mastering physics assignment 4

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Assignment 04

Due: 11:59pm on Tuesday, February 7, 2012

Note: To understand how points are aw arded, read your instructor's Grading Policy.

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Problem 1

Learning Goal:

To introduce contact forces (normal and friction forces) and to understand that, except for frictionforces under certain circumstances, these forces must be determined from: net Force = ma.

Two solid objects cannot occupy the same space at the same time. Indeed, when the objects touch,they exert repulsive normal forces on each other, as well as frictional forces that resist their slippingrelative to each other. These contact forces arise from a complex interplay between the electrostaticforces between the electrons and ions in the objects and the laws of quantum mechanics. As twosurfaces are pushed together these forces increase exponentially over an atomic distance scale,easily becoming strong enough to distort the bulk material in the objects if they approach too close.In everyday experience, contact forces are limited by the deformation or acceleration of the objects,rather than by the fundamental interatomic forces. Hence, we can conclude the following:

The magnitude of contact forces is determined by , that is, by the other forces

on, and acceleration of, the contacting bodies. The only exception is that the frictional forcescannot exceed (although they can be smaller than this or even zero).

Normal and friction forces

Two types of contact forces operate in typical mechanics problems, the normal and frictionalforces, usually designated by and (or , or something similar) respectively. These are

the components of the overall contact force: perpendicular to and parallel to the plane of

contact.

Kinetic friction when surfaces slide

When one surface is sliding past the other, experiments show three things about the frictionforce (denoted ):

The frictional force opposes the relative motion at the point of contact, is proportional to the normal force, and

the ratio of the magnitude of the frictional force to that of the normal force is fairly constantover a wide range of speeds.

The constant of proportionality is called the coefficient of kinetic friction, often designated . As

long as the sliding continues, the frictional force is then

(valid when the surfaces slide by each other).

Static friction when surfaces don't slide

When there is no relative motion of the surfaces, the frictional force can assume any valuefrom zero up to a maximum , where is the coefficient of static friction. Invariably,



is larger than , in agreement with the observation that when a force is large enough that

something breaks loose and starts to slide, it often accelerates.

The frictional force for surfaces with no relative motion is therefore

(valid when the contacting surfaces have no relative motion).

The actual magnitude and direction of the static friction force are such that it (together with otherforces on the object) causes the object to remain motionless with respect to the contacting surfaceas long as the static friction force required does not exceed

. The equation is valid only when the surfaces are on the verge of sliding.

Part A

When two objects slide by one another, which of the following statements about the force offriction between them, is true?

ANSWER:The frictional force is always equal to .

The frictional force is always less than .

The frictional force is determined by other forces on theobjects so it can be either equal to or less than .

Correct

Part B

When two objects are in contact with no relative motion, which of the following statementsabout the frictional force between them, is true?

ANSWER:The frictional force is always equal to .

The frictional force is always less than .

The frictional force is determined by other forces on theobjects so it can be either equal to or less than .

Correct

For static friction, the actual magnitude and direction of the friction force are such thatit, together with any other forces present, will cause the object to have the observedacceleration. The magnitude of the force cannot exceed . If the magnitude of static

friction needed to keep acceleration equal to zero exceeds , then the object will

slide subject to the resistance of kinetic friction. Do not automatically assume that unless you are considering a situation in which the magnitude of the static

friction force is as large as possible (i.e., when determining at what point an object willjust begin to slip). Whether the actual magnitude of the friction force is 0, less than

, or equal to depends on the magnitude of the other forces (if any) as well as the

acceleration of the object through .

Part C

When a board with a box on it is slowly tilted to larger and larger angle, common experience



shows that the box will at some point "break loose" and start to accelerate down the board.

The box begins to slide once the component of gravity acting parallel to the board equals the

force of static friction. Which of the following is the most general explanation for why the boxaccelerates down the board?

ANSWER:The force of kinetic friction is smaller than that of static friction,but remains the same.

Once the box is moving, is smaller than the force of static

friction but larger than the force of kinetic friction.

Once the box is moving, is larger than the force of static

friction.

When the box is stationary, equals the force of static

friction, but once the box starts moving, the sliding reduces thenormal force, which in turn reduces the friction.

Correct

At the point when the box finally does "break loose," you know that the component of the box'sweight that is parallel to the board is equal to (i.e., this component of gravitational force

on the box has just reached a magnitude such that the force of static friction, which has amaximum value of , can no longer oppose it.) For the box to then accelerate, there must

be a net force on the box along the board. Thus, the component of the box's weight parallel tothe board must be greater than the force of kinetic friction. Therefore the force of kineticfriction must be less than the force of static friction which implies , as

expected.

