EE1427 Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
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.
r
Solution: 1. Find the net
∑ F = ( − 40.0 N ) yˆ + ( 46.2 N ) yˆ = ( 6.2 N ) yˆ
force:
r
2. Now apply Newton’s
r ∑ F ( 6.2 N ) yˆ
r
a=
=
= (1.2 m/s 2 ) yˆ
second law to find a :
m
5.00 kg
r
F
r
W
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. Find the net force on the skier
using:
a=
v 2 − v0 2
2 ∆x
 v 2 − v0 2
F = ma = m 
 2∆x
 (12 m/s )2 − 0 

=
92
kg
 = 260 N
)
 (

 2 ( 25 m − 0 ) 
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 v y2 = v02 y + 2a∆y
2
of the parachutist:
2
2
0 − ( −3.85 m/s )
r v y − v0 y
a=
yˆ =
yˆ = 9.88 m/s 2 yˆ
2 ∆y
2(0 − 0.750 m)
2. Find the net force:
r
r
∑ F = ma = ( 42.0 kg ) ( 9.88 m/s yˆ ) = ( 415 N ) yˆ
2
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.
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Tutorial Solutions
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.
r
5
r
r F ( – 4.30 × 10 ) xˆ N
Solution: 1. (a) Find a :
a= =
= ( −1.23 m/s 2 ) xˆ
m
3.50 × 105 kg
2. Find the final speed:
(
)
v = v0 + at = 27.0 m/s + −1.23 m/s 2 ( 7.50 s ) = 17.8 m/s
3. (b) Find the distance
∆x =
travelled by the 747 as it slows down:
1
2
( v0 + v ) ∆t = 12 ( 27.0 + 17.8 m/s )( 7.50 s ) = 168 m
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/s2.
5. Picture the Problem: The free body diagram of the brick is shown at right.
r
F
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.
r
W
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:
r r r
r
r
r
∑ F = F + W = m a = 0 so that F = −W
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.
r
F1
r
−F1
r
F2
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
r
trailer there is only one force F1 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
r
r
engine F2 and the trailer −F1 . Apply Newton’s second
and third laws as appropriate to find the requested
forces.
r r
r
Solution: 1. (a) Write Newton’s Second
∑ F = F1 = m a = ( 540 kg ) (1.90 m/s2 ) xˆ = (1.0 kN ) xˆ
Law for the trailer:
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:
r
−F1 = ( −1.0 kN ) xˆ
r
r
3. (c) Write Newton’s Second Law for the ∑ F = M a = (1300 kg ) (1.90 m/s 2 ) xˆ = ( 2.5 kN ) xˆ
car:
r
Insight: The engine force F2 must be ( 3.5 kN ) xˆ 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.
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EE1427 Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
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:
F = (m1 + m2 + m3 )a ⇒ a =
a=
F
m1 + m2 + m3
7.50 N
= 0.798 m/s 2
1.30 kg + 3.20 kg + 4.90 kg
r
2. Write Newton’s Second Law for
the first box:
∑F = F − F
c12
Fc12 = F − m1 a = 7.50 N − (1.30 kg ) ( 0.798 m/s 2 ) = 6.46 N
r
3. (b) Write Newton’s Second Law
for the third box:
= m1 a
∑F = F
c23
= m3 a = ( 4.90 kg ) ( 0.798 m/s 2 ) = 3.91 N
Insight: Another way to solve the problem is to note that Fc12 accelerates boxes 2 and 3:
Fc12 = ( 3.20 + 4.90 kg ) ( 0.798 m/s 2 ) = 6.46 N , 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:
F = (m + M ) a ⇒ a =
2. Find the contact force by
writing Newton’s Second
Law for the heavy box
only:
∑F = F
r
c12
5.0 N
F
=
= 0.40 m/s 2
m + M 5.2 + 7.4 kg
= M a = ( 7.4 kg ) ( 0.40 m/s 2 ) = 3.0 N
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.
r
4. (c) Find the contact force by
F = Fc12 = m a = ( 5.2 kg ) ( 0.40 m/s 2 ) = 2.1 N
∑
writing Newton’s Second Law
for the lighter box only:
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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EE1427 Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
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:
y
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:
∑F
x
r
F
x
θ
r
mg
mg sin θ
= + F − mg sin θ = ma = 0
F = mg sin θ = ( 3900 kg ) ( 9.81 m/s 2 ) sin16° = 11, 000 N = 11 kN
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
r
pushing force F , as shown in the figure at right.
Strategy: Write Newton’s Second Law for the x
direction, where x̂ points up the incline and parallel
to it. Solve the resulting equation for the magnitude
r
of F .
Solution: The component of the
force pushing up the incline is
F cos θ and the component of the
weight pushing down the incline is
mg sin θ :
∑F
x
= F cos θ − mg sin θ = ma
2
ma + mg sin θ ( 7.5 kg ) 1.41 + ( 9.81 m/s ) sin13° 
F=
=
cos θ
cos13°
F = 28 N
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: ∑ Fy = − (130 N ) sin 34° + F sin 45° = 0
Set the sum
sin 34°
F = (130 N )
= 100 N = 0.10 kN
of
sin 45°
the forces
in the y
direction
equal to
zero:
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).
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Tutorial Solutions
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
r
direction (parallel to a ) in order to find the
acceleration of the sled.
Solution
: Write
Newton’
s Second
Law
in the x
direction
:
∑F
= 2 F cos 35° − 57 N = ( msled + mchild ) ax
2 F cos 35° − 57 N
ax =
msled + mchild
2 ( 55 N ) cos 35° − 57 N
ax =
= 1.5 m/s 2
19 + 3.7 kg
x
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 Wy ,
cancel each other. The net force on the skier is then the
only unbalanced force, Wx . Fnd the net force on the
skier.
Solution: 1. (a) The
net force
on the skier is Wx :
∑F
x
= Wx = mg sin θ
= ( 65 kg ) ( 9.81 m/s 2 ) sin 22°
∑F
x
= 0.24 kN
The force points downhill, parallel to the slope.
2. (b) As the slope becomes steeper, θ increases and the net force Wx 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.
r r
r
Solution: 1. (a) Add ∑ F = FSM + FEM
the
= ( 4.34 × 1020 N ) xˆ + (1.98 × 1020 N ) yˆ
force vectors:
r
 ∑F 
20
2. Find the direction of
N
y 
−1 
−1  1.98 × 10
=
=
θ
tan
tan
r

 = 24.5° above the +x axis
20
the


F 
 4.34 × 10 N 
∑
x
net force:


