AP-Newton's 3rd Law & Friction-Preview[1]

Version PREVIEW – Newton’s 3rd Law Friction – Johnson – (84248) 1

This print-out should have 46 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

AP M 1993 MC 5

001 10.0 pointsIf F1 is the magnitude of the force exertedby the Earth on a satellite in orbit about theEarth and F2 is the magnitude of the forceexerted by the satellite on the Earth, thenwhich of the following is true?

1. F1 ≫ F2.

2. F2 ≫ F1.

3. F1 = F2. correct

4. F2 > F1.

5. F1 > F2.

Explanation:

The forces must be equal in magnitude butopposite in direction.

Concept 05 E12

002 10.0 pointsConsider a baseball bat hitting a ball.

Which of the following is correct?

1. Both slow down.

2. The baseball bat slows down, and the ballspeeds up.

3. None of these correct

4. The baseball bat speeds up, and the ballslows down.

5. Both speed up.

Explanation:

A baseball exerts an external force on thebat (opposite to the bat’s motion) which de-celerates the oncoming bat. Whether the ballwill speed up or slow down is dependent onthe direction of its velocity.

Concept 05 E20

003 10.0 pointsKen and Joanne are astronauts floating somedistance apart in space. They are joined by asafety cord whose ends are tied around theirwaists.

If Ken starts pulling on the cord, what willhappen?

1. Joanne will move toward Ken while Kenremains stationary.

2. Neither will move.

3. The move toward each other. correct

4. Ken will move toward Joanne whileJoanne remains stationary.

Explanation:

Both will move. Ken’s pull on the rope istransmitted to Joanne, causing her to accel-erate toward him. By Newton’s third law,the rope pulls back on Ken, causing him toaccelerate toward Joanne.

Conceptual 04 07

004 (part 1 of 4) 10.0 pointsTracy (of mass 46 kg) and Tom (of mass

67 kg) are standing at rest in the center of theroller rink, facing each other, free to move.Tracy pushes off Tom with her hands and re-mains in contact with Tom’s hands, applyinga constant force for 0.45 s. Tracy moves 0.3 mduring this time. When she stops pushing offTom, she moves at a constant speed.

What is Tracy’s constant acceleration dur-ing her time of contact with Tom?Correct answer: 2.96296 m/s2.

Explanation:

Let : d = 0.3 m and

t = 0.45 s .

Since d =1

2a t2,

a =2 d

t2=

2 (0.3 m)

(0.45 s)2= 2.96296 m/s2 .


005 (part 2 of 4) 10.0 pointsWhat is Tracy’s final speed after this con-

tact?Correct answer: 1.33333 m/s.

Explanation:

Let : a = 2.96296 m/s2 and

t = 0.45 s .

∆v = a t

= (2.96296 m/s2) (0.45 s)

= 1.33333 m/s .

006 (part 3 of 4) 10.0 pointsWhat force was applied to Tracy during

this time?

Correct answer: 136.296 N.

Explanation:

Let : m = 46 kg and

a = 2.96296 m/s2 .

F = ma

= (46 kg) (2.96296 m/s2)

= 136.296 N .

007 (part 4 of 4) 10.0 pointsWhat is correct about Tom’s motion?

1. Tom’s acceleration is greater.

2. The force acting on Tom is greater thanthe force acting on Tracy.

3. Tom’s acceleration is smaller. correct

4. Tom’s final velocity is greater thanTracy’s.

Explanation:

By Newton’s third law, the forces have the

same magnitude, so a ∝ 1

m.

But ∆v = a t ∝ 1

m, so Tom has a smaller

acceleration and final velocity because he ismore massive.

Conceptual forces 02

008 (part 1 of 3) 10.0 pointsConsider the following statement made re-

garding a book at rest on a level table:The two forces exerted on the

book are the normal force directedup and the weight of the book di-rected down. These are equal andopposite to one another. By New-ton’s third law they are an action-reaction pair, so the normal force isalways equal to the weight of thebook.

Do you agree with the statement?

1. Disagree; if an additional force acts downon the book, the normal force must alsocounter this extra force. correct

2. Agree; a normal force only exerts enoughforce to keep the object from falling through.The normal force and weight are always equal,even when other forces are present.

3. Agree; for every action there is an equaland opposite reaction (Newton’s third law).Since gravity pulls down on the book, to be“equal and opposite” the force of gravity mustequal the normal force.

4. Disagree; the weight must be slightlygreater than the normal force to keep the bookin contact with the table. However, the forcesare an action-reaction force pair because theyare acting on one object.

Explanation:

If we were to put a weight on the book, thenormal force would not be simply the weightof the book as the statement implies (see part2).

Note: This force pair is not a third law


action-reaction pair. The force pairs in New-ton’s third law always act on two differentobjects! The action-reaction pair is ”Earthexerts a force on the book” and ”the bookexerts a force on the Earth.”

Caution: The explanation introduces some-thing not mentioned in the problem.

