danny/EE1427_SOL.pdfsin 7.5 kg 1.41 9.81 m/s sin13 cos cos13 28 N F F mg max ma mg F F θ θ θ θ = − = + + ° = = ° = ∑ Insight: They’d better offer double coupons at this

EE1427 Engineering Science – Dr. Daniel Nankoo Tutorial Solutions

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Solutions to Tutorial 1

1. Picture the Problem: The free body diagram for this problem is shown on

the right.

Strategy: Use the free body diagram to determine the net force on the rock,

then apply Newton’s Second Law to find the acceleration of the rock. Let

upward be the positive direction.

Solution: 1. Find the net

force: ( ) ( ) ( )ˆ ˆ ˆ40.0 N 46.2 N 6.2 N= − + =∑F y y y

r

2. Now apply Newton’s

second law to find ar

: ( )

( )2ˆ6.2 N

ˆ1.2 m/s5.00 kgm

= = =∑F y

a y

r

r

Insight: If the astronaut were to exert less than 40.0 N of upward force on the rock, it would

accelerate downward.

2. Picture the Problem: The skier is accelerated along a straight line.

Strategy: Use a constant acceleration equation of motion to determine the acceleration of the skier,

then use Newton’s Second Law to find the force exerted on the skier.

Solution: 1. Find the acceleration of

the skier from

his final velocity and the distance

covered:

2 2

0

2

v va

x

−=

∆

2. Find the net force on the skier

using: ( )

( )( )

22 2

012 m/s 0

92 kg 260 N2 2 25 m 0

v vF ma m

x

− −= = = =

∆ −

Insight: This moderate force accelerates the skier at a rate of 2.9 m/s2 or 0.29g.

3. Picture the Problem: The parachutist is moving straight downward, slowing down and coming to

rest over a distance of 0.750 m.

Strategy: Use a constant acceleration equation of motion to determine the acceleration of the

parachutist, and then use Newton’s Second Law to find the net force on her.

Solution: 1. (a) Find the acceleration

of the parachutist: ( )

2 2

0

22 2

0 2

2

0 3.85 m/sˆ ˆ ˆ9.88 m/s

2 2(0 0.750 m)

y y

y y

v v a y

v v

y

= + ∆

− − −= = =

∆ −a y y yr

2. Find the net force: ( )( ) ( )2 ˆ ˆ42.0 kg 9.88 m/s 415 Nm= = =∑F a y yr r

3. (b) If the parachutist comes to rest in a shorter distance, the acceleration will be greater and the

force will therefore be greater.

Insight: There are two other forces on the parachutist, the upward force of her parachute and the

downward force of gravity. Those two forces must cancel if she is moving downward at constant

speed (no acceleration), so in this case the net force also equals the force the ground exerts on her.

Fr

Wr


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4. Picture the Problem: The 747 is accelerated horizontally in the direction opposite its motion in

order to slow it down from its initial speed of 27.0 m/s.

Strategy: Find the acceleration from the known force and mass, then find the speed and distance

traveled.

Solution: 1. (a) Find a

r:

( )( )

5

2

5

ˆ– 4.30 10 Nˆ1.23 m/s

3.50 10 kgm

×= = = −

×

xFa x

rr

2. Find the final speed: ( )( )20 27.0 m/s 1.23 m/s 7.50 s 17.8 m/sv v at= + = + − =

3. (b) Find the distance

travelled by the 747 as it slows down: ( ) ( )( )1 1

02 227.0 17.8 m/s 7.50 s 168 mx v v t∆ = + ∆ = + =

Insight: The landing speed of a Boeing 747-200 is 71.9 m/s (161 mi/h) and it has a specified landing roll distance of 2,121 m, requiring an average landing acceleration of −1.22 m/s

2.

5. Picture the Problem: The free body diagram of the brick is shown at right.

Strategy: Use the vectors depicted in the free body diagram to answer the

questions.

Solution: 1. (a) There are two forces acting on the brick.

2. (b) The forces acting on the brick are due to gravity and your hand.

3. (c) Yes, these forces are equal and opposite, because the brick remains at rest:

0 so that m= + = = = −∑F F W a F Wr r r r rr

4. (d) No, these forces are not an action-reaction pair, because they are acting on

the same object.

Insight: Action–reaction pairs always act on different objects, and they can therefore never

“cancel” each other.

6. Picture the Problem: The free body diagram for the

car is shown at right. For the trailer there is only one

force acting on it in the forward direction, the force

exerted by the car on the trailer.

Strategy: In order to determine the forces acting on an

object, you must consider only the forces acting on that object and the motion of that object alone. For the

trailer there is only one force 1Fr

exerted on it by the

car, and it has the same acceleration (1.90 m/s2) as the

car. For the car there are two forces acting on it, the

engine 2Fr

and the trailer 1−Fr

. Apply Newton’s second

and third laws as appropriate to find the requested

forces.

Solution: 1. (a) Write Newton’s Second Law for the trailer:

( )( ) ( )2

1ˆ ˆ540 kg 1.90 m/s 1.0 kNm= = = =∑F F a x x

r r r

2. (b) Newton’s Third Law states that the

force the trailer

exerts on the car is equal and opposite to

the force the car

exerts on the trailer:

( )1ˆ1.0 kN− = −F x

r

3. (c) Write Newton’s Second Law for the

car: ( )( ) ( )2 ˆ ˆ1300 kg 1.90 m/s 2.5 kNM= = =∑F a x x

r r

Insight: The engine force 2Fr

must be ( ) ˆ3.5 kN x because it must both balance the 1.0-kN force

from the trailer and accelerate the car in the forwards, requiring an additional 2.5 kN of force.

1Fr

− 1Fr

2Fr

Fr

Wr


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7. Picture the Problem: The force pushes on box 1 in

the manner indicated by the figure at right.

Strategy: The boxes must each have the same

acceleration, but because they have different masses

the net force on each must be different. These

observations allow you to use Newton’s Second

Law for each individual box to determine the

magnitudes of the contact forces. First find the

acceleration of all the boxes and then apply

equation 5-1 to find the contact forces.

Solution: 1. (a) The 7.50 N force

accelerates all the boxes together: 1 2 3

1 2 3

2

( )

7.50 N0.798 m/s

1.30 kg 3.20 kg 4.90 kg

FF m m m a a

m m m

a

= + + ⇒ =+ +

= =+ +

2. Write Newton’s Second Law for

the first box: ( )( )

c12 1

2

c12 17.50 N 1.30 kg 0.798 m/s 6.46 N

F F m a

F F m a

= − =

= − = − =

∑Fr

3. (b) Write Newton’s Second Law

for the third box: ( )( )2

c23 3 4.90 kg 0.798 m/s 3.91 NF m a= = = =∑Fr

Insight: Another way to solve the problem is to note that c12F accelerates boxes 2 and 3:

( )( )2

c12 3.20 4.90 kg 0.798 m/s 6.46 NF = + = , the same result as in step 2.

8. Picture the Problem: The light box of mass m sits

adjacent to the heavy box of mass M as depicted in the

figure at right.

Strategy: The boxes must each have the same

acceleration, but because they have different masses

the net force on each must be different. These

observations allow you to use Newton’s Second Law

for each individual box to determine the magnitudes of

the contact forces. First find the acceleration of both

boxes and then find the contact forces.

Solution: 1. (a) The 7.50 N

force accelerates all the

boxes together:

( ) 25.0 N 0.40 m/s

5.2 7.4 kg

FF m M a a

m M= + ⇒ = = =

+ +

2. Find the contact force by

writing Newton’s Second

Law for the heavy box

only:

( ) ( )2

c12 7.4 kg 0.40 m/s 3.0 NF M a= = = =∑Fr

3. (b) When the force is applied to the heavier box the contact force between them will be less than

it was before , because the lighter box requires less force for the same acceleration, and the contact

force is the only force on the lighter box.

4. (c) Find the contact force by

writing Newton’s Second Law

for the lighter box only:

( )( )2

c12 5.2 kg 0.40 m/s 2.1 NF m a= = = =∑Fr

Insight: Another way to view the answer to (b) is to say the inertia of the heavier box shields the

lighter box from experiencing some of the pushing force. In case (a) the lighter box provides less

shielding and the contact force is greater.


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9. Picture the Problem: The trailer is pulled up the incline at

constant speed. The free body diagram of the trailer is

depicted in the diagram on the right:

Strategy: Because the trailer is moving at constant speed,

the net force on the trailer must be zero, and the force

exerted by the tractor on the trailer must equal the

component of the trailer’s weight that is pointing down the

incline and parallel to it.

Solution: Write

Newton’s Second

Law in the x

direction for the

trailer:

( )( )2

sin 0

sin 3900 kg 9.81 m/s sin16 11,000 N 11 kN

xF F mg ma

F mg

θ

θ

= + − = =

= = ° = =

∑

Insight: The steeper the hill, the larger the force the tractor must exert. If the tractor were pulling

straight upward

(θ = 90°) it would have to support the entire 38-kN weight. On level ground with no friction, the

required force is zero.

10. Picture the Problem: The trolley is pushed partly

into the incline and partly up the incline by the

pushing force Fr

, as shown in the figure at right.

Strategy: Write Newton’s Second Law for the x

direction, where x points up the incline and parallel

to it. Solve the resulting equation for the magnitude

of Fr

.

Solution: The component of the

force pushing up the incline is

cosF θ and the component of the

weight pushing down the incline is

sinmg θ :

( ) ( )2

cos sin

7.5 kg 1.41 9.81 m/s sin13sin

cos cos13

28 N

xF F mg ma

ma mgF

F

θ θ

θ

θ

= − =

+ °+ = =°

=

∑

Insight: They’d better offer double coupons at this store, because a 13° incline is a 23% slope;

danger territory for over-the-road truckers and a lot of extra work for the average grocery shopper!

11. Picture the Problem: The two men pull on the barge

in the directions indicated by the figure at right.

Strategy: Place the x-axis along the forward direction

of the boat. Use the vector sum of the forces to find

the force F such that the net force in the

y-direction is zero.

Solution: Set the sum

of

the forces

in the y

direction

equal to

zero:

( )

( )

130 N sin 34 sin 45 0

sin 34130 N 100 N 0.10 kN

sin 45

yF F

F

= − ° + ° =

°= = =

°

∑

Insight: The second crewman doesn’t have to pull as hard as the first because a larger component

of his force is pulling in the y direction. However, his force in the forward direction (73 N) is not as

large as the first crewman (110 N).

y

x Fr

θ mgr

sinmg θ


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12

.

Picture the Problem: The two teenagers pull on the

sled in the directions indicated by the figure at right.

Strategy: Write Newton’s Second Law in the x

direction (parallel to ar

) in order to find the

acceleration of the sled.

Solution: Write

Newton’

s Second

Law

in the x

direction

:

( )

( )

sled child

sled child

2

2 cos35 57 N

2 cos35 57 N

2 55 N cos35 57 N1.5 m/s

19 3.7 kg

x x

x

x

F F m m a

Fa

m m

a

= ° − = +

° −=

+

° −= =

+

∑

Insight: Some of the force exerted by the teenagers is exerted in the y direction and cancels out; only the

x components of the forces move the sled.

13. Picture the Problem: The skier glides down the incline

without friction

Strategy: Let the y direction point perpendicular to the

snow trail and the x direction point parallel to the slope

and downhill. Since the skier is not accelerating in the y

direction, the two forces in that direction, N and yW ,

cancel each other. The net force on the skier is then the

only unbalanced force, xW . Fnd the net force on the

skier.

Solution: 1. (a) The

net force

on the skier is xW :

( )( )2

sin

65 kg 9.81 m/s sin 22

0.24 kN

x x

x

F W mg

F

θ= =

= °

=

∑

∑

The force points downhill, parallel to the slope.

2. (b) As the slope becomes steeper, θ increases and the net force xW increases.

Insight: If θ were to increase to 90°, the skier would be falling straight downward and the force in

him would be maximum. If θ were to decrease to 0°, there would be no net force on the skier and

no acceleration.

14. Picture the Problem: The force vectors acting on the Moon are

depicted on the right:

Strategy: Find the vector sum of the forces by using the

component method, and use the components to find the magnitude

and direction of the net force.

Solution: 1. (a) Add

the

force vectors: ( ) ( )SM EM

20 20ˆ ˆ4.34 10 N 1.98 10 N

= +

= × + ×

∑F F F

x y

r r r

2. Find the direction of

the

net force:

( )

( )

201 1

20

1.98 10 Ntan tan 24.5 above the + axis

4.34 10 N

y

x

xθ − −

× = = = ° ×

∑

∑

F

F

r

r

3. (b) Find the

magnitude

of the net force:

( ) ( )2 2

20 20 204.34 10 N 1.98 10 N 4.77 10 N= × + × = ×∑Fr


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4. (c) Find the

acceleration

from Newton’s Second

Law:

202

22

4.77 10 N0.00649 m/s

7.35 10 kgm

×= = =

×

∑Fa

r

r

Insight: It was a calculation of the acceleration of the Moon that led Isaac Newton to conclude the

same force that causes an apple to fall is the force that keeps the Moon in its orbit.

15. Picture the Problem: The astronaut is accelerated straight upward by the force of the rocket

engine.

Strategy: There are two forces acting on the astronaut: the applied force Fr

of the seat acting

upward and the force of gravity Wr

acting downward. The force F represents the apparent weight

of the astronaut since that is both the force the seat exerts on him and the force he exerts on the

seat. Use Newton’s Second Law together with the known acceleration to determine F.

Solution: Write out Newton’s

Second Law

in the vertical direction and solve

for F :

( ) ( )( )2

a

a

88 kg 30.2 9.81 m/s

3500 N 3.5 kN

yF F mg ma

W F ma mg m a g

W

= − =

= = + = + = +

= =

∑

Insight: The astronaut’s weight on Earth is 0.86 kN, so this astronaut is experiencing 3.5/0.86 =

4.1g’s of acceleration.

16. Picture the Problem: The elevator accelerates up and down,

changing your apparent weight Wa. A free body diagram of the

situation is depicted at right.

Strategy: There are two forces acting on you: the applied force

a=F Wr r

of the scale acting upward and the force of gravity Wr

acting downward. The force Wa represents your apparent

weight because it is both the force the scale exerts on you and

the force you exert on the scale. Use Newton’s Second Law

together with the known force Wa acceleration to determine the

acceleration a.

Solution: 1. (a) The direction of acceleration is upward. An

upward acceleration results in an apparent weight greater than

the actual weight.

2. (b) Use Newton’s Second

Law together

with the known forces to

determine the acceleration a.

( )

a

2 2a a 730 610 N9.81 m/s 1.9 m/s

610 N

yF W W ma

W W W Wa

m W g

= − =

− − −= = = =

∑

3. (c) The only thing we can say about the velocity is that it is changing in the upward direction.

That means

the elevator is either speeding up if it is travelling upward, or slowing down if it is travelling

downward.

Insight: You feel the effects of apparent weight twice for each ride in an elevator, once as it

accelerates from rest and again when it slows down and comes to rest.

17. Picture the Problem: The bowling ball is lifted straight upward by the applied force.

Strategy: There are two forces acting on the bowling ball: the applied force Fr

acting upward and

the force of gravity Wr

acting downward. Write Newton’s Second Law for each case in order to

obtain two equations with two unknowns, and then use algebra to find W and a.


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Solution: 1. (a) Write Newton’s Second

Law for each

case to obtain two equations and two

unknowns:

,1 1

,2 2 2

y

y

F F mg ma

F F mg ma

= − =

= − =

∑∑

2. Divide the second equation by the first

and solve for m:

( ) ( )

22 1

1

1 2

2

22 2 2

2 82 N 92 N27.3 kg

9.81 m/s

F mg maF mg F mg

F mg ma

F Fm

g

−= = ⇒ − = −

−

−−= = =

3. Fnd the weight W: ( )( )27.3 kg 9.81 m/s 72 NW mg= = =

4. (b) Use the first Newton’s Second Law

equation to find a:

21 82 72 N1.4 m/s

7.3 kg

F mga

m

− −= = =

Insight: In the first case, 72 N of force supported the weight of the ball and 10 N accelerated it at a

rate of 1.4 m/s2. In the second case, 20 N of net force accelerated the ball at a rate of 2.8 m/s

2.

18. Picture the Problem: The free body diagram of the suitcase is

shown at right:

Strategy: Write Newton’s Second Law in the vertical direction

to determine the magnitude of the applied force F.

Solution: Use

Newton’s

Second Law in

the vertical

direction to find

F:

( )( )2

sin 0

23 kg 9.81 m/s 180 N

sin sin 25

110 N 0.11 kN

yF N W F

W NF

F

θ

θ

= − + =

−−= =

°

= =

∑

Insight: If the upward force applied by the handle were zero,

the normal force N would equal the weight W of 230 N.

19. Picture the Problem: The child sits on a chair and the

chair sits on the floor. The free body diagrams of the child

and the chair are shown at right.

Strategy: There are two forces acting on the child: the

normal force Nr

of the chair acting upward and the force of

gravity childW

r acting downward. There are three forces

acting on the chair: the normal force Nr

of the floor acting

upward, the weight of the baby acting downward, and the

force of gravity chairW

r acting downward. Write Newton’s

Second Law in the vertical direction for each case and then

use the equations to find N.

Solution: 1. (a) Write

out Newton’s

Second Law in the

vertical direction

and solve for N:

( )( )2

0

9.3 kg 9.81 m/s

91 N

yF N mg

N mg

= − =

= =

=

∑

(a) child (b) chair


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2. (b) Write out

Newton’s

Second Law in the

vertical direction

for the chair and solve

for N:

( )( )( )

baby chair

baby chair

2

g 0

9.3 3.7 kg 9.81 m/s

130 N 0.13 kN

yF N m m g

N m m g

N

= − − =

= +

= +

= =

∑

Insight: The normal force is larger in case (b) because the floor must support the weight of the

child plus the weight of the chair, whereas the chair must only support the weight of the child.

20. Picture the Problem: The free body diagram of the potatoes is shown on

the right:

Strategy: There are two forces acting on the potatoes: the normal force Nr

of the shopping cart acting upward and the force of gravity Wr

acting

downward.

Solution: 1. (a) The free body diagram for the potatoes is shown at right.

2. (b) The free-body diagram does not change. A constant velocity implies

zero acceleration and therefore zero net force on the potatoes.

Insight: Note that as far as Newton’s Second Law is concerned, zero

velocity is no different than constant, nonzero velocity. This is the essence

of Newton’s First Law.

21. Picture the Problem: The ant walks along the surface of the

bowling ball until the normal force between it and the ball becomes

too small.

Strategy: Write Newton’s Second Law for the ant along the

radial direction, remembering that the ant’s weight always acts

straight downward. The ant walks slowly enough that there is no

acceleration in any direction.

Solution: 1. Write Newton’s

Second Law

for the radial direction:

cos 0rF N mg θ= − =∑

2. Set N equal to half the weight

and

solve for θ: ( )

1

2

1 1

2

cos 0

cos 60

mg mg θ

θ −

− =

= = °

Insight: The normal force itself will not be zero until θ = 90°, at

which point nothing will stop the ant’s fall unless it has sticky feet!

22. Picture the Problem: The child slides down the incline that is tilted

31.0° above the horizontal. A free-body diagram of the situation is

depicted at right:

Strategy: Choose the x-axis along the direction of motion. Write

Newton’s Second Law in the y direction to find the normal force,

and then write Newton’s Second Law in the x direction and solve

for µk.

Solution: 1. Write

Newton’s

Second Law in the y

direction:

cos 0

cos

yF N mg

N mg

θ

θ

= − =

=

∑

2. Now write Newton’s

Second

Law in the x direction: ( )

k

k

k

sin

sin

sin cos

yF mg N ma

mg ma N

mg ma mg

θ µ

θ µ

θ µ θ

= − =

− =

− =

∑


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3. Solve the equation for

µk:

( )( )

2 2

k 2

9.81 m/s sin 31.0 1.16 m/ssin0.463

cos 9.81 m/s cos31.0

g a

g

θµ

θ

° −−= = =

°

Insight: The coefficient of friction between the child and the slide depends upon many factors,

including the clothing the child wears, the material from which the slide is constructed, and

whether the slide is wet or not.

23. Picture the Problem: The book slides in a straight line across the top of the tabletop.

Strategy: The minimum force required to get the book moving is related to the maximum

coefficient of static friction, and the force required to keep the book sliding at constant speed is

equal to the magnitude of the kinetic friction force, from which µk can be determined.

Solution: 1. When the book begins

sliding, the applied force equals the

maximum static friction force: ( )( )

app s s

app

s 2

2.25 N0.127

1.80 kg 9.81 m/s

F f mg

F

mg

µ

µ

= =

= = =

2. When the book is sliding at constant

speed, the

applied force equals the kinetic friction

force: ( )( )

app k k

app

k 2

1.50 N0.0849

1.80 kg 9.81 m/s

F f mg

F

mg

µ

µ

= =

= = =

Insight: The coefficient of kinetic friction is usually smaller than the coefficient of static friction.

This is the basic idea behind antilock brakes, which seek to keep the tire of a car rolling so that the

friction between the tire and the road remains in the static regime, where there is a greater force to

stop the car and improved handling during the stop.

24. Picture the Problem: The free-body diagram of this

situation is depicted at right.

Strategy: Write Newton’s Second Law in the vertical

direction to determine the normal force of the floor on the

crate. Then write Newton’s Second Law in the horizontal

direction to determine the minimum force necessary to

start the crate moving.

Solution: 1. Write

Newton’s

Second Law in the

vertical

direction:

sin 0

sin

y yF N mg F ma

N mg F

θ

θ

= − − = =

= +

∑

2. Write Newton’s

Second

Law in the horizontal

direction:

s

s

cos 0

cos

x xF F N ma

F N

θ µ

θ µ

= − = =

=

∑

3. Now substitute for

N and solve

for the maximum

force F that

would produce no

acceleration:

( )

( )

( )

( )( )( )( )

s

s s

2

s

s

cos sin

cos sin

0.57 32 kg 9.81 m/s

cos sin cos 21 0.57 sin 21

250 N 0.25 kN

F mg F

F mg

mgF

F

θ µ θ

θ µ θ µ

µ

θ µ θ

= +

− =

= =− ° − °

= =

Insight: The required pushing force is 250N and represents about 78% of the crate’s weight. If the

person simply pushed horizontally they would need only 180 N of force.

θ

sfr

Wr

Fr

Nr


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25. Picture the Problem: The coffee cup is accelerated in a straight line due to the static friction

between it and the roof of the car.

Strategy: The static friction between the coffee cup and the roof of the car provides the forward

force needed to accelerate the coffee cup. First write Newton’s Second Law to find the maximum

acceleration of the car, and then find the smallest amount of time in which the car can accelerate to

15 m/s.

Solution: 1. (a) Write Newton’s

Second Law for the coffee cup: s sxF f mg maµ= = =∑

2. Solve the equation for a: ( )( )2 2 2

s 0.24 9.81 m/s 2.35 m/s 2.4 m/sa gµ= = = =

3. (b) Find the minimum time: 0

2

15 m/s 06.4 s

2.35 m/s

v vt

a

− −= = =

Insight: If the person owned a Ferrari 575M Maranello capable of going from zero to 60 mi/h in

4.2 seconds (6.4 m/s2), she would need a coefficient of friction of 0.65 to prevent the cup from

slipping. Not likely, given the smooth aerodynamic finish of the sports car!

26. Picture the Problem: Your car travels in a straight line, slowing down from its initial speed v and

skidding to a stop due to the kinetic friction between the tires and the road.

Strategy: Use Newton’s Second Law to find the acceleration of the car, and then use equation 2-

12 to find the distance the car travels during the time it is slowing down.

Solution: 1. (a) Solve Newton’s Second

Law

for the acceleration of the car:

xF N mg ma

a g

µ µ

µ

= − = − =

= −∑

2. Find the travel

distance of the car: ( )

2 2 2 2 2

f 0

2 2 2

v v v vd x

a g gµ µ

− −= ∆ = = =

−

3. (b) The stopping distance depends upon the square of the initial speed, so if your speed is

doubled the stopping distance quadruples.

4. (c) If the truck has the same initial speed and coefficient of kinetic friction, its stopping distance

will be the same as your car’s stopping distance.

Insight: The stopping distance of trucks is greater than cars, but not because the coefficient of

friction for truck tires is any different. It is mainly due to the braking system of the truck, as the

brakes must safely dissipate a much larger amount of kinetic energy than car brakes. If the car and

truck each skid, the stopping distance is the same.

27. Picture the Problem: The free-body diagram

for the person in the hammock is depicted at

right.

Strategy: There are three forces acting on the

person, the two tensions and gravity. The

horizontal components of the tensions cancel

out, and the vertical components of the

tensions must balance the downward weight.

Write Newton’s Second Law in the vertical

direction to find the tension.

Solution: Write Newton’s Second Law

in the vertical direction and solve for the

tension T: ( )( )2

2 sin 0

50.0 kg 9.81 m/s948 N

2sin 2sin15.0

yF T mg

mgT

θ

θ

= − =

= = =°

∑

Insight: Notice that the tension in each rope is almost double the person’s 491 N weight. That’s

because only a small portion (sin 15.0° = 25.9%) of the tension is supporting the person’s weight.

Tr

Tr

θ θ

Wr

cosT θ

sinT θ


11 of 89

28. Picture the Problem: The spring stretches while the backpack remains at rest.

Strategy: There are two forces acting on the backpack, the spring force to the left and the friction

force to the right. The magnitudes of these two forces must be equal for the backpack to remain at

rest. Use Newton’s Second Law to relate the magnitudes of the forces and Hooke’s Law (equation

6-4) to find the spring force and hence the friction force.


Newton’s Second Law in the

horizontal direction:

0x s xF f F ma= − = =∑

2. Use Hooke’s Law to find the

spring force and solve for the

friction force. Note that the spring

stretches in the negative x direction.

( )( )150 N/m 0.0200 m 3.0 Nsf F kx= = − = − − =

3. (b) No, the answer to (a) doesn’t depend on the mass of the backpack, it only depends upon the

spring force and the fact that the backpack remains at rest.

Insight: The static friction force changes with the applied force, always matching its magnitude

until the maximum friction force is reached. If the applied force is increased even more, the

backpack will slide.

29. Picture the Problem: The spring is capable of being either stretched or compressed by the

external force.

Strategy: Use Hooke’s law to find the force required to stretch the spring to a length that is twice

its equilibrium length. In that case the stretch distance is 2x L L L= − = . Then use Hooke’s law

again to find the force required to compress the spring an amount equal to 1 1

2 2x L L L= − = − .

Solution: 1. (a) Use Hooke’s Law to

find the magnitude of the force required

to stretch the spring by the amount

x L= :

( )( )250 N/m 0.18 m 45 NF k x k L= − = = =

2. (b) Use Hooke’s Law to find the

magnitude of the force required to

compress the spring by the amount 1

2x L= − :

( ) ( ) ( )1 1

2 2250 N/m 0.18 m 23 NF k x k L= − = = =

3. Because the force depends upon the stretch distance, the magnitude of the force required to

compress the spring to half its equilibrium length is less than the force found in part (a).

