Download Exercise 7.1 Exercise 7.5

Exercise 7.1
In one day, a 75 kg mountain climber ascends from the 1500 m level on a vertical cliff to the top at 2340 m . The
next day, she descends from the top to the base of the cliff, which is at an elevation of 1340 m .
Part A
What is her change in gravitational potential energy on the first day?
ANSWER:
= 6.17×105 J ΔU
Part B
What is her change in gravitational potential energy on the second day?
ANSWER:
= −7.35×105 J ΔU
Exercise 7.5
A baseball is thrown from the roof of 23.6 m ­tall building with an initial velocity of magnitude 10.5 m/s and directed
at an angle of 51.6 ∘ above the horizontal.
Part A
What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.
ANSWER:
v2
= 23.9 m/s Part B
What is the answer for part (A) if the initial velocity is at an angle of 51.6 ∘ below the horizontal?
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Chapter 7_assignment 6
v2
= 23.9 m/s Part C
If the effects of air resistance are included, will part (A) or (B) give the higher speed?
ANSWER:
The part (A) will give the higher speed.
The part (B) will give the higher speed.
Exercise 7.11
You are testing a new amusement park roller coaster with an empty car with a mass of 100 kg . One part of the
track is a vertical loop with a radius of 12.0 m . At the bottom of the loop (point A) the car has a speed of 25.0 m/s
and at the top of the loop (point B) it has speed of 8.00 m/s .
Part A
As the car rolls from point A to point B, how much work is done by friction?
Use 9.81 m/s 2 for the acceleration due to gravity.
ANSWER:
­4510 J Exercise 7.15
A force of 400 N stretches a certain spring a distance of 0.300 m .
Part A
What is the potential energy of the spring when it is stretched a distance of 0.300 m ?
ANSWER:
U1
= 60.0 J 2/17
Chapter 7_assignment 6
Part B
What is its potential energy when it is compressed a distance of 6.00 cm ?
ANSWER:
U2
= 2.40 J Exercise 7.19
A spring of negligible mass has force constant k = 1800 N/m .
Part A
How far must the spring be compressed for an amount 3.40 J of potential energy to be stored in it?
Express your answer using two significant figures.
ANSWER:
6.1×10−2 m Part B
You place the spring vertically with one end on the floor. You then drop a book of mass 1.30 kg onto it from a
height of 0.800 m above the top of the spring. Find the maximum distance the spring will be compressed.
Take the free fall acceleration to be 9.80 m/s 2 . Express your answer using two significant figures.
ANSWER:
0.11 m Exercise 7.24
In a "worst­case" design scenario, a 2000­kg elevator with broken cables is falling at 4.00 m/s when it first
contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing
2.00 m as it does so. During the motion a safety clamp applies a constant 17000­N frictional force to the elevator.
Part A
What is the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a
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Chapter 7_assignment 6
spring?
ANSWER:
v
= 3.65 m/s Part B
When the elevator is 1.00 m below point where it first contacts a spring, what is its acceleration?
ANSWER:
a
= 4.00 m/s 2 Exercise 7.33
A 62.0 kg skier is moving at 6.50 m/s on a frictionless, horizontal snow­covered plateau when she encounters a
rough patch 3.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing
the rough patch and returning to friction­free snow, she skis down an icy, frictionless hill 2.50 m high.
Part A
How fast is the skier moving when she gets to the bottom of the hill?
ANSWER:
v
= 8.41 m/s Part B
How much internal energy was generated in crossing the rough patch?
ANSWER:
E
= 638 J Where's the Energy?
Learning Goal:
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Chapter 7_assignment 6
To understand how to apply the law of conservation of energy to situations with and without nonconservative forces
acting.
The law of conservation of energy states the following:
In an isolated system the total energy remains constant.
If the objects within the system interact through gravitational and elastic forces only, then the total mechanical
energy is conserved.
The mechanical energy of a system is defined as the sum of kinetic energy K and potential energy U . For such
systems where no forces other than the gravitational and elastic forces do work, the law of conservation of energy
can be written as
K i + Ui = K f + Uf
,
where the quantities with subscript "i" refer to the "initial" moment and those with subscript "f" refer to the final
moment. A wise choice of initial and final moments, which is not always obvious, may significantly simplify the
solution.
The kinetic energy of an object that has mass m and velocity v is given by
K=
1
2
mv
2
.
