Chapter 07 Homework

4/24/2014 Chapter 7 Homework

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2808817 1/30

Chapter 7 HomeworkDue: 10:00pm on Monday, March 17, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

Energy Required to Lift a Heavy Box

As you are trying to move a heavy box of mass , you realize that it is too heavy for you to lift by yourself. There is noone around to help, so you attach an ideal pulley to the box and a massless rope to the ceiling, which you wrap aroundthe pulley. You pull up on the rope to lift the box.

Use for the magnitude of the acceleration due to gravity andneglect friction forces.

Part A

Once you have pulled hard enough to start the box moving upward, what is the magnitude of the upward forceyou must apply to the rope to start raising the box with constant velocity?

Express the magnitude of the force in terms of , the mass of the box.

Hint 1. What force must be applied to the box to keep it moving at a constant speed?

Once you have pulled hard enough to start the box moving upward, what is the magnitude of the force thatthe pulley must exert on the box so that it moves at a constant speed?

Express your answer in terms of the mass of the box.

ANSWER:

Hint 2. What force does the pulley exert on the box?

If you take the tension in the rope to be , what is , the magnitude of the net upward force that the pulley

exerts on the box?

Express your answer in terms of .

ANSWER:

m

g

F

m

= Fp mg

T Fp

T



Hint 3. Find the tension in the rope

Find the tension in the rope in terms of , the force with which you are pulling upward.

ANSWER:

Hint 4. Putting it all together

On your own or using the previous hints, you should have found equations for he following:

1. the force needed to lift the box at constant velocity, in terms of its mass,2. the relationship between the force on the box due to the pulley and the tension in the rope, and3. the relationship between the force applied to the rope and the tension in the rope.

Use two of these equations to eliminate the force applied by the pulley and the tension in the rope. Youshould then be able to express the force applied on the rope in terms of the mass of the box.

ANSWER:

Correct

Part B

Consider lifting a box of mass to a height using two different methods: lifting the box directly or lifting the boxusing a pulley (as in the previous part).

What is , the ratio of the work done lifting the box directly to the work done lifting the box with a pulley?

Express the ratio numerically.

Hint 1. Definition of work

In each case, the amount of work you do is equal to the force you apply times the distance overwhich you apply the force:

.

Hint 2. Ratio of the forces

What is the ratio of the force needed to lift the box directly to the force needed to lift the box using thepulley?

Express your answer numerically.

= Fp 2T

F

= T F

= Fmg

2

m h

/Wd Wp

W F d

W = Fd



ANSWER:

Hint 3. Ratio of the distances

What is the ratio of the distance over which force is applied when lifting the box directly to the distance overwhich force is applied when lifting the box with the pulley?

Express the ratio of distances numerically.

Hint 1. Find the distance when using the pulley

Find , the distance over which you must apply force when lifting the box using the pulley.

Express your answer in terms of , the total height that the box is lifted.

Hint 1. Pulling the rope a short distance

Using the pully, imagine that you pull the end of the rope a short distance upward. The boxwill actually rise a distance . (Draw a picture if you have trouble visualizing this.)

ANSWER:

Hint 2. Find the distance when lifting directly

When lifting the box directly, the distance over which force is applied, , is equal to the vertical

distance that the box is raised.

ANSWER:

ANSWER:

= 2Fd

Fp

Dp

h

dx

dx/2

= Dp 2h

Dd

h

= 0.500Dd

Dp

= 1Wd

Wp



Correct

No matter which method you use to lift the box, its gravitational potential energy will increase by . So,

neglecting friction, you will always need to do an amount of work equal to to lift it.

Loop the Loop

A roller coaster car may be approximated by a block of mass . The car, which starts from rest, is released at a height

above the ground and slides along a frictionless track. The carencounters a loop of radius , as shown. Assume that theinitial height is great enough so that the car never losescontact with the track.

Part A

Find an expression for the kinetic energy of the car at the top of the loop.

Express the kinetic energy in terms of , , , and .

Hint 1. Find the potential energy at the top of the loop

What is the potential energy of the car when it is at the top of the loop? Define the gravitational potentialenergy to be zero at .