(
(
3. (b) Find the
magnitude
of the net force:
r
)
)
∑ F = ( 4.34 ×10
20
2
2
N ) + (1.98 × 1020 N ) = 4.77 × 1020 N
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EE1427 Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
r
r ∑ F 4.77 × 1020 N
4. (c) Find the
=
= 0.00649 m/s 2
a=
acceleration
m
7.35 × 10 22 kg
from Newton’s Second
Law:
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.
r
Strategy: There are two forces acting on the astronaut: the applied force F of the seat acting
r
upward and the force of gravity W 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 :
∑F
y
= F − mg = ma
Wa = F = ma + mg = m ( a + g ) = ( 88 kg ) ( 30.2 + 9.81 m/s 2 )
Wa = 3500 N = 3.5 kN
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
r r
r
F = Wa of the scale acting upward and the force of gravity W
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.
∑F
= Wa − W = ma
W − W Wa − W 730 − 610 N
a= a
=
=
(9.81 m/s2 ) = 1.9 m/s2
m
W g
610 N
y
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.
r
Strategy: There are two forces acting on the bowling ball: the applied force F acting upward and
r
the force of gravity W 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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EE1427 Engineering Science – Dr. Daniel Nankoo
Solution: 1. (a) Write Newton’s Second
Law for each
case to obtain two equations and two
unknowns:
2. Divide the second equation by the first
and solve for m:
Tutorial Solutions
∑F
∑F
y ,1
= F1 − mg = ma
y ,2
= F2 − mg = 2ma
F2 − mg 2ma
=
= 2 ⇒ F2 − mg = 2 F1 − 2mg
F1 − mg
ma
m=
2 F1 − F2 2 ( 82 N ) − ( 92 N )
=
= 7.3 kg
g
9.81 m/s 2
3. Fnd the weight W:
W = mg = ( 7.3 kg ) ( 9.81 m/s 2 ) = 72 N
4. (b) Use the first Newton’s Second Law
equation to find a:
a=
F1 − mg 82 − 72 N
=
= 1.4 m/s 2
m
7.3 kg
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/s2.
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:
∑F
y
= N − W + F sin θ = 0
2
W − N ( 23 kg ) ( 9.81 m/s ) − 180 N
=
sin θ
sin 25°
F = 110 N = 0.11 kN
F=
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.
(a) child
(b) chair
Strategy: There are two forces acting on the child: the
r
normal force N of the chair acting upward and the force of
r
gravity Wchild acting downward. There are three forces
r
acting on the chair: the normal force N of the floor acting
upward, the weight of the baby acting downward, and the
r
force of gravity Wchair 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:
∑F
y
= N − mg = 0
N = mg = ( 9.3 kg ) ( 9.81 m/s 2 )
= 91 N
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EE1427 Engineering Science – Dr. Daniel Nankoo
2. (b) Write out
Newton’s
Second Law in the
vertical direction
for the chair and solve
for N:
∑F
y
Tutorial Solutions
= N − mbaby g − mchair g = 0
N = ( mbaby + mchair ) g
= ( 9.3 + 3.7 kg ) ( 9.81 m/s 2 )
N = 130 N = 0.13 kN
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:
r
Strategy: There are two forces acting on the potatoes: the normal force N
r
of the shopping cart acting upward and the force of gravity W 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:
∑F
2. Set N equal to half the weight
and
solve for θ:
1
2
r
= N − mg cos θ = 0
mg − mg cos θ = 0
θ = cos −1 ( 12 ) = 60°
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 µ k.
Solution: 1. Write
Newton’s
Second Law in the y
direction:
2. Now write Newton’s
Second
Law in the x direction:
∑F
y
= N − mg cos θ = 0
N = mg cos θ
∑F
y
= mg sin θ − µk N = ma
mg sin θ − ma = µk N
mg sin θ − ma = µk ( mg cos θ )
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Tutorial Solutions
9.81 m/s 2 ) sin 31.0° − 1.16 m/s 2
3. Solve the equation for µ = g sin θ − a = (
= 0.463
k
g cos θ
µ k:
( 9.81 m/s2 ) cos 31.0°
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 µ k can be determined.
Solution: 1. When the book begins
sliding, the applied force equals the
maximum static friction force:
F = fs = µs mg
F
2.25 N
µs = app =
= 0.127
mg (1.80 kg ) ( 9.81 m/s 2 )
2. When the book is sliding at constant
speed, the
applied force equals the kinetic friction
force:
Fapp = f k = µk mg
F
1.50 N
µ k = app =
= 0.0849
mg (1.80 kg ) ( 9.81 m/s 2 )
app
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:
2. Write Newton’s
Second
Law in the horizontal
direction:
∑F
y
r
N
r
fs
r
W
r
F
θ
= N − mg − F sin θ = ma y = 0
N = mg + F sin θ
∑F
x
= F cos θ − µs N = max = 0
F cos θ = µs N
F cos θ = µs ( mg + F sin θ )
3. Now substitute for
N and solve
F ( cos θ − µs sin θ ) = µs mg
for the maximum
force F that
( 0.57 )( 32 kg ) ( 9.81 m/s 2 )
µs mg
F
=
=
would produce no
( cos θ − µs sin θ ) cos 21° − ( 0.57 ) sin 21°
acceleration:
F = 250 N = 0.25 kN
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.
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Tutorial Solutions
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:
∑F
2. Solve the equation for a:
a = µs g = ( 0.24 ) ( 9.81 m/s 2 ) = 2.35 m/s 2 = 2.4 m/s 2
3. (b) Find the minimum time:
x
t=
= fs = µs mg = ma
v − v0 15 m/s − 0
=
= 6.4 s
a
2.35 m/s 2
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 212 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:
∑F
2. Find the travel
distance of the car:
d = ∆x =
x = − µ N = − µ mg = ma
a = −µ g
vf2 − v 2
02 − v 2
v2
=
=
2a
2 ( −µ g ) 2µ g
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.
r
T
θ
T cos θ
r
T
θ
T sin θ
r
W
Solution: Write Newton’s Second Law ∑ Fy = 2T sin θ − mg = 0
in the vertical direction and solve for the
( 50.0 kg ) ( 9.81 m/s2 )
mg
tension T:
T=
=
= 948 N
2sin θ
2sin15.0°
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.
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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.
Solution: 1. (a) Write
Newton’s Second Law in the
horizontal direction:
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.
∑F
x
= f s − F = max = 0
f s = F = −kx = − (150 N/m )( − 0.0200 m ) = 3.0 N
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 x = 2 L − L = L . Then use Hooke’s law
again to find the force required to compress the spring an amount equal to x = 12 L − L = − 12 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:
F = −k x = k L = ( 250 N/m )( 0.18 m ) = 45 N
2. (b) Use Hooke’s Law to find the
magnitude of the force required to
compress the spring by the amount
x = − 12 L :
F = −k x = k ( 12 L ) = ( 250 N/m ) 12 ( 0.18 m ) = 23 N
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
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Tutorial Solutions
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:
∑F
x
T2 = T1
∑F
3. (c) Write Newton’s Second Law
in the vertical direction in order to
find the picture’s weight:
= −T1 cos θ1 + T2 cos θ 2 = 0
y
cos θ1
cos 65°
= (1.7 N )
= 0.85 N
cos θ 2
cos 32°
= T1 sin θ1 + T2 sin θ 2 − W = 0
W = (1.7 N ) sin 65° + ( 0.85 N ) sin 32° = 2.0 N
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.
Lower Pulley
Cupper
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.
Solution: 1. (a) Write
Newton’s Second
Law for the box:
∑F
y
F
F
F
Clower
F
Upper Pulley
= Clower − mg = 0
Clower = mg
2. Now write Newton’s Second
Law for
the lower pulley and solve for
F:
∑F
3. (b) Write Newton’s Second
Law for
the upper pulley and solve for
Cupper
∑F
y
= F + F − Clower = 0
F = 12 Clower = 12 mg =
1
2
( 52 kg ) ( 9.81 m/s2 )
= 255 N = 0.26 kN downwards
y
= Cupper − F − F = 0
Cupper = 2 F = 2 ( 0.255 kN ) = 0.51 kN
4. (c) Use the result of step 1 to Clower = mg = ( 52 kg ) ( 9.81 m/s 2 ) = 510 N = 0.51 kN
find Clower:
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.
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Solution: 1. (a)
Write Newton’s
Second Law for
block 1 and
solve for T1:
∑
Tutorial Solutions
r
T2
Free-body diagram
for the upper block
Fx = −m1 g sin θ + T1 = 0
block 1
T1 = m1 g sin θ
r
T1
= (1.0 kg ) ( 9.81 m/s 2 ) sin 31°
r
mg θ
T1 = 5.1 N
2. (b) Write
Newton’s
Second Law for
block 2 and
solve for T2:
∑
mg sin θ
Fx = −T1 − m2 g sin θ + T2 = 0
block 2
T2 = T1 + m2 g sin θ = 5.1 N + ( 2.0 kg ) ( 9.81 m/s 2 ) sin 31° = 15 N
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 T2 = ( m1 + m2 ) g = 29 N .
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
r
the magnitude of F .
Solution: 1. Find
the rope tension:
2. Write Newton’s
Second Law
in the x direction
and solve for F:
T1 = T2 = mg
∑F
x
= − F + T1 cos 30.0° + T2 cos 30.0° = 0
F = 2mg cos 30.0° = 2 ( 2.50 kg ) ( 9.81 m/s 2 ) cos 30.0° = 42.5 N = 9.54 lb
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.
Solution: 1. (a) Write
Newton’s Second
Law in the vertical direction:
2. Write Newton’s Second
Law in the
horizontal direction:
∑F
y
= f s − mg = µs N − mg = 0
N=
∑F
x
mg
µs
=F−N =F−
mg
µs
=0
mg (1.6 kg ) ( 9.81 m/s
3. Solve for the applied force F =
=
µs
( 0.79 )
F:
2
) = 20 N = 0.020 kN
= 4.5 lb
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.
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Tutorial Solutions
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).
Solution: 1. (a) Write Newton’s
Second Law for M:
2. Write Newton’s Second
Law for m and substitute the
expression for T found in
step 1:
∑F
x
= − Mg sin θ + T = M a
T = M ( a − g sin θ )
∑F
x
= −T + mg = ma
−  M ( a + g sin θ )  + mg = ma
− Mg sin θ + mg = ( m + M ) a
3. Solve for a:
 m − M sin θ
a=
 m+M
 3.2 kg − ( 5.7 kg ) sin 35° 

2
 ( 9.81 m/s )
g = 
+
3.2
5.7
kg



a = − 0.076 m/s 2
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:
3. Solve the
resulting equation
for a:
∑
Fx = T
= m1a
block 1
∑
Fx = −T + m2 g = m2 a
block 2
m2 g = ( m1 + m2 ) a
 m2
a=
 m1 + m2

2.80 kg
(9.81 m/s2 ) = 4.36 m/s2
g =
6.30
kg

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4. (c) Use the first
equation to find T:
Tutorial Solutions
T = m1a = ( 3.50 kg ) ( 4.36 m/s 2 ) = 15.3 N
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 m2 g = 27.5 N. 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.
Solution: 1. (a) Write
Newton’s Second Law for
m2 with a = 0 :
∑F
2. (b) Write Newton’s
Second Law for m1:
∑F
3. Write Newton’s Second
Law for m2 and substitute
for the acceleration from
step 2:
∑F
4. Substitute W2 = m2 g and
m2 m1 = W2 W1 :
T=
5. (c) Write
∑F
x
= 0 for
x
= −T + m2 g = 0
T = m2 g = 110 N
x
= T − m1 g = m1 a
a = T m1 − g
x
= −T + m2 g = m2 a = m2 (T m1 − g )
2m2 g = T (1 + m2 m1 )
2 (110 N )
2W2
=
= 80 N
1 + W2 W1 1+ (110 N ) ( 63 N )
∑F
x
= T − m1 g = 0 ⇒ T = m1 g = 63 N
m1 with a = 0 :
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
a = 2.7 m/s 2 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.
2
Solution: Set the static friction force
equal to
the centripetal force:
fs = Fcp = macp =
mv 2 (1200 kg )(15 m/s )
=
= 4.7 kN
r
57 m
Insight: The maximum static friction force is
µs mg = ( 0.88 )(1200 kg ) ( 9.81 m/s 2 ) = 10.4 kN which corresponds to a maximum cornering speed
(without skidding) of 22 m/s.
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Tutorial Solutions
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: v = racp = r ( 52, 000 g ) =
( 0.075 m )( 52, 000 ) ( 9.81 m/s 2 )
= 200 m/s = 0.20 km/s
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. Insert
numerical
values:
∑F
y
= N − mg = −macp = − m v 2 r
N = m ( g − v2 r )
2

(12 m/s ) 
N = ( 67 kg ) 9.81 m/s 2 −
 = 380 N = 0.38 kN
35 m 

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:
∑F
y
2. Now solve v = rg =
for v:
= N − mg = −macp = − m v 2 r
0 − mg = −m v 2 r
( 35 m ) ( 9.81 m/s2 ) = 19 m/s
Insight: The car has zero normal force in it as well, meaning it is now airborne!
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Tutorial Solutions
Solutions to Tutorial 2
1. Picture the Problem: The farmer pushes the hay
horizontally.
86 N
Strategy: Multiply the force by the distance because
in this case the two point along the same direction.
3.4 m
W = Fd = ( 86 N )( 3.4 m ) = 290 J = 0.29 kJ
Solution: Apply W = Fd directly:
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.
F
Strategy: Determine the applied force and solve for
d.
Solution: Solve for d:
d=
d
W
W
640 J
=
=
= 3.6 m
f k µk mg ( 0.26 )( 70.0 kg ) ( 9.81 m/s 2 )
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)
mg
the force and distance are antiparallel.
1.8 m
W = Fd = mgd
Solution: 1. (a) Apply W = Fd:
W = ( 3.4 kg ) ( 9.81 m/s 2 ) (1.8 m ) = 60 J
2. (b) Now the force and distance are
perpendicular:
W= 0
3. (c) Now the force and distance are W = −mgd = − 60 J = − 0.060 kJ
antiparallel:
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:
W = Fd = (120 N )( 65 m ) = 7800 J = 7.8 kJ
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.
F
θ
Strategy: Keep in mind the angle between the force
and the direction of motion.
d
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Tutorial Solutions
W = Fd cos θ = (16 N )(10.0 m ) cos 25o = 150 J = 0.15 kJ
Solution: Use W = Fdcosθ:
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.
F
θ
d
Solution: 1. (a) Use W = Fdcosθ: W = Fd cos θ = ( 50.0 N )( 0.50 m ) cos 55o = 14 J
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.
Insight: Only the component of the force along the direction of the motion does any work. The
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.
airplane
F
θ
glider
d
Strategy: Solving for the angle between the force and
the direction of motion.
W
Fd
Solution: 1. Solve W = Fdcosθ
for the angle:
W = Fd cos θ or cos θ =
2. Calculate the angle:
 2.00 × 105 J 
W 
−1
o
θ = cos −1 
 = 57.4
 = cos 
Fd
2560
N
145
m
)(
) 