009 (part 2 of 3) 10.0 pointsConsider a book on top of a level table whilethe book is being pressed straight down witha force F by your hand.

Book

F

The following figures show several attemptsat drawing free-body diagrams for the book.

Which figure has the correct directions foreach force?

Note: The magnitude of the forces are notnecessarily drawn to scale.

1.normal

gravitational hand

correct

2.

gravitational

hand normal

3.

hand

normal gravitational

4.gravitational

hand normal

5.hand

normal gravitational

6.

normal

gravitational hand

Explanation:

The normal force opposes both the gravita-tional force down and the Fhand down.

010 (part 3 of 3) 10.0 pointsWhat forces change when comparing the freebody diagram before the hand was placed onthe book to after?

1. gravitational

2. gravitational and normal

3. gravitational and Fhand

4. gravitational, normal, and Fhand


5. None of these

6. Fhand

7. normal and Fhand correct

8. normal

Explanation:

There will be a change in both the nor-mal force and the Fhand. When the handis removed, the normal force is equal to theweight of the book.

Force and Motion 07

011 (part 1 of 3) 10.0 pointsThe following 3 questions refer to the colli-

sions between a car and a truck whose weightis much heavier than the car (M ≫ m). Foreach description of a collision below, choosethe one answer from the possibilities that bestdescribes the size (or magnitude) of the forcesbetween the car and the truck.

mM

v v

They are both moving at the same speedwhen they collide.

1. Neither exerts a force on the other; thecar gets smashed simply because it is in theway of the truck.

2. The car exerts a greater amount of forceon the truck than the truck exerts on thecar.

3. The truck exerts a greater amount of forceon the car than the car exerts on the truck.

4. The truck exerts the same amount offorce on the car as the car exerts on the truck.correct

5. Not enough information is given to pickone of the answers above.

6. None of the answers above correctly de-scribes the situation.

Explanation:

By Newton’s third law, action and reactionare of the same magnitude and in oppositedirections.

012 (part 2 of 3) 10.0 pointsThe car is moving much faster than the heav-ier truck when they collide.







Explanation:

The same reason as Part 1.

013 (part 3 of 3) 10.0 pointsThe heavier truck is standing still when thecar hits it.








Explanation:

The same reason as Part 1.

Force and Motion 10

014 10.0 pointsTwo students sit in identical office chairs fac-ing each other. Bob has a mass of 95 kg, whileJim has a mass of 77 kg. Bob places his barefeet on Jim’s knees, as shown to the right.Bob then suddenly pushes outward with hisfeet, causing both chairs to move.

Bob Jim

In this situation, while Bob’s feet are incontact with Jim’s knees,1. Neither student exerts a force on the

other.

2. None of these answers is correct.

3. Each student exerts a force on the other,but Bob exerts a larger force.

4. Bob exerts a force on Jim, but Jim doesn’texert a force on Bob.

5. Each student exerts the same amount offorce on the other. correct

6. Each student exerts a force on the other,but Jim exerts a larger force.

Explanation:

By Newton’s third law, action and reactionare of the same magnitude and in the oppositedirection.

Conceptual forces 06 short

015 10.0 pointsA spherical mass rests upon two wedges, as

seen in the figure below. The sphere and thewedges are at rest and stay at rest. There is nofriction between the sphere and the wedges.

M

The following figures show several attemptsat drawing free-body diagrams for the sphere.

Which figure has the correct directions foreach force?

Note: The magnitude of the forces are notnecessarily drawn to scale.

1. normalnormal

weightweight

2.

weightfriction friction

normalnormal

3. normal

weightweight

4.

weight

normal

5. Since the sphere is not moving, no forcesact on it.


6. weight

normalnormal

7. normalnormal

weightweight

8.weight

9.

weight

normalnormal correct

Explanation:

Weight – the force of gravity – pulls thesphere down. The normal force of the leftwedge upon the sphere acts perpendicular to(normal to) their surfaces at the point of con-tact; i.e., diagonally upward and rightward.Likewise, the normal force of the right wedgeupon the sphere acts diagonally upward andleftward.

Atwood Machine 14

016 (part 1 of 2) 10.0 pointsA light, inextensible cord passes over a light,frictionless pulley with a radius of 9 cm. Ithas a(n) 19 kg mass on the left and a(n)8.2 kg mass on the right, both hanging freely.Initially their center of masses are a verticaldistance 4.8 m apart.

The acceleration of gravity is 9.8 m/s2 .

4.8 m

9 cm

ω

19 kg

8.2 kg

At what rate are the two masses accelerat-ing when they pass each other?Correct answer: 3.89118 m/s2.

Explanation:

Let : R = 9 cm ,

m1 = 8.2 kg ,

m2 = 19 kg ,

h = 4.8 m , and

v = ω R .