Express your answer in terms of and other given quantities.

ANSWER:

Correct

ANSWER:

mgh

mgh

m h

Rh

m g h R

h = 0

R

= Utop mg⋅2R



Correct

Part B

Find the minimum initial height at which the car can be released that still allows the car to stay in contact withthe track at the top of the loop.

Express the minimum height in terms of .

Hint 1. How to approach this part

Meaning of "stay in contact"

For the car to just stay in contact through the loop, without falling, the normal force that acts on the carwhen it's at the top of the loop must be zero (i.e., ).

Find the velocity at the top such that the remaining force on the car i.e. its weight provides the necessarycentripetal acceleration. If the velocity were any greater, you would additionally require some force from thetrack to provide the necessary centripetal acceleration. If the velocity were any less, the car would fall off thetrack.

Use the above described condition to find the velocity and then the result from the above part to find therequired height.

Hint 2. Acceleration at the top of the loop

Assuming that the speed of the car at the top of the loop is , and that the car stays on the track, find

the acceleration of the car. Take the positive y direction to be upward.

Express your answer in terms of and any other quantities given in the problem introduction.

ANSWER:

Hint 3. Normal force at the top of the loop

Suppose the car stays on the track and has speed at the top of the loop. Use Newton's 2nd law to find

an expression for , the magnitude of the normal force that the loop exerts on the car when the car is at thetop of the loop.

Express your answer in terms of , , , and .

Hint 1. Find the sum of forces at the top of the loop

Find the sum of the forces acting on the car at the top of the loop. Remember that the positive ydirection is upward.

= K mgh − mg2R

h

R

N = 0

vtop

vtop

= atop−vtop

2

R

vtop

N

m g R vtop

N



Express your answer in terms of , , and .

ANSWER:

ANSWER:

Hint 4. Solving for

The requirement to stay in contact results in an expression for in terms of and . Substitute this into

your expression for kinetic energy, found in Part A, to determine a relation between and .

ANSWER:

Correct

For the car will still complete the loop, though it will require some normal reaction even at the verytop.

For the car will just oscillate. Do you see this?

For , the cart will lose contact with the track at some earlier point. That is why roller coastersmust have a lot of safety features. If you like, you can check that the angle at which the cart loses contact with

the track is given by . Where is the angle measured counterclockwise from

the horizontal positive x-axis, where the origin of the x-axis is at the center of the loop.

Work and Potential Energy on a Sliding Block with Friction

A block of weight sits on a plane inclined at an angle as shown. The coefficient of kinetic friction between the planeand the block is .

N m g

= ∑ Ftop −N − mg

= N m( −g)vtop2

R

h

v2top R g

h R

= hmin 2.5R

h > 2.5 R

h < R

R < h < 2.5 R

θ = arcsin( ( −1))23

hR

θ

w θμ



A force is applied to push the block up the incline at constant speed.

Part A

What is the work done on the block by the force of friction as the block moves a distance up the incline?

Express your answer in terms of some or all of the following: , , , .

Hint 1. A formula for work

The work done by a constant force is given by the dot product of the force vector with the vector representingthe displacement over which the force is applied.

Hint 2. Find the magnitude of the frictional force

What is the magnitude of the frictional force?

Express your answer in terms of , , and .

Hint 1. Compute the normal force

Find the magnitude of the normal force on the block.

Express your answer in terms of and .

ANSWER:

ANSWER:

F

Wf L

μ w θ L

ff

μ w θ

n

w θ

= n wcos(θ)

= ff μwcos(θ)



Correct

ANSWER:

Correct

Part B

What is the work done by the applied force of magnitude ?


ANSWER:

Correct

Part C

What is the change in the potential energy of the block, , after it has been pushed a distance up the incline?


ANSWER:

Correct

Now the applied force is changed so that instead of pulling the block up the incline, the force pulls the block down theincline at a constant speed.

= Wf −wcos(θ)μL

W F

μ w θ L

= W wsin(θ)L + wcos(θ)μL

ΔU L

μ w θ L

= ΔU wLsin(θ)

F



Part D

What is the change in potential energy of the block, , as it moves a distance down the incline?