 (
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.
17 N
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.
12 N
skateboard
bicycle
r
F
r r
d = vt
r r
Solution: 1. (a) First find the d = vt = ( 4.1 m/s )( 25 s ) xˆ = 102.5 m xˆ
distance travelled:
2. Now multiply only the xcomponents:
W = Fx d x = (17 N )(102.5 m ) = 1700 J
3. (b) The force on the
bicycle is opposite that
on the skateboard by
Newton’s third law.
r
Fbike = ( −17 N ) xˆ + ( −12 N ) yˆ
4. (c) Multiply only the xcomponents:
W = Fbike,x d x = ( −17 N )(102.5 m ) = −1700 J
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.
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9.
Tutorial Solutions
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:
W = Fd cos θ ⇒
2. Calculate the angle:
θ = cos −1 
cos θ =
W
Fd


3500 J
W 
−1
o
 = 21
 = cos 
Fd
75
N
50
m
(
)(
)




Insight: Only the component of the force along the direction of the motion does any work. The
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:
K = 12 mv 2 =
1
2
(1770 kg )(120 m/s )
2
= 1.27 × 107 J = 12.7 MJ
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:
W = ∆K = mgh = 12 mv 2
v = 2 gh = 2 ( 9.81 m/s 2 ) (16 m ) = 18 m/s
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: K = 12 mv 2 =
2. (b) Solve for speed:
3. (c) Calculate W = ∆K :
v=
2K
=
m
1
2
( 0.40 kg )( 6.0 m/s )
2 ( 25 J )
0.40 kg
2
= 7.2 J
= 11 m/s
W = ∆K = K f − K i = 25 J − 7.2 J = 18 J
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.
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Tutorial Solutions
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:
W = − Fd = ∆K = 12 mvf2 − 12 mvi2 so that F = −
F =−
1
2
(1300 kg ) (152 − 182 m 2 /s 2 )
( 30.0 m )
1
2
m ( vf2 − vi2 )
d
= 2100 N = 2.1 kN
Insight: Kinetic friction always does negative work because the force is always opposite to the
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 :
W = ∆K = 12 mvf2 − 12 mvi2 = 0 − 12 ( 65 + 8.8 kg )(14 m/s )
2
= −7200 J = −7.2 kJ
2. (b) Find the distance:
∆x =
1
2
( v0 + v ) t = 12 (14 + 0 m/s )( 4.0 s ) =
28 m
W
W 7200 J
3. (c) The force can be found from W = F =
=
=
= 260 N ≅ 58 lb
d
∆x
28 m
Fd:
Insight: Kinetic friction always does negative work because the force is always opposite to the
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 12 kx 2 .
Solution W = 1 kx 2 =
2
: 1. (a)
Apply
= 44 J
equation
7-8:
1
2
( 3.5 ×10
4
N/m ) ( 0.050 m )
2. (b)
W = 12 kx 2 =
Apply
equation
= 44 J
7-8
again:
1
2
( 3.5 ×10
4
N/m ) ( − 0.050 m )
2
2
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.
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Tutorial Solutions
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 12 kx 2 . 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 Won block = ∆K block = 0 − 12 mvi2 = − 12 kx 2
and 7-8:
2
2. Now solve for k:
k =m
( 2.2 m/s )
vi2
= (1.8 kg )
= 91 N/m
2
2
x
( 0.31 m )
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. Wtotal = A = ( 0.25 m )( 0.6 + 0.4 + 0.8 N )
(a) The
= 0.45 J
total
area under
the graph:
2. (b) The W = ( 0.25 − 0.15 m )( 0.6 N ) + ( 0.25 m )( 0.4 N ) + ( 0.60 − 0.50 m )( 0.8 N )
area of the
W = 0.24 J
graph
shaded in
grey:
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.
2 (160 J )
2W
Solution: 1. (a) Solve for k = 2 =
2
x
( 0.14 m )
k:
= 1.6 × 104 N/m = 16 kN/m
2. (b) It would take more W = 1 kx 2 − 1 kx 2 =
2
1
2
2
than 160 J of
work because W is
proportional to x2:
1
2
(1.6 ×10
4
2
2
N/m ) ( 0.28 m ) − ( 0.14 m )  = 480 J


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Tutorial Solutions
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:
W=
1
2
( 0.21 m )( 4200 N ) + ( 0.30 − 0.21 m )( 4200 N ) =
2. The magnitude of the
force at x = 0.10 m can be
found from the given
formula:
F = kx = ( 2.0 × 104 N/m ) ( 0.10 m ) = 2000 N
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:
W=
1
2
0.82 kJ
( 0.21 m )( 4200 N ) + ( 0.40 − 0.21 m )( 4200 N )
− 12 ( 0.10 m )( 2000 N ) = 1.1 kJ
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:
P = Fv = ( mg ) v = (1.3 ×10-3 kg )( 9.81 m/s 2 ) ( 0.025 m/s )
P = 3.2 ×10 −4 W = 0.32 mW
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.
P
P
108 W
Solution: Solve P = Fv for v:
v= =
=
= 2.20 m/s
F mg ( 5.00 kg ) ( 9.81 m/s 2 )
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.
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Tutorial Solutions
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:
F=
P 50.0 W
=
= 33.3 N = 7.5 lb
v 1.50 m/s
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.
0.720 m


Solution: 1. (a) Apply P = Fv: P = Fv = mgv = ( 4.15 kg ) ( 9.81 m/s 2 ) 

 3.25 d × 86400 s/d 
P = 1.04 × 10−4 W = 0.104 mW
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. Now divide top and bottom by
and substitute
for the velocities:
1
2
m
t=
W ∆K 12 mvf2 − 12 mvi2
=
= 1 2
P
P
2 mvi T
t=
2
vf2 − vi2 ( 2v ) − v
=
= 3T
vi2 T
v2 T
2
3. (b) Again combine equations:
 1 mv 2 
W = ∆K = 12 mvf2 = Pt =  2
 ( 2T )
 T 
4. Now solve for vf :
1
2
mvf2 = mv 2 so vf2 = 2v 2 and thus vf = v 2
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.
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Tutorial Solutions
25. Picture the Problem: The three paths of the sliding box are depicted
at right.
Strategy: The work done by friction is W = − µ k mgd , 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:
W1 = −µk mg [ d1 + d2 + d3 + d4 + d5 ]
= −µk mg [ 4.0 + 4.0 +1.0 +1.0 +1.0 m]
W1 = −0.21( 3.2 kg) ( 9.81 m/s2 ) [11.0 m] = −73 J
2. Calculate W for path 2:
W2 = −µk mg [ d6 + d7 + d8 ]
= − 0.21( 3.2 kg ) ( 9.81 m/s 2 ) ( 2.0 m ) + ( 2.0 m ) + (1.0 m )  = −33 J
3. Calculate the work for
path 3:
W3 = −µk mg [ d9 + d10 + d11 ]
= − 0.21( 3.2 kg ) ( 9.81 m/s2 ) (1.0 m ) + ( 3.0 m ) + ( 3.0 m )  = − 46 J
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:
W1 = mg ( 0 + y1 )
2. Calculate the work for path 2:
W2 = mg ( y1 + 0 )
= ( 5.2 kg ) ( 9.81 m/s 2 ) (1.0 m ) = 51 J
= ( 2.6 kg ) ( 9.81 m/s 2 ) (1.0 m ) = 51 J
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 Wsp = 12 k ( xi2 − xf2 ) for the work
done by the spring. That way the work will
always be negative if you start out at
xi = 0 because the spring force will always be in
the opposite direction from the stretch or
compression. The work done by kinetic friction is
Wfr = − µk mgd , where d is the distance the box is
pushed irregardless of direction, because the
friction force always acts in a direction opposite
the motion.
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Tutorial Solutions
Solution: 1. (a) Sum the
Wsp = 12 k ( x12 − x42 ) + ( x42 − x32 ) 
work done by the
2
2
= 12 ( 480 N/m )  02 − ( − 0.020 m )  + ( − 0.020 m ) − ( 0.020 m 2 ) 
spring for each segment of

 

path 2:
Wsp = ( − 0.096 J ) + ( 0 J ) = − 0.096 J
{
}
2. Sum the work done by
friction for
each segment of path 2:
Wfr = − µk mg ( d1 + d 2 )
3. (b) Sum the work done
by the
spring for the direct path
from A to B:
Wsp = 12 k ( xA2 − xB2 )
= − ( 0.16 )( 2.7 kg ) ( 9.81 m/s 2 ) ( 0.020 + 0.040 m ) = − 0.25 J
=
1
2
( 480 N/m ) 02 − ( 0.020 m )
2
 = − 0.096 J

4. Sum the work done by Wfr = − µk mgd = − ( 0.16 )( 2.7 kg ) ( 9.81 m/s 2 ) ( 0.020 m ) = − 0.085 J
friction for
the direct path from A to B:
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 : U = mgy = ( 83 kg ) ( 9.81 m/s 2 ) ( 8848 m ) = 7.2 × 10 6 J = 7.2 MJ
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.
( mg x1 ) x12 = 21 mgx1
2 ( 0.962 J )
=
= 0.065 m
( 3.0 kg ) ( 9.81 m/s2 )
2. (b) Find an expression for
the stretch distance as a
function of U and m:
U1 = 12 kx12 =
3. Doubling the mass doubles
the force and therefore doubles
the stretch distance:
x1 = mg k
x2 = 2mg k = 2 x1
4. Calculate the ratio U 2 U1 :
U 2 12 kx22
( 0.13 m )
=1 2 =
=4
U1 2 kx1 ( 0.065 m )2
x1 =
2U1
mg
1
2
2
U 2 = 4U1 = 4 ( 0.962 J ) = 3.85 J
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.
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Tutorial Solutions
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:
k=
2. (b) Solve for x:
x=
2U 2 ( 0.0025 J )
=
= 200 N/m = 0.20 kN/m
x 2 ( 0.0050 m )2
2 ( 0.0084 J )
2U
=
k
200 N/m
= 0.92 cm
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:
Wreq = −Wspring = − ( −∆U ) = U 5 − U 4
= 12 kx52 − 12 kx42 = 12 k ( x52 − x42 )
k=
3. Use k to
find the new
Wreq :
2Wreq
2
5
2
4
x −x
=
2 ( 30.5 J )
2
( 0.0500 m ) − ( 0.0400 m )
Wreq = 12 k ( x2 2 − x12 ) =
1
2
( 6.78 ×10
4
2
= 6.78 × 104 N/m
2
2
N/m ) ( 0.0600 m ) − ( 0.0500 m )  = 37.3 J


