) as the positive direction, we have

. This expression can be solved for

, the normal force.

SOLUTION

(a) The magnitude of the centripetal force is

(b) The magnitude of the normal force is

42.

Pilots of high-performance fighter planes can be subjected to large centripetal accelerations during
high-speed turns. Because of these accelerations, the pilots are subjected to forces that can be much
greater than their body weight, leading to an accumulation of blood in the abdomen and legs. As a
result, the brain becomes starved for blood, and the pilot can lose consciousness (“black out”). The
pilots wear “anti-G suits” to help keep the blood from draining out of the brain. To appreciate the
forces that a fighter pilot must endure, consider the magnitude of the

normal force that the pilot's seat exerts on him at the bottom of a dive. The magnitude of the pilot's
weight is W. The plane is traveling at

on a vertical circle of radius 690 m. Determine the ratio

. For comparison, note

that blackout can occur for values of

as small as 2 if the pilot is not wearing an anti-G suit.

43.

For the normal force in Figure 5.20 to have the same magnitude at all points on the vertical track, the
stunt driver

must adjust the speed to be different at different points. Suppose, for example, that the track has a
radius of 3.0 m and that the driver goes past point 1 at the bottom with a speed of

. What speed must she have at point 3, so that

the normal force at the top has the same magnitude as it did at the bottom?

Answer:

REASONING The centripetal force is the name given to the net force pointing toward the center of
the circular path. At point 3 at the top the net force pointing toward the center of the circle consists of
the normal force and the weight, both pointing toward the center. At point 1 at the bottom the net force
consists of the normal force pointing upward toward the center and the weight pointing downward or
away from the center. In either case the centripetal force is given by Equation 5.3 as

.

SOLUTION At point 3 we have

At point 1 we have

Subtracting the second equation from the first gives

Rearranging gives

Thus, we find that

44.

A special electronic sensor is embedded in the seat of a car that takes riders around a circular loop-
the-loop ride at an amusement park. The sensor measures the magnitude of the normal force that the
seat exerts on a rider. The loop-the-loop ride is in the vertical plane and its radius is 21 m. Sitting on
the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads 770 N. At
the top of the loop, the rider is upside down and moving, and the sensor reads 350 N. What is the

speed of the rider at the top of the loop?

45.

A 0.20-kg ball on a stick is whirled on a vertical circle at a constant speed. When the ball is at the three
o'clock position, the stick tension is 16 N. Find the tensions in the stick when the ball is at the twelve
o'clock and at the six o'clock positions.

Answer:

14 N (twelve o'clock), 18 N (six o'clock)

*46.

A stone is tied to a string

and whirled in a circle at the same constant speed in two different

ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is
vertical. In the vertical case the maximum tension in the string is 15.0% larger than the tension that
exists when the circle is horizontal. Determine the speed of the stone.

*47.

A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a
circular arc with a radius of 45.0 m. Determine the maximum speed that the cycle can have while
moving over the crest without losing contact with the road.

Answer:

**48. In an automatic clothes dryer, a hollow cylinder moves the clothes on a vertical circle

, as the

drawing shows. The appliance is designed so that the clothes tumble gently as they dry. This means
that when a piece of clothing reaches an angle of above the horizontal, it loses contact with the wall of
the cylinder and falls onto the clothes below. How many revolutions per second should the cylinder
make in order that the clothes lose contact with the wall when

?

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

Problems

Section 6.1 Work Done by a Constant Force

1.

During a tug-of-war, team A pulls on team B by applying a force of 1100 N to the rope between them.
The rope remains parallel to the ground. How much work does team A do if they pull team B toward
them a distance of 2.0 m?

Answer:

REASONING AND SOLUTION We will assume that the tug-of-war rope remains parallel to the
ground, so that the force that moves team B is in the same direction as the displacement. According to
Equation 6.1, the work done by team A is

2.

You are moving into an apartment and take the elevator to the 6th floor. Suppose your weight is 685 N
and that of your belongings is 915 N.

(a) Determine the work done by the elevator in lifting you and your belongings up to the 6th floor
(15.2 m) at a constant velocity.

(b) How much work does the elevator do on you alone (without belongings) on the downward trip,
which is also made at a constant velocity?

3. The brakes of a truck cause it to slow down by applying a retarding force of

to the truck over a distance

of 850 m. What is the work done by this force on the truck? Is the work positive or negative? Why?

Answer:

(The work is negative because the retarding force points opposite to the truck's displacement.)

4. A 75.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant
velocity from ground level to the floor above, a vertical height of 4.60 m. What is the work done on
the man by (a) the gravitational force and

(b) the escalator?