Consider the free body diagrams

19 kg 8.2 kg

T T

m2g

m1g

a a

Since the larger mass will move down andthe smaller mass up, we can take motiondownward as positive for m2 and motion up-ward as positive for m1. Apply Newton’ssecond law to m1 and m2 respectively andthen combine the results:

For mass 1:∑

F1 : T − m1 g = m1 a (1)

For mass 2:∑

F2 : m2 g − T = m2 a (2)

We can add Eqs. (1) and (2) above and obtain:

m2 g − m1 g = m1 a + m2 a


a =m2 − m1

m1 + m2

g

=19 kg − 8.2 kg

19 kg + 8.2 kg(9.8 m/s2)

= 3.89118 m/s2 .

017 (part 2 of 2) 10.0 pointsWhat is the tension in the cord when theypass each other?Correct answer: 112.268 N.

Explanation:

T = m1 (g + a)

= (8.2 kg) (9.8 m/s2 + 3.89118 m/s2)

= 112.268 N .

AP M 1993 MC 9

018 10.0 pointsPretend you are on a planet similar to Earthwhere the acceleration of gravity is approxi-mately 10 m/s2.

The pulley is massless and frictionless. Amassless inextensible string is attached to themasses. The objects are initially held at rest.

7 m

0.3 m

ω

64 kg

32 kg

64 kg

If a third object with a mass of 32 kg ishung on one of the 64 kg objects as shownand the objects are released, the magnitudeof the acceleration ~a of the 32 kg object ismost nearly

1. ‖~a‖ ≈ 30 m/s2

2. ‖~a‖ ≈ 4 m/s2

3. ‖~a‖ ≈ 3 m/s2

4. ‖~a‖ ≈ 2 m/s2 correct

5. ‖~a‖ ≈ 6 m/s2

6. ‖~a‖ ≈ 8 m/s2

7. ‖~a‖ ≈ 1 m/s2

8. ‖~a‖ ≈ 10 m/s2

Explanation:

Given : m1 = 64 kg ,

m2 = 32 kg ,

R = 0.3 m ,

ℓ = 7 m , and

g ≈ 10 m/s2 .


m2 m1 m1

T1

T2

T2

m2g

T1

m1g

m1g

a a

The total unbalanced force on the system(the total mass is 320 N) is the gravitationalforce on the 32 kg object. Therefore theacceleration, according to Newton’s secondlaw of motion, is

Fnet = m2 g = (2 m1 + m2) a , and

‖~a‖ =m2 g

2 m1 + m2

≈ (32 kg)(10 m/s2)

2 (64 kg) + (32 kg)

≈ 320 N

160 kg

≈ 2 m/s2 .

Atwood Machine 11


019 (part 1 of 2) 10.0 pointsA pulley is massless and frictionless. 1 kg,2 kg, and 8 kg masses are suspended as in thefigure.

2 m

23.9 cm

ω

2 kg

1 kg

8 kg

T2

T1

T3

What is the tension T1 in the string be-tween the two blocks on the left-hand sideof the pulley? The acceleration of gravity is9.8 m/s2 .Correct answer: 14.2545 N.

Explanation:

Let : R = 23.9 cm ,

m1 = 1 kg ,

m2 = 2 kg ,

m3 = 8 kg , and

h = 2 m .


1 kg 2 kg 8 kg

T1

T2

T3

m1g

T1

m2g

m3g

a a

For each mass in the system

~Fnet = m~a .

Since the string changes direction aroundthe pulley, the forces due to the tensions T2

and T3 are in the same direction (up). Theacceleration of the system will be down tothe right (m3 > m1 + m2), and each mass inthe system accelerates at the same rate (thestring does not stretch). Let this accelerationrate be a and the tension over the pulley beT ≡ T2 = T3.

For the lower left-hand mass m1 the accel-eration is up and

T1 − m1 g = m1 a . (1)

For the upper left-hand mass m2 the acceler-ation is up and

T − T1 − m2 g = m2 a . (2)

For the right-hand mass m3 the accelerationis down and

−T + m3 g = m3 a . (3)

Adding Eqs. (1), (2), and (3), we have

(m3 − m1 − m2) g = (m1 + m2 + m3) a (4)

a =m3 − m1 − m2

m1 + m2 + m3

g (5)

=8 kg − 1 kg − 2 kg

1 kg + 2 kg + 8 kg(9.8 m/s2)

= 4.45455 m/s2 .

The tension in the string between block m1

and block m2 (on the left-hand side of thepulley) can be determined from Eq. (1).

T1 = m1(a + g) (6)

= (1 kg) (4.45455 m/s2 + 9.8 m/s2)

= 14.2545 N .

020 (part 2 of 2) 10.0 pointsWhat is the magnitude of the acceleration ofthe lower left-hand block?Correct answer: 4.45455 m/s2.

Explanation:

The acceleration is the same for every mass,since the string is inextensible.


keywords:

Atwood Machine 15

021 10.0 pointsA light, inextensible cord passes over a light,frictionless pulley with a radius of 14 cm. Ithas a(n) 18 kg mass on the left and a(n)9.5 kg mass on the right, both hanging freely.Initially their center of masses are a verticaldistance 2 m apart.