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:
∆y = L cos θ − L = L ( cos θ − 1)
2. Use ∆y to find ∆U : ∆U = mg ∆y = mgL ( cos θ − 1)
= ( 0.33 kg ) ( 9.81 m/s 2 ) (1.2 m )( cos 35° − 1)
∆U = −0.70 J
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.
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Tutorial Solutions
Solutions to Tutorial 3
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.
r
r r
r
r
p total = p d + p c = md v d + mc v c =
Solution: 1. Use the component
method of vector addition to find
= ( 20.0 kg )( 2.50 m/s yˆ ) + ( 5.00 kg )( 3.00 m/s xˆ )
the owner’s momentum:
r
p total = (15.0 kg ⋅ m/s ) xˆ + ( 50.0 kg ⋅ m/s ) yˆ
r
r
r
2. Divide the owner’s momentum p 0 = m0 v 0 = p total
r
by his mass to
p
(15.0 kg ⋅ m/s ) xˆ + ( 50.0 kg ⋅ m/s ) yˆ
v 0 = total =
get the components of the owner’s
m0
70.0 kg
velocity:
= ( 0.214 m/s ) xˆ + ( 0.714 m/s ) yˆ
3. Use the known components to
find the
direction and magnitude of the
owner’s velocity:
 0.714 
 = 73.3°
 0.214 
θ = tan −1 
v0 =
( 0.2143 m/s )
2
2
+ ( 0.7143 m/s ) = 0.746 m/s
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
for v2 . Then find the total kinetic energy of the two-cart system. Let cart 1 travel in the positive
direction.
r
r
r
r
Solution: 1. (a) Set ∑ p = 0 and ∑ p = m1 v1 + m2 v 2 = 0
solve for v2 :
v2 = −
m1v1 ( 0.35 kg )(1.2 m/s )
=
= 0.69 m/s
m2
0.61 kg
2. (b) No, kinetic energy is always greater than or equal to zero.
3. (c) Sum the kinetic
energies of the two carts:
∑K =
1
2
m1v12 + 12 m2 v2 2
=
1
2
( 0.35 kg )(1.2 m/s )
2
+ 12 ( 0.61 kg )( 0.69 m/s )
2
= 0.40 J
Insight: If cart 1 is travelling in the positive x̂ direction, then its momentum is ( 0.42 kg ⋅ m/s ) xˆ
and the momentum of cart 2 is ( − 0.42 kg ⋅ m/s ) xˆ .
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:
v=
p 0.780 kg ⋅ m/s
=
= 5.20 m/s
m
0.150 kg
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Tutorial Solutions
v 2 = v02 − 2 g ( y − y0 )
2. (b) Solve for y0 . Let y = 0
and v0 = 0 :
2
y0 =
( 5.20 m/s )
v2
=
= 1.38 m
2 g 2 ( 9.81 m/s 2 )
Insight: Another way to find the initial height is to use conservation of energy, setting
mgy0 = 12 mv 2 and solving for y0 .
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.
r r r
r r
r
Solution: 1. (a) Find ∆p : ∆p = p f − p i = m ( v f − v i )
= ( 0.220 kg ) ( 2.0 m/s ) yˆ − ( −2.5 m/s ) yˆ  = ( 0.99 kg ⋅ m/s ) yˆ
r
∆p = 0.99 kg ⋅ m/s
2. (b) Subtract the
magnitudes
of the momenta:
pf − pi = m ( vf − vi )
= ( 0.220 kg )( 2.0 m/s − 2.5 m/s )
pf − pi = − 0.11 kg ⋅ m/s
3. (c) The quantity in part (a) is more directly related to the net force acting on the ball during its
r
r
collision with the floor, first of all because ∑ F = ∆p ∆t and as we can see from above that
r
∆p ≠ pf − pi . Secondly, we expect the floor to exert an upward force on the ball but we
calculated a downward (negative) value in part (b).
r
Insight: If the ball were to rebound at 2.5 m/s upward we would find ∆p = 2mv = 1.1 kg ⋅ m/s and
pf − pi = 0 . 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 ptotal, x and
divide by v1 :
2. Find ptotal, y and divide by
v2 :
y
r
p total
r
p2
x
r
p1
p1 = ptotal, x = ptotal cos θ = (17.6 kg ⋅ m/s )( cos 66.5° ) = 7.02 kg ⋅ m/s
m1 =
p1 7.02 kg ⋅ m/s
=
= 2.51 kg
v1
2.80 m/s
m2 =
p2 ptotal sin θ (17.6 kg ⋅ m/s )( sin 66.5° )
=
=
= 5.21 kg
v2
v2
3.10 m/s
Insight: Note that object 2 has the larger momentum because the total momentum points mostly in
the ŷ 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
r
r
system to zero and solve for m2 . Let the velocity v1 point in the negative direction, v 2 in the
positive direction.
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Solution: Set ptotal = 0 and solve for
m2 :
Tutorial Solutions
p1x + p2 x = 0 = m1v1x + m2 v2 x
− m1v1x − ( 45 kg )( − 0.62 m/s )
m2 =
=
= 31 kg
0.89 m/s
v2 x
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 m1 represent the piece with the smaller
kinetic energy.
r
Solution: 1. Set ∑ p = 0 and solve p1 + p2 = 0 = m1v1 + m2 v2
for m1 m2 :
2. Set K 2 K1 = 2 :
1
m v2
K2
= 2 = 21 2 22
K1
mv
2 1 1
2
3. Combine the expressions from
steps 1 and 2:
2
 m  v 
⇒  1  = 2 
 m2   v1 
m1
v
=− 2
m2
v1
2
2
v 
m
⇒  2 =2 1
m2
 v1 
2
 v2   m1 
m1
  =
 =2
m2
 v1   m2 
⇒
m1
=2
m2
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:
pa + ps = 0 = ma va + ms vs
mv
va = − s s
ma
2. Find the distance to the space
shuttle:
d = va t = −
ms vs
(1200 kg )( − 0.14 m/s )
t=−
( 7.5 s ) = 14 m
ma
( 92 kg )
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 vL, s = speed of lumberjack relative
to the shore. Let vL, log = speed of lumberjack relative to the log, and vlog, s = speed of the log
relative to the shore.
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r
r
r
Solution: 1. (a) Write out the
v L, s = v L, log + v log, s
equation for relative motion.
v = vL, log − vlog, s
Let the log travel in the negative L, s
vlog, s = vL, log − vL, s
direction:
2. Write out the conservation of ∑ pr = 0 = mL vL, s − mlog vlog, s
momentum
with respect to the shore:
3. Substitute the expression
from step 1
into step 2 and solve for vL, s
mL vL, s = mlog vlog, s = mlog ( vL, log − vL, s )
vL, s ( mL + mlog ) = mlog vL, log
vL, s =
mlog vL, log
( mL + mlog )
=
( 380 kg )( 2.7 m/s )
=
(85 + 380 kg )
2.2 m/s
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:
vL, s =
mlog vL, log
(m
L
+ mlog )
=
( 450 kg )( 2.7 m/s )
=
( 85 + 450 kg )
2.3 m/s
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.
y
r
p2
Strategy: Since the plate falls straight down its momentum in
225°
the xy plane is zero. That means the momenta of all three
r
pieces must sum to zero. Choose the motion of the two pieces
p1
at right angles to one another to be in the x̂ and ŷ directions.
r
r
p3
Set the total momentum equal to zero and solve for v 3 .
r
r
r
Solution: 1. Set ∑ p = 0 and
∑ p = mv xˆ + mv yˆ + mv3 = 0
r
solve for v 3 :
( −mv ) xˆ + ( −mv ) yˆ
r
v3 =
= ( −v ) xˆ + ( −v ) yˆ
m
2. Find the speed v3 :
v3 =
( −v )
2
2
+ ( −v ) =
x
2v
 v3, y 
−1  − v 
 = tan   = 45° + 180° = 225°
v
 −v 
 3, x 
r
Insight: As long as the first two pieces have equal masses the direction of v 3 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 v 8 but θ would remain 225°.
3. Find the direction of v3 :
θ = tan −1 
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Solutions to Tutorial 4
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:
θ=
s 1.75 m
=
= 5.3 rad
r 0.33 m
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:
 1 rev   1 yr   1 day   1 h 
−6



 = 1.90 × 10 rev/min
 yr   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:
 1rev   2π rad   1 day   1 h 
−5

 = 7.27 ×10 rad/s


day
rev
24
h
3600s





ω =
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- ω = ∆θ = 2π rad = 2π rad = 190 rad/s
∆t
T
0.033s
3:
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.
2π
2π
=
= 31.4 rad/s
T
0.200s
Solution: 1. (a) Solve for
ω:
ω=
2. (b) Apply v = rω
directly:
vT = rω = ( 12 × 3.5 in ) ( 31.4 rad/s ) = 55 in/s × 1 m 39.4 in = 1.4 m/s
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
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Tutorial Solutions
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, θ = (125 rad/s ) t + ( 42.5 rad/s 2 ) t 2 ,with θ = θ 0 + ω0 t + 12 α t 2 , we
can identify ω0 = 125 rad/s and 12 α = 42.5 rad/s 2 .
2
1
∆θ θ − θ 0 ω0t + 2 α t  − θ 0
Solution: 1. (a) ωav =
=
=
t
t
∆t
Use equations
2

(125 rad/s )( 0.010 s ) + ( 42.5 rad/s 2 ) ( 0.010 s )  − 0
to find ωav :

=
0.010 s
ωav = 125 rad/s = 1.3 × 102 rad/s
2. (b) Use
equations
to find ωav :
3. (c) Use
equations
to find ωav :
θ = ω0 t + 12 α t 2  = (125 rad/s )(1.010 s ) + ( 42.5 rad/s 2 ) (1.010 s )  = 169.60 rad


2
θ 0 = ω0 t0 + 12 α t02  = (125 rad/s )(1.000 s ) + ( 42.5 rad/s 2 ) (1.000 s )  = 167.50 rad
2


∆θ θ − θ0 169.60 − 167.50 rad
2
=
=
= 210 rad/s = 2.1× 10 rad/s
ωav =
∆t
t − t0
1.010 − 1.000 s
2
θ = ω0t + 12 α t 2  = (125 rad/s )( 2.010 s ) + ( 42.5 rad/s 2 ) ( 2.010 s )  = 422.95 rad


2
θ 0 = ω0 t0 + 12 α t02  = (125 rad/s )( 2.000 s ) + ( 42.5 rad/s 2 ) ( 2.000 s )  = 420.00 rad