5.

Suppose in Figure 6.2 that

of work is done by the force

in moving

the suitcase a distance of 50.0 m. At what angle is the force oriented with respect to the ground?

Answer:

REASONING AND SOLUTION Solving Equation 6.1 for the angle , we obtain

6. A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of 22.0 m. She pushes
in a direction below the horizontal. A 48.0-N frictional force opposes the motion of the cart.

(a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the
pushing force,

(c) the frictional force, and
(d) the gravitational force.
7.
The drawing shows a plane diving toward the ground and then climbing back upward. During each of
these motions, the lift force
acts perpendicular to the displacement
, which has the same magnitude,
, in

each case. The engines of the plane exert a thrust
, which points in the direction of the displacement and has the same
magnitude during the dive and the climb. The weight
of the plane has a magnitude of
. In both motions,
net work is performed due to the combined action of the forces
,
, and
.
(a) Is more net work done during the dive or the climb? Explain.
Answer:
More net work is done during the dive.
REASONING AND SOLUTION
( In both cases, the lift force L is perpendicular to the displacement
a 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
o f the plane. Thus, since the thr ust T is the same fo r bo th 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,
.
(

b The work done during the dive is

,

) while the work done during the climb is

. Therefore, the difference

between the net work done during the dive and the climb is

(b) Find the difference between the net work done during the dive and the climb.

Answer:

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 thr ust T is the same fo r bo th cases, the net fo r ce in the dir ectio n o f the displacement is
greater for the case where the plane is diving. Since the displacement s is the same in both cases,

.

(b

) The work done during the dive is

, while the work done during the climb is

. Therefore, the difference between the net work done during the dive

and the climb is

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 thr ust T is the same fo r bo th cases, the net fo r ce in the dir ectio n o f the displacement is
greater for the case where the plane is diving. Since the displacement s is the same in both cases,

.

(b)

The work done during the dive is
, while the work done during the climb is
. Therefore, the difference between the net work done during the dive and the
climb is
8. A person pulls a toboggan for a distance of 35.0 m along the snow with a rope directed above the
snow. The
tension in the rope is 94.0 N.
(a) How much work is done on the toboggan by the tension force?
(b) How much work is done if the same tension is directed parallel to the snow?
*9.
As a sailboat sails 52 m due north, a breeze exerts a con-stant force
on the boat's sails. This force is
directed at an angle west of due north. A force
of the same magnitude directed due north would do the same
amount of work on the sailboat over a distance of just 47 m. What is the angle between the direction
of the force and due north?
Answer:
*10.
A 55-kg box is being pushed a distance of 7.0 m across the floor by a force
whose magnitude is 160 N. The
force
is parallel to the displacement of the box. The coefficient of kinetic friction is 0.25. Determine the
work done on the box by each of the four forces that act on the box. Be sure to include the proper plus
or minus sign for the work done by each force.
*11.
A
crate is being pushed across a horizontal floor by a force

that makes an angle of
below the horizontal. The coefficient of kinetic friction is 0.200. What should be the magnitude of
, so that the net
work done by it and the kinetic frictional force is zero?
Answer:

256 N

**12. A 1200-kg car is being driven up a
hill. The frictional force is directed opposite to the motion of the car and has
a magnitude of
. A force
is applied to the car by the road and propels the car forward. In addition to
these two forces, two other forces act on the car: its weight
and the normal force
directed perpendicular to
the road surface. The length of the road up the hill is 290 m. What should be the magnitude of
, so that the net

work done by all the forces acting on the car is
?
Sect io n 6.2 T he Wo rk–Energ y T heo rem and Kinet ic Energ y
13. A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-
powered catapult. The thrust of its engines is
. In being launched from rest it moves through a distance of 87 m and has a
kinetic energy of
at lift-off. What is the work done on the jet by the catapult?
Answer:
14.
A golf club strikes a 0.045-kg golf ball in order to launch it from the tee. For simplicity, assume that

the average net force applied to the ball acts parallel to the ball's motion, has a magnitude of 6800 N,
and is in contact with the ball for a distance of 0.010 m. With what speed does the ball leave the club?
15.
It takes 185 kJ of work to accelerate a car from
to
. What is the car's mass?
Answer:

1450 kg

REASONING The car's kinetic energy depends upon its mass and speed via

(Equation 6.2). The total

amount of work done on the car is equal to the difference between its final and initial kinetic energies:
(Equation 6.3). We will use these two relationships to determine the car's mass.