Insight: The magnitudes of the two forces would be the same if in part (a) the spring were

stretched to a length of 1.5L.

30. Picture the Problem: The free-body diagram for

the contact point between the two strings is

depicted at right.

Strategy: The horizontal components of the string

tensions must be equal because the picture is not

accelerating. The same is true of the vertical

components of the forces. Use Newton’s Second

Law in the horizontal direction to find the tension

in string 2, and in the vertical direction to find the

weight of the picture.

Solution: 1. (a) The tension in string 2 is less than

the tension in string 1, because it provides mostly a

sideways component of force that is balanced by

the horizontal component of string 1. That means

string 1 must support most of the weight of the


12 of 89

picture plus balance string 2’s horizontal

component, giving it a larger tension than string 2.

2. (b) Write Newton’s Second Law in the

horizontal direction in order to find the

tension in string 2: ( )

1 1 2 2

12 1

2

cos cos 0

cos cos 651.7 N 0.85 N

cos cos32

xF T T

T T

θ θ

θ

θ

= − + =

°= = =

°

∑

3. (c) Write Newton’s Second Law

in the vertical direction in order to

find the picture’s weight: ( ) ( )

1 1 2 2sin sin 0

1.7 N sin 65 0.85 N sin 32 2.0 N

yF T T W

W

θ θ= + − =

= ° + ° =

∑

Insight: As the angle of string 1 approaches 90° and the angle of string 2 approaches 0°, the

tension in string 2 drops to zero and the entire 2.0 N weight of the picture is supported by string 1.

31. Picture the Problem: The free-body diagram for each

pulley is depicted at right.

Strategy: The tension in the rope is the same everywhere

and is equal to F as long as the pulleys rotate without

friction. Use the free-body diagrams of the pulleys to

write Newton’s Second Law in the vertical direction for

each pulley, and use the resulting equations to find the

tensions. The acceleration is zero everywhere in the

system.


Newton’s Second

Law for the box:

lower

lower

0yF C mg

C mg

= − =

=

∑

2. Now write Newton’s Second

Law for

the lower pulley and solve for

F:

( )( )lower

21 1 1lower2 2 2

0

52 kg 9.81 m/s

255 N 0.26 kN downwards

yF F F C

F C mg

= + − =

= = =

= =

∑

3. (b) Write Newton’s Second

Law for

the upper pulley and solve for

Cupper

( )

upper

upper

0

2 2 0.255 kN 0.51 kN

yF C F F

C F

= − − =

= = =

∑

4. (c) Use the result of step 1 to

find Clower: ( ) ( )2

lower 52 kg 9.81 m/s 510 N 0.51 kNC mg= = = =

Insight: Note that there are two rope tensions pulling upward on the lower pulley. Therefore each

one supports half the weight of the crate and the tension in the rope is half that of the chains.

32. Picture the Problem: The physical situation is depicted in

the figure at right. The tension in the string between the

two blocks is T1 and the tension in the string that is tied to

the wall is T2. A free-body diagram for the upper block 2

is also shown at right..

Strategy: Set the x-axis parallel to and pointing up the

incline, and the y-axis perpendicular to the incline. Write

Newton’s Second Law in the x direction for each block

and use the resulting equations to find T1 and T2. The

acceleration is zero everywhere in the system.

F F

F F

Clower

Cupper

Lower Pulley

Upper Pulley


13 of 89


Second Law for

block 1 and

solve for T1: ( )( )

1 1

block 1

1 1

2

1

sin 0

sin

1.0 kg 9.81 m/s sin 31

5.1 N

xF m g T

T m g

T

θ

θ

= − + =

=

= °

=

∑

2. (b) Write

Newton’s

Second Law for

block 2 and

solve for T2:

( )( )

1 2 2

block 2

2

2 1 2

sin 0

sin 5.1 N 2.0 kg 9.81 m/s sin 31 15 N

xF T m g T

T T m g

θ

θ

= − − + =

= + = + ° =

∑

Insight: As expected the tension T2 is larger than T1 because it must support the weight of both

blocks. If θ were to increase so would the tension, until at θ = 90° the blocks hang straight down

and ( )2 1 2 29 NT m m g= + = .

33. Picture the Problem: The forces acting on the small

pulley are depicted at right.

Strategy: The rope tension is equal to the weight of

the 2.50 kg mass and will everywhere be the same

because the pulleys are assumed frictionless. Write

Newton’s Second Law in the x direction and solve for

the magnitude of Fr

.

Solution: 1. Find

the rope tension:

1 2T T mg= =

2. Write Newton’s

Second Law

in the x direction

and solve for F:

( ) ( )1 2

2

cos30.0 cos30.0 0

2 cos30.0 2 2.50 kg 9.81 m/s cos30.0 42.5 N 9.54 lb

xF F T T

F mg

= − + ° + ° =

= ° = ° = =

∑

Insight: The traction device is arranged to produce a force that is parallel to the leg bone so that it

can heal straight. However, the force of gravity on the leg has been ignored here and in real life

there would have to be upward component of the force exerted on the leg.

34. Picture the Problem: The free-body diagram for the wood

block is depicted at right.

Strategy: Write Newton’s Second Law in the vertical direction

for the block to find the magnitude of the normal force. Then

write Newton’s Second Law in the horizontal direction for the

block to find the magnitude of the applied force. In both cases

the acceleration of the block is zero.


Newton’s Second

Law in the vertical direction:

s s

s

0 y

F f mg N mg

mgN

µ

µ

= − = − =

=

∑

2. Write Newton’s Second

Law in the

horizontal direction: s

0x

mgF F N F

µ= − = − =∑

3. Solve for the applied force

F:

( )( )( )

2

s

1.6 kg 9.81 m/s20 N 0.020 kN 4.5 lb

0.79

mgF

µ= = = = =

4. (b) The force of static friction would remain the same if you push with a greater force because it

must exactly balance the weight of the wood board.

Insight: Pushing with a force > 20 N, the max static friction force increases.

1Tr

2Tr

sinmg θ

θ mgr

Free-body diagram

for the upper block


14 of 89

35. Picture the Problem: The free-body diagrams for

each mass are shown at right.

Strategy: The positive axis is along the line of the

string and points up the incline for the 5.7 kg mass

and in the downward direction for the hanging

mass. Let M represent the mass on the incline and

m the hanging mass. Write Newton’s Second Law

for each mass and combine the equations to find

the acceleration of the hanging mass (which is the

same as the acceleration of M because they are

connected by a string).


Second Law for M: ( )

sin

sin

xF Mg T M a

T M a g

θ

θ

= − + =

= −

∑


Law for m and substitute the

expression for T found in

step 1:

( )sin

xF T mg ma

M a g mg maθ

= − + =

− + + =

∑

3. Solve for a: ( )

( )( )2

2

sin

3.2 kg 5.7 kg sin 35sin9.81 m/s

3.2 5.7 kg

0.076 m/s

Mg mg m M a

m Ma g

m M

a

θ

θ

− + = +

− ° − = =

+ +

= −

4. The negative value means that the acceleration of the hanging mass is in the upward direction.

5. (b) The magnitude of the acceleration is 0.076 m/s2.

Insight: Verify for yourself that the angle of the incline that balances the two masses is 34° when

m = 3.2 kg.

36. Picture the Problem: Refer to the figure at right:

Strategy: Write Newton’s Second Law for each block

and add the equations to eliminate the unknown

tension T. Solve the resulting equation for the

acceleration a, and use the acceleration to find the

tension. Let x be positive in the direction of each

mass’s motion, m1 be the mass on the table, and m2 be

the hanging mass.

Solution: 1. (a) The tension in the string is less than

the weight of the hanging mass. If it were equal to the

weight, the hanging mass would not accelerate.

2. (b) Write

Newton’s Second

Law

for each block and

add the equations: ( )

1

block 1

2 2

block 2

2 1 2

x

x

F T m a

F T m g m a

m g m m a

= =

= − + =

= +

∑

∑

3. Solve the

resulting equation

for a:

( )2 22

1 2

2.80 kg9.81 m/s 4.36 m/s

6.30 kg

ma g

m m

= = =

+


15 of 89

4. (c) Use the first

equation to find T: ( )( )2

1 3.50 kg 4.36 m/s 15.3 NT m a= = =

Insight: Note that the blocks move as if they were a single block of mass 6.30 kg under the

influence of a force equal to 2 27.5 N.m g = The tension in the string would be zero if m2 fell freely,

27.5 N if m2 (and the entire system) were at rest.

37. Picture the Problem: Two buckets are attached to either end of

a rope that is hung over a pulley, as shown in the figure at right.

Strategy: Use Newton’s Second Law to determine the tension

in the rope for all three cases. For the case when the masses

accelerate, write Newton’s Second Law for each bucket and

combine the equations to find the tension. Let m1 represent the

lighter bucket and m2 the heavier bucket (on the right in the

figure). The positive x axis points upward for m1 and

downward for m2.


Newton’s Second Law for

m2 with 0a = :

2

2

0

110 N

xF T m g

T m g

= − + =

= =

∑

2. (b) Write Newton’s

Second Law for m1: 1 1

1

xF T m g m a

a T m g

= − =

= −

∑


Law for m2 and substitute

for the acceleration from

step 2:

( )

( )2 2 2 1

2 2 12 1

xF T m g m a m T m g

m g T m m

= − + = = −

= +

∑

4. Substitute 2 2W m g= and

2 1 2 1m m W W= :

( )( ) ( )

2

2 1

2 110 N280 N

1 1+ 110 N 63 N

WT

W W= = =

+

5. (c) Write 0xF =∑ for

m1 with 0a = :

1 10 63 NxF T m g T m g= − = ⇒ = =∑

Insight: Another way to solve this problem is to add the two versions of Newton’s Second Law in

steps 2 and 3 to eliminate T and solve for a, then use a to find T. Verify for yourself that 22.7 m/sa = in part (b).

38. Picture the Problem: Your car travels along a circular path at constant speed.

Strategy: Static friction between your tires and the road provides the centripetal force required to

make the car travel along a circular path. Set the static friction force equal to the centripetal force

and calculate its value.

Solution: Set the static friction force

equal to

the centripetal force:

( )( )22

s cp cp

1200 kg 15 m/s4.7 kN

57 m

mvf F ma

r= = = = =

Insight: The maximum static friction force is

( )( )( )2

s 0.88 1200 kg 9.81 m/s 10.4 kNmgµ = = which corresponds to a maximum cornering speed

(without skidding) of 22 m/s.


16 of 89

39. Picture the Problem: The test tube travels along a circular path at constant speed.

Strategy: Solve for the speed required to attain the desired acceleration.

Solution: Solve for the speed: ( ) ( )( )( )2

cp52,000 0.075 m 52,000 9.81 m/s

200 m/s 0.20 km/s

v ra r g= = =

= =

Insight: This speed corresponds to 25,000 revolutions per minute for the centrifuge, or 415

revolutions per second.

40. Picture the Problem: The car follows a circular path

at constant speed as it passes over the bump.

Strategy: The centripetal acceleration is downward,

toward the center of the circle, as the car passes over

the bump. Write Newton’s Second Law in the

vertical direction and solve for the normal force N,

which is also the apparent weight of the passenger.

Solution: 1. Write

Newton’s

Second Law for the

passenger

and solve

for N:

( )

2

cp

2

yF N mg ma mv r

N m g v r

= − = − = −

= −

∑

2. Insert numerical

values:

( )( )

2

212 m/s

67 kg 9.81 m/s 380 N 0.38 kN35 m

N

= − = =

41. Picture the Problem: The car follows a circular

path at constant speed as it passes over the bump.

Strategy: The centripetal acceleration is downward,

towards the center of the circle, as the car passes

over the bump. Write Newton’s Second Law in the

vertical direction and set the normal force N, which

is also the apparent weight of the passenger, equal to

zero. Then solve for the speed of the car.

Solution: 1. Write

Newton’s

Second Law for the

passenger, and

set N to zero:

2

cp

20

yF N mg ma mv r

mg mv r

= − = − = −

− = −

∑

2. Now solve for v:

( )( )235 m 9.81 m/s 19 m/sv rg= = =

Insight: The car has zero normal force in it as well, meaning it is now airborne!


17 of 89


1. Picture the Problem: The farmer pushes the hay

horizontally.

Strategy: Multiply the force by the distance because

in this case the two point along the same direction.

Solution: Apply W = Fd directly: ( )( )86 N 3.4 m 290 J 0.29 kJW Fd= = = =

Insight: The 23-kg mass is unneeded information unless we needed to know the amount of friction

or the acceleration of the bale.

2. Picture the Problem: The suitcase is pushed

horizontally.

Strategy: Determine the applied force and solve for

d.

Solution: Solve for d:

( )( )( )2k k

640 J3.6 m

0.26 70.0 kg 9.81 m/s

W Wd

f mgµ= = = =

Insight: The applied force equals the friction force as long as the suitcase does not accelerate.

3. Picture the Problem: The paint can is lifted vertically.

Strategy: Multiply the force by the distance because the two vectors point in the

same direction in part (a). In part (b) the distance travelled is zero, and in part (c) the force and distance are antiparallel.

Solution: 1. (a) Apply W = Fd:

( )( )( )23.4 kg 9.81 m/s 1.8 m 60 J

W Fd mgd

W

= =

= =

2. (b) Now the force and distance are

perpendicular: 0W =

3. (c) Now the force and distance are

antiparallel: 60 J 0.060 kJW mgd= − = − = −

Insight: The applied force equals the weight as long as the paint can does not accelerate. The can

gains potential energy as it is lifted and loses potential energy as it is lowered.

4. Picture the Problem: The water skier is pulled horizontally.

Strategy: Multiply the force by the distance because in this case the two point along the same direction.

Solution: 1. (a) The work is positive because the force is along the direction of motion (θ = 0°).

2. (b) Apply W = Fd directly: ( )( )120 N 65 m 7800 J 7.8 kJW Fd= = = =

Insight: While the work done by the rope is positive, the work done by friction is negative, so as

long as the skier moves at constant speed she doesn’t gain or lose kinetic energy.

5. Picture the Problem: The wagon rolls horizontally

but the force pulls upward at an angle.

Strategy: Keep in mind the angle between the force and the direction of motion.

F

d

θ

mg 1.8 m

F

d

86 N

3.4 m


18 of 89

Solution: Use W = Fdcosθ: ( )( ) ocos 16 N 10.0 m cos 25 150 J 0.15 kJW Fd θ= = = =

Insight: Only the component of the force along the direction of the motion does any work. The

vertical component of the force reduces the normal force a little.

6. Picture the Problem: The mop head is being pushed

downward into the floor.

Strategy: Keep in mind the angle between the force

and the direction of motion.

Solution: 1. (a) Use W = Fdcosθ: ( )( ) ocos 50.0 N 0.50 m cos55 14 JW Fd θ= = =

2. (b) If the angle is increased to 65°, a smaller component of the force will be along the direction

of motion and therefore the work done by the janitor will decrease.


vertical component of the force increases the normal force.

7. Picture the Problem: The plane and glider must be at

different altitudes. Since the altitudes are constant,

both are moving horizontally.

Strategy: Solving for the angle between the force and the direction of motion.

Solution: 1. Solve W = Fdcosθ

for the angle: cos or cos

WW Fd

Fdθ θ= =

2. Calculate the angle:

( )( )

51 1 o2.00 10 J

cos cos 57.42560 N 145 m

W

Fdθ − −

× = = =

Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force helps to lift the glider a little.

8. Picture the Problem: The force and distance vectors for the

woman are depicted on the right.

Strategy: Multiply the distance by the component of the

force that is parallel to the distance.

Solution: 1. (a) First find the

distance travelled: ( )( ) ˆ ˆ4.1 m/s 25 s 102.5 m t= = =d v x x

r r

2. Now multiply only the x-components:

( )( )17 N 102.5 m 1700 Jx xW F d= = =

3. (b) The force on the

bicycle is opposite that

on the skateboard by

Newton’s third law.

( ) ( )bikeˆ ˆ17 N 12 N= − + −F x y

r

4. (c) Multiply only the x-components:

( )( )bike, 17 N 102.5 m 1700 Jx xW F d= = − = −

Insight: The bicyclist must do work while pedalling or she will lose kinetic energy and come to a

stop. The force of friction on the skateboard must do −1700 J of work because her velocity (and

therefore kinetic energy) is constant.

Fr

skateboard

bicycle

t=d vr r

17 N

12 N

F

θ glider

airplane

d

F

d

θ


19 of 89

9. Picture the Problem: The boat and skier are both moving

toward the left but the rope is pulling at an angle.

Strategy: Solve for the angle between the force and the direction of motion.

Solution: 1. Solve

for the angle: cos cos

WW Fd

Fdθ θ= ⇒ =

2. Calculate the angle:

( )( )1 1 o3500 J

cos cos 2175 N 50 m

W

Fdθ − −

= = =


work the boat does on the skier is balanced by the negative work friction does on the skier, so that

the kinetic energy of the skier is constant.

10. Picture the Problem: The fragment moves at high speed in a straight line.

Strategy: Calculate the kinetic energy is using K = ½ mv2.

Solution: Apply K=½mv2

directly: ( )( )

22 71 1

2 21770 kg 120 m/s 1.27 10 J 12.7 MJK mv= = = × =

Insight: The energy came from the work the rocket motor did in order to place Skylab into orbit.

11. Picture the Problem: The pine cone falls straight down under the influence of gravity.

Strategy: The work done by gravity equals the change in kinetic energy. The work done by

gravity is always W = mgh.

Solution: 1. (a) The work done by

gravity on the pine cone equals the increase in its kinetic energy. Set the

energies equal and solve for v: ( )( )

21

2

22 2 9.81 m/s 16 m 18 m/s

W K mgh mv

v gh

= ∆ = =

= = =

2. (b) Air resistance did negative work because the speed and therefore the kinetic energy of the

pine cone when it landed was reduced. Air resistance removed energy from the pine cone.

Insight: Kinetic friction always does negative work because the force is always opposite to the

direction of motion.

12. Picture the Problem: The object falls straight down under the influence of gravity.

Strategy: Use the dependence of kinetic energy upon mass and speed to answer parts (a) and (b).

The work done by gravity can be found from the change in the kinetic energy.

Solution: 1. (a) Apply K=½mv2

directly: ( )( )221 1

2 20.40 kg 6.0 m/s 7.2 JK mv= = =

2. (b) Solve for speed:

( )2 25 J211 m/s

0.40 kg

Kv

m= = =

3. (c) Calculate W K= ∆ : f i 25 J 7.2 J 18 JW K K K= ∆ = − = − =

Insight: As an object falls, the work done by gravity increases the kinetic energy of the object.

13. Picture the Problem: The car slows down as it rolls horizontally a distance of 30.0 m through the

sand.

Strategy: The kinetic energy of the car is reduced by the amount of work done by friction. The

work done is the force times the distance, so once we know the work done and the distance, we can

find the force.


20 of 89

Solution: 1. (a) The net work done on the car must have been negative since the kinetic energy

decreased.

2. (b) The work done by

friction equals the average

force of friction times the

distance the car travelled.

Apply equations, including a

minus sign to indicate the

force and distance are in

opposite directions:

( )

( )( )( )

2 21f i22 21 1

f i2 2

2 2 2 212

so that

1300 kg 15 18 m /s2100 N 2.1 kN

30.0 m

m v vW Fd K mv mv F

d

F

−= − = ∆ = − = −

−= − = =


direction of motion. The actual force exerted on the car would be −2100 N if the distance

travelled is taken to be +30.0 m.

14. Picture the Problem: The bicycle rolls horizontally on level ground, slows down, and comes to

rest.

Strategy: The work done by the brakes equals the change in the kinetic energy of the bicycle. Find

the distance travelled, and the magnitude of the braking force.

Solution: 1. (a) Calculate W K= ∆ : ( )( )22 21 1 1

f i2 2 20 65 8.8 kg 14 m/s

7200 J 7.2 kJ

W K mv mv= ∆ = − = − +

= − = −

2. (b) Find the distance: ( ) ( )( )1 102 2

14 0 m/s 4.0 s 28 mx v v t∆ = + = + =

3. (c) The force can be found from W =

Fd:

7200 J260 N 58 lb

28 m

W WF

d x= = = = ≅

∆


direction of motion. The average velocity is half the initial velocity as long as the acceleration is

constant.

15

.

Picture the Problem: The spring is stretched to x =

0.050 m by pulling to the right or compressed to x =

−0.050 m by pushing to the left.

Strategy: The work to stretch or compress a spring

a distance x is 21

2kx .

Solution

: 1. (a) Apply

equation

7-8:

( )( )22 41 1

2 23.5 10 N/m 0.050 m

44 J

W kx= = ×

=

2. (b) Apply

equation

7-8

again:

( )( )22 41 1

2 23.5 10 N/m 0.050 m

44 J

W kx= = × −

=

Insight: The work done on the spring is the same in either case because the spring force is opposite

the applied force in each case. The applied force and distance vectors are parallel in each case, so the works are positive.


21 of 89

16. Picture the Problem: The block slides toward the

right and into the spring. After compressing the

spring the block comes to rest.

Strategy: The work to stretch or compress a spring

a distance x is 21

2kx . The work done on the block by

the spring equals the kinetic energy lost by the

block. The work done on the block is negative

because the force on the block is toward the left while the motion is toward the right.

Solution: 1. Apply equations 7-7

and 7-8:

2 21 1on block block i2 2

0W K mv kx= ∆ = − = −

2. Now solve for k: ( )

( )

( )

22

i

2 2

2.2 m/s1.8 kg 91 N/m

0.31 m

vk m

x= = =

Insight: The kinetic energy of the block is transformed into the energy stored in the spring as it is

compressed.

17. Picture the Problem: The work done by the force

is the area under the force versus position graph.

Strategy: The total work done by the force is the

total area under the graph from zero to 0.75 m. The

work done from 0.15 m to 0.60 m is the area shaded

in grey in the figure at right. Add the works done in

each of the three segments to find the total work.

Solution: 1. (a) The

total

area under

the graph:

( )( )total 0.25 m 0.6 0.4 0.8 N

0.45 J

W A= = + +

=

2. (b) The

area of the graph

shaded in

grey:

( )( ) ( )( ) ( )( )0.25 0.15 m 0.6 N 0.25 m 0.4 N 0.60 0.50 m 0.8 N

0.24 J

W

W

= − + + −

=

Insight: The work is positive as long as the object moves from left to right (from small x to large

x). Therefore the object gains energy as it moves from left to right.

18. Picture the Problem: The spring is compressed horizontally.

Strategy: The work and stretch distance can be used to find the

spring constant by applying W = ½ kx2.

Solution: 1. (a) Solve for

k :

( )

( )2 2

4

2 160 J2

0.14 m

1.6 10 N/m 16 kN/m

Wk

x= =

= × =

2. (b) It would take more

than 160 J of

work because W is

proportional to x2:

( ) ( ) ( )2 22 2 41 1 1

2 12 2 21.6 10 N/m 0.28 m 0.14 m 480 JW kx kx = − = × − =


22 of 89

Insight: The extra work required to stretch the spring an additional 0.14 m can be pictured as the

difference in area of two triangles, one with base 0.28 m and one with base 0.14 m, both bounded

by the line given by kx.

19. Picture the Problem: The work

done by the force is the area under

the force versus position graph.

Strategy: Determine the geometric

area under the graph for the various

given starting and ending positions.

The area under the graph equals the

work done on the block.

Solution: 1. (a) The work

done on the object is the

area under the graph

between 0 and 0.30 m:

( )( ) ( )( )1

20.21 m 4200 N 0.30 0.21 m 4200 N 0.82 kJW = + − =

2. The magnitude of the

force at x = 0.10 m can be

found from the given

formula:

( )( )42.0 10 N/m 0.10 m 2000 NF kx= = × =

3. (b) The work done is the

area under the graph

between 0 and 0.40 m

minus the area of the

triangle between 0 and 0.10

m:

( )( ) ( )( )

( )( )

1

2

1

2

0.21 m 4200 N 0.40 0.21 m 4200 N

0.10 m 2000 N 1.1 kJ

W = + −

− =

Insight: When the force varies as a function of x it is often useful to break the area under the

graph into simple geometric shapes to aid in the calculation of work.

20. Picture the Problem: The fly does work against gravity as it elevates its centre of mass.

Strategy: The power required is the force times the velocity, where the force is just the weight of

the fly.

Solution: Apply equation P =

Fv: ( ) ( ) ( )( )-3 2

4

1.3 10 kg 9.81 m/s 0.025 m/s

3.2 10 W 0.32 mW

P Fv mg v

P−

= = = ×

= × =

Insight: The energy and power required of the fly is higher than this because it isn’t 100%

efficient at converting food energy into mechanical energy.

21. Picture the Problem: The bucket is lifted vertically upward.

Strategy: The power required is the force times the velocity, where the force is just the weight of

the bucket.

Solution: Solve P = Fv for v:

( )( )2

108 W2.20 m/s

5.00 kg 9.81 m/s

P Pv

F mg= = = =

Insight: Lifting faster than this would require more power. If the rope’s mass were not ignored it

would require additional power since its centre of mass is also lifted. If the force exceeded the

weight, the bucket would accelerate.


23 of 89

22. Picture the Problem: The kayaker paddles horizontally in a straight line at constant speed.

Strategy: The kayaker does positive work on the kayak as she paddles, and friction does negative

work at the same time. The two works are equal because the kayak does not change its kinetic

energy. Therefore the force she exerts must be equivalent to the force of friction.

Solution: 1. (a) Solve P = Fv for F:

50.0 W33.3 N 7.5 lb

1.50 m/s

PF

v= = = =

2. (b) Since the speed of the kayak is proportional to P / F, doubling the power would double the

speed for the same F.

Insight: Newton’s Second Law F = ma states that the net force on the kayak must be zero since it

is not accelerating. That’s another way of figuring that the magnitude of the paddling force equals

the magnitude of the friction force.

23. Picture the Problem: The weight slowly descends straight down.

Strategy: The power delivered is the force (the weight) times the speed.

Solution: 1. (a) Apply P = Fv: ( ) ( )2

4

0.720 m4.15 kg 9.81 m/s

3.25 d 86400 s/d

1.04 10 W 0.104 mW

P Fv mgv

P−

= = =

×

= × =

2. (b) To increase the power delivered you must either increase the force or the velocity. In this

case, the time it takes for the mass to descend should be decreased so the velocity will increase and

so will the delivered power.

Insight: The weight delivers energy to the clock by doing work. The downward force it exerts on

the clock is parallel to its displacement, so it is doing positive work on the clock.

24. Picture the Problem: The car accelerates horizontally in a straight line.

Strategy: The power required is the work required to change the kinetic energy divided by the

time. Use ratios to easily find the desired quantities.

Solution: 1. (a) Combine equations:

2 21 1f i2 2

21i2

mv mvW Kt

P P mv T

−∆= = =

2. Now divide top and bottom by 1

2m

and substitute

for the velocities:

( )2 22 2

f i

2 2

i

23

v vv vt T

v T v T

−−= = =

3. (b) Again combine equations: ( )21

2 21f2

2mv

W K mv Pt TT

= ∆ = = =

4. Now solve for

fv : 2 2 2 21f f f2

so 2 and thus 2mv mv v v v v= = =

Insight: The assumption that the power remains constant is not realistic because car engines only

generate their rated horsepower at high engine rpm, so less power is generated when the car first

begins to accelerate.