∆θ θ − θ 0 422.95 − 420.0 rad
2
ωav =
=
=
= 295 rad/s = 3.0 × 10 rad/s
∆t
t − t0
2.010 − 2.000 s
4. (d) The angular acceleration is positive because the angular speed is positive and increasing with
time.
5. (e) Find αav: α av =
6. Find αav:
α av =
ω − ω0
∆t
ω − ω0
∆t
=
210 − 125 rad/s
= 85 rad/s 2
1.00 − 0.00 s
=
295 − 210 rad/s
= 85 rad/s 2
2.00 − 1.00 s
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 12 α = 42.5 rad/s 2 , it must be true that
α = 85.0 rad/s 2 .
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.
2
Solution: Solve for α :
α=
2
ω 2 − ω02 ( 26 rad/s ) − (12 rad/s )
=
= 17 rad/s 2
2 ∆θ
2 ( 2.5 rev × 2π rad rev )
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.
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Tutorial Solutions
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 :
∆θ =
1
2
(ω0 + ω ) t = 12 (12 + 26 rad/s )( 2.5 s ) =
48 rad = 7.5 rev
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.
2
Solution: 1. (a) Solve for α :
α=
2. (b) Solve for t:
t=
02 − ( 6.35 rad/s )
ω 2 − ω0 2
=
= − 0.226 rad/s 2
2 (θ − θ0 ) 2 (14.2 rev × 2π rad rev )
ω − ω0 0 − 6.35 rad/s
=
= 28.1 s
− 0.226 rad/s 2
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:
∆θ =
1
2
(ω + ω0 ) t = 12 ( 0 + 0.90 rev/s )( 2.2 min
2. (b) Find ∆θ:
∆θ =
1
2
(ω + ω0 ) t = 12 ( 0.45 + 0.90 rev/s )(1.1 min × 60 s
× 60 s min ) = 59 rev
min ) = 45 rev
Insight: An alternative way to solve the problem is to find α = − 0.0068 rev/s 2 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.
2
Solution: 1. (a) Solve for
∆θ :
∆θ = θ − θ 0 =
ω 2 − ω0 2 ( 6.3 rad/s ) − 02
=
= 9.0 rad × 1 rev 2π rad
2α
2 ( 2.2 rad/s 2 )
= 1.4 rev
2. (b) Solve for t:
t=
ω − ω0 6.3 − 0 rad/s
=
= 2.9 s
α
2.2 rad/s 2
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.
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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.
Solution: 1. (a) Solve for α =
α:
2. (b) Solve for ∆θ :
∆θ =
ω − ω0
t
1
2
=
0 − ( 3850 rev/min × 1 min 60 s )
10.2 s
(ω + ω0 ) t = 12 ( 0 + 3850 rev/min ×1 min
= 6.29 rev/s 2
60 s )(10.2 s ) = 327 rev
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:
∆θ =
1
2
(ω + ω0 ) t = 12 ( 310 + 0 rev/min )( 3.0 s
× 1 min 60 s ) = 7.8 rev
Insight: An alternative way to solve the problem is to find α = 1.72 rev/s 2 and use α to find
∆θ = 7.8 rev 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.
ω − ω0
Solution: 1. (a) Solve for
α:
α=
2. (b) Find ∆θ :
∆θ =
t
1
2
=
( 350, 000 rev/min ×1 min
60 s ) − 0
2.1s
= 2.8 ×103 rev/s 2
(ω + ω0 ) t = 12 ( 350, 000 + 0 rev/min )( 2.1 s ×1 min
60 s )
= 6.1×103 rev
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.
 1 rev   2π rad   1 h 
Solution: Apply v = rω: vt = rω = ( 8.2 cm ) 


 = 0.0012 cm/s = 12 µ m/s
 12 h   rev   3600 s 
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.
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Tutorial Solutions
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 :
ω=
vt
3.4 m/s
=
= 23 rad/s
r ( 0.29 m 2 )
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.
2π rad
= 0.14 rad/s
45 s
Solution: 1. (a) Find ω1 and ω2 :
ω1 = ω2 =
2. Apply v = rω directly:
vt1 = r1ω1 = ( 2.75 m )( 0.14 rad/s ) = 0.38 m/s
3. (b) Apply v = rω directly:
vt2 = r2ω2 = (1.75 m )(1.4 rad/s ) = 0.24 m/s
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 vt = rω =
directly:
1
2
( 0.120 m )( 5.25 rad/s ) =
0.315 m/s
2
= 1.65 m/s 2
1
2
( 0.315 m/s ) =
2. (b)Apply acp = rω2
directly:
acp = rω 2 =
3. (c) Use a ratio to find the
new linear speed:
v2 r2ω 12 r 1
=
=
=
⇒ v2 = 12 v1 =
v1 r1ω
r
2
4. (d) Use a ratio to find the
new acp :
a2 r2ω 2 12 r 1
=
=
=
⇒ a2 = 12 a1 =
a1 r1ω 2
r
2
1
2
( 0.120 m )( 5.25 rad/s )
1
2
0.158 m/s
(1.65 m/s ) = 0.827 m/s
2
2
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.
2
 2π rad 
2
Solution: 1. (a) Apply acp = rω2 directly: acp = rω 2 = ( 9.5 m ) 
 = 0.37 m/s
 32 s 
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
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Tutorial Solutions
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.
Top of Ferris Wheel
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.
r
at
r
v
φ
r
a cp
r
a
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.
r
2
Solution: 1. Use acp = rω2 to
find acp :
 2π rad 
2
acp = rω 2 = ( 9.5 m ) 
 = 0.37 m/s (downward)
 32 s 
2. Use equation at = rα to find
at :
at = rα = ( 9.5 m ) − 0.22 rad/s 2 = −2.1 m/s 2 (to the left)
(
3. (c) Combine the components a = acp2 + at2 =
to find a :
)
2 2
2 2
( 0.37 m/s ) + ( −2.1 m/s )
= 2.1 m/s 2
2
 acp 
−1  0.37 m/s 
= −10° or 170° below the
4. (d) Find the angle φ and relate φ = tan −1 
 = tan 
2 
at 
−2.1 m/s 


it to the
direction of motion (which is to direction of motion.
the left)
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
F = macp and
F = macp = mrω 2
solve for ω :
ω=
F
=
mr
11 N
( 0.52 kg )( 4.5 m )
= 2.2 rad/s
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.
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Tutorial Solutions
Solution: 1. (a) Apply v = rω directly: vt = rω = ( 0.00320 m ) ( 2.15 × 104 rad/s ) = 68.8 m/s
vt = rω = 2π r T
2. (b) Substitute ω = 2π T and solve
for T:
T=
2π r 2π ( 0.00320 m )
=
= 7.31× 10−5 s = 73.1 µ s
vt
275 m/s
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 ( ω0 = 0 since the wheel
starts from rest), and substitute the expression into the result of the first step. Then solve the
equation for t.
at = rα = rω 2 = acp
Solution: 1. Set at = acp :
α = ω2
2
α = (α t ) = α 2 t 2
2. (b) Substitute ω = 0 + α t and solve
for t:
1
α
= t2 ⇒ t =
1
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 α 2 .
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:
τ = r ( F sin θ )
15 N ⋅ m
τ
F=
=
= 60 N
r sin θ ( 0.25 m ) sin 90°
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:
τ = Fweed rweed
τ
1.23 N ⋅ m
Fweed =
rweed
=
0.040 m
= 31 N
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.
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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
r⊥ = r cos θ = ( 0.605 m ) cos 22.5° = 0.559 m.
Solution: 1. (a) Multiply the moment
arm by the weight:
τ = r⊥ mg = ( 0.605 m )(1.51 kg ) ( 9.81 m/s 2 ) = 8.96 N ⋅ m
2. (b) Multiply the moment arm by the
weight:
τ = r⊥ mg = ( 0.559 m )(1.51 kg ) 9.81 m/s 2 = 8.28 N ⋅ m
(
)
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:
τ = r⊥ mg = ( 0.70 m )( 3.6 kg ) ( 9.81 m/s 2 ) = 25 N ⋅ m
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 τ biceps = r⊥ F = ( 0.0275 m )(12.6 N ) = 0.347 N ⋅ m
individual
τ forearm = r⊥ mg = ( 0.170 m )(1.20 kg ) ( 9.81 m/s 2 ) = −2.00 N ⋅ m
torques and sum them:
τ ball = r⊥Wball = ( 0.340 m )(1.42 N ) = −0.483 N ⋅ m
∑τ = τ biceps + τ forearm + τ ball
= +0.347 − 2.00 − 0.483 N ⋅ m = −2.14 N ⋅ m
2. (b) Negative net torque means the clockwise direction; the forearm and hand will rotate
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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.
r⊥ adult
r⊥child
mchild g
Fadult
Solution: 1. (a) Find τ child = r⊥ child mchild g
the torque the child
= − (1.5 m )(16 kg ) ( 9.81 m/s 2 )
exerts on the teeterτ child = −235 N ⋅ m
totter.
2. Find the torque
exerted by the
parent and sum the
torques to find
the direction of
travel:
τ adult = r⊥ adult Fadult = ( 3.0 m )( 95 N ) = 285 N ⋅ m . Here τ adult + τ child > 0 so the
teeter-totter will rotate anticlockwise and the child will move up.
3. (b) Repeat step 2 τ adult = r⊥ adult Fadult = ( 2.5 m )( 95 N ) = 238 N ⋅ m . Here τ adult + τ child > 0 so the
with the new
teeter-totter will rotate anticlockwise and the child will move up.
r⊥ for the adult:
4. (c) Repeat step 2
with the new
r⊥ for the adult:
τ adult = r⊥ adult Fadult = ( 2.0 m )( 95 N ) = 190 N ⋅ m . Here τ adult + τ child < 0 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:
I=
τ
τ
τ ∆t ( − 0.120 N ⋅ m )(19.5 s )
=
=
=
= 0.883 kg ⋅ m 2
α ∆ω ∆t ∆ω
( 0 − 2.65 rad/s )
Insight: Friction converts the fan’s initial kinetic energy of
1
2
I ω02 = 3.10 J 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: I = 121 M L2 . Then find the required torque to produce the acceleration.
2
Solution: 1. Find I = 121 M L2 :
I = 121 M L2 = 121 ( 8.22 kg )( 3.15 m ) = 6.80 kg ⋅ m 2
2. Apply τ = Iα directly:
τ = I α = ( 6.80 kg ⋅ m 2 )( 0.302 rad/s 2 ) = 2.05 N ⋅ m
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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.
( 6.4 kg )( 0.71 m )
2
Solution: 1. (a) Find I = 12 MR 2 :
I = 12 MR 2 =
2. Solve for α :
α=
3. Apply τ = Iα directly:
τ = I α = (1.6 kg ⋅ m 2 )( − 0.158 rad/s 2 ) = − 0.25 N ⋅ m
1
2
= 1.6 kg ⋅ m 2
2
02 − (1.22 rad/s )
ω 2 − ω02
=
= − 0.158 rad/s 2
2∆θ
2 ( 0.75 rev × 2π rad rev )
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 I x :
I x = m1 r12 + m2 r22 + m3 r32
2. Find τ x :
τ x = I xα = ( 9.0 kg ⋅ m 2 )(1.20 rad/s 2 ) = 11 N ⋅ m
3. (b) Calculate I y :
I y = 0 + 0 + ( 2.5 kg )( 2.0 m ) = 10 kg ⋅ m 2
4. Find τ y :
τ y = I yα = (10 kg ⋅ m 2 )(1.20 rad/s 2 ) = 12 N ⋅ m
5. (c) Calculate I z :
I z = ( 9.0 kg )(1.0 m ) + 0 + ( 2.5 kg )( 2.0 m ) = 19 kg ⋅ m 2
6. Find τ z :
τ z = I zα = (19 kg ⋅ m 2 )(1.20 rad/s 2 ) = 23 N ⋅ m
2
= ( 9.0 kg )(1.0 m ) + 0 + 0
I x = 9.0 kg ⋅ m 2
2
2
2
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
( 12 MR 2 ). 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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Tutorial Solutions
( 0.84 kg )( 0.055 m )
2
Solution: 1. (a) Find I:
I = 12 MR 2 =
2. Apply τ = rF directly to find τ :
τ = r F = ( 0.055 m )( 2.1 N ) = 0.12 N ⋅ m
3. Solve τ = Ia for α :
α=
4. (b) Solve for s:
s = rθ = r ( 12 α t 2 ) = ( 0.055 m ) 12 ( 92 rad/s 2 ) ( 0.25 s )
τ
I
=
1
2
= 0.00127 kg ⋅ m 2
0.12 N ⋅ m
= 92 rad/s 2
0.0013 kg ⋅ m 2
2
= 0.16 m
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
( 12 MR 2 ). 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.
( 0.84 kg )( 0.055 m )
2
Solution: 1. (a) Find I:
I = 12 MR 2 =
2. Apply τ = rF directly to find τ :
τ = r F − C = ( 0.055 m )( 2.1 N ) − 0.047 N ⋅ m = 0.069 N ⋅ m
3. Solve τ = Iα for α :
α=
4. (b) Solve for s:
s = rθ = r ( 12 α t 2 ) = ( 0.055 m ) 12 ( 54 rad/s 2 ) ( 0.25 s )
τ
I
=
1
2
= 0.00127 kg ⋅ m 2
0.069 N ⋅ m
= 54 rad/s 2
0.00127 kg ⋅ m 2
2
= 0.093 m = 9.3 cm
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,
τ = 0.12 − 0.047 N ⋅ m = 0.07 N ⋅ m, 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:
2. (b) Multiply the period by one-quarter to
find the first zero crossing:
2π
Since cos 
T
t=


