introduction to potential energy - physics and … masteringphysics: print view with answers ......

10/17/2016 MasteringPhysics: Print View with Answers

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Chapter 7Due: 11:59pm on Sunday, October 16, 2016

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Introduction to Potential EnergyDescription: Fillinthe blank questions reviewing the WorkEnergy Theorem, then introducing the concept of PotentialEnergy.

Learning Goal:

Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energycalled potential energy that must be added to the kinetic energy to get the total mechanical energy.

The first part of this problem contains shortanswer questions that review the workenergy theorem. In the second part weintroduce the concept of potential energy. But for now, please answer in terms of the workenergy theorem.

WorkEnergy Theorem The workenergy theorem states

,where is the work done by all forces that act on the object, and and are the initial and final kinetic energies,respectively.

Part A

The workenergy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energyof the particle if the force has a component parallel to the motion.

Choose the best answer to fill in the blanks above:

ANSWER:

It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to astring and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force isalways 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 causesthe energy change.

Choose the best answer to fill in the blank above:

ANSWER:

Chapter 7 [ Edit ]

Overview Summary View Diagnostics View Print View with Answers

= +Kf Ki WallWall Ki Kf

distance / potential distance / kinetic vertical displacement / potential none of the above

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Part C

To illustrate the workenergy concept, consider the case of a stone falling from to under the influence of gravity.Using the workenergy concept, we say that work is done by the gravitational _____, resulting in an increase of the______ energy of the stone.


ANSWER:

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 workenergy theorem, enlarging the concept of energy in the most physicallyuseful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the workdone on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitationalforce 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 finalpotential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, itreplaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to becalculated.

In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which nowchanges the total energy:

,where and are the final and initial potential energies, and 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 potentialenergy rather than workenergy) say that the increased kinetic energy comes from the ______ of the _______ energy.


ANSWER:

acceleration work distance potential energy

xi xf

force / kinetic potential energy / potential force / potential potential energy / kinetic

+ = = + = + +Kf Uf Ef Wnc Ei Wnc Ki UiUf Ui Wnc



Part E

This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potentialenergies, is _______.


ANSWER:

Spring GunDescription: Simple question about potential energy in a spring being converted to gravitational potential energy of a balllaunched vertically.

A springloaded toy gun is used to shoot a ball straight up in the air.The ball reaches a maximum height , measured from theequilibrium position of the spring.

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 farbefore firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distanceby which the spring is compressed is negligible compared to .

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 springwas compressed half the distance, so the mass, when launched, has one quarter of the energy as in the first trial.

work / potential force / kinetic change / potential

sum / conserved sum / zero sum / not conserved difference / conserved

H

H



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 gravitationalpotential energy of the ball.

ANSWER:

Shooting a Block up an InclineDescription: Shoot an object up an incline (with friction) using a springgun. Use energy conservation to find the distancethe object travels up the incline.

A block of mass is placed in a smoothbored spring gun at the bottom of the incline so that it compresses the spring by anamount . The spring has spring constant . The incline makes an angle with the horizontal and the coefficient of kineticfriction between the block and the incline is . The block is released, exits the muzzle of the gun, and slides up an incline atotal distance .

Part A

Find , the distance traveled along the incline by the block after it exits the gun. Ignore friction when the block is insidethe gun. Also, assume that the uncompressed spring is just at the top of the gun (i.e., the block moves a distance while inside of the gun). Use for the magnitude of acceleration due to gravity.

Express the distance in terms of , , , , , and .

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, usethe WorkEnergy Theorem: , where is the final energy, is the initial energy,and is the work done on the system by external forces. Let the gravitational potential energy be zero beforethe spring is released. Then, is the potential energy due to the spring, is the potential energy due togravity, and is the work done by friction. Once you've set up this equation completely, solve for .

Hint 2. Find the initial energy of the block

height =

mxc k θ

μL

Lxc

g

L xc k m g μ θ

− =Efinal Einitial Wext Efinal EinitialWext

Einitial EfinalWext L



Find the initial energy of the block. Take the gravitational potential energy to be zero before the spring isreleased.