24 of 89

25. Picture the Problem: The three paths of the sliding box are depicted

at right.

Strategy: The work done by friction is kW mgdµ= − , where d is the

distance the box is pushed irregardless of direction, because the

friction force always acts in a direction opposite the motion. Sum the

work done by friction for each segment of each path..

Solution: 1. Calculate the

work for path 1: [ ][ ]

( )( )[ ]

1 k 1 2 3 4 5

k

2

1

4.0 4.0 1.0 1.0 1.0 m

0.21 3.2 kg 9.81 m/s 11.0 m 73 J

W mg d d d d d

mg

W

µ

µ

= − + + + +

= − + + + +

= − = −

2. Calculate W for path 2: [ ]( )( ) ( ) ( ) ( )

2 k 6 7 8

20.21 3.2 kg 9.81 m/s 2.0 m 2.0 m 1.0 m 33 J

W mg d d dµ= − + +

= − + + = −

3. Calculate the work for

path 3: [ ]

( )( ) ( ) ( ) ( )3 k 9 10 11

20.21 3.2 kg 9.81 m/s 1.0 m 3.0 m 3.0 m 46 J

W mg d d dµ= − + +

= − + + = −

Insight: The amount of work done depends upon the path because friction is a nonconservative

force.

26. Picture the Problem: The two paths of the object are

shown at right.

Strategy: Find the work done by gravity

W mgy= when the object is moved downward,

W mgy= − when it is moved upward, and zero

when it is moved horizontally. Sum the work done

by gravity for each segment of each path.

Solution: 1. (a) Calculate the work for

path 1: ( )

( )( )( )1 1

2

0

5.2 kg 9.81 m/s 1.0 m 51 J

W mg y= +

= =

2. Calculate the work for path 2: ( )

( )( )( )2 1

2

0

2.6 kg 9.81 m/s 1.0 m 51 J

W mg y= +

= =

3. (b) If you increase the mass of the object the work done by gravity will increase because it

depends linearly on m.

Insight: The work is path-independent because gravity is a conservative force.

27. Picture the Problem: The physical situation is

depicted at right.

Strategy: Use ( )2 21sp i f2

W k x x= − for the work

done by the spring. That way the work will

always be negative if you start out at

i 0x = because the spring force will always be in

the opposite direction from the stretch or

compression. The work done by kinetic friction is

fr kW mgdµ= − , where d is the distance the box is

pushed irregardless of direction, because the

friction force always acts in a direction opposite

the motion.


25 of 89

Solution: 1. (a) Sum the

work done by the

spring for each segment of

path 2:

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2 2 2 21sp 1 4 4 32

2 22 21

2

sp

480 N/m 0 0.020 m 0.020 m 0.020 m

0.096 J 0 J 0.096 J

W k x x x x

W

= − + −

= − − + − −

= − + = −

2. Sum the work done by

friction for

each segment of path 2:

( )

( )( )( )( )

fr k 1 2

20.16 2.7 kg 9.81 m/s 0.020 0.040 m 0.25 J

W mg d dµ= − +

= − + = −

3. (b) Sum the work done

by the

spring for the direct path

from A to B:

( )

( ) ( )

2 21sp A B2

221

2480 N/m 0 0.020 m 0.096 J

W k x x= −

= − = −

4. Sum the work done by

friction for

the direct path from A to B:

( )( )( )( )2

fr k 0.16 2.7 kg 9.81 m/s 0.020 m 0.085 JW mgdµ= − = − = −

Insight: The work done by friction is always negative, and increases in magnitude with the

distance travelled.

28. Picture the Problem: The climber stands at the top of Mt. Everest.

Strategy: Find the gravitational potential energy by using U = mgy.

Solution: Calculate U mgy= : ( )( )( )2 683 kg 9.81 m/s 8848 m 7.2 10 J 7.2 MJU mgy= = = × =

Insight: You are free to declare that the climber’s potential energy is zero at the top of Mt. Everest

and −7.2 MJ at sea level!

29. Picture the Problem: The mass is suspended from a vertical spring. As the spring is stretched it

stores potential energy.

Strategy: Doubling the mass doubles the force exerted on the spring, and therefore doubles the

stretch distance due to Hooke’s Law .F kx= − Use a ratio to find the increase in spring potential

energy when the stretch distance is doubled.

Solution: 1. (a) If the mass attached to the spring is doubled the stored potential energy in the

spring will increase by a factor of four because the stretch distance will double. A doubling of the

mass will double the extension of the spring. Doubling the extension of the spring will increase the

potential energy by a factor of 4.

2. (b) Find an expression for

the stretch distance as a

function of U and m:

( )

( )

( )( )

2 21 1 11 1 1 1 12 2 2

1

1 2

2 0.962 J20.065 m

3.0 kg 9.81 m/s

U kx mg x x mgx

Ux

mg

= = =

= = =

3. Doubling the mass doubles

the force and therefore doubles

the stretch distance:

1

2 12 2

x mg k

x mg k x

=

= =

4. Calculate the ratio

2 1U U : ( )

( )

( )

22122 2

2 211 12

2 1

0.13 m4

0.065 m

4 4 0.962 J 3.85 J

kxU

U kx

U U

= = =

= = =

Insight: Note that the change in gravitational potential energy also quadruples as the new mass is

hung on the spring. It doubles because there is twice as much mass and it doubles again because

the spring stretches twice as far.


26 of 89

30. Picture the Problem: The spring in the soap dispenser is compressed by the applied force.

Strategy: Find the spring constant using the given energy and compression distance data. Solve

the same equation for x in order to answer part (b).

Solution: 1. (a) Solve for k:

( )

( )2 2

2 0.0025 J2200 N/m 0.20 kN/m

0.0050 m

Uk

x= = = =

2. (b) Solve for x:

( )2 0.0084 J20.92 cm

200 N/m

Ux

k= = =

Insight: To compress the spring of this dispenser 0.50 cm requires 1.0 N of force.

31. Picture the Problem: A graph of the potential energy

vs. stretch distance is depicted at right.

Strategy: The work that you must do to stretch a

spring is equal to minus the work done by the spring

because the force you exert is in the opposite direction

from the force the spring exerts. Find the spring

constant and the required work to stretch the spring the

specified distance.

Solution: 1. (a) Because the stored potential energy in a

spring is proportional to the stretch distance squared,

the work required to stretch the spring from 5.00 cm to

6.00 cm will be greater than the work required to

stretch it from 4.00 cm to 5.00 cm.

2. (b) Find

k: ( )

( )( )

( ) ( )

req spring 5 4

2 2 2 21 1 15 4 5 42 2 2

req 4

2 2 2 2

5 4

2 2 30.5 J6.78 10 N/m

0.0500 m 0.0400 m

W W U U U

kx kx k x x

Wk

x x

= − = − −∆ = −

= − = −

= = = ×− −

3. Use k to

find the new

reqW :

( ) ( ) ( ) ( )2 22 2 41 1

req 2 12 26.78 10 N/m 0.0600 m 0.0500 m 37.3 JW k x x = − = × − =

Insight: Using the same procedure we discover that it would take 44.1 J to stretch the spring from

6.00 cm to 7.00 cm.

32. Picture the Problem: The pendulum bob swings from point A to point

B and loses altitude and thus gravitational potential energy. See the

figure at right.

Strategy: Use the geometry of the problem to find the change in

altitude y∆ of the pendulum bob, and then find its change in

gravitational potential energy.

Solution: 1. Find the

height change y∆ of

the pendulum bob:

( )cos cos 1y L L Lθ θ∆ = − = −

2. Use y∆ to find U∆ : ( )( )( )( )( )2

cos 1

0.33 kg 9.81 m/s 1.2 m cos35 1

0.70 J

U mg y mgL

U

θ∆ = ∆ = −

= ° −

∆ = −

Insight: Note that y∆ is negative because the pendulum swings from A to B. Likewise, y∆ is

positive and the pendulum gains potential energy if it swings from B to A.


27 of 89


1. Picture the Problem: The owner walks slowly toward the northeast while the cat runs eastward

and the dog runs northward.

Strategy: Sum the momenta of the dog and cat using the component method. Use the known

components of the total momentum to find its magnitude and direction. Let north be in the y

direction, east in the x direction.

Solution: 1. Use the component

method of vector addition to find

the owner’s momentum: ( )( ) ( )( )

( ) ( )

total d c d d c c

total

ˆ ˆ20.0 kg 2.50 m/s 5.00 kg 3.00 m/s

ˆ ˆ15.0 kg m/s 50.0 kg m/s

m m= + = + =

= +

= ⋅ + ⋅

p p p v v

y x

p x y

r r r r r

r

2. Divide the owner’s momentum

by his mass to

get the components of the owner’s

velocity:

( ) ( )

( ) ( )

0 0 0 total

total

0

0

ˆ ˆ15.0 kg m/s 50.0 kg m/s

70.0 kg

ˆ ˆ0.214 m/s 0.714 m/s

m

m

= =

⋅ + ⋅= =

= +

p v p

x ypv

x y

r rr

r

3. Use the known components to

find the

direction and magnitude of the

owner’s velocity: ( ) ( )

1

2 2

0

0.714tan 73.3

0.214

0.2143 m/s 0.7143 m/s 0.746 m/sv

θ − = = °

= + =

Insight: We bent the rules of significant figures a bit in step 3 in order to avoid rounding error. The

owner is moving much slower than either the cat or the dog because of his larger mass.

2. Picture the Problem: The two carts approach each other on a frictionless track at different speeds.

Strategy: Add the momenta of the two carts and set it equal to zero. Solve the resulting expression

for2v . Then find the total kinetic energy of the two-cart system. Let cart 1 travel in the positive

direction.

Solution: 1. (a) Set 0=∑pr

and

solve for 2 :v ( )( )

1 1 2 2

1 12

2

0

0.35 kg 1.2 m/s0.69 m/s

0.61 kg

m m

m vv

m

= + =

= − = =

∑p v vr r r

2. (b) No, kinetic energy is always greater than or equal to zero.

3. (c) Sum the kinetic

energies of the two carts: ( )( ) ( )( )

2 21 11 1 2 22 2

2 21 1

2 20.35 kg 1.2 m/s 0.61 kg 0.69 m/s

0.40 J

K m v m v= +

= +

=

∑

Insight: If cart 1 is travelling in the positive x direction, then its momentum is ( ) ˆ0.42 kg m/s⋅ x

and the momentum of cart 2 is ( ) ˆ0.42 kg m/s− ⋅ x .

3. Picture the Problem: The baseball drops straight down, gaining momentum due to the

acceleration of gravity.

Strategy: Determine the speed of the baseball before it hits the ground, then find the height from

which it was dropped.

Solution: 1. Use p = mv to find

the speed

of the ball when it lands:

0.780 kg m/s5.20 m/s

0.150 kg

pv

m

⋅= = =


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2. (b) Solve for 0y . Let 0y =

and 0 0 :v =

( )

( )

( )

2 2

0 0

22

0 2

2

5.20 m/s1.38 m

2 2 9.81 m/s

v v g y y

vy

g

= − −

= = =

Insight: Another way to find the initial height is to use conservation of energy, setting 21

0 2mgy mv= and solving for

0 .y

4. Picture the Problem: The ball falls vertically downward, landing with a speed of 2.5 m/s and

rebounding upward with a speed of 2.0 m/s.

Strategy: Find the change in momentum of the ball when it rebounds.

Solution: 1. (a) Find ∆pr

: ( )

( ) ( ) ( ) ( )f i f i

ˆ ˆ ˆ0.220 kg 2.0 m/s 2.5 m/s 0.99 kg m/s

0.99 kg m/s

m∆ = − = −

= − − = ⋅

∆ = ⋅

p p p v v

y y y

p

r r r r r

r

2. (b) Subtract the

magnitudes

of the momenta:

( )

( )( )f i f i

f i

0.220 kg 2.0 m/s 2.5 m/s

0.11 kg m/s

p p m v v

p p

− = −

= −

− = − ⋅

3. (c) The quantity in part (a) is more directly related to the net force acting on the ball during its

collision with the floor, first of all because t= ∆ ∆∑F pr r

and as we can see from above that

f ip p∆ ≠ −pr

. Secondly, we expect the floor to exert an upward force on the ball but we

calculated a downward (negative) value in part (b).

Insight: If the ball were to rebound at 2.5 m/s upward we would find 2 1.1 kg m/smv∆ = = ⋅pr

and

f i 0p p− = . Such a collision with the floor would be called elastic.

5. Picture the Problem: The individual momenta and final

momentum vectors are depicted at right.

Strategy: The momenta of the two objects are perpendicular.

Because of this we can say that the momentum of object 1 is

equal to the x-component of the total momentum and the

momentum of object 2 is equal to the y-component of the total

momentum. Find the momenta of objects 1 and 2 in this manner

and divide by their speeds to determine the masses.

Solution: 1. Find total, xp and

divide by 1v :

( )( )1 total, total

1

1

1

cos 17.6 kg m/s cos66.5 7.02 kg m/s

7.02 kg m/s2.51 kg

2.80 m/s

xp p p

pm

v

θ= = = ⋅ ° = ⋅

⋅= = =

2. Find total, yp and divide by

2v :

( )( )total2

2

2 2

17.6 kg m/s sin 66.5sin5.21 kg

3.10 m/s

ppm

v v

θ ⋅ °= = = =

Insight: Note that object 2 has the larger momentum because the total momentum points mostly in

the y direction. The two objects have similar speeds, so object 2 must have the larger mass in order

to have the larger momentum.

6. Picture the Problem: The two skaters push apart and move in opposite directions without friction.

Strategy: By applying the conservation of momentum we conclude that the total momentum of the

two skaters after the push is zero, just as it was before the push. Set the total momentum of the

system to zero and solve for 2m . Let the velocity

1vr

point in the negative direction, 2v

rin the

positive direction.

y

x

totalpr

1pr

2pr


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Solution: Set total 0p = and solve for

2 :m ( )( )1 2 1 1 2 2

1 1

2

2

0

45 kg 0.62 m/s31 kg

0.89 m/s

x x x x

x

x

p p m v m v

m vm

v

+ = = +

− −−= = =

Insight: An alternative way to find the mass is to use the equations of kinematics.

7. Picture the Problem: The two pieces fly in opposite directions at different speeds..

Strategy: As long as there is no friction the total momentum of the two pieces must remain zero,

as it was before the explosion. Combine the conservation of momentum with the given kinetic

energy ratio to determine the ratio of the masses. Let 1m represent the piece with the smaller

kinetic energy.

Solution: 1. Set 0=∑pr

and solve

for 1 2m m :

1 2 1 1 2 2

2 2

1 2 1 2

2 1 2 1

0

p p m v m v

m v m v

m v m v

+ = = +

= − ⇒ =

2. Set

2 1 2K K = :

2212 22 2 2 1

211 1 21 12

2 2m vK v m

K v mm v

= = ⇒ =

3. Combine the expressions from

steps 1 and 2:

2 2

2 1 1 1

1 2 2 2

2 2v m m m

v m m m

= = ⇒ =

4. The piece with the smaller kinetic energy has the larger mass.

Insight: The smaller mass carries the larger kinetic energy because kinetic energy increases with

the square of the velocity but is linear with mass. Its higher speed more than compensates for its

smaller mass.

8. Picture the Problem: The astronaut and the satellite move in opposite directions after the astronaut

pushes off. The astronaut travels at constant speed a distance d before coming in contact with the

space shuttle.

Strategy: As long as there is no friction the total momentum of the astronaut and the satellite must

remain zero, as it was before the astronaut pushed off. Use the conservation of momentum to

determine the speed of the astronaut, and then multiply the speed by the time to find the distance.

Assume the satellite’s motion is in the negative x-direction.

Solution: 1. Find the speed of the

astronaut using conservation of

momentum:

a s a a s s

s s

a

a

0p p m v m v

m vv

m

+ = = +

= −

2. Find the distance to the space

shuttle:

( )( )( )

( )s s

a

a

1200 kg 0.14 m/s7.5 s 14 m

92 kg

m vd v t t

m

−= = − = − =

Insight: One of the tricky things about spacewalking is that whenever you push on a satellite or

anything else, you yourself get pushed! (Newton’s Third Law). Conservation of momentum makes

it easy to predict your speed.

9. Picture the Problem: The lumberjack moves to the right while the log moves to the left.

Strategy: As long as there is no friction the total momentum of the lumberjack and the log remains

zero, as it was before the lumberjack started trotting. Combine vector addition for relative motion

with the expression from the conservation of momentum to find L, sv = speed of lumberjack relative

to the shore. Let L, logv = speed of lumberjack relative to the log, and log, sv = speed of the log

relative to the shore.


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Solution: 1. (a) Write out the

equation for relative motion.

Let the log travel in the negative

direction:

L, s L, log log, s

L, s L, log log, s

log, s L, log L, s

v v v

v v v

= +

= −

= −

v v vr r r

2. Write out the conservation of

momentum

with respect to the shore:

L L, s log log, s0 m v m v= = −∑pr

3. Substitute the expression

from step 1

into step 2 and solve for L, sv

( )

( )

( )( )( )

( )

L L, s log log, s log L, log L, s

L, s L log log L, log

log L, log

L, s

L log

380 kg 2.7 m/s2.2 m/s

85 380 kg

m v m v m v v

v m m m v

m vv

m m

= = −

+ =

= = =++

4. (b) If the mass of the log had been greater, the lumberjack’s speed relative to the shore would

have been greater than that found in part (a), because the log would have moved slower in the

negative direction.

5. (c) Use the expression from

step 3 to

find the new speed of the

lumberjack:

( )( )( )

( )log L, log

L, s

L log

450 kg 2.7 m/s2.3 m/s

85 450 kg

m vv

m m= = =

++

Insight: Taking the argument in (b) to its extreme, if the mass of the log equalled the mass of the

Earth the lumberjack’s speed would be exactly 2.7 m/s relative to the Earth (and the log). If the

mass of the log were the same as the mass of the lumberjack, the speed of each relative to the Earth

would be half the lumberjack’s walking speed.

10. Picture the Problem: The vector diagram at right indicates the

momenta of the three pieces.

Strategy: Since the plate falls straight down its momentum in

the xy plane is zero. That means the momenta of all three

pieces must sum to zero. Choose the motion of the two pieces

at right angles to one another to be in the x and y directions.

Set the total momentum equal to zero and solve for 3v

r.

Solution: 1. Set 0=∑pr

and

solve for 3v

r: ( ) ( )

( ) ( )

3

3

ˆ ˆ 0

ˆ ˆˆ ˆ

mv mv m

mv mvv v

m

= + + =

− + −= = − + −

∑p x y v

x yv x y

r r

r

2. Find the speed

3 :v ( ) ( )2 2

32v v v v= − + − =

3. Find the direction of

3 :v 3,1 1

3,

tan tan 45 180 225y

x

v v

v vθ − −

− = = = ° + ° = ° −

Insight: As long as the first two pieces have equal masses the direction of 3v

r will always be the

same. For instance, if the third piece has four times the mass of either piece 1 or 2, its speed would

be 8v but θ would remain 225°.

x

3pr

1pr

2pr

y

225°


31 of 89


1. Picture the Problem: The tire rotates about its axis through a certain angle.

Strategy: Use equation s = rθ to find the angular displacement.

Solution: Solve 10-2 for θ: 1.75 m

5.3 rad0.33 m

s

rθ = = =

Insight: This angular distance corresponds to 304° or 84% of a complete revolution.

2. Picture the Problem: The Earth travels in a nearly circular path around the Sun, completing one

revolution per year.

Strategy: Convert the known angular speed of 1 rev/yr into units of rev/min.

Solution: Convert the units: 61 rev 1 yr 1 day 1 h

1.90 10 rev/minyr 365 days 24 h 60 min

ω − = = ×

Insight: This angular speed corresponds to about 0.986 deg/day or 1.99×10−7

rad/s. A good “rule

of thumb” in astronomy is that the Sun appears to move 1°/day against the background of the

“fixed” stars.

3. Picture the Problem: The Earth rotates once on its axis every 24 hours.

Strategy: Convert the known angular speed of 1 rev/day into units of radians per second.

Solution: Convert the

units:

51rev 2 rad 1 day 1 h7.27 10 rad/s

day rev 24 h 3600s

πω −

= = ×

Insight: This angular speed corresponds to about 15° / hour. A “rule of thumb” in astronomy is

that the “fixed” stars will move across the sky at this rate (1° every 4 minutes, or 15 arcsec/s) due to

Earth’s rotation.

4. Picture the Problem: The pulsar rotates about its axis, completing 1 revolution in 0.33 s.

Strategy: Divide one revolution or 2π radians by the period in seconds to find the angular speed.

Solution: Calculate ω using equation 10-

3:

2 rad 2 rad190 rad/s

0.033st T

θ π πω

∆= = = =

∆

Insight: The rotation rate of the pulsar can also be described as 1800 rev/min.

5. Picture the Problem: The floppy disk rotates about its axis at a constant angular speed.

Strategy: Use equation ω = 2π/T to relate the period of rotation to the angular speed. Then use v =

rω to find the linear speed of a point on the disk’s rim.


ω :

2 231.4 rad/s

0.200sT

π πω = = =

2. (b) Apply v = rω

directly:

( )( )1

23.5 in 31.4 rad/s 55 in/s 1 m 39.4 in 1.4 m/sTv rω= = × = × =

3. (c) A point near the centre will have the same angular speed as a point on the rim because the

rotation periods are the same.

Insight: While the angular speed is the same everywhere on the disk, the linear speed is greatest at


32 of 89

the rim. The read/write circuitry must compensate for the different speeds at which the bits of data

will move past the head.

6. Picture the Problem: The propeller rotates about its axis with constant angular acceleration.

Strategy: Use the kinematic equations for rotating objects and the given formula to find the

average angular speed and angular acceleration during the specified time intervals. By comparison

of the formula given in the problem, ( ) ( )2 2125 rad/s 42.5 rad/st tθ = + ,with 21

0 0 2t tθ θ ω α= + + , we

can identify 0 125 rad/sω = and 21

242.5 rad/sα = .

Solution: 1. (a) Use equations

to find avω : ( )( ) ( )( )

210 020

av

22

2

av

t

125 rad/s 0.010 s 42.5 rad/s 0.010 s 0

0.010 s

125 rad/s 1.3 10 rad/s

t t

t t

ω α θθ θθω

ω

+ −−∆ = = =

∆ + − =

= = ×

2. (b) Use

equations

to find avω :

( )( ) ( )( )

( )( ) ( )( )

22 210 2

22 210 0 0 02

0

av

0

125 rad/s 1.010 s 42.5 rad/s 1.010 s 169.60 rad

125 rad/s 1.000 s 42.5 rad/s 1.000 s 167.50 rad

169.60 167.50 rad210 rad/s 2.1

1.010 1.000 s

t t

t t

t t t

θ ω α

θ ω α

θ θθω

= + = + = = + = + =

−∆ −= = = = = ×

∆ − −

210 rad/s

3. (c) Use

equations

to find avω :

( )( ) ( )( )

( )( ) ( )( )

22 210 2

22 210 0 0 02

0

av

0

125 rad/s 2.010 s 42.5 rad/s 2.010 s 422.95 rad

125 rad/s 2.000 s 42.5 rad/s 2.000 s 420.00 rad

422.95 420.0 rad295 rad/s 3.0 1

2.010 2.000 s

t t

t t

t t t

θ ω α

θ ω α

θ θθω

= + = + = = + = + =

−∆ −= = = = = ×

∆ − −

20 rad/s

4. (d) The angular acceleration is positive because the angular speed is positive and increasing with

time.

5. (e) Find αav: 20

av

210 125 rad/s85 rad/s

1.00 0.00 st

ω ωα

− −= = =

∆ −

6. Find αav: 20

av

295 210 rad/s85 rad/s

2.00 1.00 st

ω ωα

− −= = =

∆ −

Insight: We violated the rules of significant figures in order to report answers with two significant

figures. Such problems arise whenever you try to subtract two large but similar numbers to get a

small difference. The answers are only known to one significant figure, but we reported two in

order to show clearly that the angular acceleration is constant. Of course we could also have figured

from the equation given in the problem that since 21

242.5 rad/sα = , it must be true that

285.0 rad/s .α =

7. Picture the Problem: The propeller rotates about its axis, increasing its angular velocity at a

constant rate.

Strategy: Use the kinematic equations for rotation to find the angular acceleration.

Solution: Solve for α :

( ) ( )( )

22 2

2026 rad/s 12 rad/s

17 rad/s2 2 2.5 rev 2 rad rev

ω ωα

θ π

2−−

= = =∆ ×

Insight: A speed of 26 rad/s is equivalent to 250 rev/min, indicating the motor is turning pretty

slowly. A typical outboard motor is designed to operate at 5000 rev/min at full throttle.


33 of 89

8. Picture the Problem: The propeller rotates about its axis, increasing its angular velocity at a

constant rate.

Strategy: Use the kinematic equations for rotation to find the angle through which the

propeller rotated.

Solution: Solve for θ∆ : ( ) ( )( )1 102 2

12 26 rad/s 2.5 s 48 rad 7.5 revtθ ω ω∆ = + = + = =

Insight: A speed of 26 rad/s is equivalent to 250 rev/min, indicating the motor is turning pretty

slowly. A typical outboard motor is designed to operate at 5000 rev/min at full throttle.

9. Picture the Problem: The bicycle wheel rotates about its axis, slowing down with constant angular

acceleration before coming to rest.

Strategy: Use the kinematic equations for rotation to find the angular acceleration and the time

elapsed.

Solution: 1. (a) Solve for α :

( )( )

( )

222 2

20

0

0 6.35 rad/s0.226 rad/s

2 2 14.2 rev 2 rad rev

ω ωα

θ θ π

−−= = = −

− ×

2. (b) Solve for t: 0

2

0 6.35 rad/s28.1 s

0.226 rad/st

ω ω

α

− −= = =

−

Insight: The greater the friction in the axle, the larger the magnitude of the angular acceleration

and the sooner the wheel will come to rest.

10. Picture the Problem: The ceiling fan rotates about its axis, slowing down with constant angular

acceleration before coming to rest.

Strategy: Use the kinematic equations for rotation to find the number of revolutions through which

the fan rotates during the specified intervals. Because the fan slows down at a constant rate of

acceleration, it takes exactly half the time for it to slow from 0.90 rev/s to 0.45 rev/s as it does to

come to a complete stop.

Solution: 1. (a) Find ∆θ: ( ) ( )( )1 102 2

0 0.90 rev/s 2.2 min 60 s min 59 revtθ ω ω∆ = + = + × =

2. (b) Find ∆θ: ( ) ( )( )1 102 2

0.45 0.90 rev/s 1.1 min 60 s min 45 revtθ ω ω∆ = + = + × =

Insight: An alternative way to solve the problem is to find 20.0068 rev/sα = − and use α to find

θ∆ for each of the specified intervals. Note that you can stick with units of rev/s2 to find θ∆ in

units of revolutions instead of converting to radians and back again.