 2π 
t  , therefore T = 0.58 s.
 0.58 s 
t  = cos 
1
( 0.58 s ) = 0.15 s .
4
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.
 2π 
 2π 
Solution: 1. (a) Observe that in the argument
Since cos 
t  = cos 
t  , therefore T =
of
 T 
 0.68 s 
cosine the period is in the denominator:
0.68 s
2. Invert the period to obtain the frequency:
f =
1
= 1.5 Hz .
0.68 s
1
1
3. (b) The time the mass is at −7.8 cm is half a t = T = (0.68 s) = 0.34 s .
2
2
period:
Insight: This problem could also be solved by setting the motion equation equal to x = −7.8 cm
and solving for the time: t = ( 0.68 s 2π ) cos −1 ( −7.8 cm 7.8 cm ) = 0.34 s.
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 x = A cos (ωt ) . 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: A = 3.50 nm
2. Calculate the angular frequency from the ω = 2π f = 4.00π × 1014 s −1
frequency:
3. Substitute the amplitude and angular
frequency
into the cosine equation:
x=
( 3.50 nm ) cos ( 4.00π ×1014
s −1 ) t 
4. (b) It will be a sine function, x = A sin (ωt ) , 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:
 2π
x = ( 0.48 cm ) cos 
 T
2. (a) Substitute t = T/8 into the
sine equation
and evaluate:
 2π T 
π 
x = ( 0.48 cm ) cos 
 = ( 0.48 cm ) cos   = 0.34 cm
T
8


4
3. (b) Substitute t = T/4 into the
sine equation
and evaluate:
 2π T 
π 
x = ( 0.48 cm ) cos 
 = ( 0.48 cm ) cos   = 0.48 cm
 T 4
2
4. (c) Substitute t = T/2 into the
sine
equation and evaluate:
 2π T 
x = ( 0.48 cm ) cos 
 = ( 0.48 cm ) cos (π ) = 0
 T 2
5. (d) Substitute t = 3T/4 into the
sine
equation and evaluate:

t

 2π 3T 
x = ( 0.48 cm ) cos 

 T 4 
 3π 
= ( 0.48 cm ) cos  
 2 
x = − 0.48 cm
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:
 2π t 
x = ( 0.0440 m ) cos 
.
 3.15 s 
2. Insert t = 6.37 s into the general
equation and
evaluate the position:
 2π ( 6.37 s ) 
x = ( 0.0440 m ) cos 
 = 0.0436 m
 3.15 s 
3. (b) Divide the time by one period:
6.37 s
= 2.02
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:
 2π t 
vmax sin 
=
 T 
 2π t 
sin 
=
 T 
1
vmax
2
1
2
2. Take the arcsine of each side and solve for 2π t = sin −1  1 
T
2
t:
T
1
t=
sin −1  
2π
2
5π
1 π
3. Evaluate using sin −1   = and
:
2
6
6
 
t=
T
2π
T  5π
π 
  or
π  6
6
2
 
5T
 T
or
=
12
12

5T T 
 4T  2T
4. Subtract the first time from the second time 2 
−  = 2
.
=
 12 12 
 12  3
and multiply by two:
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, vmax = Aω , amax = Aω 2 . 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:
A=
2. (b) Divide the acceleration by the velocity to
determine the angular speed:
ω=
( Aω )
Aω
2
2
=
vmax 2
amax
Aω 2 amax
=
Aω
vmax
2π vmax
2π
2π
3. Divide 2 π by the angular speed to calculate the T =
=
=
ω  amax 
amax
period:


v
 max 
Insight: When two or more quantities are functions of the same variables, it is often possible to
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Tutorial Solutions
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, vmax = Aω , amax = Aω 2 . Rearrange these equations to solve for the amplitude
and angular speed. Then use the angular speed to determine the period.
2
2
( Aω ) vmax 2 ( 4.3 m/s )
Solution: 1. (a) Divide the square of the
A=
=
=
= 28 m
velocity by the acceleration to find the
amax ( 0.65 m/s 2 )
Aω 2
amplitude:
2. (b) Divide the acceleration by the
Aω
velocity to determine the angular speed: ω = Aω
2
3. Divide 2 π by the angular speed to
calculate T:
T=
2π
ω
=
=
amax
vmax
2π
 amax 


 vmax 
=
2π vmax 2π ( 4.3 m/s )
=
= 42 s
amax
( 0.65 m/s2 )
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:
 2π
vmax = Aω = A 
 T
2. Insert the amplitude and
period into the equation for
maximum speed:
 2π
vmax = A 
 T

 2π 
 = ( 0.204 m ) 
 = 0.458 m s

 2.80 s 
 2π ( 0.204 m )
= 0.458 m s
=
2.80 s

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,
amax = Aω 2 , where angular speed is 2π times the frequency.
Solution: 1. Multiply the amplitude by the square
of 2π times the frequency to get amax :
amax = Aω 2 = A ( 2π f )
2
2
= ( 0.25 mm )( 2π × 110 Hz ) = 119 m/s 2
g

2. Factor out g = 9.81 m/s2 to obtain the acceleration
amax = 119 m/s 2 
2
as a multiple of g:
 9.81 m/s

 = 12 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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Tutorial Solutions
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
C 2π r 2π ( 0.25 m )
circumference of the turntable by
T= =
=
= 2.3 s
v
v
0.67 m/s
the peg’s tangential velocity to get
the period of rotation:
2. (b) Set the amplitude equal to
the radius of the turntable:
A = r = 0.25 m
3. (c) Insert the period and
amplitude into the maximum
velocity acceleration:
vmax = Aω =
2π A Av
=
= v = 0.67 m s
T
r
2
( 0.67 m/s )
v
 2π 
v
4. (d) Insert the period and
amax = Aω 2 = A 
=
= 1.8 m/s 2
 = A  =
T
r
r
0.25 m




amplitude into the maximum
acceleration equation:
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.
2
2
2
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:
2. (a) Use the amplitude and angular
speed to solve
for maximum acceleration:
rev  2π rad  1 min 

-1
1700


 = 178 s
min

 rev  60 s 
2
amax = Aω 2 = 3.5 cm (178 s −1 ) = 1.1 km/s 2
3. (c) Use the amplitude and angular speed
vmax = Aω = 3.5 cm (178 s −1 ) = 6.2 m/s
to solve
for maximum speed:
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
A = 10.0 cm and the angular speed is ω = 2.00 rad/s . We will use these values to calculate the
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maximum velocity and acceleration.
1
1
2
Solution: 1. (a) Write the
K = mv 2 = m ( Aω )
kinetic energy in terms of mass, max 2 max
2
amplitude, and angular speed:
2
1
1
2. Insert the values of mass,
2
2
K = m Aω = ( 0.64 kg )( 0.100 m ) ( 2.00 s −1 ) = 0.13 J
amplitude and angular speed to max 2 ( )
2
calculate K max :
3. (b) Write the force equation
Fmax = mamax = m ( Aω 2 )
in terms of mass, amplitude,
and angular speed:
2
4. Insert the values of mass,
F = m ( Aω 2 ) = ( 0.64 kg )( 0.100 m ) ( 2.00 s −1 ) = 0.26 N
amplitude and angular speed to max
calculate the maximum force:
Insight: The phase shift of +π in the displacement equation has no effect on either K max or Fmax ,
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:
2
T = 2π
m
 2π 
⇒ k =
 m
k
 T 
2
2
2π 
 2π 
2. Insert the numeric values for T k = 
 m=
 ( 0.42 kg ) = 29 N m
T