Potential energy, instead, has many forms. The two forms that you will be dealing with most often in this chapter
are the gravitational and elastic potential energy. Gravitational potential energy is the energy possessed by elevated
objects. For small heights, it can be found as
Ug = mgh
,
where m is the mass of the object, g is the acceleration due to gravity, and h is the elevation of the object above
the zero level. The zero level is the elevation at which the gravitational potential energy is assumed to be (you
guessed it) zero. The choice of the zero level is dictated by convenience; typically (but not necessarily), it is
selected to coincide with the lowest position of the object during the motion explored in the problem.
Elastic potential energy is associated with stretched or compressed elastic objects such as springs. For a spring
with a force constant k , stretched or compressed a distance x, the associated elastic potential energy is
Ue =
1
2
kx
2
.
When all three types of energy change, the law of conservation of energy for an object of mass m can be written as
1
2
mv
2
i
+ mghi +
1
2
kx
2
i
=
1
2
mv
2
f
+ mghf +
1
2
kx
2
f
.
The gravitational force and the elastic force are two examples of conservative forces. What if nonconservative
forces, such as friction, also act within the system? In that case, the total mechanical energy would change. The
law of conservation of energy is then written as
1
2
mv
2
i
+ mghi +
1
2
kx
2
i
+ W nc =
1
2
mv
2
f
+ mghf +
1
2
kx
2
f
,
where W nc represents the work done by the nonconservative forces acting on the object between the initial and the
final moments. The work W nc is usually negative; that is, the nonconservative forces tend to decrease, or
dissipate, the mechanical energy of the system.
In this problem, we will consider the following situation as depicted in the diagram : A block of mass m slides at a
speed v along a horizontal, smooth table. It next slides down a smooth ramp, descending a height h, and then
slides along a horizontal rough floor, stopping eventually. Assume that the block slides slowly enough so that it
does not lose contact with the supporting surfaces (table, ramp, or floor).
You will analyze the motion of the block at different moments using the law of conservation of energy.
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Chapter 7_assignment 6
Part A
Which word in the statement of this problem allows you to assume that the table is frictionless?
ANSWER:
straight
smooth
horizontal
Although there are no truly "frictionless" surfaces, sometimes friction is small enough to be neglected. The
word "smooth" often describes such low­friction surfaces. Can you deduce what the word "rough" means?
Part B
Suppose the potential energy of the block at the table is given by mgh/3. This implies that the chosen zero
level of potential energy is __________.
Hint 1. Definition of U
Gravitational potential energy is given by
Ug = mgh
,
where h is the height relative to the zero level. Note that h > 0 when the object is above the chosen
zero level; h < 0 when the object is below the chosen zero level.
ANSWER:
a distance h/3 above the floor
a distance h/3 below the floor
a distance 2h/3 above the floor
a distance 2h/3 below the floor
on the floor
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Chapter 7_assignment 6
Part C
If the zero level is a distance 2h/3 above the floor, what is the potential energy U of the block on the floor?
Express your answer in terms of some or all the variables m, v , and h and any appropriate constants.
ANSWER:
U
= −2mgh
3
Part D
Considering that the potential energy of the block at the table is mgh/3 and that on the floor is −2mgh/3,
what is the change in potential energy ΔU of the block if it is moved from the table to the floor?
Express your answer in terms of some or all the variables m, v , and h and any appropriate constants.
Hint 1. Definition of ΔU
By definition, the change in potential energy is given by ΔU
defined as the "final" quantity minus the "initial" one.
= Uf − Ui
. In general, change is always
ANSWER:
ΔU
= −mhg
As you may have realized, this choice of the zero level was legitimate but not very convenient. Typically,
in such problems, the zero level is assumed to be on the floor. In solving this problem, we will assume just
that: the zero level of potential energy is on the floor.
Part E
Which form of the law of conservation of energy describes the motion of the block when it slides from the top of
the table to the bottom of the ramp?
Hint 1. How to approach the problem
Think about these questions:
Are there any nonconservative forces acting on the block during this part of the trip?
Are there any objects involved that can store elastic potential energy?
Is the block changing its height?
Is the block changing its speed?
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Chapter 7_assignment 6
ANSWER:
1
2
1
2
1
2
1
2
1
2
mv
2
mv
2
mv
2
mv
2
mv
2
i
i
i
i
i
+ mghi + W nc =
+
1
2
kx
2
i
=
1
2
mv
2
f
1
2
+
+ mghi = mghf +
+ mghi =
+ mghi +
1
2
1
2
mv
kx
2
i
2
f
mv
1
2
1
2
2
f
+ mghf
kx
kx
2
f
2
f
+ mghf
+ W nc =
1
2
mv
2
f
+ mghf +
1
2
kx
2
f
Part F
As the block slides down the ramp, what happens to its kinetic energy K , potential energy U , and total
mechanical energy E ?