11. Picture the Problem: The discus thrower rotates about a vertical axis through her centre of mass,

increasing her angular velocity at a constant rate.

Strategy: Use the kinematic equations for rotation to find the number of revolutions through

which the athlete rotates and the time elapsed during the specified interval.


θ∆ :

( )

( )

2 22 2

0

0 2

6.3 rad/s 09.0 rad 1 rev 2 rad

2 2 2.2 rad/s

1.4 rev

ω ωθ θ θ π

α

−−∆ = − = = = ×

=

2. (b) Solve for t: 0

2

6.3 0 rad/s2.9 s

2.2 rad/st

ω ω

α

− −= = =

Insight: Notice the athlete turns nearly one and a half times around. Therefore, she should begin

her spin with her back turned toward the range if she plans to throw the discus after reaching 6.3

rad/s. If she does let go at that point, the linear speed of the discus will be about 6.3 m/s (for a 1.0

m long arm) and will travel about 4.0 m if launched at 45° above level ground. Not that great

compared with a championship throw of over 40 m for a college woman.


34 of 89

12. Picture the Problem: The centrifuge rotates about its axis, slowing down with constant angular

acceleration and coming to rest.

Strategy: Use the kinematic equations for rotation to find the angular acceleration and the number

of revolutions through which the centrifuge rotates before coming to rest.


α :

( ) 200 3850 rev/min 1 min 60 s

6.29 rev/s10.2 st

ω ωα

− ×−= = =

2. (b) Solve for θ∆ : ( ) ( )( )1 102 2

0 3850 rev/min 1 min 60 s 10.2 s 327 revtθ ω ω∆ = + = + × =

Insight: Another way of expressing the angular acceleration is to say that it slows down at a rate of

377 rev/min/s.

13. Picture the Problem: The compact disk rotates about its axis, increasing its angular speed at a

constant rate.

Strategy: Use the kinematic equations for rotation to find the average angular speed during the

time interval and then the angle through which the disk spins during this interval.

Solution: 1. (a) Find the average angular speed over the time interval and find θ∆ .

2. (b) Find ∆θ: ( ) ( )( )1 102 2

310 0 rev/min 3.0 s 1 min 60 s 7.8 revtθ ω ω∆ = + = + × =

Insight: An alternative way to solve the problem is to find 21.72 rev/sα = and use α to find

7.8 revθ∆ = for the specified interval. Note that you can stick with units of rev/s2 to find θ∆ in

units of revolutions instead of converting to radians and back again.

14. Picture the Problem: The drill bit rotates about its axis, increasing its angular speed at a constant

rate.

Strategy: Use the kinematic equations for rotation to find the angular acceleration, average angular

speed during the time interval, and the angle through which the drill bit spins during this interval.


α :

( ) 3 20350,000 rev/min 1 min 60 s 0

2.8 10 rev/s2.1st

ω ωα

× −−= = = ×

2. (b) Find θ∆ : ( ) ( )( )1 102 2

3

350,000 0 rev/min 2.1 s 1 min 60 s

6.1 10 rev

tθ ω ω∆ = + = + ×

= ×

Insight: The angular acceleration could also be expressed as 1.7×104 rad/s2. Note that the bit spins

thousands of times during the 2.1 seconds it is coming up to speed. At full speed it spins over

12,000 times in 2.1 seconds!

15. Picture the Problem: The hour hand rotates about its axis at a constant rate.

Strategy: Convert the angular speed of the tip of the hour hand into a linear speed by

multiplying by its radius.

Solution: Apply v = rω: ( )t

1 rev 2 rad 1 h8.2 cm 0.0012 cm/s 12 m/s

12 h rev 3600 sv r

πω µ

= = = =

Insight: The tip of a minute hand travels much faster, not only because its angular speed is 12

times faster than the hour hand, but also because the minute hand is longer than the hour hand.


35 of 89

16. Picture the Problem: The Frisbee rotates at a constant rate about its central axis.

Strategy: Find the angular speed from the knowledge of the linear speed and the radius.

Solution: Solve v = rω for ω :

( )t 3.4 m/s

23 rad/s0.29 m 2

v

rω = = =

Insight: The rotation of a Frisbee produces its unique, stable flight characteristics.

17. Picture the Problem: The two horses are located at different places on the same carousel, which is

rotating about its axis at a constant rate.

Strategy: Find the angular speed of the horses by dividing 2π radians (for a complete circle) by

the time it takes to complete a revolution. Then v = rω together with the angular speed to find the

linear speed.

Solution: 1. (a) Find

1 2 and ω ω : 1 2

2 rad0.14 rad/s

45 s

πω ω= = =

2. Apply v = rω directly: ( )( )t1 1 1 2.75 m 0.14 rad/s 0.38 m/sv rω= = =

3. (b) Apply v = rω directly: ( )( )t2 2 2 1.75 m 1.4 rad/s 0.24 m/sv r ω= = =

Insight: The outer horse experiences a greater linear speed and greater centripetal acceleration

because it is at a larger radius.

18. Picture the Problem: The compact disk rotates about its central axis at a constant angular speed.

Strategy: Use v = rω to find the linear speed of a point on the outer rim of the CD, and then use

acp = rω2 to find the centripetal acceleration. Use ratios to determine the linear speed and

centripetal acceleration for a point that is half the distance to the rotation axis.

Solution: 1. (a) Apply v = rω

directly: ( )( )1

t 20.120 m 5.25 rad/s 0.315 m/sv rω= = =

2. (b)Apply acp = rω2

directly: ( )( )

22 21cp 2

0.120 m 5.25 rad/s 1.65 m/sa rω= = =

3. (c) Use a ratio to find the

new linear speed: ( )

12 2 2 1 1

2 12 2

1 1

1 0.315 m/s 0.158 m/s

2

rv rv v

v r r

ω

ω= = = ⇒ = = =

4. (d) Use a ratio to find the

new cpa : ( )

2 12 22 2 2 1 1

2 12 22

1 1

1 1.65 m/s 0.827 m/s

2

ra ra a

a rr

ω

ω= = = ⇒ = = =

Insight: The angular velocity is the same for all points on the CD regardless of the distance to the

rotation axis.

19. Picture the Problem: The Ferris wheel rotates at a constant rate, with the centripetal acceleration

of the passengers always pointing toward the axis of rotation. The acceleration of the passenger is

thus upward when they are at the bottom of the wheel and downward when they are at the top of

the wheel.

Strategy: Use acp = rω2 to find the centripetal acceleration. The centripetal acceleration remains

constant (as long as the angular speed remains the same) and points toward the axis of rotation.

Solution: 1. (a) Apply acp = rω

2 directly: ( )

2

2 2

cp

2 rad9.5 m 0.37 m/s

32 sa r

πω

= = =

2. When the passenger is at the top of the Ferris wheel, the centripetal acceleration points

downward toward the axis of rotation.

3. (b) The centripetal acceleration remains 0.37 m/s2 for a passenger at the bottom of the wheel

because the radius and angular speed remain the same, but here the acceleration points upward


36 of 89

toward the axis of rotation.

Insight: In order to double the centripetal acceleration you need to increase the angular speed by a

factor of 2 or decrease the period by a factor of 2 ; in this case a period of 23 seconds will

double the centripetal acceleration.

20. Picture the Problem: The Ferris wheel rotates

anticlockwise but is slowing down at a constant rate.

The Ferris wheel has a radius of 9.5 m and rotates once

every 32 s.

Strategy: Find the tangential acceleration of the

passenger at the top of the Ferris wheel and combine it

with the centripetal acceleration to find the total

acceleration.

Solution: 1. Use acp = rω

2 to

find cpa :

( )2

2 2

cp

2 rad9.5 m 0.37 m/s

32 sa r

πω

= = =

(downward)

2. Use equation at = rα to find

ta : ( )( )2 2

t 9.5 m 0.22 rad/s 2.1 m/sa rα= = − = − (to the left)

3. (c) Combine the components

to find a : ( ) ( )

2 22 2 2 2 2

cp t0.37 m/s 2.1 m/s 2.1 m/sa a a= + = + − =

4. (d) Find the angleφ and relate

it to the

direction of motion (which is to

the left)

2cp1 1

2

t

0.37 m/stan tan 10

2.1 m/s

a

aφ − −

= = = − ° −

or 170° below the

direction of motion.

Insight: In this case the tangential acceleration is 5.7 times greater than the centripetal

acceleration. The passengers will notice the slowing down more than they noticed the centripetal

acceleration when it was rotating at a constant rate.

21. Picture the Problem: The ball moves in a circle of constant radius at constant speed.

Strategy: The motion is approximately horizontal so we can neglect the fact that the rope would

be inclined a little bit below horizontal in order to support the weight of the ball. Set the rope

tension equal to the centripetal force required to keep the ball moving in a circle and solve for the

angular speed.

Solution: 1. (a) Set the string force

cpF ma= and

solve for ω : ( )( )

2

cp

11 N2.2 rad/s

0.52 kg 4.5 m

F ma mr

F

mr

ω

ω

= =

= = =

2. (b) Since ω is inversely proportional to r, the maximum angular velocity will increases if the

rope is shortened.

Insight: This is a fairly weak rope. Still, the problem illustrates well how the centripetal force

increases linearly with the distance from the rotation axis. Decreasing r decreases the force, or

allows a higher ω for the same amount of force.

22. Picture the Problem: The sanding disk rotates about its axis at a constant rate.

Strategy: Convert the angular speed of the disk into the linear speed of its rim by

multiplying by its radius v = rω. Use the same equation together with equation 10-5 to determine

the period of rotation for the given rim speed.

Top of Ferris Wheel

φ

ar

cpar

tar

r

vr


37 of 89

Solution: 1. (a) Apply v = rω directly: ( )( )4

t 0.00320 m 2.15 10 rad/s 68.8 m/sv rω= = × =

2. (b) Substitute 2 Tω π= and solve

for T: ( )t

5

t

2

2 0.00320 m27.31 10 s 73.1 s

275 m/s

v r r T

rT

v

ω π

ππµ−

= =

= = = × =

Insight: An angular speed of 2.15×104 rad/s is equivalent to 205,000 rev/min, or 3420 rev/s! Such

high speeds are necessary to get the linear speed of the rim of such a small tool up to a value where

it polishes well.

23. Picture the Problem: The wheel rotates about its axis, increasing its angular velocity at a constant

rate.

Strategy: Set the tangential and centripetal accelerations equal to each other for a single point on

the rim. Find an expression for the angular speed as a function of time ( 0 0ω = since the wheel

starts from rest), and substitute the expression into the result of the first step. Then solve the

equation for t.

Solution: 1. Set t cpa a= : 2

t cp

2

a r r aα ω

α ω

= = =

=

2. (b) Substitute 0 tω α= + and solve

for t:

( )2 2 2

21 1

t t

t t

α α α

α α

= =

= ⇒ =

Insight: The greater the angular acceleration, the shorter the elapsed time before the angular and

centripetal accelerations equal each other. After that the centripetal acceleration dominates

because it is proportional to 2 .α

24. Picture the Problem: The force is applied in a direction

perpendicular to the handle of the wrench and at the end of

the handle.

Strategy: Find the force from a knowledge of the torque and

the length of the wrench.

Solution: Solve for

F: ( )

( )

sin

15 N m60 N

sin 0.25 m sin 90

r F

Fr

τ θ

τ

θ

=

⋅= = =

°

Insight: A longer wrench can exert a larger torque for the same amount of force.

25. Picture the Problem: The weed is pulled by exerting a

downward force on the end of the tool handle.

Strategy: Set the torque on the tool equal to the force

exerted by the weed times the moment arm and solve for the

force.

Solution: Solve for F: weed weed

weed

weed

1.23 N m31 N

0.040 m

F r

Fr

τ

τ

=

⋅= = =

Insight: The torque must be the same everywhere on the tool. Therefore, the hand must exert a

1.23 N m 0.22 m 5.6 N⋅ = force to produce a 31-N force at the weed. The force is multiplied by a

factor of 22 4 5.5.=


38 of 89

26. Picture the Problem: The arm extends out either

horizontally or at some angle below horizontal, and the

weight of the trophy is exerted straight downward on

the hand.

Strategy: The torque equals the moment arm times the

force. In this case the moment arm is the horizontal

distance between the shoulder and the hand, and the

force is the downward weight of the trophy. Find the

horizontal distance in each case and multiply it by the

weight of the trophy to find the torque. In part (b) the

horizontal distance is

( )cos 0.605 m cos 22.5 0.559 m.r r θ⊥ = = ° =

Solution: 1. (a) Multiply the moment

arm by the weight: ( )( )( )20.605 m 1.51 kg 9.81 m/s 8.96 N mr mgτ ⊥= = = ⋅

2. (b) Multiply the moment arm by the

weight: ( )( )( )20.559 m 1.51 kg 9.81 m/s 8.28 N mr mgτ ⊥= = = ⋅

Insight: The torque on the arm is reduced as the arm is lowered. The torque is exactly zero when

the arm is vertical.

27. Picture the Problem: The arm extends out either horizontally and the weight of the crab trap is

exerted straight downward on the hand.

Strategy: The torque equals the moment arm times the force. In this case the moment arm is the

horizontal distance between the shoulder and the hand, and the force is the downward weight of

the crab trap.

Solution: Multiply the moment arm by

the weight: ( )( )( )20.70 m 3.6 kg 9.81 m/s 25 N mr mgτ ⊥= = = ⋅

Insight: If the man bent his elbow and brought his hand up next to his shoulder, the torque on the

shoulder would be zero but the force on his hand would remain 35 N.

28. Picture the Problem: The biceps muscle, the

weight of the arm, and the weight of the ball

all exert torques on the forearm as depicted at

right.

Strategy: Determine the torques produced by

the biceps muscle, the weight of the forearm,

and the weight of the ball. Sum the torques

together to find the net torque. According to

the sign convention, torques in the

anticlockwise direction are positive, and those

in the clockwise direction are negative.

Solution: 1. (a) Compute the

individual

torques and sum them:

( )( )

( )( )( )( ) ( )

2

ball

biceps

forearm

ball

biceps forearm ball

0.0275 m 12.6 N 0.347 N m

0.170 m 1.20 kg 9.81 m/s 2.00 N m

0.340 m 1.42 N 0.483 N m

0.347 2.00 0.483 N m 2.14 N m

r F

r mg

r W

τ

τ

τ

τ τ τ τ

⊥

⊥

⊥

= = = ⋅

= = = − ⋅

= = = − ⋅

= + +

= + − − ⋅ = − ⋅

∑

2. (b) Negative net torque means the clockwise direction; the forearm and hand will rotate


39 of 89

downward.

3. (c) Attaching the biceps farther from the elbow would increase the moment arm and increase the

net torque.

Insight: The biceps would need to exert a force of at least 90.3 N in order to prevent the arm from

rotating downward.

29. Picture the Problem: The adult pushes downward on the

left side of the teeter-totter and the child sits on the right

side as depicted in the figure:

Strategy: Calculate the torques exerted by the weight of

the child and the force of the parent’s hands and sum

them. The sign of the net torque indicates the direction in

which the teeter-totter will rotate.


the torque the child

exerts on the teeter-

totter.

( )( )( )child child child

2

child

1.5 m 16 kg 9.81 m/s

235 N m

r m gτ

τ

⊥=

= −

= − ⋅

2. Find the torque

exerted by the

parent and sum the

torques to find

the direction of

travel:

( )( )adult adult adult3.0 m 95 N 285 N mr Fτ ⊥= = = ⋅ . Here adult child 0τ τ+ > so the

teeter-totter will rotate anticlockwise and the child will move up.

3. (b) Repeat step 2

with the new

r⊥ for the adult:

( )( )adult adult adult2.5 m 95 N 238 N mr Fτ ⊥= = = ⋅ . Here

adult child 0τ τ+ > so the

teeter-totter will rotate anticlockwise and the child will move up.

4. (c) Repeat step 2

with the new

r⊥ for the adult:

( ) ( )adult adult adult2.0 m 95 N 190 N mr Fτ ⊥= = = ⋅ . Here adult child 0τ τ+ < so the

teeter-totter will rotate clockwise and the child will move down.

Insight: The parent would have to exert the 95-N force exactly 2.48 m from the pivot point in order

to balance the teeter-totter. We bent the rules for significant figures slightly to more easily

compare the magnitudes of the torques.

30. Picture the Problem: The ceiling fan rotates about its axis, decreasing its angular speed at a

constant rate.

Strategy: Determine the angular acceleration and then find the moment of inertia of the fan.

Solution: Solve for I:

( )( )( )

20.120 N m 19.5 s

0.883 kg m0 2.65 rad/s

tI

t

τ τ τ

α ω ω

− ⋅∆= = = = = ⋅

∆ ∆ ∆ −

Insight: Friction converts the fan’s initial kinetic energy of 2102

3.10 JIω = into heat.

31. Picture the Problem: The ladder rotates about its centre of mass, increasing its angular speed at a

constant rate.

Strategy: Find the moment of inertia of a uniform rod of mass M and length L that is rotated about

its centre of mass: 21

12I M L= . Then find the required torque to produce the acceleration.

Solution: 1. Find 21

12I M L= : ( )( )

22 21 1

12 128.22 kg 3.15 m 6.80 kg mI M L= = = ⋅

2. Apply τ = Iα directly: ( )( )2 26.80 kg m 0.302 rad/s 2.05 N mIτ α= = ⋅ = ⋅

adultr⊥childr⊥

childm gadultF


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32. Picture the Problem: The wheel rotates about its axis, decreasing its angular speed at a constant

rate, and comes to rest.

Strategy: Find the moment of inertia of a uniform disk and calculate I. Then find the angular

acceleration from the initial angular speed and the angle through which the wheel rotated. Use I

and α together to find the torque exerted on the wheel.

Solution: 1. (a) Find 21

2I MR= : ( )( )

22 21 1

2 26.4 kg 0.71 m 1.6 kg mI MR= = = ⋅

2. Solve for α :

( )( )

222 2

200 1.22 rad/s

0.158 rad/s2 2 0.75 rev 2 rad rev

ω ωα

θ π

−−= = = −

∆ ×

3. Apply τ = Iα directly: ( )( )2 21.6 kg m 0.158 rad/s 0.25 N mIτ α= = ⋅ − = − ⋅

4. (b) If the mass of the wheel is doubled and its radius is halved, the moment of inertial will be cut

in half (doubled because of the mass, cut to a fourth because of the radius). Therefore the

magnitude of the angular acceleration will increase if the frictional torque remains the same, and

the angle through which the wheel rotates before coming to rest will decrease.

Insight: If the moment of inertia is cut in half, the angular acceleration will double to −0.32 rad/s2

and the angle through which the wheel rotates will be cut in half to 0.38 rev. This is because the

wheel has less rotational inertia but the frictional torque remains the same. We bent the rules for

significant figures in step 2 to avoid rounding error in step 3.

33. Picture the Problem: The object consists of three

masses that can be rotated about any of the x, y, or z

axes, as shown in the figure at right.

Strategy: Calculate the moments of inertia about the x,

y, and z axes, and then find the required torque to give

the object an angular acceleration of 1.20 rad/s2 about

the various axes. Let m1 = 9.0 kg, m2 = 1.2 kg, and m3 =

2.5 kg.

Solution: 1. (a)

Calculate xI :

( )( )

2 2 2

1 1 2 2 3 3

2

2

9.0 kg 1.0 m 0 0

9.0 kg m

x

x

I m r m r m r

I

= + +

= + +

= ⋅

2. Find xτ : ( )( )2 29.0 kg m 1.20 rad/s 11 N mx xIτ α= = ⋅ = ⋅

3. (b) Calculate yI : ( )( )

2 20 0 2.5 kg 2.0 m 10 kg myI = + + = ⋅

4. Find yτ : ( )( )2 210 kg m 1.20 rad/s 12 N my yIτ α= = ⋅ = ⋅

5. (c) Calculate zI : ( )( ) ( )( )2 2 2

9.0 kg 1.0 m 0 2.5 kg 2.0 m 19 kg mzI = + + = ⋅

6. Find zτ : ( )( )2 2

19 kg m 1.20 rad/s 23 N mz zIτ α= = ⋅ = ⋅

Insight: When the axis of rotation passes through a particular mass, that mass does not contribute

to the moment of inertia because r = 0. The most torque is required to rotate the masses about the z

axis because that axis passes through the least amount of mass (only the 1.2-kg mass).

34. Picture the Problem: The fish exerts a torque on the fishing reel and it rotates with constant

angular acceleration.

Strategy: Determine the moment of inertia of the fishing reel assuming it is a uniform cylinder

( 21

2MR ). Find the torque the fish exerts on the reel. Then apply Newton’s Second Law for

rotation to find the angular acceleration and find the amount of line pulled from the reel.


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Solution: 1. (a) Find I: ( )( )22 21 1

2 20.84 kg 0.055 m 0.00127 kg mI MR= = = ⋅

2. Apply τ = rF directly to find τ : ( )( )0.055 m 2.1 N 0.12 N mr Fτ = = = ⋅

3. Solve τ = Ia for α :

2

2

0.12 N m92 rad/s

0.0013 kg mI

τα

⋅= = =

⋅

4. (b) Solve for s: ( ) ( ) ( )( )22 21 1

2 20.055 m 92 rad/s 0.25 s

0.16 m

s r r tθ α= = =

=

Insight: This must be a small fish because it is not pulling very hard. Or maybe the fish is tired?

35. Picture the Problem: The fish exerts a torque on the fishing reel and it rotates with constant

angular acceleration.

Strategy: Determine the moment of inertia of the fishing reel assuming it is a uniform cylinder

( 21

2MR ). Find the net torque on the reel by subtracting the torque from the friction clutch from the

torque due to the force the fish exerts. Then apply Newton’s Second Law for rotation to find the

angular acceleration and find the amount of line pulled from the reel.

Solution: 1. (a) Find I: ( )( )22 21 1

2 20.84 kg 0.055 m 0.00127 kg mI MR= = = ⋅

2. Apply τ = rF directly to find τ : ( )( )0.055 m 2.1 N 0.047 N m 0.069 N mr F Cτ = − = − ⋅ = ⋅

3. Solve τ = Iα for α :

2

2

0.069 N m54 rad/s

0.00127 kg mI

τα

⋅= = =

⋅

4. (b) Solve for s: ( ) ( ) ( )( )22 21 1

2 20.055 m 54 rad/s 0.25 s

0.093 m 9.3 cm

s r r tθ α= = =

= =

Insight: Less line is pulled because the friction clutch reduces the net torque and angular

acceleration of the reel. We bent the rules for significant figures in steps 1 and 2 to avoid rounding

errors in subsequent steps. For instance, if we follow the rules of subtraction in step 2,

0.12 0.047 N m 0.07 N m,τ = − ⋅ = ⋅ just one significant figure.


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Solutions to Tutorial 5 1. Picture the Problem: The position of the mass oscillating on a spring is given by the equation of

motion.

Strategy: The oscillation period can be obtained directly from the argument of the cosine

function. The mass is at one extreme of its motion at t = 0, when the cosine is a maximum. It then

moves toward the center as the cosine approaches zero. The first zero crossing will occur when

the cosine function first equals zero, that is, after one-quarter period.

Solution: 1. (a) Identify T with the time

0.58 s:

Since 2 2

cos cos0.58

t tT s

π π=

, therefore T = 0.58 s.

2. (b) Multiply the period by one-quarter to

find the first zero crossing: ( )1

0.58 s 0.15 s .4

t = =

Insight: A cosine function is zero at ¼ and ¾ of a period. It has its greatest magnitude at 0 and ½

of a period.

2. Picture the Problem: As the mass oscillates on the spring we can find its position at any given

time from the equation of motion.

Strategy: The oscillation period can be obtained directly from the argument of the cosine

function. The frequency is the inverse of the period. The mass is at one extreme of its motion at t

= 0, when the cosine is a maximum. The mass is at the point of interest when the cosine function

is equal to −1. This occurs one-half a period later.

Solution: 1. (a) Observe that in the argument

of

cosine the period is in the denominator:

Since 2 2

cos cos0.68 s

t tT

π π =

, therefore T =

0.68 s

2. Invert the period to obtain the frequency: 1

1.5 Hz .0.68 s

f = =

3. (b) The time the mass is at −7.8 cm is half a

period:

1 1(0.68 s) 0.34 s .

2 2t T= = =

Insight: This problem could also be solved by setting the motion equation equal to x = −7.8 cm

and solving for the time: ( ) ( )10.68 s 2 cos 7.8 cm 7.8 cm 0.34 s.t π −= − =

3. Picture the Problem: When two or more atoms are bound in a molecule they are separated by an

equilibrium distance. If the atoms get too close to each other the binding force is repulsive. When

the atoms are too far apart the binding force is attractive. The nature of the binding force therefore

is to cause the atoms to oscillate about the equilibrium distance.

Strategy: Since the mass starts at x = A at time t = 0, this is a cosine function given

by ( )cosx A tω= . From the data given we need to identify the constants A and ω . A cosine

function is at its maximum at t = 0, but a sine function equals zero at t = 0.

Solution: 1. (a) Identify the amplitude as A: 3.50 nmA =

2. Calculate the angular frequency from the

frequency:

14 12 4.00 10 sfω π π −= = ×

3. Substitute the amplitude and angular

frequency

into the cosine equation:

( ) ( )14 13.50 nm cos 4.00 10 sx tπ − = ×

4. (b) It will be a sine function, ( )sinx A tω= , since sine satisfies the initial condition of x = 0 at t

= 0.

Insight: A cos function has a maximum amplitude at t = 0. A sin function has 0 amplitude at t = 0.


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4. Picture the Problem: One period of oscillation is shown in the

figure.

Strategy: Since the mass is at x = 0 at t = 0, this will be a sine

function. Substitute the amplitude and period into the sine

equation to determine the general equation of motion. Finally

substitute in the specific times to determine the position at each

time.

Solution: 1. Write the equation

sine equation

in terms of the given amplitude

and period:

( )2

0.48 cm cosx tT

π =

2. (a) Substitute t = T/8 into the

sine equation

and evaluate:

( ) ( )2

0.48 cm cos 0.48 cm cos 0.34 cm8 4

Tx

T

π π = = =

3. (b) Substitute t = T/4 into the

sine equation

and evaluate:

( ) ( )2

0.48 cm cos 0.48 cm cos 0.48 cm4 2

Tx

T

π π = = =

4. (c) Substitute t = T/2 into the

sine

equation and evaluate:

( ) ( ) ( )2

0.48 cm cos 0.48 cm cos 02

Tx

T

ππ

= = =

5. (d) Substitute t = 3T/4 into the

sine

equation and evaluate:

( )

( )

2 30.48 cm cos

4

30.48 cm cos

2

0.48 cm

Tx

T

x

π

π

=

=

= −

6. (e) Sketch a plot with the four data points:

Insight: The sine curve has been included in the sketch of part

(e) to show that the four positions are consistent with a sine

function.