SOLUTION Combining

(Equation 6.2) and

(Equation 6.3), we obtain

Solving this expression for the car's mass , and noting that

, we find that

16. Starting from rest, a

flea springs straight upward. While the flea is pushing off from the ground, the

ground exerts an average upward force of 0.38 N on it. This force does

of work on the flea.

(a) What is the flea's speed when it leaves the ground?

(b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the
flea's weight.

17.

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle
forward, the skier accelerates. A 70.3-kg water-skier has an initial speed of

. Later, the speed increases to

.

Determine the work done by the net external force acting on the skier.

Answer:

18. As background for this problem, review Conceptual Example 6. A 7420-kg satellite has an
elliptical orbit, as in Figure 6.9 b. The point on the orbit that is farthest from the earth is called the
apogee and is at the far right side of the drawing. The point on the orbit that is closest to the earth is
called the perigee and is at the left side of the drawing.

Suppose that the speed of the satellite is
at the apogee and
at the perigee. Find the work done
by the gravitational force when the satellite moves from
(a) the apogee to the perigee and
(b) the perigee to the apogee.
19.
The hammer throw is a track-and-field event in which a 7.3-kg ball (the “hammer”), starting from
rest, is whirled around in a circle several times and released. It then moves upward on the familiar
curving path of projectile motion.
In one throw, the hammer is given a speedof
. For comparison, a .22 caliber bullet has a mass of 2.6 g and,
starting from rest, exits the barrel of a gun at a speed of
. Determine the work done to launch the motion of
(a) the hammer and
Answer:
REASONING The work done to launch either object can be found from Equation 6.3, the work-
energy theorem,
.
SOLUTION

(a) The work required to launch the hammer is
(b Similarly, the work required to launch the bullet is
)
(b) the bullet.
Answer:
REASONING The work done to launch either object can be found from Equation 6.3, the work-
energy theorem,
.
SOLUTION
(a) The work required to launch the hammer is
(b Similarly, the work required to launch the bullet is
)

REASONING The work done to launch either object can be found from Equation 6.3, the work-
energy theorem,
.
SOLUTION
(a) The work required to launch the hammer is
(b) Similarly, the work required to launch the bullet is
20.
A 16-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 24 N.
Starting from rest, the sled attains a speed of
in 8.0 m. Find the coefficient of kinetic friction between the runners of
the sled and the snow.
21.
An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and
slows it down.
The asteroid has a mass of
, and the force causes its speed to change from 7100 to
.
(a) What is the work done by the force?
Answer:
(b) If the asteroid slows down over a distance of
, determine the magnitude of the force.
Answer:
*22. The concepts in this problem are similar to those in Multiple-Concept Example 4, except that the
force doing the work in this problem is the tension in the cable. A rescue helicopter lifts a 79-kg
person straight up by means of a cable. The person has an upward acceleration of
and is lifted from rest through a distance of 11 m.
(a) What is the tension in the cable? How much work is done by
(b) the tension in the cable and



(c) the person's weight?
(d) Use the work–energy theorem and find the final speed of the person.
*23.
A 6200-kg satellite is in a circular earth orbit that has a radius of
. A net external force must act on
the satellite to make it change to a circular orbit that has a radius of
. What work W must the net external
force do? Note that the work determined here is not the work
done by the satellite's engines to change the orbit.
Instead, the work W is
, where
is the work done by the gravitational force.
Answer:
REASONING When the satellite goes from the first to the second orbit, its kinetic energy changes.
The net work that the external force must do to change the orbit can be found from the work-energy
theorem:
.The speeds and
can be obtained from Equation 5.5 for the speed of a
satellite in a circular orbit of radius . Given the speeds, the work energy theorem can be used to
obtain the work.
SOLUTION According to Equation 5.5,
. Substituting into the work-energy theorem, we have
Therefore,
*24.
Consult Multiple-Concept Example 5 for insight into solving this problem. A skier slides horizontally

along the snow for a distance of 21 m before coming to rest. The coefficient of kinetic friction
between the skier and the snow is
. Initially, how fast was the skier going?
*25.
A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force
points in the same direction as the sled's displacement, which is along the
axis. As a result, the kinetic energy of the sled
increases by 38%. By what percentage would the sled's kinetic energy have increased if this force had
pointed above the
axis?
Answer:
18%
REASONING According to the work-energy theorem, the kinetic energy of the sled increases in each
case because work is done on the sled. The work-energy theorem is given by Equation 6.3:
. The work done on the sled is given by Equation 6.1:
.The
work done in each case can, therefore, be expressed as
and
The fractional increase in the kinetic energy of the sled when
is
Therefore,
(1)

The fractional increase in the kinetic energy of the sled when

is

(2)

Equation 1 can be used to substitute for

in Equation 2.