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 of a spring with spring constant compressed a distance is .

ANSWER:

Hint 3. Find the work done by friction

Find , the work done by friction on the block.

Express in terms of , , , , and .

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 theforce times the distance over which the object moves. Work is negative if the force directly opposes themotion.

ANSWER:

Hint 4. Find the final energy of the block

Find an expression for the final energy of the block (the energy when it has traveled a distance up theincline). Assume that the gravitational potential energy of the block is zero before the spring is released and thatthe block moves a distance inside of the gun.

Your answer should contain and .

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:

ANSWER:

Einitial

U k xU = (1/2)kx2

= Einitial

Wfriction

Wfriction L m g μ θ

= Wfriction

Efinal L

xc

L xc

= Efinal



Exercise 7.1Description: 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 gravitationalpotential energy ...

In one day, a 70 mountain climber ascends from the 1440 level on a vertical cliff to the top at 2460 . The next day,she descends from the top to the base of the cliff, which is at an elevation of 1270 .

Part A

What is her change in gravitational potential energy on the first day?

ANSWER:

Part B

What is her change in gravitational potential energy on the second day?

ANSWER:

Alternative Exercise 7.116Description: A force of F stretches a certain spring a distance of x. (a) What is the potential energy of the spring when amass of m hangs vertically from it?

A force of stretches a certain spring a distance of .

Part A

What is the potential energy of the spring when a mass of hangs vertically from it?

Take the free fall acceleration to be .

ANSWER:

Exercise 7.22

= L

kg m mm

= = 7.00×105 ΔU J

= = −8.16×105 ΔU J

F x

m

g

= U J



Description: In a "worstcase" design scenario, a 2000kg elevator with broken cables is falling at 4.00 m/s when it firstcontacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 mas it does so...

In a "worstcase" design scenario, a 2000 elevator with broken cables is falling at 4.00 when it first contacts acushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 as it does so.During the motion a safety clamp applies a constant 17000 frictional force to the elevator.

Part A

What is the speed of the elevator after it has moved downward 1.00 from the point where it first contacts a spring?

ANSWER:

Part B

When the elevator is 1.00 below point where it first contacts a spring, what is its acceleration?

ANSWER:

Problem 7.40Description: A 2.00kg 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 africtionless incline with slope...

A 2.00 block is pushed against a spring with negligible mass and force constant k = 400 , compressing it 0.220 .When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0 .

Part A

What is the speed of the block as it slides along the horizontal surface after having left the spring?

kg m/sm

N

m

= 3.65 v m/s

m

= 4.00 a m/s2

kg N/m m∘



ANSWER:

Part B

How far does the block travel up the incline before starting to slide back down?

ANSWER:

Problem 7.42Description: A car in an amusement park ride rolls without friction around a track . The car starts from rest at point A at aheight 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) suchthat the...

A car in an amusement park ride rolls without friction around a track. The car starts from rest at point at a height above the bottomof the loop. Treat the car as a particle.

Part A

What is the minimum value of (in terms of ) such that the car moves around the loop without falling off at the top(point )?

Express your answer in terms of .

ANSWER:

Part B

= 3.11 v m/s

= 0.821 L m

A h

h RB

R

=

Also accepted:

hmin

=



If the car starts at height 4.10 and the radius is = 12.0 , compute the speed of the passengers when thecar is at point , which is at the end of a horizontal diameter.

Express your answer with the appropriate units.

ANSWER:

Part C

Compute the radial acceleration of the passengers when the car is at point , which is at the end of a horizontaldiameter.


ANSWER:

Part D

C ompute the tangential acceleration of the passengers when the car is at point , which is at the end of a horizontaldiameter.


ANSWER:

Problem 7.45Description: A m stone slides down a snowcovered hill (the figure ), leaving point A with a speed of v. There is nofriction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B andthe wall. After...