ANSWER:

Correct

Part E

What is the work done by the applied force of magnitude ?


ANSWER:

Correct

Part F

What is the work done on the block by the frictional force?


ANSWER:

ΔU L

μ w θ L

= ΔU −wLsin(θ)

W F

μ w θ L

= W −wsin(θ)L + wcos(θ)μL

Wf

μ w θ L



Correct

Exercise 7.5

A baseball is thrown from the roof of 20.7 -tall building with an initial velocity of magnitude 13.1 and directed at anangle of 57.1 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:

Correct

Part B

What is the answer for part (A) if the initial velocity is at an angle of 57.1 below the horizontal?

ANSWER:

Correct

Part C

If the effects of air resistance are included, will part (A) or (B) give the higher speed?

ANSWER:

Correct

Bungee Jumping

= Wf −wcos(θ)μL

m m/s∘

= 24.0 v2 m/s

∘

= 24.0 v2 m/s

The part (A) will give the higher speed.

The part (B) will give the higher speed.



Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass , and thesurface of the bridge is a height above the water. The bungee cord, which has length when unstretched, will firststraighten and then stretch as Kate falls.

Assume the following:

The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant .Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward.Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as apoint particle.

Use for the magnitude of the acceleration due to gravity.

Part A

How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest?Assume that she doesn't touch the water.

Express the distance in terms of quantities given in the problem introduction.

Hint 1. Decide how to approach the problem

Here are three possible methods for solving this problem:

1. No nonconservative forces are acting, so mechanical energy is conserved. Set Kate'sgravitational potential energy at the top of the bridge equal to the spring potential energy in thebungee cord (which depends on the cord's final length ) and solve for .

2. Since nonconservative forces are acting, mechanical energy is not conserved. Set the springpotential energy in the bungee cord (which depends on ) equal to Kate's gravitational potentialenergy plus the work done by dissipative forces. Eliminate the unknown work, and solve for .

3. When Kate comes to rest she has zero acceleration, so the net force acting on her must bezero. Set the spring force due to the bungee cord (which depends on ) equal to the force ofgravity and solve for .

Which of these options is the simplest, most accurate way to find given the information available?

ANSWER:

Correct

Hint 2. Compute the force due to the bungee cord

When Kate is at rest, what is the magnitude of the upward force the bungee cord exerts on her?

Express your answer in terms of the cord's final stretched length and quantities given in theproblem introduction. Your answer should not depend on Kate's mass .

mh L

k

g

d d

d

d

d

d

d

a

b

c

Fb

d

m



Hint 1. Find the extension of the bungee cord

The upward force on Kate is due to the extension of the bungee cord. What is this extension?

Express your answer in terms of the cord's final (stretched) length and .

ANSWER:

Hint 2. Formula for the force due to a stretched cord

The formula for the force due to a stretched cord is,

where is the spring constant of the cord and is the extension of the cord.

ANSWER:

Incorrect; Try Again; 5 attempts remaining

ANSWER:

Correct

Part B

If Kate just touches the surface of the river on her first downward trip (i.e., before the first bounce), what is the springconstant ? Ignore all dissipative forces.

Express in terms of , , , and .

Hint 1. Decide how to approach the problem

Here are three possible methods for solving this problem:

1. Since nonconservative forces are ignored, mechanical energy is conserved. Set Kate'sgravitational potential energy at the top of the bridge equal to the spring potential energy in thebungee cord at the lowest point (which depends on ) and solve for .

2. Nonconservative forces can be ignored, so mechanical energy is conserved. Set the springpotential energy in the bungee cord (which depends on ) equal to Kate's gravitational potentialenergy at the top of the bridge plus the work done by gravity as Kate falls. Compute the workdone by gravity, then solve for .

d L

Extension = d − L

F = −kΔx

k Δx

= Fb k(d − L)

= d L+ mg

k

k

k L h m g

k k

k

k



3. When Kate is being held just above the water she has zero acceleration, so the net forceacting on her must be zero. Set the spring force due to the bungee cord (which depends on )equal to the force of gravity and solve for .

Which of these options is the simplest, most accurate way to find given the information available?

