Formal Homework Assignment 4

10/26/14 10:39 PMFormal Homework Assignment 4

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Formal Homework Assignment 4Due: 11:15am on Tuesday, October 21, 2014

To understand how points are awarded, read the Grading Policy for this assignment.

A message from your instructor...

General Comment: In this assignment you will be asked to draw a number of free-body diagrams. Be very careful whendrawing these diagrams. Mastering Physics can be exceptionally picky about the relative lengths and directions in whichthese vectors are drawn. As a result, I suggest spacing out the problems you do so that you don't get too frustrated with it.


For Problem 3.30, take the origin as the initial position of the clock's hand.

Problem 3.30

The minute hand on a watch is 3 in length. What is the displacement vector of the tip of the minute hand

Part A

From 8:00 to 8:20 a.m.?

Express vector in the form , , where the x and y components are separated by a comma.

ANSWER:

Correct

Part B

From 8:00 to 9:00 a.m.?

Express vector in the form , , where the x and y components are separated by a comma.

ANSWER:

Correct

cm

r rx ry

= 2.61,-4.50 cm r

r rx ry

= 0,0 cm r



Problem 3.32

Jim's dog Sparky runs 50.0 northeast to a tree, then 75.0 west to a second tree, and finally 25.0 south to a third tree.

Part A

Calculate Sparky's net displacement vector.

Enter the east and north components of the net displacement, separated by commas.

ANSWER:

Correct

Part B

Calculate the magnitude of Sparky's net displacement.

Express your answer in meters.

ANSWER:

Correct

Part C

Calculate the direction of Sparky's net displacement.

Express your answer in degrees.

ANSWER:

Correct

Reading Question 5.01

m m m

, = -39.6,10.4 (D net )East (D net )North m

= 41.0 Dnet m

= 14.6 north of west net



Part A

What is a net force?

ANSWER:

Correct

Two Forces Acting at a Point

Two forces, and , act at a point. has a magnitude of 9.00 and is directed at an angle of 61.0 above thenegative x axis in the second quadrant. has a magnitude of 6.20 and is directed at an angle of 53.2 below thenegative x axis in the third quadrant.

Part A

What is the x component of the resultant force?

Express your answer in newtons.

Hint 1. How to approach the problem

The resultant force is defined as the vector sum of all forces. Thus, its x component is the sum of the xcomponents of the forces, and its y component is the sum of the y components of the forces.

Hint 2. Find the x component of

Find the x component of .


Hint 1. Components of a vector

Consider a vector that forms an angle with the positive x axis. The x and y components of are,respectively,

and ,

The weight excluding the container

The vector sum of all forces in a problem

The vector sum of all forces acting on an object

The vector force applied by a net

The vector sum of all forces that add up to zero

F 1 F 2 F 1 N

F 2 N

F 1

F 1

A A

= AcosAx = A sinAy

A



where is the magnitude of the vector. Note that

and if ,

and if .

Hint 2. Find the direction of

is directed at an angle of 61.0 above the x axis in the second quadrant. When you calculate the

components of , however, the direction of the force is commonly expressed in terms of the angle that

the vector representing the force forms with the positive x axis. What is the angle that forms with thepositive x axis? Select an answer from the following list, where 61.0 .

ANSWER:

ANSWER:

Hint 3. Find the x component of

Find the x component of .




and ,


and if ,

A

< 0Ax > 0Ay < < pipi2

< 0Ax < 0Ay pi < < 3pi2

F 1

F 1

F 1F 1

=

180

+ 180

+ 90

-4.36 N

F 2

F 2

A A

= AcosAx = A sinAy

A

< 0Ax > 0Ay < < pipi2

pi < < 3pi



and if .

Hint 2. Find the direction of

is directed at an angle of 53.2 below the x axis in the third quadrant. When you calculate the

components of , however, the direction of the force is commonly expressed in terms of the angle that

the vector representing the force forms with the positive x axis. What is the angle that forms with thepositive x axis? Select an answer from the following list, where 53.2 .

ANSWER:

ANSWER:

ANSWER:

Correct

Part B

What is the y component of the resultant force?



Follow the same procedure that you used in Part A to find the x component of the resultant force, though nowcalculate the y components of the two forces.

Hint 2. Find the y component of

Find the y component of .

< 0Ax






and ,


and if ,

and if .

ANSWER:

Hint 3. Find the y component of

Find the y component of .




and ,


and if ,

and if .

ANSWER:

A A

= AcosAx = A sinAy

A

< 0Ax > 0Ay < < pipi2

< 0Ax < 0Ay pi < < 3pi2

7.87 N

F 2

F 2

A A

= AcosAx = A sinAy

A

< 0Ax > 0Ay < < pipi2

< 0Ax < 0Ay pi < < 3pi2

-4.96 N



ANSWER:

Correct

Part C

What is the magnitude of the resultant force?


Hint 1. Magnitude of a vector

Consider a vector , whose components are and . The magnitude of is

.

ANSWER:

Correct

Problem 5.4

A baseball player is sliding into second base.

Part A

Identify the forces on the baseball player.

ANSWER:

2.91 N

A Ax Ay A

A = +A2x A2y

8.58 N

Normal force ; Kinetic friction

Thrust ; Normal force

Gravity ; Normal force ; Kinetic friction

Gravity ; Normal force ; Static friction

n fk

Fthrust

n

F G n fk

F G n fs



Correct

Problem 5.18

The figure shows two of the three forces acting on an object inequilibrium.

Part A

Redraw the diagram, showing all three forces. Label the third force .

Draw the force vector starting at the black dot. The location and orientation of the vector will be graded. Thelength of the vector will not be graded.

