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Week 7: Chapter 7
Due: 11:00pm on Sunday, October 13, 2013
To understand how points are awarded, read the Grading Policy for this assignment.
Introduction to Potential Energy
Description: Fill-in-the blank questions reviewing the Work-Energy Theorem, then introducing the concept of Potential Energy.
Learning Goal:
Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that
must be added to the kinetic energy to get the total mechanical energy.
The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential
energy. But for now, please answer in terms of the work-energy theorem.
Work -Energy Theorem
The work-energy theorem states
K f = K i + Wall ,
where Wall is the work done by all forces that act on the object, and K i and K f are the initial and final kinetic energies, respectively.
Part A
The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a
component parallel to the motion.
Choose the best answer to fill in the blanks above:
ANSWER:
distance / potential
distance / kinetic
vertical displacement / potential
none of the above
It is important
Typesetting
math: 69%
that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform
circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and
cannot change the kinetic energy of the ball.
Part B
To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change.
Choose the best answer to fill in the blank above:
ANSWER:
acceleration
work
distance
potential energy
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Part C
To illustrate the work-energy concept, consider the case of a stone falling from
xi
to xf under the influence of gravity.
Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone.
Choose the best answer to fill in the blanks above:
ANSWER:
force / kinetic
potential energy / potential
force / potential
potential energy / kinetic
Potential Energy You should read about potential energy in your text before answering the following questions.
Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect
that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the
object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not.
The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is
equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative
force. Then only the work due to nonconservative forces needs to be calculated.
In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy:
K f + U f = E f = Wnc + E i = Wnc + K i + U i ,
where U f and U i are the final and initial potential energies, and Wnc is the work due only to nonconservative forces.
Now, we will revisit the falling stone example using the concept of potential energy.
Part D
Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy)
say that the increased kinetic energy comes from the ______ of the _______ energy.
Choose the best answer to fill in the blanks above:
ANSWER:
work / potential
force / kinetic
change / potential
Part E
This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______.
Choose the best answer to fill in the blanks above:
ANSWER:
sum / conserved
sum / zero
sum / not conserved
difference / conserved
Spring Gun
Description: Simple question about potential energy in a spring being converted to gravitational potential energy of a ball launched vertically.
A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a
maximum height H , measured from the equilibrium position of the spring.
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Part A
The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up
does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible
compared to H .
Hint 1. Potential energy of the spring
The potential energy of a spring is proportional to the square of the distance the spring is compressed. The spring was compressed half the
distance, so the mass, when launched, has one quarter of the energy as in the first trial.
Hint 2. Potential energy of the ball
At the highest point in the ball's trajectory, all of the spring's potential energy has been converted into gravitational potential energy of the ball.
ANSWER:
height =
Shooting a Block up an Incline
Description: Shoot an object up an incline (with friction) using a spring-gun. Use energy conservation to find the distance the object travels up the
incline.
A block of mass m is placed in a smooth-bored spring gun at the bottom of the incline so that it compresses the spring by an amount xc . The spring has
spring constant k. The incline makes an angle θ with the horizontal and the coefficient of kinetic friction between the block and the incline is μ. The block
is released, exits the muzzle of the gun, and slides up an incline a total distance L.
Part A
Find L, the distance traveled along the incline by the block after it exits the gun. Ignore friction when the block is inside the gun. Also, assume that the
uncompressed spring is just at the top of the gun (i.e., the block moves a distance xc while inside of the gun). Use g for the magnitude of acceleration
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due to gravity.
Express the distance
L
in terms of xc , k,
, ,
m g
μ
, and θ .
Hint 1. How to approach the problem
This is an example of a problem that would be very difficult using only Newton's laws and calculus. Instead, use the Work-Energy Theorem:
Efinal − E initial = Wext , where Efinal is the final energy, Einitial is the initial energy, and Wext is the work done on the system by external
forces. Let the gravitational potential energy be zero before the spring is released. Then,
Einitial
is the potential energy due to the spring,
Efinal
is the potential energy due to gravity, and Wext is the work done by friction. Once you've set up this equation completely, solve for L.
Hint 2. Find the initial energy of the block
Find the initial energy
Einitial
of the block. Take the gravitational potential energy to be zero before the spring is released.
Express your answer in terms of parameters given in the problem introduction.
Hint 1. Potential energy of a compressed spring
Recall that the potential energy
U
of a spring with spring constant
k
compressed a distance x is
2
U = (1/2)kx
.
ANSWER:
Einitial
=
Hint 3. Find the work done by friction
Find Wfriction , the work done by friction on the block.
Express Wfriction in terms of L,
, ,
m g
μ
, and θ .
Hint 1. How to compute work
The work done by a force acting along the direction of motion of an object is equal to the magnitude of the force times the distance over
which the object moves. Work is negative if the force directly opposes the motion.