 0.75 s 
and m:
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
F mg
F = ky ⇒ k = =
force equation
y
y
to solve for the spring constant:
2. Insert numeric values to
obtain k:
k=
0.50 kg ( 9.81 m/s 2 )
15 × 10 −2 m
= 32.7 N/m
2
3. Solve the period equation for T = 2π
the mass:
m
 T 
⇒ m=
 k
k
 2π 
2
 0.75 s 
4. Insert numeric values to
m=
 ( 32.7 N/m ) = 0.47 kg
 2π 
obtain the mass:
Insight: Since the period is proportional to the square root of the mass, increasing the mass will
increase the period.
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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 = ky ⇒ k =
2. Enter numeric values for the spring
constant:
k=
3. (a) Solve the period equation for the total
load (M+m):
T = 2π
F mg
=
y
y
125 kg ( 9.81 m/s 2 )
0.0800 m
= 1.53 × 104 N/m
2
M +m
 T 
⇒ M +m=
 k
k
 2π 
2
4. Enter numeric values for the total mass:
5. (b) Subtract the mass of the two people to
obtain M:
 1.65 s 
4
M +m=
 (1.53 × 10 N/m ) = 1060 kg
 2π 
M = ( M + m ) − m = 1057 kg − 125 kg = 932 kg
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
m
0.85 kg
T = 2π
= 2π
= 0.47 s
and spring
k
150 N/m
constant into the period
equation:
Tvmax 0.4730 s ( 0.35 m/s )
=
= 2.6 cm
2π
2π
2. (b) Solve the maximum
velocity equation
for the amplitude:
A=
3. (c) Insert the amplitude and
period into the maximum
acceleration equation:
 2π 
 2π 
2
amax = Aω 2 = A 
 = 0.02635 m 
 = 4.6 m s
 T 
 0.4730 s 
vmax
ω
=
2
2
Insight: Another way to solve this problem is to calculate the angular speed ω = k m instead of
the period. Then A = vmax ω and amax = vmax 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:
T = 2π
m
k
3. Write the equation for the period of the
motorcycle with the rider:
T2 = 2π
m+M
k
4. Calculate the percent difference between the
two periods:
T2 − T
=
T
5. Simplify by factoring out 2π m k from the
numerator
and denominator:
T2 − T
=
T
=
2π
m+M
m
− 2π
k
k
m
2π
k
m+M
−1
m
511 + 122 kg
− 1 = 0.104 = 10.4%
511 kg
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
T ∝1 k .
1
2. (b) Write the combined spring constant as onek′ = k
half of the single spring constant:
2
m
= 2π
k′
m
( k / 2)
3. Replace k’ with k/2 in the period equation:
T ′ = 2π
4. Rearrange the equation:

m
m
T ′ = 2π 2   = 2  2π

k
k 
 


m
T ′ = 2  2π
 = 2T
k 

Insight: If the spring had been replaced by 3 identical springs the resulting spring constant would
be k ′ = k / 3 , giving T ′ = 3T . In general the period will be proportional to the square-root of the
length of the spring.
5. Replace the term in parentheses with the
original period:
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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.
2
L
 2π 
⇒ g =
Solution: 1. Solve the period equation for T = 2π
 L
g
 T 
gravity:
2
2. Insert numeric values:
 2π 
2
g = 1
 ( 2.5 m ) = 9.6 m/s
 5 (16 s ) 
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.
2
Solution: 1. Solve the period equation for T = 2π L ⇒ L =  T  g
g
 2π 
length:
2
2. Insert numeric values:
 1.00 s 
2
L=
 ( 9.81 m/s ) = 24.8 cm
2
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, ( g Moon = 16 g Earth ).
Solution: 1. Write the
period on the Moon in terms
of the period on Earth:
2. Calculate the period on the Moon:
TMoon = 2π
1
6
L
L
= 6 ⋅ 2π
= 6 ⋅ TEarth
g Earth
g Earth
TMoon = 6 (1.00 s ) = 2.45 s
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
L
T = 2π
equation for the period of a pendulum:
g+a
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2. (b) ) Replace “g” by “g − a” in the equation
L
T = 2π
for
g −a
the period of a pendulum:
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.
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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-totrough distance. Set the amplitude equal to the vertical
crest-to-midline distance, or half the vertical crest-totrough distance.
Solution: 1. (a) Double the horizontal crest-totrough distance:
λ = 2 ( 26 cm ) = 52 cm
2. (b) Halve the vertical crest-to-trough
distance:
A=
1
2
(11 cm ) =
5.5 cm
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.
 1 min 
v = λ f = ( 34 m )(14 /min ) 
 = 7.9 m/s
 60 sec 
Solution: Multiply wavelength by
frequency:
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.
Solution: 1. Calculate the
frequency:
f =
2. Calculate the new wavelength:
λ2 =
v1
λ1
=
2.0 m/s
= 1.333 Hz
1.5 m
v2
1.6 m/s
=
= 1.2 m
f 1.333 Hz
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:
f =
v
λ
=
( 750 km/h ) 
1h 
−4

 = 6.7 × 10 Hz
310 km  3600 s 
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. (b) Calculate the vertical
distance:
d w = vt = ( λ f ) t = ( 27 ×10 −2 m ) ( 4.5 Hz )( 0.50 s ) = 0.61 m
t
dk = ( 4 A) 
T

−2
 = 4 Aft = 4 (12 × 10 m ) ( 4.5 Hz )( 0.50 s ) = 1.1 m

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 v = g λ 2π 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:
v = g λ / ( 2π ) =
(9.81 m/s ) ( 4.5 m ) / ( 2π ) =
2
2.65 m s
2.651 m/s
= 0.59 Hz
4.5 m
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.
2. (b) Solve for the frequency:
7.
f =
v
λ
=
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:
v = gd =
( 9.81 m/s ) ( 0.026 m ) =
2
0.51 m s
0.505 m/s
= 67 Hz
λ 0.0075 m
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.
2. (b) Calculate the wave frequency:
8.
f =
v
=
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:
v = F µ ⇒ F = v2 µ
2. Divide the tension at higher velocity by the
initial tension:
F2 v22 µ  v2   32 m/s 
=
=
=
 =4
F v 2 µ  v   16 m/s 
2
2
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.
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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:
2. Substitute µ = m d and insert
numerical values:
t=
d
µ
=d
v
F
t=d
m/d
=
F
md
=
F
( 0.032 kg )( 9.5 m )
8.6 N
= 0.19 s
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:
t=
( 0.032 kg )( 9.5 m ) / ( 9.0 N ) =
3. (c) Set the tension equal to 10.0 N:
t=
( 0.032 kg )( 9.5 m ) / (10.0 N ) =
0.18 s
0.17 s
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:
v=
F
µ
2
F = µ v2 =
m L
mL
  = 2
Lt 
t
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2. Insert the given mass, length and time:
Tutorial Solutions
F=
( 0.085 kg )( 7.3 m )
=
2
( 0.94 s )
0.70 N
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.
F=
( 0.095 kg )( 7.3 m )
=
2
( 0.94 s )
0.78 N
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:
vA
=
vB
FA / µA
=
FB / µB
FA µ B
FB µ A
2. Write the linear mass density in terms of density
and area:
vA
=
vB
FA ρ AB
=
FB ρ AA
FA AB
FB AA
3. Write the area in terms of the diameter:
vA
=
vB
FA π ( d B 2 )
=
FB π ( d A 2 )2
4. Insert the given tensions and diameters:
vA
410 N  1.0 mm 
=

 = 2 = 1.4
vB
820 N  0.50 mm 
2
FA d B
FB d A
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. Use the dimensions of time to calculate
α:
α
[ L] =  [ M ][ L ]  L β  [ M ] 
[ ]  3
[T ]  [T ]2 
 [ L] 
1
α +γ
= [M ]
α + β − 3γ
[ L]
[T ]
−2α
[T ]−1 = [T ]−2α
−2α = −1
α=
3. Use the dimensions of mass to
determine γ:
−1
[ L ] [T ]
γ
1
2
[ M ]0 = [ M ]α +γ
α +γ = 0
γ = −α = − 12
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[ L ]1 = [ L ]α + β −3γ
4. Use the dimensions of length to
determine β:
1 = α + β − 3γ
β = 1 − α + 3γ = 1 − ( 12 ) + 3(− 12 ) = −1
5. Use the dimensions to write a
dimensionally correct equation for
velocity:
1
v ∝ T 2 R −1 ρ
− 12
∝
T
R2 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.
d=
Solution: Calculate the distance to
the cliff:
1
1
vt = ( 343 m/s )(1.85 s ) = 317 m
2
2
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:
t=
2. (b) Solve for the wavelength:
λ=
2d 2 ( 75 m )
=
= 0.098 s
v 1530 m/s
v 1530 m/s
=
= 28 ×10 −3 m = 28 mm
f
55 kHz
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:
λ=
v 343 m/s
=
= 0.807 m
f
425 Hz
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Tutorial Solutions
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:
λ=
343 m/s
= 0.722 m
475 Hz
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.
2d
g
Solution: 1. (a) Solve for the falling time:
t1 =
2. Solve for the time for the sound to travel up
the well:
t2 =
3. Sum the two times to equal the total time:
t = t1 + t2 =
4. Rewrite as a quadratic
equation in terms of the
variable d :
0=
d
vs
1
vs
( d)
2d d
+
g vs
2
1


0=

 343 m/s 
5. Solve for d using the
quadratic formula and square
the result:
2
d −t
g
+
( d)
2

2
+ 
2
 9.81 m/s

 d − 1.2s

d = 3.2537 m ⇒ d = 10.587 m = 11 m
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.
d 8.80 m
=
= 0.0257 s
v 343 m/s
Solution: 1. Calculate ts :
ts =
2. Subtract ts from the total
time:
tf = 1.2 s − 0.0257 s = 1.1743 s
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3. Solve for v0 :
Tutorial Solutions
1
y = y0 + v0 tf + atf 2
2
y − y0 1
v0 =
− atf
tf
2
−8.8 m 1
− ( −9.81 m/s 2 ) (1.1743 s ) = −1.7 m s
1.1743 s 2
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.
=
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Tutorial Solutions
Solutions to Tutorial 7
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.
m
0.014 g
=
= 6.4 g cm3
V 0.0022 cm3
Solution: 1. Divide the volume by the
mass:
ρ=
2. Compare with the density of gold from
the table:
ρgold = 19.3 g cm3 . 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.
m
0.347 kg
=
= 1.05 × 104 kg m3
V 3.21 cm ( 1 m )  3
100 cm 

Solution: 1. Calculate the density of
the cube:
ρ=
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.
2
Solution: 1. (a) Convert the pressure to
pascals:
2. (b) Convert to atmospheres:
−5
 1 dyne   10 N   100 cm 
–1