ANSWER:

F 3



Correct

Problem 5.14

The figure shows an object's acceleration-versus-force graph.

Part A



What is the object's mass?

Express your answer with the appropriate units.

ANSWER:

Correct

Problem 5.28

For each part, look at the motion diagrams shown and choose the direction in which the net force is acting on the object.

Part A

ANSWER:

Correct

Part B

0.250 kg

down

up

right

left



ANSWER:

Correct

Problem 5.32

A net-force with -component acts on a object as it moves along the -axis. A graph of versus is shown inthe figure .

down

up

right

left

x Fx 2.0kg x Fx t



Part A

Draw an acceleration graph( versus ) for this object.

ANSWER:

Correct

Relating Graphs and Free-Body Diagrams

Two forces are exerted on an object of mass in the x direction as illustrated in the free-body diagram. Assume that theseare the only forces acting on the object.

ax t

m



Part A

Which of the curves labeled A to D on the graph could be aplot of , the velocity of the object in the x direction as afunction of time?


Analyze the free-body diagram to determine whether there is a net force acting on the object along the x axis. Ifthe object is experiencing a net force, then its velocity must be changing in that direction.

Hint 2. Relate force, acceleration, and velocity

If a constant nonzero net force is applied to an object, what will the object's acceleration and velocity be?

Hint 1. Relating force and acceleration

Recall that Newton's 2nd law applied in the x direction gives

(t)vx

= mFx x



,where is the mass of the object and is the acceleration of the object along the x axis.

Because the object's mass is constant, is proportional to . This means that if increases, must also increase.

Hint 2. Relating acceleration and velocity

The average acceleration of an object along the x direction is defined as the rate of change of

velocity,

,

where time occurs after time .

It may also help to recall that, on a graph of velocity versus time, the slope of the velocity curve is theaverage acceleration.

ANSWER:

ANSWER:

CorrectThe net force on the object in the x direction indicates that the object is accelerating in the x direction. Butaccelerating doesn't necessarily mean speeding up. As depicted by curve B, at the time the net force was applied tothe object, the object had already been moving with nonzero velocity in the +x direction. The effect of theacceleration in the x direction on the object was to

1. slow down the object,2. bring the object to an instantaneous stop (which occurs when line B intersects the horizontal t axis),

and3. speed up the object in the x direction.

= mFx axm ax

Fx ax Fxax

aavg, x

= =aavg, x vxtv2x v1xt2 t1

t2 t1

Both acceleration and velocity will be constant.

Acceleration will not be constant and velocity will change at a nonconstant rate.

Acceleration will be constant and velocity will change at a constant rate.

Acceleration will be constant and velocity will change at a nonconstant rate.

A

B

C

D



Part B

Which of the curves labeled A to D on the graph could be aplot of , the position of the object along the x axis as afunction of time?


The average velocity of an object along the x direction is defined as the rate of change of position,

,

where time occurs after time . On a graph of position versus time, the slope of the position curve is theaverage velocity. Determine what kind of position graph will yield the average velocity found in Part A.

ANSWER:

x(t)

vavg, x

= =vavg, x xtx2 x1t2 t1

t2 t1

A

B

C

D



Correct

Notice that the correct graph shows the object

first, moving in the +x direction with decreasing speed,then, stopping momentarily (at the top of the curve), andfinally, moving in the x direction with increasing speed.

The graphs in Parts A and B were not the only possible velocity and position graphs for the given net force.However, all graphs illustrating motion under the influence of a constant force will have the same characteristics.

In this particular problem, the acceleration of the object was constant. This caused the velocity graph to be a linearcurve of constant slope (i.e., a straight line). When velocity obeys a linear relationship, the position of the objectfollows a curve whose shape is quadratic (also called parabolic). You saw examples of this type of motion when youstudied motion under constant velocity in an earlier chapter.


For Problem 5.35, all of the parts are independent of one another. As a result, when answering each part go back to theoriginal problem statement.

Problem 5.35

A constant force is applied to an object, causing the object to accelerate at 11.0 .

Part A

What will the acceleration be if the force is halved?


ANSWER:

Correct

Part B

What will the acceleration be if the object's mass is halved?


ANSWER:

x(t)

m/s2

= 5.50 a ms2



Correct

Part C

What will the acceleration be if the force and the object's mass are both halved?


ANSWER:

Correct

Part D

What will the acceleration be if the force is halved and the object's mass is doubled?


ANSWER:

Correct

Problem 5.7

Two rubber bands pulling on an object cause it to accelerate at 2.2 .

Part A

What will be the object's acceleration if it is pulled by four rubber bands?


ANSWER:

= 22.0 a ms2

= 11.0 a ms2

= 2.75 a ms2

m/s2

= 4.40 a ms2



Correct

Part B

What will be the acceleration of two of these objects glued together if they are pulled by two rubber bands?


ANSWER:

Correct

A Push or a Pull?

Learning Goal:

To understand the concept of force as a push or a pull and to become familiar with everyday forces.

A force can be simply defined as a push or a pull exerted by one object upon another.

Although such a definition may not sound too scientific, it does capture three essential properties of forces:

Each force is created by some object.Each force acts upon some other object.The action of a force can be visualized as a push or a pull.

Since each force is created by one object and acts upon another, forces must be described as interactions. The proper wordsdescribing the force interaction between objects A and B may be any of the following:

"Object A acts upon object B with force ."

"Object A exerts force upon object B."

"Force is applied to object B by object A."

"Force due to object A is acting upon object B."

One of the biggest mistakes you may make is to think of a force as "something an object has." In fact, at least two objects arealways required for a force to exist.

Each force has a direction: Forces are vectors. The main result of such interactions is that the objects involved change theirvelocities: Forces cause acceleration. However, in this problem, we will not concern ourselves with acceleration--not yet.