ANSWER:

Hint 2. Find the initial gravitational potential energy

What is Kate's gravitational potential energy at the moment she steps off the bridge? (Define the zero of

gravitational potential to be at the surface of the water.)

Express your answer in terms of quantities given in the problem introduction.

ANSWER:

Hint 3. Find the elastic potential energy in the bungee cord

What is the elastic potential energy stored in the bungee cord when Kate is at the lowest point of her

first downward trip?

Express your answer in terms of quantities given in the problem introduction.

Hint 1. Formula for elastic potential energy

The elastic potential energy of the bungee cord (which we are treating as an ideal spring) is

,

where is the amount by which the cord is stretched beyond its unstretched length.

Hint 2. How much is the bungee cord stretched?

By how much is the bungee cord stretched when Kate is at a depth below the bridge?

Express your answer in terms of and .

ANSWER:

k

k

k

a

b

c

Ug

= Ug mgh

Uel

= k(ΔxUel12

)2

Δx

d1

d1 L

= Δx − Ld1



ANSWER:

ANSWER:

Correct

Dancing Balls

Four balls, each of mass , are connected by four identical relaxed springs with spring constant . The balls aresimultaneously given equal initial speeds directed away from the center of symmetry of the system.

Part A

As the balls reach their maximum displacement, their kinetic energy reaches __________.

ANSWER:

Correct

= Uel k12 (h−L)

2

= k2mgh

(h−L)2

m kv

a maximum

zero

neither a maximum nor zero



Part B

Use geometry to find , the distance each of the springs has stretched from its equilibrium position. (It may help todraw the initial and the final states of the system.)

Express your answer in terms of , the maximum displacement of each ball from its initial position.

ANSWER:

Correct

Part C

Find the maximum displacement of any one of the balls from its initial position.

Express in terms of some or all of the given quantities , , and .

Hint 1. A useful equation

The equation

could be useful. If you are familiar with this equation, you most likely have seen the expression applied to asingle mass on a single spring. For the situation with four balls and four masses, you will need to considercarefully which quantities to use in this expression.

ANSWER:

Correct

Spring Gun

A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height , measuredfrom the equilibrium position of the spring.

x

d

= x d 2√

d

d k v m

m = k12

v2 12

x2

= d v m2k

−−−√

H



Part A

The same ball is shot straight up a second time from the same gun, but this time the spring is compressed onlyhalf as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal andthat the distance by 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. Thespring was compressed half the distance, so the mass, when launched, has one quarter of the energy as inthe 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 intogravitational potential energy of the ball.

ANSWER:

Correct

Stretching a Spring

As illustrated in the figure, a spring with spring constant is stretched from to , where is theequilibrium position of the spring.

H

height = H14

k x = 0 x = 3d x = 0



Part A

During which interval is the largest amount of energy required to stretch the spring?

Hint 1. How to approach the problem

The force exerted on a spring to stretch or compress it from equilibrium is given by Hooke's law:

,

where is the displacement of the spring from equilibrium. Notice that this force varies in magnitude: as increases so does the magnitude of the force. On a graph of force as a function of position, the total workdone by the force is represented by the area under the curve between the initial and final positions. Plot agraph of force versus displacement and compare the areas under the curve from to , to

, and to .

ANSWER:

= kxFon spring

x x

x = 0 x = d x = d

x = 2d x = 2d x = 3d

From to

From to

From to

The energy required is the same in all three intervals.

x = 0 x = d

x = d x = 2d

x = 2d x = 3d



Correct

A graph of the force exerted on the spring versus the displacement of the spring is shown in the figure.

Recall that on a graph of force as a function ofposition, the work done by the force is represented by the area under the curve. The work done by the hand inthe first segment to pull the spring from to is represented by a single triangle. The area under thesecond segment from to is three times larger than the first segment, and the area under thethird segment from to is five times larger than in the first segment. So more energy is requiredto pull the spring through the third segment.

Part B

A spring is stretched from to , where is the equilibrium position of the spring. It is thencompressed from to . What can be said about the energy required to stretch or compress thespring?