2 m

14 cm

ω

18 kg

9.5 kg

At what rate are the two masses accelerat-ing when they pass each other?Correct answer: 3.02909 m/s2.

Explanation:

Let : R = 14 cm ,

m1 = 9.5 kg ,

m2 = 18 kg ,

h = 2 m , and

v = ω R .


18 kg 9.5 kg

T T

m2g

m1g

a a

Since the larger mass will move down andthe smaller mass up, we can take motiondownward as positive for m2 and motion up-ward as positive for m1. Apply Newton’ssecond law to m1 and m2 respectively andthen combine the results:

For mass 1:∑

F1 : T − m1 g = m1 a (1)

For mass 2:∑

F2 : m2 g − T = m2 a (2)

We can add Eqs. (1) and (2) above and obtain:

m2 g − m1 g = m1 a + m2 a

a =m2 − m1

m1 + m2

g

=18 kg − 9.5 kg

18 kg + 9.5 kg(9.8 m/s2)

= 3.02909 m/s2 .

T = m1 (g + a)

= (9.5 kg) (9.8 m/s2 + 3.02909 m/s2)

= 121.876 N .

Exam Accelerated System

022 10.0 pointsA mass of 2.7 kg lies on a frictionless table,pulled by another mass of 3.8 kg under theinfluence of Earth’s gravity.


2.7 kg

3.8 kg

What is the magnitude of the accelerationa of the two masses?Correct answer: 5.72923 m/s2.


Explanation:

Given : m1 = 2.7 kg and

m2 = 3.8 kg .

m1 m2

a

T

N m1 g

a

T

m2 g

Let the direction of acceleration as indi-cated in the figure be positive. The net forceon the system is simply the weight of m2.

Fnet = m2 g = 37.24 N .

From Newton’s second law,

Fnet = m2 g = (m1 + m2) a .

Solving for a,

a =m2

m1 + m2

g

=3.8 kg

2.7 kg + 3.8 kg

= 5.72923 m/s2 .

Accelerated System 01

023 (part 1 of 2) 10.0 pointsA block of mass 3.22 kg lies on a frictionlesshorizontal surface. The block is connectedby a cord passing over a pulley to anotherblock of mass 7.52 kg which hangs in the air,as shown on the following picture. Assumethe cord to be light (massless and weightless)and unstretchable and the pulley to have nofriction and no rotational inertia.


3.22 kg

7.52 kg

Calculate the acceleration of the firstblock.Correct answer: 6.86182 m/s2.

Explanation:

Given : m1 = 3.22 kg , and

m2 = 7.52 kg .

m1 m2

a

T

N m1 g

a

T

m2 g

Since the cord is unstretchable, the firstblock accelerates to the right at exactly thesame rate a as the second (hanging) block ac-celerates downward. Also, the cord’s tensionpulls the first block to the right with exactlythe same tension T as it pulls the second blockupward.

The only horizontal force acting on the firstblock is the cord’s tension T , hence by New-ton’s Second Law

m1 a = F net→1 = T .

The second block feels two vertical forces:The cord’s tension T (upward) and the block’sown weight W2 = m2 g (downward). Conse-quently,

m2 a = F net↓2

= m2 g − T .

Adding the two equations together, we arriveat

(m1 + m2) a = m2 g ,

and hence

a =m2

m1 + m2

g

=7.52 kg

3.22 kg + 7.52 kg(9.8 m/s2)

= 6.86182 m/s2 .

024 (part 2 of 2) 10.0 points


Calculate the tension in the cord.Correct answer: 22.0951 N.

Explanation:

T = m1 a

= (3.22 kg) (6.86182 m/s2)

= 22.0951 N .


025 10.0 pointsA block of mass 4.51 kg lies on a frictionlesshorizontal surface. The block is connected bya cord passing over a pulley to another blockof mass 4.94 kg which hangs in the air, asshown on the following picture. Assume thecord to be light (massless and weightless) andunstretchable and the the pulley to have nofriction and no rotational inertia.


4.51 kg

4.94 kg

Calculate the acceleration of the firstblock.Correct answer: 5.12296 m/s2.

Explanation:

Let : m1 = 4.51 kg and

m2 = 4.94 kg .

m1 m2

a

T

N m1 g

a

T

m2 g

Since the cord is unstretchable, the firstblock accelerates to the right at exactly thesame rate a as the second (hanging) block ac-celerates downward. Also, the cord’s tension

pulls the first block to the right with exactlythe same tension T as it pulls the second blockupward.

The only horizontal force acting on the firstblock is the cord’s tension T , hence by New-ton’s Second Law

m1 a = F net→1 = T .

The second block feels two vertical forces:The cord’s tension T (upward) and the block’sown weight W2 = m2 g (downward). Conse-quently,

m2 a = F net↓2

= m2 g − T .