Some common types of forces that you will be dealing with include the gravitational force (weight), the force of tension, the

= 1.10 a ms2

F

F

F

F



force of friction, and the normal force.

It is sometimes convenient to classify forces as either contact forces between two objects that are touching or as long-rangeforces between two objects that are some distance apart. Contact forces include tension, friction, and the normal force. Long-range forces include gravity and electromagnetic forces. Note that such a distinction is useful but not really fundamental: Forinstance, on a microscopic scale the force of friction is really an electromagnetic force.

In this problem, you will identify the types of forces acting on objects in various situations.

First, consider a book resting on a horizontal table.

Part A

Which object exerts a downward force on the book?

ANSWER:

Correct

Part B

The downward force acting on the book is __________.

ANSWER:

Correct

Part C

What is the downward force acting on the book called?

ANSWER:

the book itself

the earth

the surface of the table

a contact force

a long-range force



Correct

Part D

Which object exerts an upward force on the book?

ANSWER:

Correct

Part E

The upward force acting on the book is __________.

ANSWER:

Correct

Part F

What is the upward force acting on the book called?

ANSWER:

tension

normal force

weight

friction

the book itself

the earth


a contact force

a long-range force



Correct

Now consider a different situation. A string is attached to a heavy block. The string is used to pull the block to the right alonga rough horizontal table.

Part G

Which object exerts a force on the block that is directed toward the right?

ANSWER:

Correct

Part H

The force acting on the block and directed to the right is __________.

ANSWER:

CorrectTo exert a tension force, the string must be connected to (i.e., touching) the block.

Part I

What is the force acting on the block and directed to the right called?

tension

normal force

weight

friction

the block itself

the earth


the string

a contact force

a long-range force



ANSWER:

Correct

Part J

Which object exerts a force on the block that is directed toward the left?

ANSWER:

Correct

Part K

The force acting on the block and directed to the left is __________.

ANSWER:

Correct

Part L

What is the force acting on the block and directed to the left called?

ANSWER:

tension

normal force

weight

friction

the block itself

the earth


the string

a contact force

a long-range force



Correct

Now consider a slightly different situation. The same block is placed on the same rough table. However, this time, the stringis disconnected and the block is given a quick push to the right. The block slides to the right and eventually stops. Thefollowing questions refer to the motion of the block after it is pushed but before it stops.

Part M

How many forces are acting on the block in the horizontal direction?

ANSWER:

CorrectOnce the push has ended, there is no force acting to the right: The block is moving to the right because it was givena velocity in this direction by some force that is no longer applied to the block (probably, the normal force exerted bya student's hand or some spring launcher).

Once the contact with the launching object has been lost, the only horizontal force acting on the block is directed tothe left--which is why the block eventually stops.

Part N

What is the force acting on the block that is directed to the left called?

ANSWER:

tension

normal force

weight

friction

0

1

2

3



CorrectThe force of friction does not disappear as long as the block is moving. Once the block stops, fricion becomes zero(assuming the table is perfectly horizontal).

Free-Body Diagrams

Learning Goal:

To gain practice drawing free-body diagrams

Whenever you face a problem involving forces, always start with a free-body diagram.To draw a free-body diagram use the following steps:

1. Isolate the object of interest. It is customary to represent the object of interest as a point in your diagram.2. Identify all the forces acting on the object and their directions. Do not include forces acting on other objects in

the problem. Also, do not include quantities, such as velocities and accelerations, that are not forces.3. Draw the vectors for each force acting on your object of interest. When possible, the length of the force vectors

you draw should represent the relative magnitudes of the forces acting on the object.

In most problems, after you have drawn the free-body diagrams, you will explicitly label your coordinate axes and directions.Always make the object of interest the origin of your coordinate system. Then you will need to divide the forces into x and ycomponents, sum the x and y forces, and apply Newton's first or second law.

In this problem you will only draw the free-body diagram.

Suppose that you are asked to solve the following problem:Chadwick is pushing a piano across a level floor (see the figure). The piano can slide across the floor without friction. IfChadwick applies a horizontal force to the piano, what is the piano's acceleration?To solve this problem you should start by drawing a free-body diagram.

tension

normal force

weight

friction



Part A

Determine the object of interest for the situation described in the problem introduction.


You should first think about the question you are trying to answer: What is the acceleration of the piano? Theobject of interest in this situation will be the object whose acceleration you are asked to find.

ANSWER:

Correct

Part B

Identify the forces acting on the object of interest. From the list below, select the forces that act on the piano.

Check all that apply.

ANSWER:

For this situation you should draw a free-body diagram for the floor.

Chadwick.

the piano.



Correct

Now that you have identified the forces acting on the piano, you should draw the free-body diagram. Draw the length of yourvectors to represent the relative magnitudes of the forces, but you don't need to worry about the exact scale. You won't havethe exact value of all of the forces until you finish solving the problem. To maximize your learning, you should draw thediagram yourself before looking at the choices in the next part. You are on your honor to do so.

Part C

Select the choice that best matches the free-body diagram you have drawn for the piano.

Hint 1. Determine the directions and relative magnitudes of the forces

Which of the following statements best describes the correct directions and relative magnitudes of the forcesinvolved?

ANSWER:

ANSWER:

acceleration of the piano

gravitational force acting on the piano (piano's weight)

speed of the piano

gravitational force acting on Chadwick (Chadwick's weight)

force of the floor on the piano (normal force)

force of the piano on the floor

force of Chadwick on the piano

force of the piano pushing on Chadwick

The normal force and weight are both upward and the pushing force is horizontal.

The normal force and weight are both downward and the pushing force is horizontal.