5. Picture the Problem: A mass is attached to a spring. The mass is displaced from equilibrium and

released from rest. The spring force causes the mass to oscillate about the equilibrium position in

harmonic motion.

Strategy: Since the mass starts from rest at t = 0, the harmonic equation will be a sine function.

We will use the amplitude and period to determine the general equation, which can be evaluated

for any specific time. During the first half of each period the mass will be moving in the negative

x-direction toward the minimum and during the second half the mass will move in the positive

direction, back toward the maximum. Therefore we can determine the direction of motion by

finding in which half of a period the time is located.

Solution: 1. (a) Insert the period and

amplitude to

create the general harmonic equation:

( )2

0.0440 m cos . 3.15 s

tx

π =

2. Insert t = 6.37 s into the general

equation and

evaluate the position:

( )( )2 6.37 s

0.0440 m cos 0.0436 m3.15 s

xπ

= =

3. (b) Divide the time by one period:

6.37 s2.02

3.15 s=

4. Since this is slightly greater than two full periods, the mass is headed toward the origin from its

maximum displacement and is moving in the negative x direction.


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Insight: This problem could also be solved by inserting a time slightly later than t = 6.37 s (such as

t = 6.38 s) and evaluating the position (x = 0.0434 m). Since this result is smaller than 0.0436 m,

the mass is moving in the negative

x direction.

6. Picture the Problem: One velocity over one period is shown in the

figure. The region for which the speed (|v|) is greater than vmax/2 is

shaded gray.

Strategy: Since speed is the magnitude of the velocity it will be

greater than vmax/2 twice during each period. The two regions are

symmetric; the total time is double the time interval over which the

velocity is positive. The end points of the regions are found when the

sine function is equal to one-half.

Solution: 1. Set v = -Aωsin(ωt) equal to

vmax/2 and divide out vmax: max max

2 1sin

2

2 1sin

2

tv v

T

t

T

π

π

=

=

2. Take the arcsine of each side and solve for

t:

1

1

2 1sin

2

1sin

2 2

t

T

Tt

π

π

−

−

=

=

3. Evaluate using

1 1 5sin and

2 6 6

π π− =

:

5 5 or or

2 6 2 6 12 12

T T T Tt

π π

π π

= =

4. Subtract the first time from the second time

and multiply by two:

5 4 22 2 .

12 12 12 3

T T T T − = =

The mass’s speed is greater than vmax/2 for

two-thirds of a cycle.

Insight: If the problem had asked for the time that the velocity was greater than +vmax/2, then only

the crest of the cycle would have been included and the time would have been one-third of a cycle.

7. Picture the Problem: When an object is oscillating in simple harmonic motion it experiences a

maximum acceleration when it is displaced at its maximum amplitude. As the object moves

toward the equilibrium position, the acceleration decreases and the velocity of the object increases.

The object experiences its maximum velocity as it passes through the equilibrium position.

Strategy: The maximum velocity and acceleration can both be written in terms of the amplitude

and angular speed, 2

max max,v A a Aω ω= = . We can rearrange these equations to solve for the

amplitude and angular speed. Then we can use the angular speed to determine the period.

Solution: 1. (a) Divide the square of the velocity

by the acceleration to find the amplitude:

( )2 2

max

2

max

A vA

aA

ω

ω= =

2. (b) Divide the acceleration by the velocity to

determine the angular speed:

2

max

max

aA

A v

ωω

ω= =

3. Divide 2 π by the angular speed to calculate the

period:

max

maxmax

max

22 2 vT

aa

v

ππ π

ω= = =

Insight: When two or more quantities are functions of the same variables, it is often possible to


45 of 89

rearrange the equations to isolate one or more of the variables. This can be a useful mathematical

procedure.

8. Picture the Problem: When an object is oscillating in simple harmonic motion it experiences a

maximum acceleration when it is displaced at its maximum amplitude. As the object moves

toward the equilibrium position the acceleration decreases and the velocity of the object increases.

The object experiences its maximum velocity as it passes through the equilibrium position.

Strategy: The maximum velocity and acceleration can both be written in terms of the amplitude

and angular speed, 2

max max,v A a Aω ω= = . Rearrange these equations to solve for the amplitude

and angular speed. Then use the angular speed to determine the period.

Solution: 1. (a) Divide the square of the

velocity by the acceleration to find the

amplitude:

( ) ( )

( )

2 22

max

2 2max

4.3 m/s28 m

0.65 m/s

A vA

aA

ω

ω= = = =

2. (b) Divide the acceleration by the

velocity to determine the angular speed:

2

max

max

A

A

a

v

ωω

ω= =

3. Divide 2 π by the angular speed to

calculate T:

( )

( )max

2maxmax

max

2 4.3 m/s22 242 s

0.65 m/s

vT

aa

v

πππ π

ω= = = = =

Insight: When two or more quantities are functions of the same variables, it is often possible to

rearrange the equations to uniquely determine those variables.

9. Picture the Problem: As the child rocks back and forth on the swing, her speed increases as she

approaches the equilibrium of the swing and then decreases back to zero at the end of the swing.

The maximum speed occurs when the swing is vertical.

Strategy: The maximum velocity equals the amplitude times the angular speed, which in turn

depends upon the period.

Solution: 1. Write the

maximum velocity in terms

of amplitude and period:

( )max

2 20.204 m 0.458 m s

2.80 sv A A

T

π πω

= = = =

2. Insert the amplitude and

period into the equation for

maximum speed:

( )max

2 0.204 m20.458 m s

2.80 sv A

T

ππ = = =

Insight: The girl’s motion has an amplitude of 20.34cm and a maximum speed of 1.64 km/h. Not

a very exciting swing.

10. Picture the Problem: A structural beam is a metal rod that is necessary to maintain the shape

and integrity of the spacecraft. Large forces, which could be caused by small, but rapid

oscillations, could damage the support beam, jeopardizing the integrity of the spacecraft.

Strategy: The maximum acceleration can be written in terms of the amplitude and angular speed, 2

maxa Aω= , where angular speed is 2π times the frequency.

Solution: 1. Multiply the amplitude by the square

of 2π times the frequency to get maxa :

( )

( )( )

22

max

2 2

2

0.25 mm 2 110 Hz = 119 m/s

a A A fω π

π

= =

= ×

2. Factor out g = 9.81 m/s2 to obtain the acceleration

as a multiple of g: 2

max 2119 m/s 12

9.81 m/s

ga g

= =

Insight: Since the acceleration is proportional to the square of the frequency, a large frequency

will result in a very large acceleration. This is true even for small amplitude oscillations.


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11. Picture the Problem: The figure shows a turntable rotating

with a peg on its outer rim. The table is illuminated on one

side. The shadow of the peg moves with simple harmonic

motion along the wall.

Strategy: The shadow of the peg moves along the wall with

simple harmonic motion. The period is the time for the peg

(and as such the shadow) to complete a full revolution. The

amplitude is the same as the radius of the circle, and is also the

maximum distance from the center. Knowing the period and

amplitude we can calculate the maximum velocity,

remembering that the angular frequency is inversely related to

the period. Finally, use the amplitude and period to calculate

the maximum acceleration.

Solution: 1. (a) Divide the

circumference of the turntable by

the peg’s tangential velocity to get

the period of rotation:

( )2 0.25 m22.3 s

0.67 m/s

C rT

v v

ππ= = = =

2. (b) Set the amplitude equal to

the radius of the turntable: 0.25 mA r= =

3. (c) Insert the period and

amplitude into the maximum

velocity acceleration:

max

20.67 m s

A Avv A v

T r

πω= = = = =

4. (d) Insert the period and

amplitude into the maximum

acceleration equation:

( )22 2 2

2 2

max

0.67 m/s21.8 m/s

0.25 m

v va A A A

T r r

πω

= = = = = =

Insight: The maximum speed of the shadow is equal to the tangential speed of the peg. The

shadow and the peg travel at the same speed when the peg travels perpendicular to the light

source. The shadow travels slower than the peg when a component of the peg’s velocity is

parallel to the light.

12. Picture the Problem: In an engine the moving pistons compress the fuel in the chamber and

expand after the fuel has been ignited. This motion provides the power to the car. The frequency

of the piston motion is measured by the number of revolutions of the crankshaft per minute

(rev/min).

Strategy: To solve for the maximum acceleration and speed we must first convert the angular

speed from rev/min to rad/s. Then we can use the amplitude and angular speed in the equations

for maximum acceleration and maximum speed.

Solution: 1. Convert the angular

speed to rad/s:

-1rev 2 rad 1 min1700 178 s

min rev 60 s

π =

2. (a) Use the amplitude and angular

speed to solve

for maximum acceleration:

( )2

2 1 2

max 3.5 cm 178 s 1.1 km/sa Aω −= = =

3. (c) Use the amplitude and angular speed

to solve

for maximum speed:

( )1

max 3.5 cm 178 s 6.2 m/sv Aω −= = =

Insight: The maximum acceleration of the pistons is over 100 times the acceleration due to

gravity. Therefore the gravitational force has negligible effect on the motion of a working piston.

13. Picture the Problem: An air cart is attached to the end of a spring and pulled slightly away from

equilibrium and released. It oscillates about the equilibrium position at a frequency determined by

the mass of the cart and the stiffness of the spring.

Strategy: The maximum kinetic energy can be obtained by inserting the maximum velocity into

the kinetic energy equation. The maximum force can be obtained by substituting the maximum

acceleration into Newton’s Second Law. From the position equation we see that the amplitude is

10.0 cmA = and the angular speed is 2.00 rad/sω = . We will use these values to calculate the


47 of 89

maximum velocity and acceleration.

Solution: 1. (a) Write the

kinetic energy in terms of mass,

amplitude, and angular speed:

( )22

max max

1 1

2 2K mv m Aω= =

2. Insert the values of mass,

amplitude and angular speed to

calculate maxK :

( ) ( )( ) ( )22 2 1

max

1 10.64 kg 0.100 m 2.00 s 0.13 J

2 2K m Aω −= = =

3. (b) Write the force equation

in terms of mass, amplitude,

and angular speed:

( )2

max maxF ma m Aω= =

4. Insert the values of mass,

amplitude and angular speed to

calculate the maximum force:

( ) ( )( )( )2

2 1

max0.64 kg 0.100 m 2.00s 0.26 NF m Aω −= = =

Insight: The phase shift of +π in the displacement equation has no effect on either maxK or maxF ,

but it reverses the sign of the displacement. Therefore, at t = 0 the cart starts x = −10 cm instead of

at x = 10 cm.

14. Picture the Problem: A mass attached to a spring is pulled slightly away from equilibrium and

released. The mass then oscillates about the equilibrium position at a frequency determined by the

stiffness of the spring.

Strategy: We can determine the spring constant by solving the equation for the period of a mass

on a spring for the spring constant, and substituting in the given period and mass.

Solution: 1. Solve the period

equation for

the spring constant:

22

2 m

T k mk T

ππ

= ⇒ =

2. Insert the numeric values for T

and m: ( )

2 22 2

= 0.42 kg 29 N m0.75 s

k mT

π π = =

Insight: Measuring the period of oscillation is in many cases the most accurate way of measuring

a spring constant.

15. Picture the Problem: The picture shows the unstretched

spring and the spring with a 0.50-kg mass attached to it.

Strategy: We can use the displacement of the spring to

calculate the spring constant. The spring constant and period

can then be inserted into the period equation to solve for the

necessary mass.

Solution: 1. Use the spring

force equation

to solve for the spring constant:

F mg

F ky ky y

= ⇒ = =

2. Insert numeric values to

obtain k:

( )2

2

0.50 kg 9.81 m/s32.7 N/m

15 10 mk

−= =

×

3. Solve the period equation for

the mass:

2

2 2

m TT m k

kπ

π

= ⇒ =

4. Insert numeric values to

obtain the mass: ( )

20.75 s

32.7 N/m 0.47 kg2

mπ

= =

Insight: Since the period is proportional to the square root of the mass, increasing the mass will

increase the period.


48 of 89

16. Picture the Problem: When the two people enter the car they compress the springs. The distance

that the springs are compressed is regulated by their mass and the stiffness of the spring. When the

car hits a bump in the road the car begins to oscillate up and down at a frequency determined by

the total mass of the car and riders and the stiffness of the springs.

Strategy: Using the mass of the two people and the amount the springs compressed, we can

calculate the spring constant. The total load can be obtained by solving the period of oscillation

equation for the mass. The mass of the car is found by subtracting the mass of the two people

from the total mass.

Solution: 1. Solve the force equation for the

spring constant:

F mg

F ky ky y

= ⇒ = =

2. Enter numeric values for the spring

constant:

( )2

4125 kg 9.81 m/s

1.53 10 N/m0.0800 m

k = = ×

3. (a) Solve the period equation for the total

load (M+m):

2

2 2

M m TT M m k

kπ

π

+ = ⇒ + =

4. Enter numeric values for the total mass: ( )

2

41.65 s1.53 10 N/m 1060 kg

2M m

π

+ = × =

5. (b) Subtract the mass of the two people to

obtain M: ( ) 1057 kg 125 kg 932 kgM M m m= + − = − =

Insight: The period of oscillation is an excellent method of determining the mass of an object.

This is especially useful in orbit, where conventional scales do not work.

17. Picture the Problem: A mass attached to a vertical spring, pulled slightly down from the

equilibrium position and released will oscillate in simple harmonic motion. The acceleration of the

mass will be a maximum when the spring is at maximum displacement. As the mass moves back

to the equilibrium position the speed increases and the acceleration decreases. The maximum

speed is at equilibrium position. As the mass moves away from equilibrium the velocity decreases

as the deceleration increases until the mass stops at the opposite amplitude.

Strategy: The period can be found from the spring constant and mass. From the maximum speed

and the period we can calculate the amplitude. We can calculate the maximum acceleration from

the period and the amplitude.

Solution: 1. (a) Insert the mass

and spring

constant into the period

equation:

0.85 kg2 2 0.47 s

150 N/m

mT

kπ π= = =

2. (b) Solve the maximum

velocity equation

for the amplitude:

( )max max

0.4730 s 0.35 m/s2.6 cm

2 2

v TvA

ω π π= = = =

3. (c) Insert the amplitude and

period into the maximum

acceleration equation:

2 2

2 2

max

2 20.02635 m 4.6 m s

0.4730 sa A A

T

π πω

= = = =

Insight: Another way to solve this problem is to calculate the angular speed k mω = instead of

the period. Then maxA v ω= and max maxa v ω= .

18. Picture the Problem: If the motorcycle is pushed down slightly on its springs it will oscillate up

and down in harmonic motion. A rider sitting on the motorcycle effectively increases the mass of

the motorcycle and oscillates also.

Strategy: We can use the equation for the period of a mass on a spring. Writing this equation for


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the motorcycle without rider and again for the motorcycle with rider we can calculate the percent

difference in the periods.

Solution: 1. (a) The period increases, because the person’s mass is added to the system and

.T m∝

2. (b) Write the equation for the period of the

motorcycle

without the rider:

2m

Tk

π=

3. Write the equation for the period of the

motorcycle with the rider: 2 2

m MT

kπ

+=

4. Calculate the percent difference between the

two periods:

2

2 2

2

m M m

T T k k

T m

k

π π

π

+−

−=

5. Simplify by factoring out 2 m kπ from the

numerator

and denominator:

2 1

511 122 kg1 0.104 10.4%

511 kg

T T m M

T m

− += −

+= − = =

Insight: The percent change in the period does not depend on the spring constant. It only depends

on the fractional increase in mass.

19. Picture the Problem: The mass on the left is hung from one

spring. The mass on the right is hung from two identical springs.

Strategy: When a mass is attached to a single spring it stretches

by a distance x. When two identical springs are connected end to

end and the same mass is attached, each spring will stretch by the

same amount and the total stretch will be 2x. Since the force has

not changed, but the total stretch is twice as much, the effective

spring constant will be half the single spring constant. Use this

information to find a relationship between the two periods.

Solution:(a) The period is more than the period of a single

spring because the effective spring constant is smaller and

1 .T k∝

2. (b) Write the combined spring constant as one-

half of the single spring constant:

1

2k k′ =

3. Replace k’ with k/2 in the period equation:

( )2 2

/ 2

m mT

k kπ π′ = =

′

4. Rearrange the equation: 2 2 2 2

m mT

k kπ π

′ = =

5. Replace the term in parentheses with the

original period: 2 2 2

mT T

kπ

′ = =

Insight: If the spring had been replaced by 3 identical springs the resulting spring constant would

be / 3k k′ = , giving 3T T′ = . In general the period will be proportional to the square-root of the

length of the spring.


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20. Picture the Problem: A mass is attached to the end of a 2.5-meter-long string, displaced slightly

from the vertical and released. The mass then swings back and forth through the vertical with a

period determined by the length of the string.

Strategy: Use the period of the pendulum and its length to calculate the acceleration of gravity.

Solution: 1. Solve the period equation for

gravity:

22

2 L

T g Lg T

ππ

= ⇒ =

2. Insert numeric values: ( )( )

2

2

15

22.5 m 9.6 m/s

16 sg

π = =

Insight: The small variations in gravity around the surface of the Earth are measured using the

period of a pendulum.

21. Picture the Problem: A simple pendulum is a mass attached to a string. The mass is displaced so

the string is slightly away from the vertical and released. The mass then oscillates about the

vertical with a period determined by the length of the string and gravity.

Strategy: Calculate the length of the pendulum from its period.

Solution: 1. Solve the period equation for

length:

2

2 2

L TT L g

gπ

π

= ⇒ =

2. Insert numeric values: ( )

2

21.00 s9.81 m/s 24.8 cm

2L

π

= =

Insight: This is the length of the pendulum in many older clocks. Larger clocks, such as a

grandfather clock, have pendulums about a meter long with a period of 2 seconds.

22. Picture the Problem: The pendulum on the Moon is the same length string and mass, with the

mass displaced from the vertical and released. The period is determined by the length of the string

and the acceleration due to gravity.

Strategy: Since the period of a pendulum is inversely proportional to the square-root of gravity,

the smaller gravitational pull on the Moon would increase the period of the pendulum. The period

of the pendulum on the Moon can be calculated by replacing the acceleration of gravity on the

Earth with the acceleration of gravity on the Moon, ( 1Moon Earth6

g g= ).


period on the Moon in terms

of the period on Earth:

Moon Earth1Earth Earth6

2 6 2 6L L

T Tg g

π π= = ⋅ = ⋅

2. Calculate the period on the Moon: ( )Moon 6 1.00 s 2.45 sT = =

Insight: A grandfather clock taken to the Moon would run 2.45 times slower than one on the

Earth. To run properly, the pendulum in the clock would need to be shortened to one-sixth of its

original length.

23. Picture the Problem: A pendulum is made by attaching a mass to the end of a string. The

opposite end of the string is attached to the ceiling of an elevator. The mass is displaced slightly

from vertical and released. The pendulum oscillated back and forth with a period determined by

the length of the string and the effective gravity experienced in the elevator.

Strategy: The acceleration of gravity is that felt by the pendulum. As an object accelerates

upward it experiences an effective gravity g + a. When the elevator accelerates downward the

effective gravity is or g – a. To solve for the accelerated period we can substitute the effective

gravity into the period equation.

Solution: 1. (a) Replace “g” by “g + a” in the

equation for the period of a pendulum: 2L

Tg a

π=+


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2. (b) ) Replace “g” by “g − a” in the equation

for

the period of a pendulum:

2L

Tg a

π=−

Insight: Consider the effect on the pendulum if the elevator were to be in free fall. According to

our answer to part (b), as the downward acceleration approaches g, the period increases. In the

limit that a g→ , T → ∞ . If the elevator were in free fall the tension in the pendulum string

would be zero and the pendulum would not oscillate.


52 of 89

Solutions to Tutorial 6 1. Picture the Problem: The image shows a wave with

the given wave dimensions.

Strategy: Set the wavelength equal to the horizontal

crest-to-crest distance, or double the horizontal crest-to-

trough distance. Set the amplitude equal to the vertical

crest-to-midline distance, or half the vertical crest-to-

trough distance.

Solution: 1. (a) Double the horizontal crest-to-

trough distance: ( )2 26 cm 52 cmλ = =

2. (b) Halve the vertical crest-to-trough

distance: ( )1

211 cm 5.5 cmA = =

Insight: Note the difference in wavelength and amplitude. The wavelength is the entire distance

from crest to crest, but amplitude is only from the equilibrium point to the crest.

2. Picture the Problem: A surfer measures the frequency and length of the waves that pass her.

From this information we wish to calculate the wave speed.

Strategy: Write the wave speed as the product of the wavelength and frequency.

Solution: Multiply wavelength by

frequency: ( )( )

1 min34 m 14 /min 7.9 m/s

60 secv fλ

= = =

Insight: The wave speed can increase by either an increase in wavelength or an increase in

frequency.

3. Picture the Problem: The image shows

water waves passing to a shallow region

where the speed decreases. We need to

calculate the wavelength in the shallow

area.

Strategy: Calculate the frequency in the deep water. Then use the constant frequency and the speed

in the shallow water to calculate the new wavelength.


frequency:

1

1

2.0 m/s1.333 Hz

1.5 m

vf

λ= = =

2. Calculate the new wavelength: 2

2

1.6 m/s1.2 m

1.333 Hz

v

fλ = = =

Insight: Note that decreasing the speed, with constant frequency, will decrease the wavelength.

4. Picture the Problem: The speed and wavelength of a tsunami are given and we wish to calculate

the frequency.

Strategy: Solve for the frequency.

Solution: Calculate the frequency:

( ) 4750 km/h 1 h

6.7 10 Hz310 km 3600 s

vf

λ−

= = = ×

Insight: Although the tsunami has a very high speed, the long wavelength gives the tsunami a low

frequency.


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5. Picture the Problem: A wave of known amplitude, frequency, and wavelength travels along a string.

We wish to calculate the distance travelled horizontally by the wave in 0.5 s and the distance

travelled by a point on the string in the same time period.

Strategy: Multiply the time by the wave speed to calculate the horizontal distance travelled by the

wave. A point on the string travels up and down a distance four times the amplitude during each

period. Calculate the fraction of a period by dividing the time by the time of a full period. Set the

period equal to the inverse of the period and multiply by four times the amplitude to calculate the

distance traveled by a point on the string.

Solution: 1. (a) Calculate the

horizontal distance: ( ) ( )( )( )2

w 27 10 m 4.5 Hz 0.50 s 0.61 md vt f tλ −= = = × =

2. (b) Calculate the vertical

distance: ( ) ( )( )( )2

k 4 4 4 12 10 m 4.5 Hz 0.50 s 1.1 mt

d A AftT

− = = = × =

3. (c) The distance travelled by a wave peak is independent of the amplitude, so the answer in part (a)

is unchanged. The distance travelled by the knot varies directly with the amplitude, so the answer in

part (b) is halved.

Insight: A point on the string travels four times the wave amplitude in the same time that the crest

travels one wavelength.

6. Picture the Problem: Using the equation for the speed of deep water waves given in the problem we

want to calculate the speed and frequency of the waves.

Strategy: Insert the given data into the equation 2v gλ π= to solve for the speed of the waves.

Then calculate the wave frequency.

Solution: 1. (a) Insert the frequency

into the deep water velocity equation: ( ) ( )( ) ( )2/ 2 9.81 m/s 4.5 m / 2 2.65 m sv gλ π π= = =

2. (b) Solve for the frequency:

2.651 m/s0.59 Hz

4.5 m

vf

λ= = =

Insight: Since the velocity is proportional to the square-root of the wavelength, the frequency is

inversely proportional to the square-root of the wavelength. Increasing the wavelength by a factor of

four will double the wave speed and cut the frequency in half.

7. Picture the Problem: The speed of shallow water waves is proportional to the square-root of the

water depth. We wish to calculate the speed and frequency of some shallow water waves.

Strategy: Use the speed equation v gd= given in the problem, where d is the water depth, to

calculate the wave speed. Then calculate the wave frequency.

Solution: 1. (a) Calculate the wave

speed: ( )( )29.81 m/s 0.026 m 0.51 m sv gd= = =

2. (b) Calculate the wave frequency: 0.505 m/s

67 Hz0.0075 m

vf

λ= = =

Insight: As the wave approaches shallower water, with constant frequency, its wavelength

decreases. In this problem, if the depth drops to 1.3 cm, the wavelength will decrease to 0.59 cm.

8. Picture the Problem: The string tension is changed until the wave speed doubles.

Strategy: The speed of a wave on a string is given by v = √(F/µ). Solve the equation for the

tension in the string. Then use a ratio to find the factor by which the tension increases.

Solution: 1. Solve for the tension: 2 v F F vµ µ= ⇒ =

2. Divide the tension at higher velocity by the

initial tension:

2 22

2 2 2

2

32 m/s4

16 m/s

F v v

F vv

µ

µ

= = = =

The tension increases by a factor of 4.

Insight: The tension increases by a factor equal to the square of the fractional increase in velocity.


54 of 89

9. Picture the Problem: The image shows two

people talking on tin can telephone. The cans

are connected by a 9.5-meter-long string

weighing 32 grams. We wish to calculate the

time it takes for a message to travel across the

string.

Strategy: Set the time equal to the distance

divided by the velocity. The linear mass

density is the total mass divided by the length.

Solution: 1. Set

the time equal to

the distance

divided by

velocity:

dt d

v F

µ= =

2. Substitute m dµ = and insert

numerical values:

( )( )0.032 kg 9.5 m/0.19 s

8.6 N

m d mdt d

F F= = = =

Insight: The message travels the same distance in the air in 0.028 seconds, about 7 times faster.

10. Picture the Problem: The image shows two

people talking on tin can telephone. The cans

are connected by a 9.5-meter-long string

weighing 32 grams. We wish to determine

how the tension in the string affects the time

for the message to travel across the string.

Strategy: In problem 9, we found that the

travel time across the string is given by

/t md F= . Use this equation to calculate

the time for the different tensions.

Solution: 1. (a) Since the time is inversely related to the tension, increasing the tension will result

in less time.

2. (b) Set the tension equal to 9.0 N: ( )( ) ( )0.032 kg 9.5 m / 9.0 N 0.18 st = =

3. (c) Set the tension equal to 10.0 N: ( ) ( ) ( )0.032 kg 9.5 m / 10.0 N 0.17 st = =

Insight: As predicted, increasing the tension decreases the time for the message to travel the

string.

11. Picture the Problem: Sound takes 0.94 seconds to travel across a wire of known length and

density. We want to calculate the tension in the wire.

Strategy: Solve for the tension in the wire. The velocity is given by the length of the wire

divided by the time for the sound to travel across it. The linear mass density is the mass divided by

the length.

Solution: 1. (a) Solve for the tension:

2

2

2

Fv

m L mLF v

L t t

µ

µ

=

= = =


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2. Insert the given mass, length and time:

( )( )

( )2

0.085 kg 7.3 m0.70 N

0.94 sF = =

3. (b) The mass is proportional to the tension (if L and t remain constant). So increased mass

means increased tension.

4. (c) Solve with a mass of 0.095 kg.

( )( )

( )2

0.095 kg 7.3 m0.78 N

0.94 sF = =

Insight: A heavier string requires greater tension for a wave to travel across it in the same time.