Adding the two equations together, we arriveat

(m1 + m2) a = m2 g ,

and hence

a =m2

m1 + m2

g

=4.94 kg

4.51 kg + 4.94 kg(9.8 m/s2)

= 5.12296 m/s2 .


026 (part 1 of 3) 10.0 pointsA block of mass 2.6792 kg lies on a friction-less table, pulled by another mass 4.97091 kgunder the influence of Earth’s gravity.


2.6792 kg

4.97091 kg

What is the magnitude of the net externalforce F acting on the two mass system con-nected by the string?Correct answer: 48.7149 N.

Explanation:

Given : m1 = 2.6792 kg and

m2 = 4.97091 kg .


m1 m2

a

T

N m1 g

a

T

m2 g

The net force on the system is simply theweight of m2.

Fnet = m2 g

= (4.97091 kg) (9.8 m/s2)

= 48.7149 N .

027 (part 2 of 3) 10.0 pointsWhat is the magnitude of the acceleration aof the two masses?Correct answer: 6.36787 m/s2.

Explanation:


Fnet = m2 g = (m1 + m2) a .

Solving for a,

a =m2

m1 + m2

g

=4.97091 kg

2.6792 kg + 4.97091 kg(9.8 m/s2)

= 6.36787 m/s2 .

028 (part 3 of 3) 10.0 pointsWhat is the magnitude of the tension T of therope between the two masses?Correct answer: 17.0608 N.

Explanation:

Analyzing the horizontal forces on blockm1, we have

∑

Fx : T = m1 a

= (2.6792 kg) (6.36787 m/s2)

= 17.0608 N .


029 (part 1 of 2) 10.0 pointsA mass of 2 kg lies on a frictionless table,

pulled by another mass of 4.6 kg under theinfluence of Earth’s gravity.


2 kg

4.6 kg

What is the magnitude of the accelerationa of the two masses?Correct answer: 6.8303 m/s2.

Explanation:

Let : m1 = 2 kg and

m2 = 4.6 kg .

m1 m2

a

T

N m1 g

a

T

m2 g

Let the direction of acceleration as indi-cated in the figure be positive. The net forceon the system is simply the weight of m2.

Fnet = m2 g .


Fnet = m2 g = (m1 + m2) a .

a =m2

m1 + m2

g

=4.6 kg

2 kg + 4.6 kg(9.8 m/s2)

= 6.8303 m/s2 .

030 (part 2 of 2) 10.0 pointsT , a and g represent positive quantities.

Which equation is correct?

1. (4.6 kg) g − T = (2 kg) a


2. T − (4.6 kg) g = (4.6 kg) a

3. T − (4.6 kg) g = (2 kg + 4.6 kg) a

4. T + (2 kg) g = (2 kg) a

5. T − (2 kg) g = (2 kg) a

6. T − (4.6 kg) g = (2 kg) a

7. T + (2 kg) g = (4.6 kg) a

8. (4.6 kg) g − T = (4.6 kg) a correct

9. (4.6 kg) g − T = (2 kg + 4.6 kg) a

10. T − (2 kg) g = (4.6 kg) a

Explanation:

Applying Newton’s second law (and thefree-body diagram above) to the mass m2

yields

Fnet = m2 g − T = m2 a

(4.6 kg) g − T = (4.6 kg) a

Static Friction and Pulley

031 10.0 pointsA 75 N block rests on a table. The sus-

pended mass has a weight of 24 N.

75 N

24 N

µs

What frictional force is required to keep theblocks from moving?Correct answer: 24 N.

Explanation:

Given : W1 = 75 N and

W2 = 24 N .

a

T

N W1µs N

a

T

W2

Consider the forces then the blocks wouldfirst start to move: For block W2 vertically,

Fy,net = T −W2 = 0

For block W1 horizontally,

Fx,net = T − µN = 0

µN = T = W1 = 24 N .

Accelerated Blocks

032 10.0 pointsTwo blocks are arranged at the ends of a mass-less string as shown in the figure. The systemstarts from rest. When the 2.12 kg mass hasfallen through 0.377 m, its downward speed is1.28 m/s.


2.12 kg

3.72 kg

µ

a

What is the frictional force between the3.72 kg mass and the table?Correct answer: 8.08601 N.

Explanation:

Given : m1 = 2.12 kg ,

m2 = 3.72 kg ,

v0 = 0 m/s , and

v = 1.28 m/s .

Basic Concept: Newton’s Second Law

F = M a


Solution: The acceleration of m1 is obtainedfrom the equation

v2 − v2

0 = 2 a (s − s0)

a =v2 − v2

0

2 h

=(1.28 m/s)2 − (0 m/s)2

2 (0.377 m)

= 2.17294 m/s2 .

Consider free body diagrams for the twomasses

T

m1 g

a

T

N

µN

a

m2 g

Because m1 and m2 are tied together withstring, they have same the speed and the sameacceleration, so the net force exerted on m2 is

F2 = m2 a

The net force on m1 is m1 a = m1 g − T , sothat T = m1 g − m1 a.