The normal force is upward, the weight is downward, and the pushing force is horizontal. The normalforce has a greater magnitude than the weight.

The normal force is upward, the weight is downward, and the pushing force is horizontal. The normalforce and weight have the same magnitude.

The normal force is upward, the weight is downward, and the pushing force is horizontal. The normalforce has a smaller magnitude than the weight.



CorrectIf you were actually going to solve this problem rather than just draw the free-body diagram, you would need todefine the coordinate system. Choose the position of the piano as the origin. In this case it is simplest to let the yaxis point vertically upward and the x axis point horizontally to the right, in the direction of the acceleration.

Chadwick now needs to push the piano up a ramp and into a moving van. at left. The ramp is frictionless. Is Chadwick strongenough to push the piano up the ramp alone or must he get help?To solve this problem you should start by drawing a free-bodydiagram.

Part D

Determine the object of interest for this situation.

ANSWER:

Correct

Now draw the free-body diagram of the piano in this new situation. Follow the same sequence of steps that you followed forthe first situation. Again draw your diagram before you look at the choices below.

Part E

Which diagram accurately represents the free-body diagram for the piano?

ANSWER:

For this situation, you should draw a free-body diagram for the ramp.

Chadwick.

the piano.



CorrectIn working problems like this one that involve an incline, it is most often easiest to select a coordinate system that isnot vertical and horizontal. Instead, choose the x axis so that it is parallel to the incline and choose the y axis so thatit is perpendicular to the incline.

Tactics Box 5.2 Identifying Forces

Learning Goal:

To practice Tactics Box 5.2 Identifying Forces.

The first basic step in solving force and motion problems generally involves identifying all of the forces acting on an object.This tactics box provides a step-by-step method for identifying each force in a problem.

TACTICS BOX 5.2 Identifying forces

1. Identify the object of interest. This is the object whose motion you wish to study.2. Draw a picture of the situation. Show the object of interest and all other objectssuch as ropes, springs, or

surfacesthat touch it.3. Draw a closed curve around the object. Only the object of interest is inside the curve; everything else is outside.4. Locate every point on the boundary of this curve where other objects touch the object of interest. These are the

points where contact forces are exerted on the object.5. Name and label each contact force acting on the object. There is at least one force at each point of contact;

there may be more than one. When necessary, use subscripts to distinguish forces of the same type.6. Name and label each long-range force acting on the object. For now, the only long-range force is the

gravitational force.

Apply these steps to the following problem: A crate is pulled up a rough inclined wood board by a tow rope. Identify the forceson the crate.

Part A

Which of the following objects are of interest?




ANSWER:

CorrectNow that you have identified the object of interest, draw a sketch of the situation and draw a closed curve aroundthe object, as shown in the figure below.

Part B

Identify the contact forces exerted on the crate.


ANSWER:

rope

wood board

earth

crate

static friction

kinetic friction

drag

gravitational force

spring force

thrust

tension

normal force



Correct

Part C

Identify the long-range forces acting on the crate.


ANSWER:

CorrectNow that you have identified all the forces acting on the system, your final sketch describing the situation might looklike this:

Problem 5.25

An ice hockey puck glides across frictionless ice.

kinetic friction

gravitational force

normal force

static friction

spring force

drag

tension

thrust



Part A

Identify all forces acting on the object.

ANSWER:

Correct

Part B

Draw a free-body diagram of the ice hockey puck.

Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact lengthof your vectors will not be graded but the relative length of one to the other will be graded.

ANSWER:

Normal force ; Gravity

Normal force ; Gravity ; Kinetic friction

Tension ; Weight

Thrust ; Gravity

n F G

n F G fk

T w

Fthrust

F G



Correct

Problem 5.26

Your physics textbook is sliding to the right across the table.

Part A

Identify all forces acting on the object.

ANSWER:

Correct

Part B

Draw a free-body diagram of the object.


ANSWER:

Weight ; Kinetic friction

Thrust ; Kinetic friction

Normal force ; Weight ; Kinetic friction

Normal force ; Weight ; Static friction

w fk

Fthrust

fk

n w fk

n w fs



Correct

Free-Body Diagrams: Introduction

Learning Goal:

To learn to draw free-body diagrams for various real-life situations.

Imagine that you are given a description of a real-life situation and are asked to analyze the motion of the objects involved.Frequently, that analysis involves finding the acceleration of the objects, which, in turn, requires that you find the net force.

To find the net force, you must first identify all of the forces acting on the object and then add them as vectors. Such aprocedure is not always trivial. It is helpful to replace the sketch of the situation by a drawing of the object (represented as aparticle) and all the forces applied to it. Such a drawing is called a free-body diagram. This problem will walk you throughseveral examples of free-body diagrams and will demonstrate some of the possible pitfalls.Here is the general strategy for drawing free-body diagrams:

Identify the object of interest. This may not always be easy: A sketch of the situation may contain many objects,each of which has a different set of forces acting on it. Including forces acting on different objects in the samediagram will lead to confusion and a wrong solution.Draw the object as a dot. Draw and clearly label all the forces acting on the object of interest. The forces shouldbe shown as vectors originating from the dot representing the object of interest. There are two possibledifficulties here: omitting some forces and drawing the forces that either don't exist at all or are applied to otherobjects. To avoid these two pitfalls, remember that every force must be applied to the object of interest by someother object.



Once all of the forces are drawn, draw the coordinate system. The origin should coincide with the dotrepresenting the object of interest and the axes should be chosen so that the subsequent calculations of vectorcomponents of the forces will be relatively simple. That is, as many forces as possible must be either parallel orperpendicular to one of the axes.