Part D

Consider a problem in which a car of mass is on a road tilted at an angle . The normal force

Select the best answer.

ANSWER:

is found using

Correct

The key point is that contact forces must be determined from Newton's equation. In theproblem described above, there is not enough information given to determine the normalforce. Each of the answer options is valid under some conditions, but in fact none is likely tobe correct if there are other forces on the car or if the car is accelerating. Do not memorize



values for the normal force valid in different problems -- you must determine from

.

Problem 2

A small box of mass is sitting on a board of mass and length . The board rests on a

frictionless horizontal surface. Thecoefficient of static friction between theboard and the box is . The coefficient of

kinetic friction between the board and thebox is, as usual, less than .

Throughout the problem, use for the

magnitude of the acceleration due togravity. In the hints, use for the

magnitude of the friction force betweenthe board and the box.

Part A

Find , the constant force with the least magnitude that must be applied to the board in

order to pull the board out from under the the box (which will then fall off of the opposite endof the board).

Hint A.1 Condition for the board sliding out from under the box

The board will slide out from under the box if the magnitude of the board's accelerationexceeds the magnitude of the maximum acceleration that friction can give to the box.

Hint A.2 Find the acceleration of the box in terms of

Assume that the coefficient of static friction between the board and the box is not knownat this point. What is the magnitude of the acceleration of the box in terms of the frictionforce ?

Express your answer in terms of and .

ANSWER:

=

Correct

Hint A.3 Find the largest acceleration of the box

Now take the coefficient of static friction between the board and the box to be . What is

the largest possible magnitude of the acceleration of the box?

Hint A.3.1 Maximum force on the box

Hint not displayed



Express your answer in terms of some or all of the variables , , and .

ANSWER: =

Correct

Hint A.4 Find the sum of horizontal forces on the board

Write down the sum of all the horizontal forces acting on the board. Take the positive xdirection to be to the right.

Hint A.4.1 Friction and Newton's 3rd law

Hint not displayed

Give your answer in terms of , , and any constants necessary.

ANSWER: =

Correct

Hint A.5 Find the acceleration of the board for large

In Hint 4 you found the net horizontal force on the board. Now, find the acceleration of theboard when the force of static friction reaches its maximum possible value.

Express your answer in terms of , , , , and .

ANSWER:

=

Correct

Hint A.6 Putting it all together

Reread Hint 1. In Hint 3, you found the largest possible acceleration of the box, . In

Hint 5, you found the acceleration of the board, . What is the minimum value of the

constant force, , so that ?

Express your answer in terms of some or all of the variables , , , , and .

Do not include in your answer.

ANSWER: =

Correct

Problem 3

A sky diver of mass 80.0 (including parachute) jumps off a plane and begins her descent.



Throughout this problem use 9.80 for the magnitude of the acceleration due to gravity.

Part A

At the beginning of her fall, does the sky diver have an acceleration?

Hint A.1 Free fall

Hint not displayed

ANSWER:No; the sky diver falls at constant speed.

Yes and her acceleration is directed upward.

Yes and her acceleration is directed downward.

Correct

This applet shows the sky diver (not to scale) with her position, speed, and accelerationgraphed as functions of time. You can see how her acceleration drops to zero overtime, giving constant speed after a long time.

Part B

At some point during her free fall, the sky diver reaches her terminal speed. What is themagnitude of the drag force due to air resistance that acts on the sky diver when she

has reached terminal speed?

Hint B.1 Dynamic equilibrium

Hint not displayed

Express your answer in newtons.

ANSWER: = 784

Correct

Part C

For an object falling through air at a high speed , the drag force acting on it due to air resistance

can be expressed as

,

where the coefficient depends on the shape and size of the falling object and on the density of

air. For a human body, the numerical value for is about 0.250 .

Using this value for , what is the terminal speed of the sky diver?

Hint C.1 Terminal speed

Hint not displayed



Express you answer in meters per second.

ANSWER: = 56.0

Correct

Recreational sky divers can control their terminal speed to some extent by changingtheir body posture. When oriented in a headfirst dive, a sky diver can reach speeds ofabout 54 meters per second (120 miles per hour). For maximum drag and stability, skydivers often will orient themselves "belly-first." In this position, their terminal speed istypically around 45 meters per second (100 miles per hour).

Part D

When the sky diver descends to a certain height from the ground, she deploys herparachute to ensure a safe landing. (Usually the parachute is deployed when the sky diverreaches an altitude of about 900 --3000 .) Immediately after deploying the parachute,

does the skydiver have a nonzero acceleration?

Hint D.1 How to approach the problem

Hint not displayed

ANSWER:No; the sky diver keeps falling at constant speed.