Thus

F2 = T − fk ,

fk = T − F2

= m1 g − (m1 + m2) a

= (2.12 kg) (9.8 m/s2)

− (2.12 kg + 3.72 kg)

× (2.17294 m/s2)

= 8.08601 N .

Falling Mass 01

033 (part 1 of 2) 10.0 pointsTwo blocks are arranged at the ends of a mass-less cord over a frictionless massless pulley asshown in the figure. Assume the system startsfrom rest. When the masses have moved a dis-tance of 0.374 m, their speed is 1.31 m/s.


3.6 kg

2.2 kg

µ

What is the coefficient of friction betweenm2 and the table?Correct answer: 0.233938.

Explanation:

Given : m1 = 2.2 kg ,

m2 = 3.6 kg ,

s = 0.374 m , and

v0 = 0 m/s .

Basic Concept: Newton’s Second Law

F = M a

Solution: The acceleration of m1 is obtainedfrom the equation

v2 − v2

0 = 2 a (s − s0)

a =v2 − v2

0

2 h

=(1.31 m/s)2 − (0 m/s)2

2 (0.374 m)

= 2.29425 m/s2 .

Consider free body diagrams for the twomasses

m2 m1

a

T

Nm2 g

a

T

m1 gµ m2 g

which leads to∑

F1y: m1 a = m1 g − T (1)

∑

F2x: m2 a = T − fk (2)

∑

F2y: N = m2 g , (3)


where T is the tension in the cord and fromEq. 3, fk ≡ µN = µ m2 g.

Because m1 and m2 are tied together with acord, they have same the speed and the sameacceleration. Adding Eqs. 1 & 2 we have

(m1 + m2) a = m1 g − fk = m1 g − µ m2 g

so that

µ m2 g = m1 g − (m1 + m2) a .

Thus

µ =m1

m2

− (m1 + m2)

m2

a

g

=2.2 kg

3.6 kg− 2.2 kg + 3.6 kg

3.6 kg

× 2.29425 m/s2

9.8 m/s2

= 0.233938 .

034 (part 2 of 2) 10.0 pointsWhat is the magnitude of the tension in thecord?Correct answer: 16.5126 N.

Explanation:

Using Eq. 1 the tension T is

T = m1 (g − a)

= (2.2 kg) (9.8 m/s2 − 2.29425 m/s2)

= 16.5126 N

or, using Eq. 2 and µ from Part 1, the tensionT is

T = m2 [a + µ g]

= (3.6 kg) [(2.29425 m/s2)

+ (0.233938) (9.8 m/s2)]

= 16.5126 N .

Since the T is the same using Eqs. 1 & 2: Part1, µ = 0.233938, is verified.

Accelerating Down a Plane 03

035 10.0 pointsA block accelerates 3 m/s2 down a plane in-clined at angle 27.0◦.


m

µk

3 m/s2

27◦

Find µk between the block and the inclinedplane.

1. 0.148

2. 0.166 correct

3. 1.289

4. 0.853

5. 0.510

Explanation:

Given : a = 3 m/s2 ,

θ = 27◦ , and

g = 9.81 m/s2 .

Consider the free body diagram for theblock

mg sinθ N = mg cos

θµN

a

mg

Basic Concepts:

~Fnet = m~a

Parallel to the ramp:

Fx,net = max = Fg,x − Fk

Fg,x = mg sin θ

Fk = µkFn


Perpendicular to the ramp:

Fy,net = Fn − Fg,y = 0

Fg,y = mg cos θ

Solution:

Consider the forces parallel to the ramp:

Fk = Fg,x − max

= mg sin θ − max

Consider the forces perpendicular to theramp:

Fn = Fg,y = mg cos θ

Thus the coefficient of friction is

µk =Fk

Fn

=mg sin θ − ma

mg cos θ

=g sin θ − a

g cos θ

=9.81 m/s2 sin 27◦ − 3 m/s2

9.81 m/s2 cos 27◦

= 0.166306 .

Accelerating Down a Plane

036 10.0 pointsA block accelerates 3 m/s2 down a plane in-clined at angle 27.0◦.


m

µk

3 m/s2

27◦

Find µk between the block and the inclinedplane.

Correct answer: 0.166306.

Explanation:

Given : a = 3 m/s2 ,

θ = 27◦ , and

g = 9.81 m/s2 .


mg sinθ N = mg cos

θµN

a

mg

Basic Concepts:

~Fnet = m~a



Fg,x = mg sin θ

Fk = µkFn



Fg,y = mg cos θ

Solution:


Fk = Fg,x − max

= mg sin θ − max




µk =Fk

Fn

=mg sin θ − ma

mg cos θ

=g sin θ − a

g cos θ

=9.81 m/s2 sin 27◦ − 3 m/s2

9.81 m/s2 cos 27◦

= 0.166306 .