A 10.0 stone slides down a snowcovered hill (the figure ), leaving point with a speed of 12.0 . There is no frictionon the hill between points and , but there is friction on the level ground at the bottom of the hill, between and the wall.After entering the rough horizontal region, the stone travels 100 and then runs into a very long, light spring with forceconstant 2.40 . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and0.80, respectively.

h = R R1 mC

= = 27.0

Also accepted: = 27.0 , = 27.0

vC

C

= = 60.8

Also accepted: = 60.8 , = 60.8

arad

C

= 9.80

Also accepted: 9.81 , 9.80

atan

kg A m/sA B B

mN/m



Part A

What is the speed of the stone when it reaches point ?

ANSWER:

Part B

How far will the stone compress the spring?

ANSWER:

Part C

Will the stone move again after it has been stopped by the spring?

ANSWER:

Problem 7.54Description: A ##kg skier starts from rest at the top of a ski slope of height ## m. (a) If frictional forces do ## J of workon her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crossesa patch of...

B

= = 23.2 v2 m/s

= = 17.7 x m

yes no



A 60.0kg skier starts from rest at the top of a ski slope of height 62.0 .

Part A

If frictional forces do −1.06×104 of work on her as she descends, how fast is she going at the bottom of the slope?

Take free fall acceleration to be = 9.80 .

ANSWER:

Part B

Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.25. If the patch is ofwidth 70.0 and the average force of air resistance on the skier is 140 , how fast is she going after crossing thepatch?

ANSWER:

Part C

After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.3 into it before coming to astop. What is the average force exerted on her by the snowdrift as it stops her?

ANSWER:

Problem 7.58Description: A truck with mass m has a brake failure while going down an icy mountain road of constant downward slopeangle alpha . Initially the truck is moving downhill at speed v_0. After careening downhill a distance L with negligiblefriction, the truck...

A truck with mass has a brake failure while going down an icy mountain road of constant downward slope angle . Initiallythe truck is moving downhill at speed . After careening downhill a distance with negligible friction, the truck driver steersthe runaway vehicle onto a runaway truck ramp of constant upward slope angle . The truck ramp has a soft sand surface forwhich the coefficient of rolling friction is .

m

J

g m/s2

= = 29.4 v m/s

m N

= = 13.9 v m/s

m

= = 2510 F N

m αv0 L

βμr



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 , , , , , and .

ANSWER:

Problem 7.59Description: 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 itis 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 if it is stretched orcompressed, where and . The mass of the spring is negligible.

Part A

Calculate the potential energy function for this spring. Let when .

Express your answer in terms of .

ANSWER:

Part B

An object with mass on a frictionless, horizontal surface is attached to this spring, pulled a distance tothe right (the + direction) to stretch the spring, and released. What is the speed of the object when it is to theright of the equilibrium position?

m α v0 L g β μr

= x

(x) = −αx − βFx x2

α = 60.0 N/m β = 18.0 N/m2

U(x) U = 0 x = 0

x

= U(x)

0.900 kg 1.00 mx 0.50 m

x = 0



ANSWER:

Problem 7.72Description: A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. Thependulum 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 is fastened to a massless string with length 0.80 to form a pendulum. The pendulum isswinging so as to make a maximum angle of 45 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:

Part B

What is the tension in the string when it makes an angle of with the vertical?


ANSWER:

Part C

What is the tension in the string as it passes through the vertical?


ANSWER:

Problem 7.73Description: A wooden block with mass ## kg is placed against a compressed spring at the bottom of a slope inclined atan angle of ## degree(s) (point A). When the spring is released, it projects the block up the incline. At point B, a distanceof ## m up the...

A wooden block with mass 1.75 is placed against a compressed spring at the bottom of a slope inclined at an angle of 25.0 (point ). When the spring is released, it projects the block up the incline. At point , a distance of 7.35 up the incline

from , the block is moving up the incline at a speed of 5.10 and is no longer in contact with the spring. The coefficient

= 7.85 v2 m/s

kg m∘

= 2.1 v m/s

45∘

0.83 N

1.9 N

kg∘ A B m

m/s



from , the block is moving up the incline at a speed of 5.10 and is no longer in contact with the spring. The coefficientof kinetic friction between the block and incline is = 0.55. 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 .

ANSWER:

A m/sμk

m/s2

= = 139 U1 J

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introduction to potential energy - physics and … masteringphysics: print view with answers ......

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