Even though real life can present us with a wide variety of situations, we will be mostly dealing with a very small number offorces. Here are the principal ones of interest:

Weight, or the force due to gravity. Weight acts on every object and is directed straight down unless we areconsidering a problem involving the nonflat earth (e.g., satellites).Normal force. The normal force exists between two surfaces that are pressed against each other; it is alwaysperpendicular to the surfaces.Force of tension. Tension exists in strings, springs, and other objects of finite length. It is directed along thestring or a spring. Keep in mind that a spring can be either compressed or stretched whereas a string can onlybe stretched.Force of friction. A friction force exists between two surfaces that either move or have a tendency to moverelative to each other. Sometimes, the force of air drag, similar in some ways to the force of friction, may comeinto play. These forces are directed so that they resist the relative motion of the surfaces. To simplify problemsyou often assume that friction is negligible on smooth surfaces and can be ignored. In addition, the word frictioncommonly refers to resistive forces other than air drag that are caused by contact between surfaces, so youcan ignore air drag in problems unless you are explicitly told to consider its effects.

The following examples should help you learn to draw free-body diagrams. We will start with relatively simple situations inwhich the object of interest is either explicitly suggested or fairly obvious.

Part A

A hockey puck slides along a horizontal, smooth icy surface at a constant velocity as shown. Which of the followingforces act on the puck?


ANSWER:



CorrectThere is no such thing as "the force of velocity." If the puck is not being pushed, there are no horizontal forcesacting on it. Of course, some horizontal force must have acted on it before, to impart the velocity--however, in thesituation described, no such "force of push" exists. Also, the air drag in such cases is assumed to be negligible.Finally, the word "smooth" usually implies negligible surface friction. Your free-body diagram should look like theone shown here.

Part B

Consider a block pulled by a horizontal rope along a horizontal surface at a constant velocity as shown. There is tensionin the rope. Which of the following forces act on the block?


normal force

weight

acceleration

friction

force of velocity

air drag

force of push



ANSWER:

CorrectBecause the velocity is constant, there must be a force of friction opposing the force of tension. Since the block ismoving, it is kinetic friction. Your free-body diagram should look like that shown here.

Part C

A block is resting on an slope. Which of the following forces act on the block?


ANSWER:

force of tension

weight

normal force

force of velocity

friction

acceleration

air drag



Correct

Part D

Draw the free-body diagram for the block resting on a slope.

Draw the force vectors such that their tails align with the center of the block (indicated by the black dot). Theorientations of your vectors will be graded but not the lengths.

ANSWER:

Correct

Part E

force of push

weight

static friction

normal force

kinetic friction



Now consider a block sliding up a rough slope after having been given a quick push as shown . Which of the followingforces act on the block?


ANSWER:

CorrectThe word "rough" implies the presence of friction. Since the block is in motion, it is kinetic friction. Once again, thereis no such thing as "the force of velocity." However, it seems a tempting choice to some students since the block isgoing up.

Part F

Draw the free-body diagram for the block sliding up a rough slope after having been given a quick push.

Draw the force vectors such that their tails align with the center of the block (indicated by the black dot). Theorientations of your vectors will be graded but not the lengths.

ANSWER:

weight

kinetic friction

static friction

force of push

normal force

the force of velocity



Correct

Part G

Now consider a block being pushed up a smooth slope. The force pushing the block is parallel to the slope. Which of thefollowing forces are acting on the block?


ANSWER:



CorrectYour free-body diagram should look like the one shown here.

The force of push is the normal force exerted, possibly, by the palm of the hand of the person pushing the block.

In all the previous situations just described, the object of interest was explicitly given. In the remaining parts of the problem,consider a situation where choosing the objects for which to draw the free-body diagrams is up to you.Two blocks of masses and are connected by a light string that goes over a light frictionless pulley. The block ofmass is sliding to the right on a rough horizontal surface of a lab table.

Part H

To solve for the acceleration of the blocks, you will have to draw the free-body diagrams for which objects?

weight

kinetic friction

static friction

force of push

normal force

m1 m2m1




ANSWER:

Correct

Part I

Draw the free-body diagram for the block of mass and draw a free-body diagram for the block of mass .

Draw the force vectors acting on such that their tails align with the center of the block labeled (indicated by the black dot). Draw the force vectors acting on with their tails aligned with the center of theblock labeled . The orientations of your vectors will be graded but not the lengths.

ANSWER:

the block of mass

the block of mass

the connecting string

the pulley

the table

the earth

m1m2

m1 m2

m1 m1m2

m2



Correct

Problem 5.44

A rocket is being launched straight up. Air resistance is not negligible.

Part A

Which of the following is the correct motion diagram for the situation described above?

Enter the letter that corresponds with the best answer.

ANSWER:

Correct

Part B

Draw a free-body diagram.


ANSWER:

b



Incorrect; Try Again

Problem 5.46

You've slammed on the brakes and your car is skidding to a stop while going down a hill.

Part A

Which of the following is the correct motion diagram for the system described above?

20



ANSWER:

Correct

Part B

Draw a free-body diagram.


ANSWER:

figure a

figure b

figure c

figure d



Correct

Problem 5.12

Answer the following true/false questions on the topic "Force and Motion."

Part A

An object will not accelerate unless there is a net force acting on it.

ANSWER:

Correct

Part B

An object's acceleration vector always points in the same direction as its net force vector .

true

false

a F net



ANSWER:

Correct

Part C

An object cannot be in motion unless there is a net force acting on it.

ANSWER:

Correct

Part D

When the same force is applied to two objects, the more massive object will experience a greater acceleration.

ANSWER:

Correct

Part E

The magnitude of the net force acting on an object is equal to the sum of the magnitudes of the individual forces actingon that object.