ANSWER:
Wfriction
=
Hint 4. Find the final energy of the block
Find an expression for the final energy
Efinal
of the block (the energy when it has traveled a distance L up the incline). Assume that the
gravitational potential energy of the block is zero before the spring is released and that the block moves a distance xc inside of the gun.
Your answer should contain
L
and
xc
.
Hint 1. What form does the energy take?
When the block stops sliding up the ramp, all of its energy is in the form of gravitational potential energy.
ANSWER:
Efinal
=
ANSWER:
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L
=
Exercise 7.1
Description: In one day, a m mountain climber ascends from the y level on a vertical cliff to the top at y1. The next day, she descends from the top to
the base of the cliff, which is at an elevation of y2. (a) What is her change in gravitational potential energy ...
In one day, a 90kg mountain climber ascends from the 1600m level on a vertical cliff to the top at 2420m . The next day, she descends from the top to the
base of the cliff, which is at an elevation of 1310m .
Part A
What is her change in gravitational potential energy on the first day?
ANSWER:
ΔU
= 7.23×105
=
J
Part B
What is her change in gravitational potential energy on the second day?
ANSWER:
ΔU
= −9.79×105
=
J
Exercise 7.15
Description: A force of ## N stretches a certain spring a distance of ## m. (a) What is the potential energy of the spring when it is stretched a
distance of ## m? (b) What is its potential energy when it is compressed a distance of ## cm?
A force of 800N stretches a certain spring a distance of 0.200m .
Part A
What is the potential energy of the spring when it is stretched a distance of 0.200m ?
ANSWER:
U1
=
= 80.0
J
Part B
What is its potential energy when it is compressed a distance of 4.00cm ?
ANSWER:
U2
=
= 3.20
J
Exercise 7.24
Description: In a "worst-case" design scenario, a 2000-kg 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...
In a "worst-case" design scenario, a 2000-kg 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 17000-N
frictional force to the elevator.
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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 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
2
m/s
Exercise 7.29 - Copy
Description: A ##-kg book slides on a horizontal table. The kinetic friction force on the book has magnitude ## N. (a) How much work is done on the
book by friction during a displacement of ## m to the left? (b) The book now slides a distance ## m to the right, ...
A 0.58−kg book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.4N .
Part A
How much work is done on the book by friction during a displacement of 2.1m to the left?
Express your answer using two significant figures.
ANSWER:
W
=
= -2.9
J
Part B
The book now slides a distance 2.1m to the right, returning to its starting point. During this second displacement of 2.1m , how much work is done on
the book by friction?
Express your answer using two significant figures.
ANSWER:
W
=
= -2.9
J
Part C
What is the total work done on the book by friction during the complete round trip?
Express your answer using two significant figures.
ANSWER:
W
=
= -5.9
J
Part D
On the basis of your answer to part (C), would you say that the friction force is conservative or nonconservative?
ANSWER:
conservative
nonconservative
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Exercise 7.38
Description: A marble moves along the x-axis. The potential-energy function is shown in the figure . (a) At which of the labeled x-coordinates is the
force on the marble zero? (b) Which of the labeled x-coordinates is a position of stable equilibrium? (c) Which ...
A marble moves along the x-axis. The potential-energy function is shown in the figure .
Part A
At which of the labeled x-coordinates is the force on the marble zero?
ANSWER:
a and c
b and d
Part B
Which of the labeled x-coordinates is a position of stable equilibrium?
ANSWER:
a
b
c
d
Part C
Which of the labeled x-coordinates is a position of unstable equilibrium?
ANSWER:
a
b
c
d
Problem 7.42
Description: A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the
block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope...
A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the block is released, it
moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0∘ .
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Part A
What is the speed of the block as it slides along the horizontal surface after having left the spring?
ANSWER:
v
= 3.11
m/s
Part B
How far does the block travel up the incline before starting to slide back down?
ANSWER:
L
= 0.821
m
Problem 7.46
Description: 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. (a) What is the minimum value of h (in terms of R)...
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.
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
=
Also accepted:
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Part B
If the car starts at height h = 5.00 R and the radius is
end of a horizontal diameter.
R
= 15.0m , compute the speed of the passengers when the car is at point
R
= 15.0m , compute the radial acceleration of the passengers when the car is at point
C
, which is at the
ANSWER:
vC
=
= 34.3
m/s
Part C
If the car starts at height h = 5.00 R and the radius is
which is at the end of a horizontal diameter.
C
,
ANSWER:
arad
=
= 78.4 {\rm m}/{\rm s}^{2}
Part D
If the car starts at height h= 5.00 R and the radius is \texttip{R}{R_1} = 15.0{\rm 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_{\rm tan}| = 9.80 {\rm m}/{\rm s}^{2}
Problem 7.49
Description: A m stone slides down a snow-covered hill (the figure ), leaving point A with a speed of v. There is no friction on the hill between points A
and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After...