12. Picture the Problem: Waves travel down two strings, made of the same material and having the

same length, but having different diameters and tensions. We wish to calculate the ratio of the

wave speeds on these two strings.

Strategy: Calculate the ratio of the velocities. Set the linear mass densities equal to the density

of steel times the cross-sectional area of the wires.

Solution: 1. (a) Write the ratio of the velocities: A A A A

B B B B

/

/

B

A

v F F

v F F

µ µ

µ µ= =

2. Write the linear mass density in terms of density

and area:

A A A

B B B

B B

A A

v F A F A

v F A F A

ρ

ρ= =

3. Write the area in terms of the diameter: ( )

( )

2

A A A

2

B B B

2

2

B B

AA

dv F F d

v F F dd

π

π= =

4. Insert the given tensions and diameters: A

B

410 N 1.0 mm2 1.4

820 N 0.50 mm

v

v

= = =

Insight: The ratio of the velocities is proportional to the square root of the tensions and inversely

proportional to the diameters.

13. Picture the Problem: The speed of a wave on a string depends on the tension, radius and density

of the string. We wish to use dimensional analysis to create an equation relating the speed to these

parameters.

Strategy: Set the dimensions of speed, [L]/[T] equal to the dimensions of tension, [M][L]/[T]2;

radius, [L]; and density [M]/[L]3 each raised to the powers α, β, and γ respectively and solve for

the powers.

Solution: 1. Write the dimensions

of speed in terms of the product of

powers of the dimensions of force,

diameter, and density:

[ ][ ]

[ ][ ]

[ ][ ]

[ ]

[ ]

[ ] [ ] [ ] [ ] [ ]

2 3

1 1 3 2

L M L ML

T T L

L T M L T

α γ

β

α γ α β γ α− + + − −

=

=

2. Use the dimensions of time to calculate

α: [ ] [ ]

1 2

1

2

2 1

T Tα

α

α

− −=

− = −

=

3. Use the dimensions of mass to

determine γ: [ ] [ ]

0

1

2

0

M Mα γ

α γ

γ α

+=

+ =

= − = −


56 of 89

4. Use the dimensions of length to

determine β: [ ] [ ]

( )

1 3

1 1

2 2

1 3

1 3 1 3( ) 1

L Lα β γ

α β γ

β α γ

+ −=

= + −

= − + = − + − = −

5. Use the dimensions to write a

dimensionally correct equation for

velocity:

1 12 21

2

Tv T R

Rρ

ρ

−−∝ ∝

Insight: The exact velocity equation cannot be derived from dimensional analysis because of

nondimensional constants. However, from this analysis we can determine that doubling the radius

would cut the velocity in half. Doubling the tension, increases the velocity by a factor of 2 .

14. Picture the Problem: The picture depicts a

person shouting toward a distant cliff and

hearing her echo. We want to calculate the

distance to the cliff based on the time to

hear the echo.

Strategy: Between the shout and hearing

the echo the sound has travelled to the cliff

and back, or twice the distance to the cliff.

Multiply the speed of sound by the time

lapse to calculate the distance the sound has

travelled. The cliff will be one-half of this

distance away.

Solution: Calculate the distance to

the cliff: ( )( )

1 1343 m/s 1.85 s 317 m

2 2d vt= = =

Insight: On a cold morning, when the speed of sound is only 320 m/s, it would take the echo 1.98

seconds to be heard.

15. Picture the Problem: The dolphin sends a signal to the ocean floor and hears its

echo.

Strategy: We want to calculate the time the elapses before the dolphin hears the

echo and the wavelength of the sound in the ocean. The wave must travel to the

ocean floor and back before it is heard. So the distance travelled is twice the

distance to the floor. Divide this distance by the speed of sound in water to

calculate the time. Calculate the wavelength.

Solution: 1. (a) Divide the

distance by the

speed of sound in water:

( )2 75 m20.098 s

1530 m/s

dt

v= = =

2. (b) Solve for the wavelength: 31530 m/s28 10 m 28 mm

55 kHz

v

fλ −= = = × =

Insight: In air the wavelength would be 6.2 mm. The wavelength is longer in the water because

the wave travels faster in water, while the frequency is the same.

16. Picture the Problem: We need to calculate the wavelength of sound in air from its frequency.

Strategy: Solve for the wavelength, using 343 m/s for the speed of sound in air.

Solution: 1. (a) Solve for the

wavelength:

343 m/s0.807 m

425 Hz

v

fλ = = =


57 of 89

2. (b) Examine the relationship between

wavelength and frequency:

Wavelength is inversely related to frequency so,

if the frequency increases the wavelength

decreases.

3. (c) Calculate the wavelength at 450

Hz:

343 m/s0.722 m

475 Hzλ = =

Insight: As predicted, an increase in frequency corresponds to a decrease in wavelength.

17. Picture the Problem: The figure represents you dropping a rock

down a well and listening for the splash. From the time lapse

between dropping the rock and hearing the splash we want to

calculate the depth of the well.

Strategy: The time to hear the splash, t = 1.5 s, is the sum of the

time for the rock to fall to the water, t1, and the time for the sound

of the splash to reach you, t2. Solve the free-fall equation for the

time to fall and displacement at constant velocity to calculate the

time for the sound to return. Set the sum of these times equal to

the time to hear the splash and solve for the distance.

Solution: 1. (a) Solve for the falling time: 1

2dt

g=

2. Solve for the time for the sound to travel up

the well: 2

s

dt

v=

3. Sum the two times to equal the total time:

1 2

2

s

d dt t t

g v= + = +

4. Rewrite as a quadratic

equation in terms of the

variable d :

( )

( )

2

2

2

1 20

1 20 1.2s

343 m/s 9.81 m/s

s

d d tv g

d d

= + −

= + −

5. Solve for d using the

quadratic formula and square

the result:

3.2537 m 10.587 m 11 md d= ⇒ = =

6. (b) The time to hear the sound would be less then 3.0 seconds because, although the sound

travel time would double, the fall time would less than double.

Insight: The time to hear the sound for a 21-meter-deep well is 2.1 s, which is indeed less than 3.0

s.

18. Picture the Problem: The figure shows a person throwing a rock

down an 8.80-m deep well. The sound of the splash reaches the

person’s ear 1.20 seconds after the rock is thrown. We want to

calculate the speed of the rock.

Strategy: Solve for the initial velocity of the rock, where the fall

time is equal to the total time minus the sound travel time. The sound

travel time is the depth of the well divided by the speed of sound.

Solution: 1. Calculate st : s

8.80 m0.0257 s

343m/s

dt

v= = =

2. Subtract st from the total

time:

f 1.2 s 0.0257 s 1.1743 st = − =


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3. Solve for

0v :

( )( )

2

0 0 f f

0

0 f

f

2

1

2

1

2

8.8 m 19.81 m/s 1.1743 s 1.7 m s

1.1743 s 2

y y v t at

y yv at

t

= + +

−= −

−= − − = −

The initial velocity of the rock is 1.7 m/s downward

Insight: Even though the speed of sound is much larger than the speed of the rock, the time for the

sound to travel up the well is significant. If the sound travel time was not included, the initial

velocity of the rock would incorrectly be calculated as 1.4 m/s, which is 15% off of the actual

velocity.


59 of 89


1. Picture the Problem: A gold ring has a density equal to its mass divided by its volume.

Strategy: If the ring is pure gold, its density will be equal to the density of gold. Since the mass

and volume of the ring are known, use ρ = M/V to calculate the density. Compare the result with

the density of gold given in tables.

Solution: 1. Divide the volume by the

mass:

3

3

0.014 g6.4 g cm

0.0022 cm

m

Vρ = = =

2. Compare with the density of gold from

the table:

3

gold 19.3 g cmρ = . Therefore, the ring is not

solid gold.

Insight: If the ring were pure gold of the same volume given in the problem, its mass would be

42.5 g.

2. Picture the Problem: A cube has a mass of 0.347 kg and sides of 3.21 cm each.

Strategy: Calculate the density of the cube. Compare the resulting density with the densities

given determine the likely composition.

Solution: 1. Calculate the density of

the cube: ( )4 3

31 m

100 cm

0.347 kg1.05 10 kg m

3.21 cm

m

Vρ = = = ×

2. Compare with the densities: The cube has the density of silver.

Insight: Cubes made of different materials could have considerably different masses. For

example, a cube of gold (with the same volume as the silver cube) would have a mass 0.638 kg,

while a cube of aluminium would have a mass of 0.089 kg.

3. Picture the Problem: A pressure of 1 dyne per square centimetre needs to be converted to the

units of pascals and atmospheres.

Strategy: The pressure is given as a force divided by area. Convert the units of force and area to

the standard SI units of Newton and square meter to write the pressure in pascals. Then convert to

atmospheres.

Solution: 1. (a) Convert the pressure to

pascals:

25–1

2

1 dyne 10 N 100 cm10 Pa

dyne mcm

− =

2. (b) Convert to atmospheres: –1 –6

3

1 atm10 Pa 10 atm

101.3 10 Pa

=

×

Insight: A dyne/cm2 is the standard unit of pressure in the cgs system of units, which is

commonly used in chemistry. The pascal is the standard unit of pressure in the mks system.

4. Picture the Problem: When a person sits in a four-legged chair the weight of the person and chair

is distributed over each leg of the chair, increasing the pressure each leg exerts on the ground.

Strategy: Calculate the pressure each leg exerts on the floor. Set the force equal to the sum of the

weights of the person and chair and the area equal to four times the cross-sectional area of each

leg.

Solution: 1. Set the pressure equal to the

weight divided by area: 22

42

F mg mgP

A dd ππ

= = =


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2. Insert given data:

( )( )

2

6

2

(72 kg 3.8 kg) 9.81 m/s1.4 10 Pa

0.013 mP

π

+= = ×

Insight: Leaning back in the chair, so that is rests on only two legs, doubles the pressure those

legs exert on the floor.

5. Picture the Problem: When walking with crutches, a person supports a large portion of her weight

on the crutch. If the end of the crutch did not have a rubber tip, the entire weight would be

supported over the small area of the crutch. The rubber tip increases the area over which the

weight is distributed, thus decreasing the pressure.

Strategy: Let Pwo represent the pressure without the rubber tip and Pw represent the pressure with

the rubber tip. Calculate the ratio of Pw to Pwo . The force on the crutches is the same and the

cross-sectional area is the area of a circle.

Solution: 1. Find the ratio of the

pressure with the tip to the pressure

without:

( )

( )w

wo

22

w wo wo

2 2

wo w w

1.2 cm0.23

2.5 cm

FA

FA

P A r

P A r

π

π= = = = =

2. Invert the ratio to find the factor

by which the pressure decreases:

14.3

0.23=

Insight: Since area is proportional to the square of the radius, the pressure is decreased by the

square of the fractional increase in radius.

6. Picture the Problem: When you ride a bicycle, your weight and the weight of the bicycle are

supported by the air pressure in both tires spread out over the area of contact between the tires and

the road.

Strategy: To calculate your weight, first solve for the supporting force of the air pressure on the

tires. Set this force equal to the sum of your weight and the weight of the bicycle. Subtract the

weight of the bicycle to determine your weight.

Solution: 1. Multiply

the tire pressure by the

contact area to calculate

the supporting force on

the bicycle:

( ) ( )25

2 2

2

1.01 10 Pa 1 m70.5 lb/in 2 7.13 cm 690.7 N

100 cm14.7 lb/in

F PA=

× = × =

2. Set the supporting

force equal to the sum of

your weight and the

weight of the bicycle:

you bicycle you bicycleF W W W m g= + = +

3. Solve for your weight: ( )2

you bicycle 690.7 7.7kg 9.8m/s 615 NW F m g N= − = − =

Insight: When “popping a wheelie” on the bicycle, such that only one wheel is touching the

ground, that wheel must support the entire weight of the bicycle and rider. Therefore, since the tire

pressure has not changed, the area of contact for the single tire would double. In this problem the

area would increase to 14.26 cm2.

7. Picture the Problem: The weight of the car is supported by the air pressure in all four tires spread

out over the area of contact between the tires and the road.

Strategy: Find the necessary contact area to support the weight of the car for the given tire

pressure. Divide the area by four to calculate the area of contact for each tire. Since the weight of

the car does not change significantly as the tire pressure is increased, the tire pressure and contact

area are inversely proportional to each other. Finally, solve for the air pressure.

Solution: 1. (a) Solve

equation 15-2

for the total contact area:

F mg

AP P

= =


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2. Solve for the contact

area on one tire: tire tire4

4

mg mgA A A

P P= = ⇒ =

3. Insert given values: ( )

( )( )5

2

2

2

tire1.01 10 Pa2

14.7 lb/in

1320 kg 9.81 m/s0.0135 m

4 35.0 lb/inA

×= =

4. (b) Since the area and pressure are inversely proportional, as the pressure increases the area of

contact decreases.

5. (c) Solve for the

pressure: tire4

mgP

A=

6. Insert the given values:

( )

( ) ( )( )

2 25 2

2 52 1 m

100 cm

1320 kg 9.81 m/s 14.7 lb/in2.79 10 Pa 40.6 lb/in

1.01 10 Pa4 116cmP

= = × =

×

Insight: The pressure in part (c) was greater than the pressure given in part (a). As predicted, the

increase in pressure resulted in a decrease in contact area.

8. Picture the Problem: The soft drink can has zero pressure inside and atmospheric pressure

pushing inward from the outside.

Strategy: Calculate the net force, with the pressure being atmospheric pressure and the area the

area of a cylinder, .A D hπ=

Solution: 1. Calculate

vertical area: ( ) ( ) 20.065 m 0.12 m 0.0245 mA D hπ π= = =

2. Solve for inward force: ( )5 2 2

at 1.01 10 N/m 0.0245 m 2.5 kNF P A= = × =

Insight: This force is equal to over 500 lbs, which will easily crush the can.

9. Picture the Problem: Atmospheric pressure will cause a column of mercury to rise 760 mm into a

vacuum. Changes in air pressure are measured by the changes in height of the barometer.

Atmospheric pressure will cause a column of water to rise much higher because of is lower density.

Strategy: Convert the height of the mercury column to pascals by using the relation for

atmospheric pressure 51 atm 1.01 10 Pa 760 mmHg= × = . To calculate the height of the water

column, use equation 15-7 with P1 equal to zero and P2 equal to the answer to part (a).

Solution: 1. (a) Convert the height of

the mercury column to pascals:

51.01 10 Pa(736 mmHg) 97.8kPa

760 mmHgP

×= =

2. (b) Solve for the height of the

water:

( )( )

2 1

4

2 1

3 2

9.78 10 Pa 09.97 m

1000 kg/m 9.81 m/s

P P gh

P Ph

g

ρ

ρ

= +

− × −= = =

Insight: The water column rises over 13 times further than the mercury column because its density

is over 13 times smaller than the density of mercury. The height of the water column makes it

impractical to use as a barometer.


62 of 89

10. Picture the Problem: Two pistons are supported

by a fluid, as shown in the figure. The pressure in

the fluid at the bottom of the left piston is equal to

the pressure in the right piston at the same vertical

level, which is a distance h below the right piston.

Strategy: Set the pressures in the two columns

equal at the depth of the left piston. Calculate the

pressure due to the pistons and calculate the

increase in pressure due to the fluid in the right-

hand column.

Solution: 1. Set the pressures

equal: L R

L R

L R

P P

m g m ggh

A Aρ

=

= +

2. Solve for the height h:

L R L R L R

2 2 2 2

L R L R L R4 4

1 1 4m m m m m mh

A A D D D Dπ πρ ρ πρ

= − = − = −

3. Insert the given values:

( ) ( ) ( )2 23

4 1.7 kg 3.2 kg1.0 m

750kg/m 0.045 m 0.12 mh

π

= − =

Insight: The height difference does not depend on the height of the fluid in the left column.

11. Picture the Problem: As water is poured into the tube shown in

the figure, the pressure inside the barrel increases. When the

upward force on the barrel lid exceeds 643 N, the barrel will burst.

Strategy: The force on the barrel top is the pressure at the surface

times the area of the top. Calculate the height of the water column

when the barrel will burst. Calculate the weight of the water

column from the height, cross-sectional area, and density of water.


bursting pressure of

the lid:

( )2

643 N1460 Pa

0.75 m 2

FP

A π∆ = = =

2. Solve for the

height of the water

column: ( )3 2

1455 Pa0.148 m

1000 kg/m 9.81 m/s

P gh

Ph

g

ρ

ρ

∆ =

∆= = =

3. Solve for the

weight:

( )( )

2

2

3 2

2

0.010 m1000 kg/m 0.148 m 9.81 m/s 0.11 N

2

dW mg hgρ π

π

= =

= =

Insight: A very short column of water is able to increase the pressure sufficiently to burst the

barrel. This is one reason why rain barrels always have a hole at the top to allow excess water to

flow out.


63 of 89

12. Picture the Problem: A cylinder is filled with a fluid, as

shown in the diagram. The pressure at the bottom of the fluid

is greater than the atmospheric pressure at the top. We wish to

find the depth of the fluid that will result in a pressure at the

bottom of 116 kPa. Adding additional fluid to the container

will increase the pressure at the bottom. We wish to calculate

the increase in pressure when

2.05 × 10-3 m3 are added.

Strategy: Calculate the depth of the fluid. To calculate the

pressure when additional fluid has been added, divide the

volume of the fluid by the cross-sectional area to find the

additional height of the fluid. Then insert the total height into

P = Pat + ρgh for the total pressure.

Solution: 1. (a) Solve for h: at

at

P P gh

P Ph

g

ρ

ρ

= +

−=

2. Insert given values:

( )( )

3 5

3 2

116 10 Pa 1.01 10 Pa1.90 m

806 kg/m 9.81 m/sh

× − ×= =

3. (b) Divide the volume by

A:

3 3

2 4 2

2.05 10 m0.314 m

65.2 10 mh

−

−

×= =

×

4. Add the heights: total 2

1.897 m 0.314 m 2.211 mh h h= + = + =

5. Insert data:

( )( )at

5 3 21.01 10 Pa 806 kg/m 9.81 m/s (2.211 m) 118 kPa

P P ghρ= +

= × + =

Insight: Part (b) could also have been solved by adding the additional depth to the bottom of the

cylinder, such as: ( ) ( )( )5 3 21.16 10 Pa 806 kg/m 9.81 m/s 0.314 m 118 kPaP = × + = .

13. Picture the Problem: As a submarine dives, the pressure difference between the interior and

exterior increases. To be safe, this pressure difference cannot exceed 10.0 N/mm2. We need to

solve for the maximum depth that the submarine can dive.

Strategy: Solve for the depth to which the submarine can descend. Use the density of sea water

from the table. Examine the resulting equation to determine how the density affects the maximum

depth.

Solution: 1. (a) Solve for h: atm w

atm

w

P P gh

P Ph

g

ρ

ρ

= +

−=

2. Insert the given values: ( )( )( )

3 22 10 mm

m

3 2

10.0 N/mm995 m

1025 kg/m 9.81 m/sh = =

3. (b) Fresh water is less dense than sea water, so the maximum safe depth in fresh water is greater

than in salt water.

Insight: The maximum depth in fresh water is 1020 m.


64 of 89

14. Picture the Problem: A water tower is filled with water. The pressure in the tank increases as the

water depth increases. We wish to calculate the pressure at specific depths.

Strategy: Solve for the pressure at the given depths. Use the density of water given in the table

15-1.

Solution: 1. (a) Apply P =

Pat + ρgh directly: ( )( )

at

5 3 2 51.01 10 Pa 1000 kg/m 9.81 m/s 4.5 m 1.45 10 Pa

P P ghρ= +

= × + = ×

2. (b) Repeat for a depth of

5.5 m: ( )( )5 3 2 51.01 10 Pa 1000 kg/m 9.81 m/s 5.5 m 1.54 10 PaP = × + = ×

3. (c) The bands are closer together near to bottom because pressure increases with depth. A

greater confining force is needed near the bottom than near the surface of the water.

Insight: The pressure at the bottom of the tank reaches a maximum of 1.64×105 Pa.

15. Picture the Problem: Water is held in a glass on an elevator that is accelerating

upward. A free body diagram for the water is shown at right. We want to

calculate the additional pressure at the bottom of the glass due to the

acceleration of the elevator.

Strategy: Calculate the acceleration of the water from its change in speed and

time. Use Newton’s Second Law to write the additional force necessary to

accelerate the water. Divide this force by the cross-sectional area of the glass to

find the added pressure in the water.

Solution: 1. (a) The force that the glass exerts upward on the water is greater

than the weight of the water in order to provide upward acceleration. By

Newton’s Third Law, the water exerts an equal force downward on the glass

bottom, so the pressure is greater than it was before the elevator began to move.

2. (b) Calculate the

acceleration of the

elevator:

22.2 m/s 00.710 m/s

3.1 s

va

t

∆ −= = =

∆

3. Write the change in

pressure as the added

force divided by the

area of the glass:

F maP

A A∆ = =

4. Write the mass as

density times volume:

( )( )( )( )3 21000 kg/m 0.065 m 0.710 m/s 46 Pa

AhP a ha

A

ρρ∆ = = = =

Insight: If the elevator were accelerating downward at 0.710 m/s2 the pressure in the glass would

decrease by 46 Pa.

16. Picture the Problem: A 12-cm-tall column of water lies under a 7.2-cm-tall

column of olive oil, as shown in the figure. We wish to calculate the

pressure at the bottom of the water.

Strategy: Establish the relationship between pressure and depth within a

fluid of known density. The pressure under the oil, P1, can be calculated

with the density of the oil and height of the oil. To find the pressure at the

bottom of the water, insert the pressure P1 and add the pressure change from

the density of water and height of the water.


pressure at the

bottom of the

oil:

( )( )( )1 atm

5 3 2

5

1.013 10 Pa 920kg/m 9.81m/s 0.072m

1.0195 10 Pa

P P ghρ= +

= × +

= ×


65 of 89

2. Calculate the

pressure at the

bottom of the

water:

( )( )( )

1 water water

5 3 2 51.0195 10 Pa 1000 kg/m 9.81 m/s 0.12 m 1.03 10 Pa

P P ghρ= +

= × + = ×

Insight: Since the density of oil is less than the density of water, the pressure at the bottom of the

water is slightly less than the pressure would be if the entire column were water.

17. Picture the Problem: A straw sits in a glass of water. When you suck on the

straw, the water rises in the water. We want to know, theoretically, what is the

highest the water can rise in the straw.

Strategy: The minimum pressure that you could cause inside the straw would be a

pure vacuum. The pressure outside the straw is atmospheric pressure. Solve for the

height of the water in the straw with no pressure above the water and atmospheric

pressure at the bottom.

Solution: 1. (a) The atmospheric pressure that is exerted on the surface of the water

creates an upward force on the water column in the straw that overcomes the force

of gravity.

2. (b) Set the pressure above

the fluid equal to zero: atP P gh ghρ ρ= + =

3. Solve for the height of the

water column: ( )( )

5

at

3 2

1.01 10 Pa10.3 m

1000 kg/m 9.81 m/s

Ph

gρ

×= = =

Insight: No amount of suction can cause the water to rise higher than 10.3 meters in the straw.

18. Picture the Problem: As shown in the figure, an IV solution is elevated

above the injection point on a patient. The pressure in the bag is

atmospheric pressure, while the pressure at the injection point is 109 kPa.

We need to calculate the height of the bag.

Strategy: Solve for the height of the solution above the injection point.

Solution: 1. (a) Solve for h: at

at

g

P PP P gh hρ

ρ

−= + ⇒ =

2. Insert the given

data: ( )( )3 2

109 kPa 101.3 kPa0.770 m

1020 kg/m 9.81 m/sh

−= =

3. (b) From the equation in step 1, we see that the height is inversely proportional to the density of

the fluid. Therefore if a less dense fluid is used, the height must be increased.

Insight: If the density of the fluid were reduced to 920 kg/m3, the bag would need to be suspended

at a height of 0.853 m, which is, as predicted, higher than the 0.770 meters.


66 of 89

19. Picture the Problem: A cylinder is filled with mercury up to a depth d,

and then filled the rest of the way with water, as shown in the figure.

The pressure at the bottom of the cylinder is two atmospheres.

Strategy: Set the pressure at the bottom of the cylinder equal to the

pressure at the top (atmospheric) plus the pressure increases due to the

water and the mercury. Calculate the pressure increases. The height of

the water is one meter minus the height of the mercury, 1.0 mwh d= − .

Solution: 1. Write the pressure at the

bottom of the cylinder: at w w Hg Hg

at at w Hg2 (1.0 m )

P P gh gh

P P g d gd

ρ ρ

ρ ρ

= + +

= + − +

2. Solve for d:

atw

Hg w

(1.0 m)P

gd

ρ

ρ ρ

−

=−

3. Insert the given values: ( )

53

2

4 3 3

1.01 10 Pa1000 kg/m 1.0 m

9.81 m/s 0.74 m1.36 10 kg/m 1000 kg/m

d

×−

= =× −

Insight: Atmospheric pressure is 760 mmHg. Since the density of mercury is much greater than

the density of water, the height of the mercury is almost the same as if it were a vacuum above the

mercury column.


67 of 89


1. Picture the Problem: When the SR-71 Blackbird is in flight, its surface heats up significantly.

This increase in temperature causes the plane to expand in length.

Strategy: We want to calculate the temperature of the plane using the known expansion amount.

Establish the relationship between the expansion of the plane and its change in temperature. Solve

for the final temperature of the plane.

Solution: Solve for T: 0

0 –6 1 50 12

0.50 ft23 C 220 C

(24 10 K )(107 ft)

L L T

LT T

L

α

α −

∆ = ∆

∆= + = ° + = °

×

Insight: Note that in this problem it was not necessary to convert the lengths to metric units.

2. Picture the Problem: The Akashi Kaikyo Bridge is Japan is made of steel. When steel is heated it

expands and when it is cooled it contracts.

Strategy: In this problem we wish to find the change in length of the bridge between a cold winter

day and a warm summer day. Determine the change in length. The coefficient of linear expansion

for steel is given in the table.

Solution: Insert

the given

values: ( )[ ]0

5 11.2 10 (C ) 3910 m 30.0 C ( 5.00 C) 1.6 m

L L Tα

− −

∆ = ∆

= × ° ° − − ° =

Insight: This change in length is about the height of a person. If there were no expansion joints in

the bridge this increase in length would be sufficient to buckle the bridge.

3. Picture the Problem: An aluminium plate has a hole cut in its

centre. The plate expands as it is heated.

Strategy: We want to find the size of the hole after the

temperature has increased to 199.0°C. The hole will expand at

the same rate as the aluminium. Since the diameter of the hole is

a unit of length, calculate the diameter as a function of the

increase in temperature. The coefficient of linear expansion is

given in the able 16-1.