Recall that on a graph of force as a function of position, the work done by the force is represented by thearea "under" the curve, or more accurately, the area between the curve and the horizontal axis. Plot a graphof force versus displacement and compare the areas "under" the curve from to and to

.

ANSWER:

x = 0 x = d

x = d x = 2d

x = 2d x = 3d

x = 0 x = d x = 0x = 0 x = −d

x = 0 x = d x = 0x = −d

More energy is required to stretch the spring than to compress it.

The same amount of energy is required to either stretch or compress the spring.

Less energy is required to stretch the spring than to compress it.



Correct

The work done to stretch or compress a spring from equilibrium is given by,

where is the distance away from equilibrium that the spring moves. Since is squared in the equation forwork, stretching ( ) or compressing ( ) a spring by the same distance requires the same positiveamount of work.

Part C

Now consider two springs A and B that are attached to a wall. Spring A has a spring constant that is four times thatof the spring constant of spring B. If the same amount of energy is required to stretch both springs, what can besaid about the distance each spring is stretched?

Hint 1. How to approach this problem

The work done to stretch or compress a spring is given by

,

where is the distance away from equilibrium that the spring is displaced.Use this expression to relate the information provided about the work done on each spring and the springconstants to the distance each spring stretches.

Hint 2. Use proportional reasoning to find a relationship between the springs

From the problem statement you know that,

where

and

.

Use this information to find an expression for .

ANSWER:

ANSWER:

= kWon spring12 x2

x x

x > 0 x < 0

= kWon spring12 x2

x

=WA WB

W = k12

x2

= 4kA kB

(xA )2

(xB)2

= 0.25(xA )2

(xB)2



Correct

The energy required to stretch a spring is proportional to and to . If is four times , must be half

that of , so the energy required is the same for both springs.

Part D

Two identical springs are attached to two different masses, and , where is greater than . The

masses lie on a frictionless surface. Both springs are compressed the same distance, , as shown in the figure.Which of the following statements descibes the energy required to compress spring A and spring B?

ANSWER:

Correct

Good job; you have realized an important fact. The work done on a spring to compress it a distance is given

by . The amount of mass attached to the spring does not affect the work required to stretch or

compress the spring.

Spring A must stretch 4 times as far as spring B

Spring A must stretch 2 times as far as spring B.

Spring A must stretch the same distance as spring B.

Spring A must stretch half the distance spring B stretches.

Spring A must stretch one-quarter of the distance spring B stretches.

k x2 kA kB xA

xB

MA MB MA MB

d

Spring A requires more energy than spring B.

Spring A requires the same amount of energy as spring B.

Spring A requires less energy than spring B.

Not enough information is provided to answer the question.

d

k12

d 2



Exercise 7.15

A force of 700 stretches a certain spring a distance of 0.400 .

Part A

What is the potential energy of the spring when it is stretched a distance of 0.400 ?

ANSWER:

Correct

Part B

What is its potential energy when it is compressed a distance of 5.00 ?

ANSWER:

Correct

Exercise 7.25

You are asked to design a spring that will give a 1300 satellite a speed of 3.45 relative to an orbiting spaceshuttle. Your spring is to give the satellite a maximum acceleration of . The spring's mass, the recoil kineticenergy 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 = 9.80 .

ANSWER:

Correct

Part B

What distance must the spring be compressed?

N m

m

= 140 U1 J

cm

= 2.19 U2 J

kg m/s5.00g

g m/s2

= 2.62×105 k N/m



ANSWER:

Correct

Sliding In Socks

Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowingthis, Zak decides to get a running start and then slide across the floor.

Part A

If Zak's speed is 3.00 when he starts to slide, what distance will he slide before stopping?

Express your answer in meters.

ANSWER:

Correct

Part B

Now, suppose that Zak's younger cousin, Greta, sees him sliding and takes off her shoes so that she can slide aswell (assume her socks have the same coefficient of kinetic friction as Zak's). Instead of getting a running start, sheasks Zak to give her a push. So, Zak pushes her with a force of 125 over a distance of 1.00 . If her mass is20.0 , what distance 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.


This problem can be solved using work and energy. Pick the moment just before the push starts as theinitial time, and pick the point at which she stops sliding as the final time. What is the change inenergy between these two times?