A 15.0{\rm kg} stone slides down a snow-covered hill (the figure ), leaving point \texttip{A}{A} with
a speed of 10.0{\rm m/s} . There is no friction on the hill between points \texttip{A}{A} and
\texttip{B}{B}, but there is friction on the level ground at the bottom of the hill, between \texttip{B}
{B} and the wall. After entering the rough horizontal region, the stone travels 100 {\rm m} and then
runs into a very long, light spring with force constant 2.50{\rm N/m} . The coefficients of kinetic
and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.
Part A
What is the speed of the stone when it reaches point \texttip{B}{B}?
ANSWER:
v_{\rm 2} =
= 22.2 {\rm m/s}
Part B
How far will the stone compress the spring?
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ANSWER:
= 15.4 {\rm m}
x=
Part C
Will the stone move again after it has been stopped by the spring?
ANSWER:
yes
no
Problem 7.62
Description: A ##-kg skier starts from rest at the top of a ski slope of height ## m. (a) If frictional forces do ## J of work on her as she descends, how
fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of...
A 64.0-kg skier starts from rest at the top of a ski slope of height 65.0{\rm m} .
Part A
If frictional forces do −1.09×104{\rm 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 \texttip{g}{g} = 9.80{\rm m/s^2} .
ANSWER:
v=
= 30.6 \rm m/s
Part B
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.15. If the patch is of width 62.0{\rm m} and the
average force of air resistance on the skier is 170{\rm N} , how fast is she going after crossing the patch?
ANSWER:
v=
= 20.5 \rm m/s
Part C
After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.1{\rm 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=
= 6430 \rm N
Problem 7.66
Description: A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle alpha . Initially the
truck is moving downhill at speed v_0. After careening downhill a distance L with negligible friction, the truck...
A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle \alpha . Initially the truck is moving
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downhill at speed v_0. After careening downhill a distance L with negligible friction, the truck
driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle
\beta. The truck ramp has a soft sand surface for which the coefficient of rolling friction is \mu_r.
Part A
What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.
Express your answer in terms of m, \alpha, v_0, L, g, \beta and \mu_r.
ANSWER:
x=
Problem 7.67
Description: A certain spring is found not to obey Hooke's law; it exerts a restoring force F_x(x)= - alpha x- beta x^2 if it is stretched or compressed,
where alpha = 60.0 (N/m) and beta = 18.0 (N/m^2). The mass of the spring is negligible. (a) Calculate the ...
A certain spring is found not to obey Hooke's law; it exerts a restoring force F_{x}(x)= - \alpha x- \beta x^{2} if it is stretched or compressed, where \alpha =
60.0 \; {\rm N/m} and \beta = 18.0 \; {\rm N/m^2}. The mass of the spring is negligible.
Part A
Calculate the potential energy function U(x) for this spring. Let U=0 when x=0.
Express your answer in terms of x.
ANSWER:
U(x) =
Part B
An object with mass 0.900 \: {\rm kg} on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \; {\rm m} to the right (the
+ x - direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \; {\rm m} to the right of the x=0 equilibrium
position?
ANSWER:
v_{2} = 7.85 {\rm m}/{\rm s}
Problem 7.81
Description: A wooden block with mass ## kg is placed against a compressed spring at the bottom of a slope inclined at an angle of ## degree(s)
(point A). When the spring is released, it projects the block up the incline. At point B, a distance of ## m up the...
A wooden block with mass 1.45{\rm kg} is placed against a compressed spring at the bottom of a slope inclined at an angle of 26.0{\rm ^\circ} (point A).
When the spring is released, it projects the block up the incline. At point B, a distance of 4.30{\rm m} up the incline from A, the block is moving up the
incline at a speed of 6.65{\rm m/s} and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is
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\texttip{\mu_k}{mu_k} = 0.50. The mass of the spring is negligible.
Part A
Calculate the amount of potential energy that was initially stored in the spring.
Take free fall acceleration to be 9.80{\rm m/s^2} .
ANSWER:
U_{1} =
= 86.3 {\rm J}
Problem 7.82
Description: A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as
to make a maximum angle of 45 degree(s) with the vertical. Air resistance is negligible. (a) What is the speed of...
A small rock with mass 0.12 \rm kg is fastened to a massless string with length 0.80 \rm m to form a pendulum. The pendulum is swinging so as to make
a maximum angle of 45 ^\circ with the vertical. Air resistance is negligible.
Part A
What is the speed of the rock when the string passes through the vertical position?
Express your answer using two significant figures.
ANSWER:
v = 2.1 \rm m/s
Part B
What is the tension in the string when it makes an angle of 45 ^\circ with the vertical?
Express your answer using two significant figures.
ANSWER:
0.83 \rm N
Part C
What is the tension in the string as it passes through the vertical?
Express your answer using two significant figures.
ANSWER:
1.9 N
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