ANSWER:
K
decreases;U increases;E stays the same
K
decreases;U increases;E increases
K
increases;U increases;E increases
K
increases;U decreases;E stays the same
Part G
Using conservation of energy, find the speed v b of the block at the bottom of the ramp.
Express your answer in terms of some or all the variables m, v , and h and any appropriate constants.
Hint 1. How to approach the problem
Use the equation for the law of conservation of energy that describes the motion of the block as it slides
down the ramp. Then substitute in all known values and solve for the unknown.
ANSWER:
vb
= −
−
−
−
−−
−
2
√ v + 2gh
Part H
Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from
8/17
Chapter 7_assignment 6
the bottom of the ramp to the moment it stops?
Hint 1. How to approach the problem
Think about these questions:
Are there any nonconservative forces acting on the block during this part of the trip?
Are there any objects involved that can store elastic potential energy?
Is the block changing its height?
Is the block changing its speed?
ANSWER:
1
2
1
2
1
2
1
2
1
2
mv
mv
mv
mv
mv
2
i
2
i
2
i
2
i
2
i
+ mghi + W nc =
=
1
2
mv
1
2
mv
2
f
+ mghf
2
f
+ W nc =
1
2
+ mghi =
+ mghi +
mv
1
2
1
2
2
f
mv
kx
2
i
2
f
+ mghf
+ W nc =
1
2
mv
2
f
+ mghf +
1
2
kx
2
f
Part I
As the block slides across the floor, what happens to its kinetic energy K , potential energy U , and total
mechanical energy E ?
ANSWER:
K
decreases;U increases;E decreases
K
increases;U decreases;E decreases
K
decreases;U stays the same;E decreases
K
increases;U stays the same;E decreases
K
decreases;U increases;E stays the same
K
increases;U decreases;E stays the same
Part J
What force is responsible for the decrease in the mechanical energy of the block?
ANSWER:
9/17
Chapter 7_assignment 6
tension
gravity
friction
normal force
Part K
Find the amount of energy E dissipated by friction by the time the block stops.
Express your answer in terms of some or all the variables m, v , and h and any appropriate constants.
Hint 1. How to approach the problem
Use the equation for the law of conservation of energy that you selected as the most appropriate for the
block sliding on the floor. Then substitute in all known values and solve for the unknown. You will need
to use the value for v b that you found earlier in Part G, as your initial speed.
ANSWER:
E
= 0.5mv
2
+ mgh
Conceptual Question 7.09
Part A
A ball drops some distance and gains 30 J of kinetic energy. Do NOT ignore air resistance. How much
gravitational potential energy did the ball lose?
ANSWER:
less than 30 J
exactly 30 J
more than 30 J
Conceptual Question 7.11
10/17
Chapter 7_assignment 6
Part A
Block 1 and block 2 have the same mass, m, and are released from the top of two inclined planes of the same
height making 30° and 60° angles with the horizontal direction, respectively. If the coefficient of friction is the
same in both cases, which of the blocks is going faster when it reaches the bottom of its respective incline?
ANSWER:
Both blocks have the same speed at the bottom.
Block 2 is faster.
We must know the actual masses of the blocks to answer.
Block 1 is faster.
There is not enough information to answer the question because we do not know the value of the
coefficient of kinetic friction.
Prelecture Concept Question 7.01
Part A
Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the
mass of the first object. They both slide up the same frictionless incline plane. Which object rises to a greater
height?
ANSWER:
Object 1 rises to the greater height because it weighs less.
Object 2 rises to the greater height because it contains more mass.
Object 1 rises to the greater height because it possesses a smaller amount of kinetic energy.
Object 2 rises to the greater height because it possesses a larger amount of kinetic energy.
The two objects rise to the same height.
Problem 7.04
Part A
A tennis ball bounces on the floor three times. If each time it loses 11% of its energy due to heating, how high
does it rise after the third bounce, provided we released it 1.1 m from the floor?
ANSWER:
11/17
Chapter 7_assignment 6
78 cm
7.8 cm
87 cm
78 mm
Problem 7.31
Part A
A 0.85­kg block is held in place against the spring by a 65­N horizontal external force (see the figure). The
external force is removed, and the block is projected with a velocity v 1 = 1.2 m/s upon separation from the
spring. The block descends a ramp and has a velocity v 2 = 1.5 m/s at the bottom. The track is frictionless
between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic
friction over this section is 0.28. The velocity of the block is v 3 = 1.4 m/s at C. The block moves on to D,
where it stops. The height h of the ramp is closest to
ANSWER:
9.6
11
4.1
7.8
7.3
Problem 7.46
A car in an amusement park ride rolls without friction around the track shown in the figure . It starts from rest at
point A at a height h above the bottom of the loop. Treat the car as a particle.