Block on Incline 01

037 (part 1 of 2) 10.0 pointsYour teacher placed a 2.5 kg block at the posi-tion marked with a “ +” (horizontally, 0.5 mfrom the origin) on a large incline outlined onthe graph below and let it slide, starting fromrest.

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

10

Ver

tica

lH

eight,

y(m

)

Horizontal Distance, x (m)

Figure: Drawn to scale.Calculate the speed of the block when it

reaches the bottom edge (right-hand edge) ofthe incline. The coefficient of kinetic frictionfor the block on the incline is 0.83 and theacceleration of gravity is 9.8 m/s2 .Correct answer: 4.49856 N.

Explanation:

Let : µ = 0.83 ,

xb = 4 m ,

yb = 4 m ,

xt = 0 m ,

yt = 8.5 m ,

θ = arctanyt − yb

xb − xt,

= arctan8.5 m − 4 m

4 m − 0 m,

= 48.3665◦ , and

x = 0.5 m .

The graph below shows the trajectory of

the block while it is on the incline and after itslides off the incline where it finally hits thefloor.

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

10

Ver

tica

lH

eight,

y(m

)

Horizontal Distance, x (m)

θ

The free-body diagram when the block ison the incline is

mg

mg sin

θµm

g cos θ

mgcos θ

N

y = yt −yt − yb

xb

= 8.5 m − 8.5 m − 4 m

4 m= 7.9375 m .

a = g (sin θ − µ cos θ)

= (9.8 m/s2)[

sin 48.3665◦

− (0.83) cos 48.3665◦]

= 1.92068 m/s2 .

xd = xb − x = 4 m − 0.5 m = 3.5 m .


yd = y − yb = 7.9375 m − 4 m = 3.9375 m .

d =√

x2

d + y2

d

=√

(3.5 m)2 + (3.9375 m)2

= 5.2682 m .

vb =√

2 a d

=√

2 (1.92068 m/s2) × (5.2682 m)

= 4.49856 m/s .

038 (part 2 of 2) 10.0 pointsCalculate the horizontal distance from thebottom edge (right-hand edge) of the inclineto where the block hits the floor (i.e., y = 0).Correct answer: 1.86304 m/s.

Explanation:

vy = −vb sin θ

= −(4.49856 m/s) sin 48.3665◦

= −3.36226 m/s .

vx = vb cos θ

= (4.49856 m/s) cos 48.3665◦

= 2.98868 m/s .

∆y = yb = 4 m

= vy t +1

2g t2 , so

g

2t2 + vy t − ∆y = 0 and

v2

y + 2 g ∆y = (−3.36226 m/s)2

+ 2 (9.8 m/s2) (4 m)

= 89.7048 m2/s2,

t =1

g

(

vy ±√

v2y + 2 g ∆y

)

=−3.36226 m/s +

√

89.7048 m2/s2

9.8 m/s2

= 0.623367 s .

∆x = vx t

= (2.98868 m/s) (0.623367 s)

= 1.86304 m .

Block Sliding Down 01

039 (part 1 of 2) 10.0 pointsHint: This problem requires a train of logic.

(1) Analyze force diagram,(2) use Newton’s Laws, and(3) solve the equations of motion.

A block starts from rest at a height of 8.8 mon a fixed inclined plane.


7 kg

µ = 0.27

25◦

What is the speed of the block at the bot-tom of the ramp?Correct answer: 8.52122 m/s.

Explanation:

Let : h = 8.8 m ,

m = 7 kg ,

µ = 0.27 ,

θ = 25◦ , and

vf = final speed .


Ff

N

mg

25◦

dh 25◦

The normal force to the inclined plane isN = mg cos θ. The sum of the forces par-allel to the inclined plane is

Fnet = ma = mg sin θ − µ mg cos θ

a = g sin θ − µ g cos θ

Sincev2

f = v2

0 + 2 a x = 2 a d (1)

along the plane, and neglecting the dimensionof the block, the distance, d, to the end of theramp is d sin θ = h

d =h

sin θ(2)

therefore

vf =

√

2 a h

sin θ=

√

2 g h (sin θ − µ cos θ)

sin θ

=√

2 g h (1 − µ cot θ) (3)

=√

2 (9.8 m/s2) (8.8 m) [1 − (0.27) cot 25◦]

= 8.52122 m/s .

040 (part 2 of 2) 10.0 pointsIf the block continues to slide on the groundwith the same coefficient of friction, how farwill the block slide on the ground until comingto rest?Correct answer: 13.7209 m.

Explanation:

Ff

N

mg

Using Newton’s Law,

ma = ff = µ mg

a = µ g

If v0 is speed of block when it leaves incline,and as the final speed vf = 0, we have

v2

f = v2

0 + 2 a x

v2

0 = −2 a x = −2 µ g x

then

x =−v2

0

2 g µ

|x| =(8.52122 m/s)2

2 (9.8 m/s2) (0.27)= 13.7209 m .

Friction on an Incline 02

041 (part 1 of 2) 10.0 pointsA block is at rest on an inclined plane.