Solution: 1. (a) Solve

for the final diameter: ( )1

d d d d T

d d d T d T

α

α α

′∆ = − = ∆

′ = + ∆ = + ∆

2. Insert the given

data: ( )( )–6 11.178 cm 1+ 24 10 K 199.0 C 23.00 C

1.183 cm

d− ′ = × ° − °

=

3. (b) Solve for the

change in temperature:

0

d d d d T

d dT T T

d

α

α

′∆ = − = ∆

′ −∆ = − =

4. Solve for the final

temperature:

( )( )

0

–6 1

1.176 cm 1.178 cm23.00 C 48 C

24 10 K 1.178 cm

d dT T

dα

−

′ −= +

−= ° + = − °

×

Insight: Since the final diameter (1.176 cm) is smaller than the diameter at 23°C we would expect

that the final temperature would be below 23°C. The calculations show that this is the case.


68 of 89

4. Picture the Problem: A steel bar has a diameter that

is 0.040 cm larger than the inner diameter of an

aluminium ring that you would like to slip over the

bar.

Strategy: Calculate the temperature at which the

ring’s inner diameter will equal the diameter of the

bar. The coefficient of thermal expansion is given in

the table.

Solution: 1. (a) The ring should be heated. Imagine

that the ring is cut and “unrolled.” It would be a

rectangle. If the rectangle is heated, it will expand

along its length and width. Its length is the

circumference of the ring. Since the length of the

rectangle increases, the circumference of the circle

increases, and therefore, so does its diameter.

2. (b) Solve for the change in temperature:

0

d d T

dT T T

d

α

α

∆ = ∆

∆∆ = − =

3. Solve for the final temperature:

( )( )( )

0

0

15

4.040 cm – 4.000 cm10.00 C

2.4 10 C 4.000 cm

430 C

dT T

d

T

α

−−

∆= +

= ° +× °

= °

Insight: When the aluminium is heated to 430°C it will slip over the steel rod. As it cools back

down it will shrink to form a tight bond with the steel.

5. Picture the Problem A brass sleeve has an inner

diameter slightly smaller than the diameter of a steel

bar. To shrink-fit the sleeve over the bar, you must

either heat the sleeve or cool the bar.

Strategy: Solve for the temperature at which the

change in diameter is equal to the difference in

diameters of the brass sleeve and the steel rod. For the

case of heating the brass sleeve use the coefficient of

thermal expansion of brass, the initial inner diameter

of the brass and a positive change in diameter. For

the case of cooling the steel rod, use the coefficient of

thermal expansion of steel, the diameter of the steel

rod, and a negative change in diameter. The

coefficients of thermal expansion of brass and steel

aluminium are given in the table.

Solution: 1. Solve for the final

temperature: ( )0

0

L L T L T T

LT T

L

α α

α

∆ = ∆ = −

∆= +

2. (a) Insert the data for heating the

brass sleeve: ( )( )5 1

2.19893 cm 2.19625 cm12.25 C 76 C

1.9 10 K 2.19625 cmT

− −

−= ° + = °

×

3. (b) Insert the data for cooling the

steel: ( )( )–6 1

2.19625 cm – 2.19893 cm12.25 C 89 C

12 10 K 2.19893 cmT

−= ° + = − °

×

Insight: Since the coefficient of thermal expansion for brass is greater than the coefficient of


69 of 89

thermal expansion for steel, the brass does not have to be heated through as large of a temperature

difference as the steel has to be cooled to achieve the same change in diameter.

6. Picture the Problem: A steel gasoline tank is completely filled with gasoline, such that the gasoline

and the tank have the same initial volumes. When the gas and tank are heated, the gas expands

more than the tank, causing some of the gas to spill out of the tank.

Strategy: Since the initial volumes of the gas and tank are equal, the amount that will spill out is the

difference in the increase in volume of the gas and tank, namely: The volume of spilled gasoline

spill gas tank .V V V= ∆ − ∆ Calculate the changes in volume for the gas and tank. The coefficient of

volume expansion for steel is 3 times the coefficient of linear expansion, which is given in the table.

The coefficient of volume expansion for gas is given in the table.

Solution: 1. Write

the volume

difference:

( ) ( )spill gas tank gas 0 tank 0 gas tank 03 3V V V V V T V Tβ α β α= ∆ − ∆ = − ∆ = − ∆

2. Insert the given

data: ( )( ) ( )( )

14 5

spill9.5 10 3 1.2 10 C 51 L 25 5.0 C 0.93 LV

−− −= × − × × ° − ° =

Insight: 0.93 L is about a quarter of a gallon. Most commercial gas pumps shut off before your

car’s tank is completely filled to prevent spillover due to the expansion of gas.

7. Picture the Problem: When at room temperature, a stainless steel pot

has the same diameter as the pot’s copper bottom. When the pot is

heated, the copper expands faster than the steel, causing a difference

in diameters.

Strategy: Calculate the difference in diameters of the steel and copper

when the temperature is 610°C. The coefficients of expansion are

given in the problem.


equation for the increases in

diameters:

Cu Cu 0

st st 0

d d T

d d T

α

α

∆ = ∆

∆ = ∆

2. Subtract the two

differences: ( )

( )( ) ( )( )

st Cu st Cu 0

5 5 11.73 10 1.70 10 C 8.0 in. 610 22 C

0.0014 in.

d d d Tα α

− − −

∆ − ∆ = − ∆

= × − × ° − °

=

Insight: Since the coefficients of expansion between stainless steel and copper are similar (less

than 2% difference) the difference in expansion is small. If instead of stainless steel, normal steel

were used, the difference in diameters would be 0.23 in; more than sufficient to the break the pan

apart.

8. Picture the Problem: Two cubes are constructed of aluminium and copper wire. The cubes

initially enclose equal volumes. When the two cubes are heated they expand at different rates,

resulting in different enclosed volumes.

Strategy: Calculate the increase in volume of each wire cube. The coefficients of volume

expansion are three times the coefficients of linear expansion given in the table.

Solution: 1. (a) Aluminium has the larger coefficient of volumetric expansion, therefore, the

aluminium cube will enclose a greater volume.

2. (b) Write the changes in

volume for each cube: Al Al 0 Al 0

Cu Cu 0 Cu 0

3

3

V V T V T

V V T V T

β α

β α

∆ = ∆ = ∆

∆ = ∆ = ∆


70 of 89

3. Subtract the changes in

volume to calculate the

difference in volumes of

the two cubes:

( )

( ) ( ) ( )( )

Al Cu Al Cu 0

15 3

5 3

3

3 2.4 1.7 10 C 0.016m 97 21 C

2.6 10 m

V V V Tα α

−−

−

∆ − ∆ = − ∆

= − × ° − °

= ×

Insight: This difference in volume is 26 mL, about two tablespoons, and is about 0.16% of the

initial volume of 4.2 gallons.

9. Picture the Problem: A copper ball expands as it is heated, and in proportion to its increase in

temperature.

Strategy: The copper ball expands equally in all directions when heated. Relate the increase in

diameter to the change in temperature. Solve the equation for the final temperature. The

coefficient of linear expansion for copper is given in the table.

Solution: 1. Solve for the

final temperature: 0 0

0

L

L L T T TL

αα

∆∆ = ∆ ⇒ = +

2. Insert the given data:

( ) ( )

3

15 2

0.18 10 m22°C 680 C

1.7 10 C 1.6 10 mT

−

−− −

×= + = °

× ° ×

Insight: This problem could equivalently be solved by relating the change in volume of the

sphere to the temperature.

10. Picture the Problem: The volume of an aluminium saucepan expands as the temperature of the

pan increases. Water, which initially fills the saucepan to the brim, also increases in temperature

and expands. If the water expands more than the saucepan, the water will spill over the top. If the

saucepan expands more that the water, the water level will drop from the brim of the pan.

Strategy: Calculate the change in volumes of the saucepan and of the water. Subtract the change

in volume of the saucepan from the change in volume of the water to determine the volume of

water that overflows the saucepan. The coefficient of volume expansion for water is given in

Table 16-1. The coefficient of volume of expansion of aluminium is three times its coefficient of

linear expansion, which is also given in the table.

Solution: 1. (a) Because water has a larger coefficient of volumetric expansion, its volume will

increase more than the volume of the aluminium sauce pan. Therefore, water will overflow from

the pan.

2. (b) Calculate the

initial volumes of the

saucepan and water:

( )2

2 3

0

23 cm6.0 cm = 2493c m

2V r hπ π

= =

3. Write the changes in

volume: w w 0

Al Al 03

V V T

V V T

β

α

∆ = ∆

∆ = ∆

4. Subtract the change

on volume of the pan

from the water to

calculate the volume of

water spilled:

( ) ( ) ( ) ( )

spill w Al w Al 0

15 3

3

spill

( 3 )

21 3 2.4 10 C 2493 cm 88 19 C

24 cm

V V V V T

V

β α

−−

= ∆ − ∆ = − ∆

= − × × ° − °

=

Insight: This is the same principle that enables a mercury thermometer to work. The mercury

expands faster than the surrounding glass, causing the mercury column to rise.


71 of 89

11. Picture the Problem: Heat is added to a glass ball, resulting in an increase in temperature.

Strategy: Solve for the heat necessary to increase the temperature. The specific heat of glass is

given in the table.

Solution: Solve for the heat:

( ) ( ) ( )0.055 kg 837 J/ kg K 15C 0.69 kJ

Q mc T= ∆

= ⋅ ° =

Insight: The change in temperature is proportional to the heat added. Doubling the heat added

would result in a temperature change of 15 C°.

12. Picture the Problem: A lead bullet travelling at 250 m/s has kinetic energy. As the bullet

encounters a fence post it slows to a stop, converting the kinetic energy to heat. Half of the energy

heats the bullet resulting in an increase in bullet temperature.

Strategy: Solve for the change in temperature. Set the heat equal to one half of the initial kinetic

energy of the bullet. The specific heat of lead is given in the table.

Solution: 1. Solve equation for :T∆

QT

mc∆ =

2. Set the heat equal to half the initial kinetic

energy:

( )

( )( )

21 1 212 22

2

4

250 m/s120 K

4 128 J/ kg K

mvKQ vT

mc mc mc c∆ = = = =

= =⋅

Insight: The relatively small specific heat of lead leads to this large increase in temperature. A

silver bullet travelling at the same speed would only heat up by 68 K.

13. Picture the Problem: As the hot silver pellets are dropped into the cool water, heat transfers from

the pellets to the water. This results in a decrease in the temperature of the pellets and an increase

in the temperature of the water.

Strategy: Use conservation of energy, setting the sum of the heat lost by the silver and the heat

gained by the water equal to zero. Solve the resulting equation for the mass of the silver which

gives a final temperature of 25°C. Divide the resulting mass by the mass of each silver pellet to

calculate the number of pellets needed. For the copper pellets, repeat the same calculation, but with

the specific heat of copper. The specific heats of water, silver, and copper are found in the table.

Solution: 1. (a) Sum the heat transfers

to zero: ( ) ( )

Ag w

Ag Ag Ag w w w

0

0

Q Q

m c T T m c T T

+ =

− + − =

2. Solve for the mass of silver:

( )

( )( ) ( )

( ) ( )

w w w

Ag

Ag Ag

0.220 kg 4186 J/ kg K 14 25 C0.722 kg

234 J/ kg K 25 85 C

m c T Tm

c T T

−=

−

⋅ − ° = =

⋅ − °

3. Divide by the mass of one pellet: Ag 2

pellet

0.722 kg = 7.2 10 pellets

0.001 kg

mn

m= = ×

4. (b) Copper has a higher specific heat, so the required number of pellets decreases.

5. (c) Solve the

conservation of

energy equation for

the mass of copper:

( )( )

( ) ( )

( ) ( )

w w w

Cu

Cu Cu

(0.220 kg) 4186 J/ kg K 14 25 C0.436 kg

387 J/ kg K 25 85 C

m c T Tm

c T T

−=

−

⋅ − ° = =⋅ − °


72 of 89

6. Divide by the mass of one

pellet:

2Cu

pellet

0.436 kg = 4.4 10 pellets

0.001 kg

mn

m= = ×

Insight: The amount of heat needed to increase the water’s temperature does not depend on

whether silver or copper pellets provide the heat. Since the copper has a higher specific heat, each

pellet is able to transfer more heat to the water, so fewer copper pellets are needed.

14. Picture the Problem: Heat transfers from the hot lead ball to the cool water, causing the lead to

cool and the water to heat up. Eventually the water and lead will come to the same equilibrium

temperature.

Strategy: Calculate the equilibrium temperature. The specific heats of water and lead are given in

the table.

Solution: Insert given data:

( )

( )

( ) ( )

Pb Pb Pb w w w

Pb Pb w w

0.235 kg 128 J/ kg K 84.2 C

0.177 kg 4186 J/ kg K 21.5 C

0.235 kg 128 J/ kg K 0.177 kg 4186 J/ kg K

23.9 C

m c T m c TT

m c m c

T

+=

+

⋅ ° + ⋅ °

=⋅ + ⋅

= °

Insight: Since the specific heat of water is greater than the specific heat of lead, the final

temperature is much closer to the initial temperature of the water.

15. Picture the Problem: As heat is added to an object its temperature rises. The ratio of the heat to

the change in temperature is the heat capacity. The specific heat is the heat capacity per unit mass.

Strategy: Calculate the heat capacity by dividing the heat by the change in temperature. Calculate

the specific heat.

Solution: 1. (a) Divide the heat

by the change in temperature: 2200 J

0.18 kJ C12C

QC

T= = = °

∆ °

2. (b) Divide the heat capacity by

the mass:

( ) ( )

0.183 kJ/ C°

0.190 kg

0.96 kJ kg C° 0.96 kJ kg K

Q Cc

m T m= = =

∆

= ⋅ = ⋅

Insight: When the heat capacity is known, you can calculate the amount of heat necessary to

produce a specific temperature change. For example, to increase the temperature of the object by

30 C°, 5.4 kJ of heat should be added.

16. Picture the Problem: A lead ball is dropped from the specified height. During the fall the

gravitational potential energy is converted to kinetic energy and finally to heat. The resulting heat

increases the temperature of the ball.

Strategy: Use conservation of energy to calculate the kinetic energy of the ball just before it hits

the ground. Set the heat entering the lead ball equal to the kinetic energy. Solve for the change in

temperature

Solution: 1. Set the initial and

final energies equal and solve for

the kinetic energy:

0 0

0 0

U K U K

mgh K

K mgh Q

+ = +

+ = +

= =

2. Solve for the change in

temperature:

( )( )

( )

29.81 m/s 5.43 m

128 J/ kg K

0.416 K 0.416 C

Q mgh ghT

mc mc c∆ = = = =

⋅

= = °


73 of 89

Insight: The mass of the ball did not affect the change in temperature. A ball with a larger mass

would have a greater amount of heat available, but would require the additional heat to increase the

temperature of the additional mass.

17. Picture the Problem: A hot object is immersed in water in a calorimeter cup. Heat transfers from

the hot object to the cold water and cup, causing the temperature of the object to decrease and the

temperature of the water and cup to increase.

Strategy: Since the heat only transfers between the water, cup, and object, we can use

conservation of energy to calculate the heat given off by the object by summing the heats absorbed

by the water and cup. Use the heat given off by the object and its change in temperature to

calculate its specific heat.

Solution: 1.

Let 0Q =∑

and solve for

ObQ :

( ) ( )Ob w Al

Ob w Al w w Al Al w

0 Q Q Q

Q Q Q m c m c T

= + +

= − + = − + ∆

2. Solve for

the specific

heat:

( )

( )( )( )

( ) ( ) ( )

( )

Ob

Ob Ob

w w Al A1 w

Ob

Ob Ob

Ob

0.103 kg 4186 J/ kg K 0.155 kg 900 J/ kg K 20 22 C

0.0380 kg 22.0 100 C

385 J (kg C ) 385 J (kg K)

obQc

m T T

m c m c T Tc

m T T

c

=−

+ −=

−

⋅ + ⋅ − ° =

− °

= ⋅ ° = ⋅

3. Look up the

specific heat in

the table:

The object is made of copper.

Insight: It is important to include the effect of the aluminium cup in this calculation. If the

contribution of the cup were excluded, the specific heat of the object would have been calculated as

291 J/(kg K).

18. Picture the Problem: Heat transfers from a hot horseshoe to the cold water. This decreases the

temperature of the horseshoe and increases the temperature of the water until the water and

horseshoe are at the same equilibrium temperature.

Strategy: Since only two objects are transferring heat, calculate the equilibrium temperature. To

determine which object would cause a larger final temperature, you should compare the heat

capacities heats of the two objects. The object with the higher heat capacity will have more heat to

transfer to the water, causing the final temperature to be greater.

Solution: 1. (a) Insert given data:

( ) ( )

( ) ( ) ( ) ( )

h h h w w w

h h w w

0.50 kg 448 J/ kg K 450 C 25 kg 4186 J/ kg K 23 C

0.50 kg 448 J/ kg K 25 kg 4186 J/ kg K

24 C

m c T m c TT

m c m c

T

+=

+

⋅ ° + ⋅ ° =⋅ + ⋅

= °

2. (b) Write the heat

capacities of the

lead and iron:

( )

( )

Pb

Fe

1 kg 128J/ kg K 128 J/K

0.50 kg 8 J/ kg K 224 J/K

C

C

= × ⋅ =

= × 44 ⋅ =

3. Compare the heat

capacities:

Since the heat capacity of the lead is less than the heat capacity of the

iron, the final temperature will be less than 24°C.

Insight: Even though the lead had twice the mass of the iron, the specific heat of lead is small

enough that the heat capacity of the iron was larger than the heat capacity of the lead.


74 of 89

19 Picture the Problem: As coffee and cream are

poured and mixed in a ceramic cup, heat exchanges

between the three objects until they come to the

same equilibrium temperature.

Strategy: Set the sum of the heat exchanges

between the coffee, cream, and cup equal to zero

since no heat leaves the system. Then using the

specific heat equation, solve for the equilibrium

temperature. The specific heat of ceramic is given

in the problem. Use the specific heat of water for

the specific heat of the coffee and cream.

Solution: 1. Set the sum

of the heats

equal to

zero:

( ) ( ) ( )

( ) ( )

cup cof crm

cup cup cup cof w cof crm w crm

cup cup cof crm w cup cup cup cof cof crm crm w

0

0

Q Q Q

m c T T m c T T m c T T

T m c m m c m c T m T m T c

= + +

= − + − + −

= + + − + +

2. Solve for

the equilib-

rium

temperature:

( )( )

( ) ( ) ( )

( )( ) ( )( ) ( )

( ) ( ) ( ) ( )

cup cup cup cof cof crm crm w

cup cup cof crm w

0.116 kg 1090 J/ kg K 24.0 C

0.225 kg 80.3 C 0.0122 kg 5.00 C 4186 J/ kg K70

0.116 kg 1090 J/ kg K 0.225 kg 0.0122 kg 4186 J/ kg K

m c T m T m T cT

m c m m c

+ +=

+ +

⋅ °

+ ° + ° ⋅ = =

⋅ + + ⋅ .5 C°

Insight: The comparatively large heat capacity of the coffee, compared with the heat capacities of the

cream and cup, causes the equilibrium temperature to be much closer to the initial temperature of the

coffee than to the initial temperature of the cream or cup.

20. Picture the Problem: Heat conducts through a lead brick from the warm end to the cooler end.

Strategy: Calculate the heat flow through the brick. The thermal conductivity of lead is found in

the table.

Solution: Apply the

equation directly: ( )( ) ( )3 2 8.5 C

34.3 W/ m K 1.2 10 m 1.0 s 2.3 J0.15 m

TQ kA t

L

−∆ ° = = ⋅ × =

Insight: An identical copper brick would transfer 27 J because of its higher thermal conductivity.

21. Picture the Problem: Two metal bars, of equal

lengths and diameters, are connected in parallel. Heat

transfers across both bars. We want to know what

length is necessary for the rate of heat transfer to be

27.5 J/s.

Strategy: Set the total rate of heat transfer equal to

the sum of the heat transfers through each bar. Solve

for the length of the bars.

Solution: 1. (a) Sum the heat

transfer rates

for each bar:

( )total stlAl st Al st

AQ QQ T T T

k A k A A k kt t t L L L

∆ ∆ ∆ = + = + = +

2. Solve for the

length of the bars: ( )

( ) ( ) ( ) ( )

( )

Al st

total

214

( )

0.0350 m 118 20 C 217 16.3 W/ m K0.800 m

27.5 J/s

A T k kL

Q t

π

∆ +=

− ° + ⋅ = =


75 of 89

3. (b) Part (a) shows that the rate of heat transfer is inversely proportional to the length of the bar.

Therefore, doubling the length of the bars causes the rate to decrease by a factor of 2.

Insight: The thermal conductivity of aluminium is much greater than the thermal conductivity of

stainless steel, so most of the heat transfers through the aluminium; 25.6 J sAlQ t = and

1.9 J sstQ t = .

22. Picture the Problem: Two metal bars, of equal

lengths, are connected in parallel. Heat transfers

across both bars. The diameter of the lead rod is

known. We want to know what diameter of the copper

rod is necessary for the rate of heat transfer through

both rods to be 33.2 J/s.

Strategy: Set the total rate of heat transfer equal to the

sum of the heat transfers through each rod. Solve for

the diameter of the copper rod.

Solution: 1. Sum the heat

transfer rates

of the two

rods:

total Cu Pb

Cu Cu Pb Pb

2 2 2 2

Cu Cu Pb Pb Cu Cu Pb Pb4 4 4

Q Q Q T Tk A k A

t t t L L

T Tk d k d k d k d

L L

π π π

∆ ∆ = + = +

∆ ∆ = + = +

2. Solve for

the diameter

of the

copper rod: ( )( )

( )( ) ( )

( )

2totalPb Pb

Cu

Cu

2

4

4 0.650 m33.2 J/s 34.3W/ m K 0.0276 m

112 21 C2.64 cm

395 W/ m K

Q Lk d

t Td

k

π

π

− ∆ =

− ⋅ − °= =

⋅

Insight: The diameters of both rods are about the same. However, since the thermal conductivity of

the copper is over 10 times the thermal conductivity of lead, over 90% of the heat passes through

the copper.

23. Picture the Problem: Two metal rods are connected in

series. Heat flows from the high temperature source

through the copper rod and then through the lead rod

before reaching the cold temperature end.

Strategy: Use the heat flow through the rods to

calculate the temperature at the junction of the rods.

Solution: 1. (a) The temperature of the junction is

greater than 54 °C. Since lead has a smaller thermal

conductivity than copper, it must have a greater

temperature difference across it to have the same heat

flow.

2. (b) Solve for

the junction

temperature: ( )( )

( ) ( ) ( )

2 1

1 2 2

1.41 J 0.525 m106 °C 98°C

395W/ m K 0.015 m 1 s ec

T TQ k A t

L

QLT T

k At

−=

= − = − =⋅

Insight: This problem can also be solved by finding the temperature difference across the lead:

( )( )

( ) ( ) ( )2 1 2

1.41 J 0.525 m2°C 98°C

34.3W/ m K 0.015 m 1 sec

QLT T

k At= + = + =

⋅


76 of 89

24. Picture the Problem: Two metal rods are connected

end to end as shown in the diagram. Heat flows from

the hot temperature end through the aluminium and

then through the lead to reach the cold side.

Strategy: To calculate the length of the aluminum rod,

set the heat flow through the aluminum rod equal to the

heat flow through the lead rod and solve for the length.

Solution: 1. (a) Since the heat must flow through one

rod before it passes through the other, the heat flow

rate through both rods must be the same.

2. (b) Set

the rate of

heat flow

through the

rods equal:

PbAl

Al Pb

Al Pb

TTk A k A

L L

∆∆⋅ =

3. Solve for

the length of

the

aluminum

rod:

( )( )

( )Al Al

Al Pb

Pb Pb

217 W/ m K 80.0 C 50.0 C14 cm 89 cm

34.3 W/ m K 50.0 C 20.0 C

k TL L

k T

⋅ ∆ ° − °= = =

∆ ⋅ ° − °

Insight: Since heat flows more easily through the aluminium rod, and both rods experience the

same change in temperature, the aluminium rod must be longer than the lead rod.

25. Picture the Problem: A copper poker is kept hot at

one end and cool at the other as shown in the figure.

Strategy: We wish to find the temperature 23 cm

from the cold end. The rate of heat flow through the

copper rod is constant throughout the rod. Set the

heat flow through the entire rod equal to the heat flow

through the final 23 cm and solve for the temperature

23 cm from the end.

Solution: 1. Set the heat flow

through the rod equal to the

heat flow through the last 23

cm:

23QQ

t t=

2. Substitute for the heat flows: 23 21 C105 C 21 C

85 cm 23 cm

TkA kA

− °° − ° =

3. Now solve for T23: ( )23

23 cm21 C 84 C 21 C 23 C 44 C

85 cmT

= ° + ° = ° + ° = °

Insight: Note that the thermal conductivity of copper cancelled out of this problem. The

temperature would be 44°C at 23 cm from the end for any type of thermally conductive poker.

26. Picture the Problem: Two identical objects at different temperatures radiate heat into a room,

which is at a lower temperature than the object.

Strategy: We wish to find the ratio of power radiated by the hotter object to power radiated by the

colder object. Divide the net power radiated by the hotter object by the power radiated by the

cooler object.

Solution: Write the

ratio:

( )( )

( ) ( )

( ) ( )

4 44 4

1 s1

4 44 42 2 s

273.15 95 K 273.15 21 K26

273.15 25 K 273.15 21 K

e A T TP

P e A T T

σ

σ

− + − + = = =− + − +

Insight: Since the cooler object’s temperature is close to the room temperature, its net power


77 of 89

radiated is much smaller than the power radiated by the warmer object. If the room temperature

were increased by only 2 degrees (to 23°C), the ratio would increase to 51.

27. Picture the Problem: A rectangular cube has sides L, 2L, and 3L,

as shown in the figure. When the L × 2L faces are held at fixed

temperature, the cube is a conductor with cross-sectional area A = 2

L2 and length 3L. Changing the sides that are held at fixed

temperature changes the cross-sectional area and length of the

conductor.

Strategy: Write P in terms of the thermal conductivity, length L, and change in temperature. For

each rotation of the cube, use the appropriate the cross-sectional area and length to write the rate of

heat flow in terms of P.

Solution: 1. Set P equal to

the heat flow through the

length 3L and 2A L L= × :

( )2 22

3 3

T TP kA k L k L T

L L

∆ ∆ = = = ∆ ′

2. (a) Set Pa equal to the heat

flow through the length 2L

and 3A L L= × :

( )2 3 93

2 2 4a

TP k L k L T P

L

∆ = = ∆ =

3. (b) Set Pb equal to the heat

flow through the length L and

3 2A L L= × :

( )2

b 6 6 9T

P k L k L T PL

∆ = = ∆ =

Insight: Note that the heat flow is greatest when the length is the shortest and the cross-sectional

area is the greatest. The heat flow is least when the length is the greatest.

3 L

L

2 L


78 of 89

Solutions to Tutorial 9 1. Picture the Problem: A proton moves in a magnetic field that is directed at right angles to its

velocity.

Strategy: Combine Newton's Second Law with the magnetic force (equation 22-1) to find the

acceleration of the particle.