ANSWER:

true

false

true

false

true

false

true

false



Correct

Tactics Box 5.3 Drawing a Free-Body Diagram

Learning Goal:

To practice Tactics Box 5.3 Drawing a Free-Body Diagram.

A free-body diagram is a diagram that represents the object as a particle and shows all of the forces acting on the object.Learning how to draw such a diagram is a very important skill in solving physics problems. This tactics box explains theessential steps to construct a correct free-body diagram.

TACTICS BOX 5.3 Drawing a free-body diagram

1. Identify all forces acting on the object. This step was described in Tactics Box 5.2.2. Draw a coordinate system. Use the axes defined in your pictorial representation. If those axes are tilted, for

motion along an incline, then the axes of the free-body diagram should be similarly tilted.3. Represent the object as a dot at the origin of the coordinate axes. This is the particle model.4. Draw vectors representing each of the identified forces. This was described in Tactics Box 5.1. Be sure to label

each force vector.5. Draw and label the net force vector . Draw this vector beside the diagram, not on the particle. Or, if

appropriate, write . Then, check that points in the same direction as the acceleration vector on your motion diagram.

Apply these steps to the following problem: Your physics book is sliding on the carpet. Draw a free-body diagram.

Part A

Which forces are acting on the book?


F net=F net 0 F net a



Hint 1. How to identify all forces acting on the object

These are the steps outlined in Tactics Box 5.2 that will help you to identify all forces acting on the object whosemotion you wish to study:

1. Identify the object of interest. This is the object whose motion you wish to study.2. Draw a picture of the situation. Show the object of interest and all other objectssuch as ropes,

springs, or surfacesthat touch it.3. Draw a closed curve around the object. Only the object of interest is inside the curve; everything

else is outside.4. Locate every point on the boundary of this curve where other objects touch the object of interest.

These are the points where contact forces are exerted on the object.5. Name and label each contact force acting on the object. There is at least one force at each point of

contact; there may be more than one. When necessary, use subscripts to distinguish forces of thesame type.

6. Name and label each long-range force acting on the object. For now, the only long-range force isthe gravitational force.

ANSWER:

CorrectAs explained in Tactics Box 5.2, to identify all the forces acting on the object it helps to sketch the situation. So,draw a closed curve around the object of interest, and mark and label each contact force and long-range forceacting on it. In this problem, your sketch might look like this:

Part B

Draw the most appropriate set of coordinate axes for this problem.

drag

spring force

kinetic friction

tension

gravity

normal force

static friction



The orientation of your vectors will be graded.

ANSWER:

Correct

Part C

In the diagram below, the book is represented by a black dot at the origin of the coordinate axes, in accord with theparticle model. Use this diagram to draw a free-body diagram for this problem that shows all of the forces identified inPart A. Make certain all vectors have the correct orientation, and choose their magnitudes consistent with the expecteddirection of the net force.The net force vector, , has been provided for you in this item. When you are finished drawing the force vectors

identified in Part A, should point in the correct direction.

Draw each force vector with its tail centered at the black dot. The location and orientation of your vectors will begraded.

Hint 1. The relationship between the net force and the acceleration

If you apply Newton's second law to this problem, you will find that the net force should point in the samedirection as the book's acceleration.

Hint 2. Find the direction of the book's acceleration

F netF net

F net



What is the direction of the book's acceleration? Recall that the book is moving to the right on a surface wherefriction cannot be neglected.

ANSWER:

ANSWER:

Correct

Video Tutor: Cart with Fan and Sail

First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment.Then, close the video window and answer the question on the right. You can watch the video again at any point.

to the right

to the left

upward

downward



Part A

Which of the force diagrams in the figure correctly displays all of the horizontal forces exerted on the cart by thesurrounding air?


First, what does the acceleration of the cart imply about the net force acting on the cart? (Only two of the choicesare compatible with the carts behavior.)Next, decide which direction is correct for the force vectors.

When the fan pushes air to the right, in which direction does the air push the fan? (Remember how the cartmoved in the video when only the fan was attached to it.)

When air moving to the right strikes the sail, in which direction does the air push the sail?

ANSWER:



CorrectThe net force on the cart is zero, since the air is the only thing acting on the cart in the horizontal direction.

Video Tutor: Tension in String between Hanging Weights

First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment.Then, close the video window and answer the question at right. You can watch the video again at any point.

Part A

Consider the video tutorial you just watched. Suppose that we duplicate this experimental setup in an elevator. What willthe spring scale read if the elevator is moving upward at constant speed?


What does the phrase "at constant speed" imply about the acceleration of the system?

ANSWER:

A

B

C

D

Less than 18 but greater than 0

More than 18

18

0

N NN

NN



CorrectSince the elevator is not accelerating, the reading on the scale is the same as in the video.


Parts C and E of "Understanding Newton's Laws" can be difficult - be careful and think about it fully.

Understanding Newton's Laws

Part A

An object cannot remain at rest unless which of the following holds?


This problem describes a situation of static equilibrium (i.e., a body that remains at rest). Hence, it is appropriateto apply Newton's 1st law.

Hint 2. Newton's 1st law: a body at rest

According to Newton's 1st law, a body at rest remains at rest if the net force acting on it is zero.

ANSWER:

CorrectIf there is a net force acting on a body, regardless of whether it is a constant force, the body accelerates. If the bodyis at rest and the net force acting on it is zero, then it will remain at rest. The net force could be zero either becausethere are no forces acting on the body at all or because several forces are acting on the body but they all cancelout.

Part B

If a block is moving to the left at a constant velocity, what can one conclude?

The net force acting on it is zero.

The net force acting on it is constant and nonzero.

There are no forces at all acting on it.

There is only one force acting on it.