Express your answer in joules.

ANSWER:

ANSWER:

= 0.243 x m

m/s d

1.84 m

N mkg d2

ΔE

= 0 ΔE J



Correct

Exercise 7.32

While a roofer is working on a roof that slants at 40.0 above the horizontal, he accidentally nudges his 90.0 toolbox,causing it to start sliding downward, starting from rest.

Part A

If it starts 4.25 from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge ofthe roof if the kinetic friction force on it is 18.0 ?

ANSWER:

Correct

Exercise 7.36

An object moving in the -plane is acted on by a conservative force described by the potential-energy function , where is a positive constant.

Part A

Derive an expression for the force expressed in terms of the unit vectors and .

Express your answer in terms of the given quantities.

ANSWER:

Correct

Potential Energy Graphs and Motion

Learning Goal:

To be able to interpret potential energy diagrams and predict the corresponding motion of a particle.

= 1.55 d2 m

∘ N

mN

= 6.07 v m/s

xyU(x,y) = α(1/ + 1/ )x2 y2 α

F i j

= F +2αx3 i 2α

y 3 j



Potential energy diagrams for a particle are useful in predicting the motion of that particle. These diagrams allow one todetermine the direction of the force acting on the particle at any point, the points of stable and unstable equilibrium, theparticle's kinetic energy, etc.Consider the potential energy diagram shown. The curverepresents the value of potential energy as a function of theparticle's coordinate . The horizontal line above the curverepresents the constant value of the total energy of theparticle . The total energy is the sum of kinetic ( ) andpotential ( ) energies of the particle.

The key idea in interpreting the graph can be expressed in theequation

where is the x component of the net force as functionof the particle's coordinate . Note the negative sign: It meansthat the x component of the net force is negative when thederivative is positive and vice versa. For instance, if theparticle is moving to the right, and its potential energy isincreasing, the net force would be pulling the particle to theleft.

If you are still having trouble visualizing this, consider the following: If a massive particle is increasing its gravitationalpotential energy (that is, moving upward), the force of gravity is pulling in the opposite direction (that is, downward).

If the x component of the net force is zero, the particle is said to be in equilibrium. There are two kinds of equilibrium:

Stable equilibrium means that small deviations from the equilibrium point create a net force thataccelerates the particle back toward the equilibrium point (think of a ball rolling between two hills).Unstable equilibrium means that small deviations from the equilibrium point create a net force thataccelerates the particle further away from the equilibrium point (think of a ball on top of a hill).

In answering the following questions, we will assume that there is a single varying force acting on the particle alongthe x axis. Therefore, we will use the term force instead of the cumbersome x component of the net force.

Part A

The force acting on the particle at point A is __________.

Hint 1. Sign of the derivative

If a function increases (as increases) in a certain region, then the derivative of the function in that region ispositive.

Hint 2. Sign of the component

If increases to the right, as in the graph shown, then a (one-dimensional) vector with a positive xcomponent points to the right, and vice versa.

ANSWER:

Ux

E E KU

(x) = − ,FxdU(x)

dx(x)Fx

x

F

x

x



Correct

Consider the graph in the region of point A. If the particle is moving to the right, it would be "climbing the hill,"and the force would "pull it down," that is, pull the particle back to the left. Another, more abstract way ofthinking about this is to say that the slope of the graph at point A is positive; therefore, the direction of isnegative.

Part B

The force acting on the particle at point C is __________.

Hint 1. Sign of the derivative

If a function increases (as increases) in a certain region, then the derivative of the function in that region ispositive, and vice versa.

Hint 2. Sign of the component

If increases to the right, as in the graph shown, then a (one-dimensional) vector with a positive xcomponent points to the right, and vice versa.

ANSWER:

Correct

Part C

The force acting on the particle at point B is __________.

Hint 1. Derivative of a function at a local maximum

At a local maximum, the derivative of a function is equal to zero.

ANSWER:

directed to the right

directed to the left

equal to zero

F

x

x



equal to zero



Correct

The slope of the graph is zero; therefore, the derivative , and .

Part D

The acceleration of the particle at point B is __________.