Solution: Set the magnetic force

equal to the mass multiplied by the

acceleration and solve for a: ( )( )( )19

9 2

27

sin 90

1.6 10 C 9.5 m/s 1.6 T1.5 10 m/s

1.673 10 kg

F ma evB

evBa

m

−

−

= = °

×= = = ×

×

Insight: If the magnetic field were parallel to the velocity, the angle θ = 0° and the force and

acceleration would be zero.

2. Picture the Problem: An electron moves at right angles to a magnetic field and experiences a

magnetic force.

Strategy: Solve for the speed of the electron that would produce the specified force.

Solution: Solve for v:

( )( )

155

19

8.9 10 N4.6 10 m/s

sin 90 1.6 10 C 0.12 T

Fv

e B

−

−

×= = = ×

° ×

Insight: Doubling the magnetic field would cut the required speed in half in order for the electron

to experience the same magnetic force.

3. Picture the Problem: A negatively charged ion moves due north while immersed in Earth’s

magnetic field at the equator.

Strategy: The magnetic field of Earth points due north at the equator. Therefore, the velocity of

the particle is parallel to the magnetic field, which means that the angle θ = 0 and the force on the

particle is zero.

Solution: Solve directly: sin 0 0F qvB= ° =

Insight: If the ion had the same charge as an electron, and moved due east instead of north, and

the magnetic field of Earth pointed north and had a magnitude of 5.0×10−5

T at that location, the

magnetic force on the ion would be 1.2×10−17 N in the downwards direction (remember the ion is

negatively charged!).

4. Picture the Problem: A proton moves straight downwards from a location above the equator,

moving at right angles to the magnetic field that is horizontal and points due north.

Strategy: Combine Newton's Second Law with the magnetic force to find the acceleration of the

proton.

Solution: Set the magnetic force

equal to the mass multiplied by

the acceleration and solve for a:

( )( )( )19 5

27

6 2

1.60 10 C 355 m/s 4.05 10 Tsin 90

1.673 10 kg

1.38 10 m/s

evBa

m

− −

−

× ×°= =

×

= ×

Insight: The proton would experience the very same acceleration if it were travelling due east or

due west, or any other direction that is perpendicular to the horizontal magnetic field that points due

north.

5. Picture the Problem: A charged particle moves in a region in which a magnetic field exists.

Strategy: Solve the magnetic force equation for the angle θ that would produce the specified force.

Solution: 1. (a) Solve for θ : 1

sinF

qvBθ −

=


79 of 89

2. Insert the numerical values for F =

4.8 µN: ( )( ) ( )

61

6

4.8 10 Nsin 81

0.32 10 C 16 m/s 0.95 Tθ

−−

−

× = = °

×

3. (b) Insert the numerical values for F

= 3.0 µN: ( )( )( )

61

6

3.0 10 Nsin

0.32 10 C 16 m/s 0.95 Tθ

−−

−

× = = 38°

×

4. (c) Insert the numerical values for F

= 0.10 µN: ( )( )( )

71

6

1.0 10 Nsin 1.2

0.32 10 C 16 m/s 0.95 Tθ

−−

−

× = = °

×

Insight: The magnetic force is the largest when the velocity and the magnetic field are

perpendicular, and it is zero when vr

and Br

are parallel.

6. Picture the Problem: A charged particle moves in a region in which a magnetic field exists.

Strategy: Use a ratio to determine the force the particle experiences after changing its speed and

the angle its velocity makes with the magnetic field.

Solution: 1. Make a ratio:

( )( )

new new new new new

old old old old old

6.3 m/s sin 25sin sin0.099

sin sin 27 m/s sin 90

F qv B v

F qv B v

θ θ

θ θ

°= = = =

°

2. Now solve for

newF : ( )( )4 5

new old0.099 0.099 2.2 10 N 2.2 10 NF F− −= = × = ×

Insight: The speed decreased by a factor of 4.3 and the angle changed to 25°, decreasing the force

by another factor of 2.4. Together these two effects decreased the force by a factor of 10.

7. Picture the Problem: An ion moves with constant speed in a magnetic

field.

Strategy: The ion experiences no magnetic force when it is moving in

the y direction, so we conclude that the magnetic field also points in the

y direction. When the ion travels in the xy plane and along the line y = x,

it moves at an angle of 45° with respect to .Br

When it moves in the x

direction, it experiences the maximum magnetic force.

Solution: Letting maxF q vB= : ( )16 16

maxsin 45 6.2 10 N sin 45 4.4 10 NF F

− −= ° = × ° = ×

Insight: The ion won’t travel along the line y = x for very long, because the magnetic force will

cause the charged particle to travel in a circle.

8. Picture the Problem: An electron moves in a region in which a magnetic field exists.

Strategy: The ion experiences no magnetic force when it is moving in the x direction, so we

conclude that the magnetic field also points along either the ˆ ˆ or + −x x direction. We can use either

the Left-Hand Rule for negative charges, or use the Right-Hand Rule and remember to reverse the

direction of the force because the charge on the electron is negative. Try pointing your left thumb

downwards ( ˆ−z direction) and the fingers of your left hand in the y direction to find that the

magnetic field must point in the ˆ−x direction. Then use the force experienced by the electron when

it moves in the ˆ+y direction to find the magnitude of .Br

Solution: Find :Br

( )

( ) ( )( )

13

19 5

2.0 10 N sin 90sinˆ1.4 T 1.4 T

1.6 10 C 9.1 10 m/s

FB

ev

θ−

−

× °= = = ⇒ = −

× ×B xr

Insight: If instead the electron were to travel in the ˆ+z direction, it would experience a force

( )13 ˆ2.0 10 N .−= ×F yr


80 of 89

9. Picture the Problem: Two charged particles travel in a magnetic field along the same direction

and experience the same magnetic force, but they travel at different speeds.

Strategy: Calculate the ratio of the speeds of the particles.

Solution: 1. (a) Since the magnetic force is directly proportional to both the charge and the speed

of the particles, and since the particles experience the same force, particle 2 must have a greater

speed because particle 1 has the greater charge.

2. (b) Find the ratio: 1 1 2 2

2 2 1 2

sin 1

sin 4 4

v F q B q q

v F q B q q

θ

θ= = = =

Insight: Suppose both charges were allowed to travel in circles as described in section 22-3.

Assuming both charges have the same mass, we find that 1 1 1 1 2 1 2

2 2 2 2 1 1 2

1

4 4 16

r mv q B v q v q

r mv q B v q v q= = = = ;

that is, the slower particle 1 with the larger charge has a much smaller radius of motion in the

magnetic field.

10. Picture the Problem: The magnetic force on an electron travelling at constant speed causes it to

move in a circle.

Strategy: Find the radius of the electron’s circular orbit.

Solution: Apply equation

directly:

( )( )( )( )

31 5

6

19

9.11 10 kg 6.27 10 m/s7.9 10 m 7.9 m

1.6 10 C 0.45 T

mvr

eBµ

−

−

−

× ×= = = × =

×

Insight: A proton has the same magnitude charge but a much larger mass, which causes it to orbit

in a circle of much larger radius than an electron if the two particles have the same speed.

11. Picture the Problem: The magnetic force on a proton travelling at constant speed causes it to

move in a circle.

Strategy: Find the radius of the proton’s circular orbit.

Solution: Apply equation

directly:

( )( )( )( )

27 5

2

19

1.673 10 kg 6.27 10 m/s1.5 10 m 1.5 cm

1.6 10 C 0.45 T

mvr

eB

−

−

−

× ×= = = × =

×

Insight: An electron has the same magnitude charge but a much smaller mass, which causes it to

orbit in a circle of much smaller radius than a proton if the two particles have the same speed.

12 Picture the Problem: The magnetic force on an electron

travelling at constant speed causes it to move in a circle.

Strategy: The magnetic force provides the centripetal force

required to keep the electron moving in a circle. The radius of the

circle in terms of m, v, q, and B is given. We must first find the

speed v of the electron by using conservation of energy, then solve

for B.

Solution: 1. Find

v:

( ) ( )

( )

19

31

7

2 1.60 10 C 410 V2

9.11 10 kg

1.2 10 m/s

e Vv

m

−

−

×∆= =

×

= ×

2. Solve for B: ( ) ( )

( )( )

31 7

4

19

9.11 10 kg 1.2 10 m/s4.0 10 T 0.40 mT

1.60 10 C 0.17 m

mvB

er

−

−

−

× ×= = = × =

×

Insight: The electron’s speed is only 4.0% of the speed of light, so that we can safely neglect

relativistic effects. The magnetic field of 0.40 mT is about eight times larger than Earth’s magnetic

field (0.50 G = 0.050 mT).


81 of 89

13. Picture the Problem: The magnetic force on a charged particle



required to keep the particle moving in a circle. The radius of

the circle in terms of m, v, q, and B is given. We must first solve

for the speed v of the particle, then find the time it takes the

particle to complete one orbit by dividing the circumference by

the speed.


v:

( )( )( )6

5

12.5 10 C 26.8 m 1.01 T

2.80 10 kg

12.1 m/s

qrBv

m

−

−

×= =

×

=

2. (b) Divide the

circum-

ference by

the speed:

( )( )

2 26.8 m2 213.9 s

12.1 m/s

r mt

v qB

ππ π= = = =

Insight: A fairly large field (1.01 T is 20,200 times stronger than Earth’s field) is required to keep

this particle travelling in a circle of large radius (26.8 m) because the charge to mass ratio q / m is

relatively small.

14. Picture the Problem: The magnetic force on a charged particle


Strategy: Apply the Right-Hand Rule to the diagram at the

right in order to determine whether the particle is positively or

negatively charged. Then find the radius of the circle in terms

of m, v, q, and B, in order to find the mass m of the particle.

Solution: 1. (a) According to the RHR, a positively charged

particle would experience a force to the left. Since the particle

is experiencing a force to the right, it must be negatively

charged.

2. (b) Solve for m:

( )( )( )

( )

19

276

1.60 10 C 0.520 m 0.180 T 1.0 u1.5 u

1.67 10 kg6.0 10 m/s

erBm

v

−

−

×= = × =

××

Insight: Another way to answer part (a) is to point the fingers of your hand along vr

and curl them

into the page in the direction of .Br

For which hand does the thumb (and therefore Fr

) point to the

right? Since it is the left hand that works, the particle must be negatively charged, since negative

particles follow the Left-Hand Rule.

15. Picture the Problem: The magnetic force on a proton


Strategy: The magnetic force provides the centripetal

force required to keep the electron moving in a circle.

The radius of the circle in terms of m, v, q, and B . We

must first find the speed v of the proton by using the

definition of kinetic energy, then find the radius r.

Solution: 1. Solve for

v:

21 2

2

KK mv v

m= ⇒ =

2. Find r: ( )( )

( )( )

27 16

19

2 1.673 10 kg 4.9 10 J2 23.1 cm

1.6 10 C 0.26 T

mv m K mKr

eB eB m eB

− −

−

× ×= = = = =

×

Insight: The magnetic field must be increased in order to decrease the radius of the proton’s path.


82 of 89

16. Picture the Problem: The magnetic force on an alpha particle



required to keep the particle moving in a circle. The radius of

the circle in terms of m, v, q, and B is given. We must first find

the time it takes the particle to travel halfway through a

complete circle by dividing half the circumference by the

speed, then substitute for the speed v of the particle in the

expression.


half of the orbit

period, then substitute

for v:

1 2

2 2

T r r mv mt

v v v qB qB

π π π π = = = = =

2. Insert numerical

values:

( )( )( )

27

7

19

6.64 10 kg4.21 10 s 421 ns

1.60 10 C 0.155 Tt

π −

−

−

×= = × =

×

3. (b) The time does not depend upon the speed of the particle, so the answer to part (a) will stay

the same .

4. (c) The time stays

the same:

74.21 10 s 421 nst−= × =

Insight: Since the charge-to-mass ratio is a constant for alpha particles, the only way to change the

period of its orbit (or 1 ,T f= which is sometimes called the cyclotron frequency) is to change the

magnitude of B.

17. Picture the Problem: The magnetic force causes both an electron and a proton to move in circles

at constant speed.

Strategy: Form a ratio electron protonr r for each of two cases: (a) the particles have equal momenta,

and (b) the particles have equal kinetic energies.

Solution: 1. (a) Calculate e pr r when

e p :p p=

e e ee e

p pp p p

1.00m v q Br p e B

r p e Bm v q B= = =

2. (b) Calculate e pr r when e p :K K=

e e e e e ee e e e

p p p pp p p p p p

31

27

2

2

9.11 10 kg0.0233

1.673 10 kg

m v q B m K m e Br m K m

r m K mm v q B m K m e B

−

−

= = = =

×= =

×

Insight: If the kinetic energies are the same, the proton has a much larger radius. If the momenta

are the same, the two radii are equal.

18. Picture the Problem: A current-carrying wire experiences a force due to the presence of a

magnetic field.

Strategy: Find the magnetic force that is exerted on the wire.

Solution: Find F: ( )( )( )( )sin 90 0.695 A 2.15 m 0.720 T 1 1.08 NF I L B= ° = =

Insight: This force amounts to 0.50 N or 1.8 ounces of force per meter of wire. If the current were

increased to 15 A, the force would increase to 11 N per meter of wire.


83 of 89


magnetic field.

Strategy: Find the magnetic force that is exerted on the wire.

Solution: Find F: ( ) ( )( )sin 2.8 A 2.25 m 0.88 T sin 36.0 3.3 NF I L B θ= = ° =

Insight: The maximum force occurs at θ = 90°, at which angle the force amounts to 2.5 N of force

per meter of wire.


magnetic field.

Strategy: Solve for the angle θ that would produce the given magnetic force on the wire.

Solution: Solve for θ:

( )( )( )1 1

sin

1.6 Nsin sin 63

3.0 A 1.2 m 0.50 T

F

I L B

F

I L B

θ

θ − −

=

= = = °

Insight: If the angle between the wire and the magnetic field were θ = 90°, the force on the wire

would be 1.8 N.

21. Picture the Problem: A magnetic field exerts forces on the four

sides of a square loop of current-carrying wire.

Strategy: Find the force on each of the four sides of the square

current loop.

Solution: 1. The top

and bottom wires

run parallel to the

field:

top bottom

0 orsin 0

180F F I L B

° = = =

°

2. The left and right wires

are perpendicular to the

field:

( )( )( )left right sin 90 9.5 A 0.46 m 0.34 T 1.5 NF F I L B= = ° = =

Insight: The force on the left wire points out of the page and the force on the right wire points into

the page.

22. Picture the Problem: A wire carries a current in the

positive x direction while immersed in a magnetic field that

exerts an upward force on the wire.

Strategy: Set the magnitude of the magnetic force equal to

the weight of the wire to find the magnitude of .Br

Then

use the Right-Hand Rule to determine the direction that Br

must point in order for the magnetic force to be exerted in

the upward direction on the wire.

Solution: 1. Set

the magnetic

force equal to

the weight:

sin 90F mg I L B I L B= = ° =

2. Solve for B:

( )( )( )( )

20.17 kg 9.81 m/s

0.34 T11 A 0.45 m

mgB

I L= = =

3. Let upward in the figure be the positive y-direction, and the positive x-direction be to the right.

The RHR stipulates that Br

must point into the page, or in the ˆ−z direction. Therefore,


84 of 89

( ) ˆ0.34 T .= −B zr

Insight: This physics setup is related to that used for magnetic levitation trains. A pretty

substantial field (0.34 T is 6800 times stronger than Earth’s magnetic field) is required to lift a

pretty small wire (0.17 kg) in this arrangement.

23. Picture the Problem: A current-carrying wire experiences a

force due to the presence of a magnetic field.

Strategy: Solve for B to answer the question of part (a), and

solve the same equation for θ to answer the question of part

(b).


equation in terms of

force per unit length:

sinF

I BL

θ=

2. Solve for B:

( )0.033 N/m

0.041 T 41 mTsin 6.2 A sin 7.5

F LB

I θ= = = =

°

2. (b) Solve for θ :

( )( )1 1 0.015 N/m

sin sin 3.46.2 A 0.041 T

F L

I Bθ − −= = = °

Insight: At small angles the sinθ function is roughly linear with θ, so that cutting the force per

unit length approximately in half requires cutting the angle in half.

24. Picture the Problem: A wire carries a current in a region

where the magnetic field exerts an upward force on the

wire.

Strategy: Set the magnitude of the magnetic force equal to

the weight of the wire to find the current required to

levitate the wire. The force is maximum when the current

is perpendicular to the field. Therefore, the minimum

required current occurs in the perpendicular configuration.

Solution: 1. Set

the magnetic

force equal to

the weight:

sin 90F mg I L B I L B= = ° =

2. Solve for I:

( )( )( )( )

20.75 kg 9.81 m/s

3.2 A2.7 m 0.84 T

mgI

LB= = =

Insight: This physics setup is related to that used for magnetic levitation trains. A pretty

substantial field (0.84 T is 16,800 times stronger than Earth’s magnetic field) is required to lift a

pretty small wire (0.75 kg) in this arrangement.

25

.

Picture the Problem: Two current-carrying wires experience a force due

to Earth’s magnetic field.

Strategy: Use the Right-Hand Rule to determine the direction and

magnitude of the magnetic force on the currents. Let x point east, y point

north, and z point upward. Converting the magnetic field from gauss to

tesla, we find that 4 50.59 G 1.0 10 T/G 5.9 10 T− −× × = × .

Solution: 1. (a) According to the RHR, the magnetic force points toward

north, 18° above the horizontal.


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2. Calculate

F: ( )( )( )5

sin

110 A 250 m 5.9 10 T sin 90 1.6 N

F I L B θ

−

=

= × ° =

3. (b) According to the RHR, the magnetic force points toward the east.

4. Calculate F: ( )( )( )5sin 110 A 250 m 5.9 10 T sin 72 1.5 NF I L B θ −= = × ° =

Insight: These are the forces if the wires carry direct current. In reality, the electrical current in high-

voltage wires is alternating current, so that the net force on the wire due to the magnetic field of Earth is

zero.

26. Picture the Problem: A metal bar is suspended from

two conducting wires and immersed in a magnetic

field that points straight downward.

Strategy: Looking down the bar from the left, so that

the current Ir

points into the page, the magnetic force

I L B points to the left, mg points downward, and the

tension in the wires points up and to the right. Use

Newton's Second Law in the vertical and horizontal

directions to find the angle θ at which the rod is in

equilibrium.

Solution: 1. Set the net forces equal to

zero:

0 cos

0 sin

y

x

F T mg

F T I LB

θ

θ

= = −

= = −

∑∑

2. Solve each expression from step 1 for T:

cos sin

mg I L BT

θ θ= =

3. Rearrange to solve for θ :

1

tan

tan

I LB

mg

I LB

mg

θ

θ −

=

=

Insight: The angle θ can be increased by increasing I or B and decreased by increasing m.


86 of 89


1. Picture the Problem: The image show a ring of radius 3.1 cm

oriented at an angle of θ = 16º from a B = 0.055 T magnetic field.

Strategy: Solve for the flux.

Solution: Calculate the flux:

( ) ( )2

4

cos

0.055 T 0.031 m cos16

1.6 10 Wb

BA θ

π−

Φ =

= °

Φ = ×

Insight: The maximum flux through this coil, 1.66×10− 4 Wb, occurs when the angle θ is zero.

2. Picture the Problem: The image shows a

box immersed in a vertical magnetic field.

Strategy: Calculate the flux through each

side.

Solution: 1. The sides of the box are parallel

to the field, so the magnetic flux through the

sides is zero .

2. Calculate the flux

through the bottom: ( )( )( ) 4cos 0.0250 T 0.325 m 0.120 m cos0 9.75 10 Wb .BA θ −Φ = = ° = ×

Insight: The height of the box is not important in this problem.

3. Picture the Problem: The image shows a rectangular loop oriented 42

degrees from a magnetic field.

Strategy: Solve for the magnetic field.

Solution: Calculate the

magnetic field: ( )( )

5 24.8 10 Tm17 mT

cos 0.055 m 0.068 m cos 42B

A θ

−Φ ×= = =

°

Insight: The minimum magnetic field that would produce this flux would occur when the rectangle

is parallel to the magnetic field.

4. Picture the Problem: A house has a floor of dimensions 22 m by 18 m. The local magnetic field

due to Earth has a horizontal component 2.6×10-5

T and a downward vertical component 4.2×10-5

T.

Strategy: The horizontal component of the magnetic field is parallel to the floor, so it does not

contribute to the flux. Calculate the flux using the vertical component.


magnetic flux: ( )( )( )5 2cos 4.2 10 T 22 m 18 m 1.7 10 WbBA B Aθ − −

⊥Φ = = = × = ×

Insight: The flux through the vertical walls of the house is determined by the horizontal component

of the magnetic field instead of the vertical component.

5. Picture the Problem: A solenoid of diameter 1.2 m produces a magnetic field of 1.7 T.

Strategy: Multiply the magnetic field by the cross-sectional area of the solenoid to calculate the

magnetic flux.

Solution: Calculate the magnetic flux: ( )

21.2 m

cos 1.7 T cos 0 1.9 Wb2

BA θ π

Φ = = ° =

Insight: Note that the length of the solenoid does not affect the flux through the solenoid.


87 of 89

6. Picture the Problem: A magnetic field of magnitude 5.9×10−5

T is directed 72° below the

horizontal and passes through a horizontal region 130 cm by 82 cm.

Strategy: Calculate the flux, where the angle from the vertical is θ = 90° − 72°.


flux:

( )( )( ) ( )5 5cos 5.9 10 T 1.30 m 0.82 m cos 90 72 6.0 10 WbBA θ − −Φ = = × ° − ° = ×

Insight: Increasing the angle from the horizontal increases the flux through the desk top. For

example, if the angle were increased to 80° from the horizontal the total flux would increase to

6.2×10-5 Wb.

7. Picture the Problem: A coil of radius 15 cm and 53 turns is oriented perpendicular to a magnetic

field. The magnetic field changes from 0.25 T to zero in 0.12 s.

Strategy: Calculate the induced emf, with the flux.


emf:

( ) ( )2

0.25 T 0.15 m053 7.8 V

0.12 s

BAN N

t t

πε ∆Φ −

= = = =∆ ∆

Insight: Note that the emf is inversely proportional to the time it takes for the magnetic field to

change. In this case, if it dropped to zero in only 0.060 seconds (half the time), the average emf

would be twice as large, or 15.6 V.

8. Picture the Problem: The figure shows the flux through

a single loop coil as a function of time.

Strategy: Calculate the emf at the times

t = 0.05 s, 0.15 s, and 0.50 s. Use the graph to find the

change in flux.

Solution: 1. (a) Calculate

the emf at t =

0.05 s:

10 Wb 00.1 kV

0.1 sN

tε ∆Φ −

= − = − = −∆

2. (b Calculate

the emf at

t = 0.15 s:

0ε =

3. (c) Calculate the

emf at

t = 0.50 s:

5Wb 10 Wb0.04 kV

0.6 s 0.2 sε − −

= − =−

Insight: When the slope of the flux is constant, the emf is constant. The emf is 0.04 kV from t = 0.2

s to t = 0.6 s.

9. Picture the Problem: The image shows the emf

through a single loop as a function of time.

Strategy: Calculate the emf at the times t = 0.25 s and

t = 0.55 s

Solution: 1. (a) The flux at t = 0.25 s is about 8 Wb.

This is greater than the flux at t = 0.55 s, which is

about −3 Wb.

2. (b) The two emf values are the same, because at

those two times the flux is changing at the same rate.

3. (c) Calculate the slope of the flux between

0.2 s and 0.6 s:

5 Wb 10 Wb37.5 Wb/s

0.6 s 0.2 st

∆Φ − −= = −

∆ −


88 of 89

4. Calculate the induced emf : ( )1 37.5 Wb/s 0.04 kVN

tε ∆Φ

= − = − − =∆

Insight: Note that the emf is zero for times 0.1 s < t < 0.2 s and t > 0.6 s. The voltage is not

determined by the magnitude of the flux but by the slope of the flux vs. time graph. For these two

time periods the slope is zero.

10. Picture the Problem: The image shows a single loop of area

7.4×10−2

m2 and resistance 110 Ω. The loop is perpendicular to a

magnetic field.

Strategy: Solve Ohm’s Law for the necessary emf . Then insert

the emf to calculate the rate of change in the magnetic field.

Solution: 1. Calculate

the emf : ( )( )0.32 A 110 35.2 VIRε = = Ω =

2. Solve for the

change in

magnetic field:

( )2

2 2

35.2 V4.8 10 T/s

1 7.4 10 m

A BN N

t t

B

t ΝΑ

ε

ε−

∆Φ ∆= =

∆ ∆

∆= = = ×

∆ ×

Insight: The magnitude of the magnetic field (0.28 T) is not important in this problem, only the

change in the field.

11. Picture the Problem: A coil with 120 loops is oriented perpendicular to a changing magnetic

field.

Strategy: Solve for the emf , where the magnetic flux is given.

Solution: Calculate the emf

:

( )

( )( )2

2

2 0.20 T 0.050 m120 7.1 V

0.34 s

BA BA BAN N N

t t tε

− −∆Φ= = =

∆ ∆ ∆

= =

Insight: This emf is double the value the coil would experience if the magnetic field simply

dropped to zero in the same time period.

12

.

Picture the Problem: The image shows a square loop of wire with

circumference 1.12 m that is in a 0.105 T magnetic field. The

square is changed into a circle with the same perimeter in 4.25 s.

Strategy: Calculate the flux through the circle and the rectangle.

Then insert the fluxes to calculate the induced emf .

Solution:

1. Calculate

the radius

of the

circle:

2

0.1783 m2

C r

Cr

π

π

=

= =

2. Calculat

e the

flux

through

the

circle:

( ) ( )22

circle 0.105 T 0.1783 m

0.0105 Wb

BA B rπ πΦ = = =

=


89 of 89

3. Calculate

the side

of the

square:

4

1.12 m0.28 m

4 4

P s

Ps

=

= = =

4. Calculate

the flux

through

the

square:

( )( )22

square 0.105 T 0.28 m 0.00823 WbBA BsΦ = = = =

5. Calculate

the emf :

circle square 40.0105 0.00823 Wb1 5.3 10 V

4.25 sN

tε −

Φ − Φ −= = = ×

∆

Insight: The induced current will be clockwise because the flux through the coil is increasing.

13. Picture the Problem: A number of circular loops of wire are oriented perpendicular to a changing

magnetic field.

Strategy: Solve for the number of coils, with the flux given. The radius of the loops is

( )1 1

2 20.12 m 0.060 m.r d= = =


number of coils:

( ) ( )1

21 1

30.20 T 0.060 m

6.0 V 4.0 101.5 s

Nt

BAN

t t

π

ε

ε ε−

− −

∆Φ=

∆

∆Φ= = = = ×

∆ ∆

Insight: Note that the number of coils is proportional to the desired emf . To obtain an emf of 3.0

V, only 2.0×103 coils would be needed.

Documents

danny/EE1427_SOL.pdfsin 7.5 kg 1.41 9.81 m/s sin13 cos cos13 28 N F F mg max ma mg F F θ θ θ θ = − = + + ° = = ° = ∑ Insight: They’d better offer double coupons at this