SOLUTION Combining Equations 1 and 2, we have

Thus, the sled's kinetic energy would increase by

.

*26.

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a
horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. The
snowmobile and its rider have a mass of 136 kg. Under the influence of a drive force of 205 N, it is
moving at a constant velocity whose magnitude is

. The drive force is then shut off. Find

(a) the distance in which the snowmobile coasts to a halt and

(b) the time required to do so.

**27. The model airplane in Figure 5.6 is flying at a speed of

on a horizontal circle of radius 16 m. The

mass of the plane is 0.90 kg. The person holding the guideline pulls it in until the radius becomes 14
m. The plane speeds up, and the tension in the guideline becomes four times greater. What is the net
work done on the plane?

Answer:

**28. Multiple-Concept Example 5 reviews many of the concepts that play roles in this problem. An
extreme skier, starting from rest, coasts down a mountain slope that makes an angle of

with the horizontal. The

coefficient of kinetic friction between her skis and the snow is 0.200. She coasts down a distance of
10.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands
downhill at a point whose vertical distance is 3.50 m below the edge. How fast is she going just before
she lands?

Section 6.3

Gravitational

Potential

Energy

Section 6.4 Conservative Versus Nonconservative Forces

29. A 75.0-kg skier rides a 2830-m-long lift to the top of a mountain. The lift makes an angle of with
the

horizontal. What is the change in the skier's gravitational potential energy?

Answer:

30. Juggles and Bangles are clowns. Juggles stands on one end of a teeter-totter at rest on the ground.
Bangles jumps off a platform 2.5 m above the ground and lands on the other end of the teeter-totter,
launching Juggles into the air. Juggles rises to a height of 3.3 m above the ground, at which point he
has the same amount of gravitational potential energy as Bangles had before he jumped, assuming
both potential energies are measured using the ground as the reference level.

Bangles' mass is 86 kg. What is Juggles' mass?

31. A 0.60-kg basketball is dropped out of a window that is 6.1 m above the ground. The ball is caught
by a person whose hands are 1.5 m above the ground.

(a) How much work is done on the ball by its weight? What is the gravitational potential energy of the
basketball, relative to the ground, when it is

Answer:

27 J

(b) released and
Answer:

36 J

(c) caught?
Answer:
8.8 J
(d) How is the change
in the ball's gravitational potential energy related to the work done by its
weight?

Answer:

The change in gravitational potential energy is

, where W is the work done by the weight.

32. A pole-vaulter just clears the bar at 5.80 m and falls back to the ground. The change in the
vaulter's potential energy during the fall is

. What is his weight?

33.

A bicyclist rides 5.0 km due east, while the resistive force from the air has a magnitude of 3.0 N and
points due west. The rider then turns around and rides 5.0 km due west, back to her starting point. The
resistive force from the air on the return trip has a magnitude of 3.0 N and points due east.

(a) Find the work done by the resistive force during the round trip.

Answer:

REASONING During each portion of the trip, the work done by the resistive force is given by
Equation 6.1,

.Since the resistive force always points opposite to the displacement of the

bicyclist,

; hence, on each part of the trip,

. The work done by the

resistive force during the round trip is the algebraic sum of the work done during each portion of the
trip.

SOLUTION

(a) The work done during the round trip is, therefore,

(b) Since the work done by the resistive force over the closed path is not zero, we can conclude that
.
(b) Based on your answer to part (a), is the resistive force a conservative force? Explain.
Answer:
The resistive force is not a conservative force.
REASONING During each portion of the trip, the work done by the resistive force is given by
Equation 6.1,
.Since the resistive force always points opposite to the displacement of the
bicyclist,
; hence, on each part of the trip,
. The work done by the
resistive force during the round trip is the algebraic sum of the work done during each portion of the
trip.
SOLUTION
(a) The work done during the round trip is, therefore,
(b) Since the work done by the resistive force over the closed path is not zero, we can conclude that
.
REASONING During each portion of the trip, the work done by the resistive force is given by
Equation 6.1,
.Since the resistive force always points opposite to the displacement of the bicyclist,
;
hence, on each part of the trip,
. The work done by the resistive force during the round
trip is the algebraic sum of the work done during each portion of the trip.
SOLUTION
(a) The work done during the round trip is, therefore,
(b) Since the work done by the resistive force over the closed path is not zero, we can conclude that

.