12/17
Chapter 7_assignment 6
Part A
What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at
the top (point B)?
ANSWER:
hmin
= 5R
2
Part B
If the car starts at height h = 4.00 R and the radius is R = 20.0 m , compute the speed of the passengers
when the car is at point C , which is at the end of a horizontal diameter.
ANSWER:
vC
= 34.3 m/s Part C
If the car starts at height h = 4.00 R and the radius is R = 20.0 m , compute the radial acceleration of the
passengers when the car is at point C , which is at the end of a horizontal diameter.
ANSWER:
a rad
= 58.8 m/s 2 13/17
Chapter 7_assignment 6
Part D
If the car starts at height h = 4.00 R and the radius is R = 20.0 m , compute the tangential acceleration of
the passengers when the car is at point C , which is at the end of a horizontal diameter.
ANSWER:
|a tan |
= 9.80 m/s 2 Problem 7.50
A 2.8­kg block slides over the smooth, icy hill shown in the figure . The top of the hill is horizontal and 70 m higher
than its base.
Part A
What minimum speed must the block have at the base of the hill so that it will not fall into the pit on the far side
of the hill?
Express your answer using two significant figures.
ANSWER:
v min
= 42 m/s Problem 7.56
A 1510 kg rocket is to be launched with an initial upward speed of 54.0 m/s . In order to assist its engines, the
engineers will start it from rest on a ramp that rises 53 ∘ above the horizontal (the figure ). At the bottom, the ramp
turns upward and launches the rocket vertically. The engines provide a constant forward thrust of 2000 N, and
friction with the ramp surface is a constant 500 N.
14/17
Chapter 7_assignment 6
Part A
How far from the base of the ramp should the rocket start, as measured along the surface of the ramp?
ANSWER:
d
= 165 m Problem 7.62
A 61.0­kg skier starts from rest at the top of a ski slope of height 67.0 m .
Part A
If frictional forces do −1.05×104 J of work on her as she descends, how fast is she going at the bottom of the
slope?
Take free fall acceleration to be g = 9.80 m/s 2 .
ANSWER:
v
= 31.1 m/s Part B
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.16. If the
patch is of width 68.0 m and the average force of air resistance on the skier is 160 N , how fast is she going
after crossing the patch?
ANSWER:
15/17
Chapter 7_assignment 6
v
= 20.0 m/s Part C
After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.0 m into it before
coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?
ANSWER:
F
= 6080 N Sliding In Socks
Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250.
Knowing this, Zak decides to get a running start and then slide across the floor.
Part A
If Zak's speed is 3.00 m/s when he starts to slide, what distance d will he slide before stopping?
Express your answer in meters.
ANSWER:
1.84 m Part B
Now, suppose that Zak's younger cousin, Greta, sees him sliding and takes off her shoes so that she can slide
as well (assume her socks have the same coefficient of kinetic friction as Zak's). Instead of getting a running
start, she asks Zak to give her a push. So, Zak pushes her with a force of 125 N over a distance of 1.00 m. If
her mass is 20.0 kg , what distance d 2 does she slide after Zak's push ends?
Remember that the frictional force acts on Greta during Zak's push and while she is sliding after the push.
Express your answer in meters.
Hint 1. How to approach the problem
This problem can be solved using work and energy. Pick the moment just before the push starts as the
initial time, and pick the point at which she stops sliding as the final time. What is the change ΔE in
energy between these two times?
Express your answer in joules.
16/17
Chapter 7_assignment 6
ANSWER:
ΔE
= 0 J ANSWER:
d2
= 1.55 m Exercise 7.25
You are asked to design a spring that will give a 1120 kg satellite a speed of 2.05 m/s relative to an orbiting space
shuttle. Your spring is to give the satellite a maximum acceleration of 5.00g . The spring's mass, the recoil kinetic
energy of the shuttle, and changes in gravitational potential energy will all be negligible.
Part A
What must the force constant of the spring be?
Take the free fall acceleration to be g = 9.80 m/s 2 .
ANSWER:
k
= 6.40×105 N/m Part B
What distance must the spring be compressed?
ANSWER:
x
= 8.58×10−2 m 17/17