Hint 1. Relation between acceleration and force

The relation between acceleration and force is given by Newton's 2nd law,.

ANSWER:

Correct

If the net force is zero, so is the acceleration. The particle is said to be in a state of equilibrium.

Part E

If the particle is located slightly to the left of point B, its acceleration is __________.

Hint 1. The force on such a particle

To the left of B, is an increasing function and so its derivative is positive. This implies that the x

component of the force on a particle at this location is negative, or that the force is directed to the left, justlike at A. What can you say now about the acceleration?

ANSWER:



equal to zero

dU/dx = 0 | | = 0F

F = ma



equal to zero

U(x)



Correct

Part F

If the particle is located slightly to the right of point B, its acceleration is __________.

Hint 1. The force on such a particle

To the right of B, is a decreasing function and so its derivative is negative. This implies that the x

component of the force on a particle at this location is positive, or that the force is directed to the right, justlike at C. What can you now say about the acceleration?

ANSWER:

Correct

As you can see, small deviations from equilibrium at point B cause a force that accelerates the particle furtheraway; hence the particle is in unstable equilibrium.

Part G

Name all labeled points on the graph corresponding to unstable equilibrium.

List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, andE, type your answer as BDE.

Hint 1. Definition of unstable equilibrium

Unstable equilibrium means that small deviations from the equilibrium point create a net force thataccelerates the particle further away from the equilibrium point (think of a ball on top of a hill).

ANSWER:



equal to zero

U(x)



equal to zero

BF



Correct

Part H

Name all labeled points on the graph corresponding to stable equilibrium.


Hint 1. Definition of stable equilibrium

Stable equilibrium means that small deviations from the equilibrium point create a net force that acceleratesthe particle back toward the equilibrium point. (Think of a ball rolling between two hills.)

ANSWER:

Correct

Part I

Name all labeled points on the graph where the acceleration of the particle is zero.


Hint 1. Relation between acceleration and force

The relation between acceleration and force is given by Newton's 2nd law,.

ANSWER:

Correct

Your answer, of course, includes the locations of both stable and unstable equilibrium.

Part J

Name all labeled points such that when a particle is released from rest there, it would accelerate to the left.


DH

F = ma

BDFH



Hint 1. Determine the sign of the x component of force

If the acceleration is to the left, so is the force. This means that the x component of the force is __________.

ANSWER:

Hint 2. What is the behavior of ?

If the x component of the force at a point is negative, then the derivative of at that point is positive. This

means that in the region around the point is __________.

ANSWER:

ANSWER:

Correct

Part K

Consider points A, E, and G. Of these three points, which one corresponds to the greatest magnitude ofacceleration of the particle?

Hint 1. Acceleration and force

The greatest acceleration corresponds to the greatest magnitude of the net force, represented on the graphby the magnitude of the slope.

ANSWER:

positive

negative

U(x)

U(x)U(x)

increasing

decreasing

AE

A

E

G



Correct

Kinetic energy

If the total energy of the particle is known, one can also use the graph of to draw conclusions about the kineticenergy of the particle since

.

As a reminder, on this graph, the total energy is shown by the horizontal line.

Part L

What point on the graph corresponds to the maximum kinetic energy of the moving particle?

Hint 1. , , and

Since the total energy does not change, the maximum kinetic energy corresponds to the minimum potentialenergy.

ANSWER:

Correct

It makes sense that the kinetic energy of the particle is maximum at one of the (force) equilibrium points. Forexample, think of a pendulum (which has only one force equilibrium point--at the very bottom).

Part M

At what point on the graph does the particle have the lowest speed?

ANSWER:

Correct

As you can see, many different conclusions can be made about the particle's motion merely by looking at thegraph. It is helpful to understand the character of motion qualitatively before you attempt quantitative problems.This problem should prove useful in improving such an understanding.

Score Summary:Your score on this assignment is 101%.You received 14.07 out of a possible total of 14 points.

E U(t)

K = E − U

E

K U E

D

B

Date post:	18-Jan-2016
Category:	Documents
Upload:	fatboy91
View:	4,152 times
Download:	1 times

Chapter 07 Homework

Documents