Yes and her acceleration is directed downward.

Yes and her acceleration is directed upward.

Correct

Part E

When the parachute is fully open, the effective drag coefficient of the sky diver plusparachute increases to 60.0 . What is the drag force acting on the sky diver

immediately after she has opened the parachute?

Hint E.1 How to approach the problem

Hint not displayed

Hint E.2 Find the speed of the sky diver when the parachute is deployed

Hint not displayed

Express your answer in newtons.

ANSWER: = 1.88×105

Correct

Part F



What is the terminal speed of the sky diver when the parachute is opened?

Hint F.1 How to approach the problem

Hint not displayed

Express your answer in meters per second.

ANSWER: = 3.61

Correct

A typical "student" parachute for recreational skydiving has a drag coefficient that givesa terminal speed for landing of about 2 meters per second (5 miles per hour). If thisseems slow based on video or real-life sky divers you have seen, that may be becausethe sky divers you saw were using high-performance parachutes; these offer the skydivers more maneuverability in the air but increase the terminal speed up to 4 metersper second (10 miles per hour).

Problem 4

A small button placed on a horizontal rotating platform with diameter 0.320 will revolve with

the platform when it is brought up to a rotational speed of 40.0 , provided the button is

a distance no more than 0.140 from the axis.

Part A

What is the coefficient of static friction between the button and the platform?

ANSWER: 0.251Correct

Part B

How far from the axis can the button be placed, without slipping, if the platform rotates at65.0 ?

ANSWER: 5.30×10−2

Correct

Problem 5

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.1×106 , one

an angle 18 west of north and the other an angle 18 east of north, as they pull the tanker a

distance 0.82 toward the north.

Part A

What is the total work they do on the supertanker?

Express your answer using two significant figures.

ANSWER: 3.3×109



= Correct

Problem 6

You are a member of an alpine rescue team and must project a box of supplies, with mass ,

up an incline of constant slope angle so that it reaches a stranded skier who is a vertical

distance above the bottom of the incline. The incline is slippery, but there is some friction

present, with kinetic friction coefficient .

Part A

Use the work-energy theorem to calculate the minimum speed that you must give the box

at the bottom of the incline so that it will reach the skier.

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Find the total work done on the box

Hint not displayed

Hint A.3 What is the initial kinetic energy?

Hint not displayed

Hint A.4 What is the final kinetic energy?

Hint not displayed

Express your answer in terms of some or all of the variables , , , , and .

ANSWER: =

Correct

Problem 7

A block of weight sits on an inclined plane as shown. A force of magnitude is applied to

pull the block up the incline at constant speed. The coefficient of kinetic friction between theplane and the block is .



Part A

What is the total work done on the block by the force of friction as the block moves a

distance up the incline?

Hint A.1 How to start

Hint not displayed

Hint A.2 Find the magnitude of the friction force

Hint not displayed

Express the work done by friction in terms of any or all of the variables , , , ,

, and .

ANSWER: =

Correct

Part B

What is the total work done on the block by the applied force as the block moves a

distance up the incline?

Express your answer in terms of any or all of the variables , , , , , and .

ANSWER: =

Correct

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



Part C

What is the total work done on the block by the force of friction as the block moves a

distance down the incline?


ANSWER: =

Correct

Part D

What is the total work done on the box by the appled force in this case?


ANSWER: =

Correct

Problem 8

A particle of mass moves along a straight line with initial speed . A force of magnitude

pushes the particle a distance along the direction of its motion.

Part A

Find , the final speed of the particle after it has traveled a distance .

Hint A.1 Find the final kinetic energy

Hint not displayed

Express the final speed in terms of , , , and .



ANSWER:

=

Correct

Increase in mass

For the next two parts, assume that the particle's mass is increased to , while all other

parameters in the problem introduction remain the same.

Part B

By what multiplicative factor does the initial kinetic energy increase, and by what multiplicative

factor does the work done by the force increase (with respect to the case when the particle

had a mass )?

If one of the quantities doubles, for instance, it would increase by a factor of 2. If a quantity staysthe same, then the multiplicative factor would be 1.

Hint B.1 Find the work done

Hint not displayed

You should enter the two factors separated by a comma.

ANSWER: , = 3,1

Correct

Part C

The particle's change in speed over the distance will be ______ the change in speed

when it had a mass equal to .

ANSWER:less than

Correct

Increase in initial velocity

For the final two parts, assume that the initial speed of the particle is increased to , with

the particle's mass once again equal to .

Part D

By what factor does the initial kinetic energy increase (with respect to the first situation,

with mass and speed ), and by what factor does the work done by the force

increase?