This problem describes a situation of dynamic equilibrium (i.e., a body that moves at a constant velocity). Hence,it is appropriate to apply Newton's 1st law.

Hint 2. Newton's 1st law: a body in motion

According to Newton's 1st law, a body initially in motion continues to move with constant velocity if the net forceacting on it is zero.

ANSWER:

CorrectIf there is a net force acting on a body, regardless of whether the body is already moving, the body accelerates. If abody is moving with constant velocity, then it is not accelerating and the net force acting on it is zero. The net forcecould be zero either because there are no forces acting on the body at all or because several forces are acting onthe body but they all cancel out.

Part C

A block of mass is acted upon by two forces: (directed to the left) and (directed to the right). What canyou say about the block's motion?


This problem describes a situation of dynamic motion (i.e., a body that is acted on by a net force). Hence, it isappropriate to apply Newton's 2nd law, which allows you to relate the net force acting on a body to theacceleration of the body.

Hint 2. Newton's 2nd law

Newton's 2nd law states that a body accelerates if a net force acts on it. The net force is proportional to theacceleration of the body and the constant of proportionality is equal to the mass of the body. In other words,

,

where is the net force acting on the body, and and are the mass and the acceleration of the body,respectively.

Hint 3. Relating acceleration to velocity

There is exactly one force applied to the block.

The net force applied to the block is directed to the left.

The net force applied to the block is zero.

There must be no forces at all applied to the block.

2 kg 3 N 4 N

F = ma

F m a



Acceleration is defined as the change in velocity per unit time. Keep in mind that both acceleration and velocityare vector quantities.

ANSWER:

CorrectThe acceleration of an object tells you nothing about its velocity--the direction and speed at which it is moving. Inthis case, the net force on (and therefore the acceleration of) the block is to the right, but the block could be movingleft, right, or in any other direction.

Part D

A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be__________.


This problem describes a situation of dynamic motion (i.e., a body that is acted on by a net force). Hence, it isappropriate to apply Newton's 2nd law, which allows you to relate the net force acting on a body to theacceleration of the body.

Hint 2. Newton's 2nd law

Newton's 2nd law states that a body accelerates if a net force acts on it. The net force is proportional to theacceleration of the body and the constant of proportionality is equal to the mass of the body. In other words,

,

where is the net force acting on the body, and and are the mass and the acceleration of the body,respectively.

ANSWER:

It must be moving to the left.

It must be moving to the right.

It must be at rest.

It could be moving to the left, moving to the right, or be instantaneously at rest.

F = ma

F m a

continuously changing direction

moving at constant velocity

moving with a constant nonzero acceleration

moving with continuously increasing acceleration



CorrectSince there is a net force acting, the body does not move at a constant velocity, but it accelerates instead.However, the force acting on the body is constant. Hence, according to Newton's 2nd law of motion, theacceleration of the body is also constant.

Part E

Two forces, of magnitude and , are applied to an object. The relative direction of the forces is unknown. Thenet force acting on the object __________.



By definition, the net force is the vector sum of all forces acting on the object. To find the magnitude of the netforce you need to add the components of the two forces acting. Try adding the two forces graphically (byconnecting the head of one force to the tail of the other). The directions of the two forces are arbitrary, but bytrying different possibilities you should be able to determine the maximum and minimum net forces that could acton the object.

Hint 2. Find the net force when the two forces act on the object in opposite directions

Find the magnitude of the net force if both the forces acting on the object are horizontal and the 10-N force isdirected to the right, while the 4-N force is directed to the left.


Hint 1. Vector addition

The magnitude of the vector sum of two parallel forces is the sum of the magnitudes of the forces. Themagnitude of the vector sum of two antiparallel forces is the absolute value of the difference in magnitudesof the forces.

ANSWER:

Hint 3. Find the direction of the net force when the two forces act in opposite directions

If both the forces acting on the object are horizontal and the 10-N force is directed to the right, while the 4-N forceis directed to the left, the net force is horizontal and directed __________.

ANSWER:

4 N 10 N

6.0 N



ANSWER:

Correct

A World-Class Sprinter

World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and hasmagnitude .

Part A

How much horizontal force must a sprinter of mass 55 exert on the starting blocks to produce this acceleration?

Express your answer in newtons using two significant figures.

Hint 1. Newton's 2nd law of motion

According to Newton's 2nd law of motion, if a net external force acts on a body, the body accelerates, andthe net force is equal to the mass of the body times the acceleration of the body:

.

ANSWER:

Correct

Part B

Which body exerts the force that propels the sprinter, the blocks or the sprinter?

in the same direction as the 10-N force

in the opposite direction to the 10-N force

cannot have a magnitude equal to

cannot have a magnitude equal to

cannot have the same direction as the force with magnitude

must have a magnitude greater than

5 N10 N

10 N10 N

15 m/s2

F kg

Fnetm a

= maFnet

= 830 F N



Hint 1. How to approach the question

To start moving forward, sprinters push backward on the starting blocks with their feet. Newton's 3rd law tells youthat the blocks exert a force on the sprinter of the same magnitude, but opposite in direction.

ANSWER:

CorrectTo start moving forward, sprinters push backward on the starting blocks with their feet. As a reaction, the blockspush forward on their feet with a force of the same magnitude. This external force accelerates the sprinter forward.

Enhanced EOC: Problem 5.9

The figure shows acceleration-versus-force graphs for two objectspulled by rubber bands.

You may want to review ( pages 127 - 130) .

For help with math skills, you may want to review:

Finding the Slope of a Line from a Graph

Part A

What is the mass ratio ?

Express your answer using two significant figures.


How are the acceleration and the force on an object related to its mass? How is the slope of each line in the

the blocks

the sprinter

m1m2



figure related to each object's mass?