80kg

µs= 0.4

8θc

Find the critical angle, θc, at which theblock just begins to slide.Correct answer: 25.641◦.

Explanation:

Basic Concepts: Friction: fs ≤ µsN ,fk = µkN

Let : m = 80 kg ,

µs = 0.48 ,

θc = 25.641◦ , and

vf = final speed .


Ff

N

mg

25.641◦

Solution: At the critical angle θc, the mag-nitude of the static frictional force attains itsmaximum value µs N . Since the block is atrest, the component of the gravitational forceparallel to the incline is equal in magnitude tothe frictional force

mg sin θc = fs = µs N . (1)

Also the component of the gravitational forceperpendicular to the incline is equal in mag-nitude to the normal force

N = mg cos θc . (2)

Combining (1) and (2) yields

mg sin θc = µs mg cos θc ,

ortan θc = µs

θc = tan−1 µs

= arctan(0.48)

= 25.641◦ .

042 (part 2 of 2) 10.0 pointsFind the magnitude of the static frictionalforce when θ < θc.

1. fs = 0

2. fs = mg

3. fs = mg cos θ

4. fs =1

2µs mg sin θ

5. fs =1

2µs mg cos θ

6. fs =1

2mg cos θ

7. fs = mg sin θ correct

8. fs =1

2mg sin θ

9. fs = µs mg cos θ

10. fs = µs mg sin θ

Explanation:

Below the critical angle, the block is stillat rest. So by Newton’s second law, it stillremains true that the parallel component ofthe gravitational force is equal in magnitudeto the static frictional force

fs = mg sin θ .

Holt SF 04D 03

043 (part 1 of 2) 10.0 pointsA 76.0 kg box slides down a 24.0◦ ramp withan acceleration of 3.10 m/s2.


76kg

µk

3.1m/s

2

24◦

a) Find µk between the box and the ramp.Correct answer: 0.0993192.

Explanation:

Given : m = 76 kg ,

a = 3.1 m/s2 ,

θ = 24◦ , and

g = 9.81 m/s2 .



mg sinθ N = mg cos

θµN

a

mg

Basic Concepts:

~Fnet = m~a



Fg,x = mg sin θ

Fk = µkFn



Fg,y = mg cos θ

Solution:


Fk = Fg,x − max

= mg sin θ − max

= (76 kg)(9.81 m/s2) sin 24◦

− (76 kg)(3.1 m/s2)

= 67.6466 N



= (76 kg)(9.81 m/s2) cos 24◦

= 681.103 N


µk =Fk

Fn=

67.6466 N

681.103 N= 0.0993192 .

044 (part 2 of 2) 10.0 pointsb) What acceleration would a 161 kg masshave down this ramp?Correct answer: 3.1 m/s2.

Explanation:

Basic Concepts: Parallel to the ramp:

Fx,net = max = Fg,x − Fk = mg sin θ − µkFn



Fg,y = mg cos θ

Given : m = 161 kg andµk= 0.0993192 .

Solution:

Fx,net = mg sin θ − µkmg cos θ

= (161 kg)(9.81 m/s2) sin 24◦

− (0.0993192)(161 kg)

· (9.81 m/s2) cos 24◦

= 499.1 N

Thus the acceleration is

ax =Fx,net

m=

499.1 N

161 kg= 3.1 m/s2

down the ramp.

Pushing a Block Upward 02

045 (part 1 of 2) 10.0 pointsAs shown in the figure, a block is pushed upagainst the wall. Let the mass of the block bem = 2.2 kg, the coefficient of kinetic frictionbetween the block and the wall be µ = 0.66,and θ = 48◦. Suppose F = 67 N.


2.2 kg

F

48◦

µk

=0.6

6

Find the force of friction.Correct answer: 29.5889 N.

Explanation:


Recall that

f = µ N .

From∑

F = 0 in the horizontal direction,

one can see that

N = F cos θ .

Hence the force of friction is

f = µ N = µ F cos θ = 29.5889 N .

046 (part 2 of 2) 10.0 pointsThe force, F , which keeps the block mov-ing upwards with a constant velocity satisfieswhich equations?

1. F cos θ = mg + µ F sin θ

2. F cos θ = mg + µ F cos θ

3. F sin θ = mg + µ F sin θ

4. F sin θ = µ mg

5. F sin θ = mg − µ F cos θ

6. F cos θ = µ mg

7. F cos θ = mg − µ F sin θ

8. F cos θ = mg − µ F cos θ

9. F sin θ = mg − µ F sin θ

10. F sin θ = mg + µ F cos θ correct

Explanation:

m

F

θ

v

mgfµ

For constant velocity, acceleration is zero.Hence

∑

Fy = 0 = F sin θ − mg − µ F cos θ .

The first term is the applied upward force, thesecond term is the weight of the block, andthe third term is the frictional force.

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AP-Newton's 3rd Law & Friction-Preview[1]