Again, enter the two factors, separated by a comma.

ANSWER: , = 9,1

Correct

Part E

The particle's change in speed over the distance will be ______ the change in speed

when it had an initial velocity equal to .

Hint E.1 Some math help

Hint not displayed

ANSWER:less than

Correct

Make sure you understand this result:

The amount of energy needed to increase an object's velocity by a certain fixed amountincreases the faster the object is already moving. An elegant way to see this is to take thederivative of kinetic energy with respect to velocity:

,

or

.

This says that when the velocity of an object increases by an amount , its kinetic energy

increases by , so if you have a very fast moving object (large ), it takes a large

amount of work or energy to increase its velocity even just a little bit!

Problem 9

A luggage handler pulls a suitcase of mass 16.9 up a ramp inclined at an angle 24.0

above the horizontal by a force of magnitude 150 that acts parallel to the ramp. The

coefficient of kinetic friction between the ramp and the incline is 0.254. The suitcase travels adistance 3.90 along the ramp.

Part A

Calculate the work done on the suitcase by the force .

ANSWER: = 585

Correct

Part B



Calculate the work done on the suitcase by the gravitational force.

ANSWER: = -263

Correct

Part C

Calculate the work done on the suitcase by the normal force.

ANSWER: = 0

Correct

Part D

Calculate the work done on the suitcase by the friction force.

ANSWER: = -150

Correct

Part E

Calculate the total work done on the suitcase.

ANSWER: = 172

Correct

Part F

If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it hastraveled 3.90 along the ramp?

ANSWER: = 4.52

Correct

Problem 10

Two blocks are connected by a very light string passing over a massless and frictionless pulley(Figure ). The 20.0- block moves 75.0 to the right and the 12.0- block moves 75.0

downward.



Part A

Find the total work done on 20.0- block if there is no friction between the table and the

20.0- block.

ANSWER: = 5.62

Correct

Part B

Find the total work done on 12.0- block if there is no friction between the table and the

20.0- block.

ANSWER: = 3.38

Correct

Part C

Find the total work done on 20.0- block if =0.500 and =0.325 between the table and

the 20.0- block.

ANSWER: = 2.58

Correct

Part D

Find the total work done on 12.0- block if and between the table

and the 20.0- block.

ANSWER: = 1.54

Correct



Problem 11

A box is sliding with a speed of 4.50 on a horizontal surface when, at point P, it

encounters a rough section. On the rough section, the coefficient of friction is not constant, butstarts at 0.100 at P and increases linearly with distance past P, reaching a value of 0.600 at12.5 past point P.

Part A

Use the work-energy theorem to find how far this box slides before stopping.

ANSWER: = 5.11

Correct past point P

Part B

What is the coefficient of friction at the stopping point?

ANSWER: = 0.304

Correct

Part C

How far would the box have slid if the friction coefficient didn't increase, but instead had theconstant value of 0.100?

ANSWER: = 10.3

Correct past point P

Problem 12

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed.To keep the spring stretched or compressed an amount , a force along the x-axis with

x-component must be applied to the free end. Here ,

, and . Note that when the spring is stretched and

when it is compressed.

Part A

How much work must be done to stretch this spring by 0.050 from its unstretched

length?


ANSWER: = 0.11

Correct

Part B

How much work must be done to compress this spring by 0.050 from its unstretched

length?




ANSWER: = 0.17

Correct

Part C

Is it easier to stretch or compress this spring?

ANSWER:to stretch

to compress

Correct

Problem 13

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1300 car moving at 0.66 is to compress the spring no more than 8.1×10−2 before

stopping.

Part A

What should be the force constant of the spring? Assume that the spring has negligiblemass.


ANSWER: = 8.6×104

Correct

Problem 14

A truck engine transmits 28.0 (37.5 ) to the driving wheels when the truck is traveling at

a constant velocity of magnitude 60.0 ( 37.3 ) on a level road.

Part A

What is the resisting force acting on the truck?

ANSWER: = 1680

Correct

Part B

Assume that 65% of the resisting force is due to rolling friction and the remainder is due toair resistance. If the force of rolling friction is independent of speed, and the force of airresistance is proportional to the square of the speed, what power will drive the truck at 30.0

? Give your answer in kilowatts .

ANSWER: 10.3



= Correct

Part C


? Give your answer in horsepower.

ANSWER: = 13.8

Correct

Part D


? Give your answer in kilowatts.

ANSWER: = 115

Correct

Part E


? Give your answer in horsepower.

ANSWER: = 154

Correct

Score Summary:

Your score on this assignment is 92.9%.You received 130.08 out of a possible total of 140 points.

mastering physics assignment 4

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