For each line, what two points are easy to measure accurately to determine the slope of line? How is the slopedetermined from the x and y coordinates of the two points you chose for each line?

ANSWER:

Correct

Problem 6.10

A horizontal rope is tied to a 51.0 box on frictionless ice. What is the tension in the rope if:

Part A

The box is at rest?

Express your answer as an integer and include the appropriate units.

ANSWER:

Correct

Part B

The box moves at a steady = 5.20 ?


ANSWER:

Correct

Part C

The box = 5.20 and = 5.20 ?

= 0.36m1m2

kg

= 0 T N

vx m/s

= 0 T N

vx m/s ax m/s2



Express your answer to three significant figures and include the appropriate units.

ANSWER:

Correct

Problem 6.13

A woman has a mass of .

Part A

What is her weight while standing on earth?

Express your answer to two significant figures and include the appropriate units.

ANSWER:

Correct

Part B

What is her mass on the moon, where


ANSWER:

Correct

Part C

What is her weight on the moon?

Express your answer to two significant figures and include the appropriate units.

ANSWER:

= 265 T N

55 kg

= 540 wEarth N

g = 1.62 m/s2 ?

= 55 mMoon kg



Correct

PSS 6.1 Equilibrium Problems - Copy

Learning Goal:

To practice Problem-Solving Strategy 6.1 for equilibrium problems.

A pair of students are lifting a heavy trunk on move-in day. Usingtwo ropes tied to a small ring at the center of the top of the trunk,they pull the trunk straight up at a constant velocity . Each ropemakes an angle with respect to the vertical. The gravitationalforce acting on the trunk has magnitude .

Find the tension in each rope.

PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems

MODEL: Make simplifying assumptions.

VISUALIZE:

Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is theprocess of translating words into symbols.Identify all forces acting on the object, and show them on a free-body diagram.These elements form the pictorial representation of the problem.

SOLVE: The mathematical representation is based on Newton's first law:

.The vector sum of the forces is found directly from the free-body diagram.

ASSESS: Check if your result has the correct units, is reasonable, and answers the question.

Model

= 89 wMoon N

v

FG

T

= =F net i F i 0



The trunk is moving at a constant velocity. This means that you can model it as a particle in dynamic equilibrium and applythe strategy above. Furthermore, you can ignore the masses of the ropes and the ring because it is reasonable to assumethat their combined weight is much less than the weight of the trunk.

Visualize

Part A

The most convenient coordinate system for this problem is one in which the y axis is vertical and the ropes both lie in thexy plane, as shown below.

Identify the forces acting on the trunk, and then draw a free-body diagram of the trunk in the diagram below. The blackdot represents the trunk as it is lifted by the students.

Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The lengthof the vectors will not be graded.

ANSWER:

Correct

Part B

In the free-body diagram drawn in the previous part, different symbols are used to represent the tensions in the tworopes. This notation could be simplified by identifying a useful relationship between these two forces. Which of the

T1



following statements properly describes the relationship between the magnitude of the tension force in rope 1 and themagnitude of the tension force in rope 2?

ANSWER:

CorrectThis is a type of reasoning, used often in physics, called a symmetry argument. Since the ropes are in identicalsituations, except for one being the mirror image of the other, they have to possess identical tensions. Since the twotension forces have equal magnitude, just use to denote the magnitude of the tension force in either rope. Withthe information you have gathered here, you can build a complete pictorial representation:

Solve

Part C

To assess whether your results make sense, sort the following situations according to whether the tension in the ropesincreases, decreases, or is unchanged as a result of the change mentioned in each picture. In each case, assume thatall the conditions, other than those mentioned in each picture, remain the same as in the situation described in theproblem introduction. Use your intuition, not your math skills, to find your answers.

Drag the appropriate items to their respective bins.

T1T2

, because the first rope attached must hold the full weight of the trunk before the second rope isattached.

, because rope 1 is shorter than rope 2.

, because two ropes attached to the same object should have the same tension.

, because the ropes attach to the trunk at the same point and at the same angle.

>T1 T2



ANSWER:

CorrectNow, use your math skills. Look at your expression for from Part C. How does change if increases ordecreases? How does change if the gravitational force on the trunk has a larger magnitude, that is, if the trunk isheavier? Your answer from Part C says that is directly proportional to and inversely proportional to .This means your mathematical expression for correctly predicts what your intuition has suggested. Yourcalculations do make sense!

Problem 6.2

The three ropes in the figure are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles withthe tensions shown in the figure.

T T T

T FG cosT



Part A

What is the magnitude of the tension in the third rope?


ANSWER:

Correct

Part B

What is the direction of the tension in the third rope?


ANSWER:

Correct

Lifting a Bucket

A 6- bucket of water is being pulled straight up by a string at a constant speed.

T 3

= 94 T3 N

T 3

= 58 below horizontal

kg



Part A

What is the tension in the rope?

ANSWER:

Correct

Part B

At a certain point the speed of the bucket begins to change. The bucket now has an upward constant acceleration ofmagnitude 3 . What is the tension in the rope now?

ANSWER:

Correct

Now assume that the bucket has a downward acceleration, with a constant acceleration of magnitude 3 .

Part C

Now what is the tension in the rope?

ANSWER:

about 42

about 60

about 78

0 because the bucket has no acceleration.

NNN

N

m/s2

about 42

about 60

about 78

It is increasing as the speed increases.

NNN

m/s2

about 42

about 60

about 78

It is decreasing as the speed increases.

NNN



CorrectNote that the force of tension is not related to the direction or the magnitude of the bucket's velocity: Only theacceleration matters.

Score Summary:Your score on this assignment is 96.7%.You received 61.88 out of a possible total of 64 points.

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Formal Homework Assignment 4