 = 10 Pa
2 
 cm   dyne   m 
1 atm


–6
10 –1 Pa 
 = 10 atm
3
 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:
P=
F
mg
mg
=
=
2
A
  d   πd2
4 π   
  2  
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2. Insert given data:
Tutorial Solutions
P=
(72 kg + 3.8 kg) ( 9.81 m/s 2 )
π ( 0.013 m )
2
= 1.4 × 106 Pa
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.
F
Aw
2
Solution: 1. Find the ratio of the
pressure with the tip to the pressure
without:
Pw
=
Pwo
2. Invert the ratio to find the factor
by which the pressure decreases:
1
= 4.3
0.23
F
Awo
=
Awo π rwo 2 (1.2 cm )
=
=
= 0.23
Aw
π rw 2 ( 2.5 cm )2
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:
F = PA
2. Set the supporting
force equal to the sum of
your weight and the
weight of the bicycle:
F = Wyou + Wbicycle = Wyou + mbicycle g
3. Solve for your weight:
Wyou = F − mbicycle g = 690.7 N − 7.7kg ( 9.8m/s 2 ) = 615 N
2
 1.01× 105 Pa 
 1m 
= ( 70.5 lb/in 2 ) 
2 × 7.13 cm 2 ) 
 = 690.7 N
2 (
 100 cm 
 14.7 lb/in 
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.
F
mg
Solution: 1. (a) Solve
A=
=
equation 15-2
P
P
for the total contact area:
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2. Solve for the contact
area on one tire:
A = 4 Atire =
3. Insert given values:
Atire =
Tutorial Solutions
mg
P
⇒ Atire =
mg
4P
1320 kg ( 9.81 m/s 2 )
4 ( 35.0 lb/in
2
)(
1.01×105 Pa
14.7 lb/in 2
)
= 0.0135 m 2
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:
P=
6. Insert the given values: P =
mg
4 Atire
1320 kg ( 9.81 m/s 2 )
4 (116 cm 2 )
(
1m
100 cm
)
2
 14.7 lb/in 2 
2
= ( 2.79 × 105 Pa )
 = 40.6 lb/in
5
 1.01× 10 Pa 
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 = Dπ h.
Solution: 1. Calculate
vertical area:
A = Dπ h = ( 0.065 m ) π ( 0.12 m ) = 0.0245 m 2
2. Solve for inward force:
F = Pat A = 1.01× 105 N/m 2 ( 0.0245 m 2 ) = 2.5 kN
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 1 atm = 1.01× 105 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:
 1.01× 105 Pa 
P = (736 mmHg) 
 = 97.8 kPa
 760 mmHg 
2. (b) Solve for the height of the
water:
P2 = P1 + ρ gh
h=
P2 − P1
9.78 × 104 Pa − 0
=
= 9.97 m
ρg
(1000 kg/m3 )( 9.81 m/s2 )
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.
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Tutorial Solutions
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 righthand column.
PL = PR
Solution: 1. Set the pressures
equal:
mL g mR g
=
+ ρ gh
AL
AR
2. Solve for the height h:
h=
3. Insert the given values:
h=
1  mL mR
−

AR
ρ  AL
 1  mL
mR  4  mL mR 
 2− 2
= π 2 −π 2 =
D
ρ

 4 L 4 DR  πρ  DL DR 
 1.7 kg
3.2 kg 
−

 = 1.0 m
2
π ( 750kg/m )  ( 0.045 m ) ( 0.12 m )2 
4
3
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.
Solution: 1.
Calculate the
bursting pressure of
the lid:
2. Solve for the
height of the water
column:
3. Solve for the
weight:
∆P =
643 N
F
=
= 1460 Pa
A π ( 0.75 m 2 )2
∆P = ρ gh
h=
∆P
1455 Pa
=
= 0.148 m
ρ g 1000 kg/m3 ( 9.81 m/s 2 )
  d 2 
W = mg = ρ π    hg
  2  
  0.010 m  2 
2
= 1000 kg/m3 π 
  ( 0.148 m ) ( 9.81 m/s ) = 0.11 N
2
 
 
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.
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Tutorial Solutions
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: P = Pat + ρ gh
h=
2. Insert given values:
h=
P − Pat
ρg
116 × 103 Pa − 1.01× 105 Pa
(806 kg/m )( 9.81 m/s )
3
2
= 1.90 m
2.05 × 10−3 m3
3. (b) Divide the volume by h2 =
= 0.314 m
65.2 × 10−4 m 2
A:
4. Add the heights:
htotal = h + h2 = 1.897 m + 0.314 m = 2.211 m
5. Insert data:
P = Pat + ρ gh
= 1.01×105 Pa + (806 kg/m3 )( 9.81 m/s 2 ) (2.211 m) = 118 kPa
Insight: Part (b) could also have been solved by adding the additional depth to the bottom of the
cylinder, such as: P = 1.16 × 105 Pa + ( 806 kg/m 3 )( 9.81 m/s 2 ) ( 0.314 m ) = 118 kPa .
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:
P = Patm + ρ w gh
h=
2. Insert the given values:
h=
P − Patm
ρw g
10.0 N/mm 2
(
103 mm
m
)
2
(1025 kg/m )( 9.81 m/s )
3
2
= 995 m
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.
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Tutorial Solutions
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:
P = Pat + ρ gh
2. (b) Repeat for a depth of
5.5 m:
P = 1.01×105 Pa + (1000 kg/m3 )( 9.81 m/s 2 ) 5.5 m = 1.54 × 105 Pa
= 1.01× 105 Pa + (1000 kg/m3 )( 9.81 m/s 2 ) 4.5 m = 1.45 × 105 Pa
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.
∆v 2.2 m/s − 0
2. (b) Calculate the
a=
=
= 0.710 m/s 2
acceleration of the
∆t
3.1 s
elevator:
3. Write the change in
pressure as the added
force divided by the
area of the glass:
4. Write the mass as
density times volume:
∆P =
∆P =
F ma
=
A
A
( ρ Ah )
A
a = ρ ha = (1000 kg/m3 ) ( 0.065 m ) ( 0.710 m/s 2 ) = 46 Pa
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.
Solution: 1.
Calculate the
pressure at the
bottom of the
oil:
P1 = Patm + ρ gh
= 1.013 × 105 Pa + ( 920kg/m3 )( 9.81m/s 2 ) ( 0.072m )
= 1.0195 ×105 Pa
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2. Calculate the
pressure at the
bottom of the
water:
Tutorial Solutions
P = P1 + ρ water ghwater
= 1.0195 × 105 Pa + (1000 kg/m3 )( 9.81 m/s 2 ) ( 0.12 m ) = 1.03 ×105 Pa
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:
Pat = P + ρ gh = ρ gh
3. Solve for the height of the
water column:
h=
Pat
1.01× 105 Pa
=
= 10.3 m
ρ g (1000 kg/m3 )( 9.81 m/s 2 )
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: P = Pat + ρ gh ⇒ h =
2. Insert the given
data:
h=
P − Pat
ρg
109 kPa − 101.3 kPa
(1020 kg/m )(9.81 m/s )
3
2
= 0.770 m
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.
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Tutorial Solutions
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, hw = 1.0 m − d .
Solution: 1. Write the pressure at the
bottom of the cylinder:
P = Pat + ρ w ghw + ρ Hg ghHg
2 Pat = Pat + ρ w g (1.0 m − d ) + ρ Hg gd
2. Solve for d:
Pat
− ρ w (1.0 m)
g
d=
ρ Hg − ρ w
3. Insert the given values:
1.01× 105 Pa
− (1000 kg/m3 )1.0 m
2
9.81
m/s
d=
= 0.74 m
1.36 × 104 kg/m3 − 1000 kg/m3
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.
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Tutorial Solutions
Solutions to Tutorial 8
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:
∆L = α L0 ∆T
T = T0 +
∆L
α L0
= 23 °C +
0.50 ft
= 220 °C
(24 ×10 –6 K −1 )(107 125 ft)
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:
∆L = α L0 ∆T
= 1.2 × 10−5 (C°) −1  ( 3910 m ) [30.0 °C − (−5.00 °C) ] = 1.6 m
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:
∆d = d ′ − d = α d ∆T
2. Insert the given
data:
d ′ = 1.178 cm 1+ ( 24 × 10 –6 K −1 ) (199.0 °C − 23.00 °C ) 
d ′ = d + α d ∆T = d (1 + α∆T )
= 1.183 cm
3. (b) Solve for the
change in temperature:
∆d = d ′ − d = α d ∆T
d′ − d
∆T = T − T0 =
αd
4. Solve for the final
temperature:
T = T0 +
d′ − d
αd
= 23.00 °C +
1.176 cm − 1.178 cm
( 24 ×10–6 K −1 ) (1.178 cm )
= − 48 °C
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.
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Tutorial Solutions
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:
∆d = α d ∆T
∆T = T − T0 =
3. Solve for the final temperature:
T = T0 +
∆d
αd
∆d
α d0
= 10.00 °C +
4.040 cm – 4.000 cm
( 2.4 ×10
−5
( C°) ) ( 4.000 cm )
−1
T = 430°C
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:
∆L = α L∆T = α L (T − T0 )
T = T0 +
∆L
αL
2. (a) Insert the data for heating the
brass sleeve:
T = 12.25 °C +
3. (b) Insert the data for cooling the
steel:
T = 12.25 °C +
2.19893 cm − 2.19625 cm
(1.9 ×10
−5
K −1 ) ( 2.19625 cm )
2.19625 cm – 2.19893 cm
(12 ×10
–6
K −1 ) ( 2.19893 cm )
= 76 °C
= −89 °C
Insight: Since the coefficient of thermal expansion for brass is greater than the coefficient of
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Tutorial Solutions
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
Vspill = ∆Vgas − ∆Vtank . 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:
Vspill = ∆Vgas − ∆Vtank = ( β gasV0 − 3α tankV0 ) ∆T = ( β gas − 3α tank ) V0 ∆T
2. Insert the given
data:
Vspill = ( 9.5 × 10−4 − 3 × 1.2 ×10 −5 ) ( C° )
−1
( 51 L )( 25 − 5.0 ) °C =
0.93 L
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.
Solution: 1. Write the
∆d Cu = α Cu d 0 ∆T
equation for the increases in ∆d = α d ∆T
st
st 0
diameters:
2. Subtract the two
differences:
∆d st − ∆d Cu = (α st − α Cu ) d 0 ∆T
= (1.73 × 10−5 − 1.70 × 10−5 ) ( C° ) −1 ( 8.0 in.)( 610 − 22 ) °C
= 0.0014 in.
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:
∆VAl = β AlV0 ∆T = 3α AlV0 ∆T
∆VCu = β CuV0 ∆T = 3α CuV0 ∆T
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3. Subtract the changes in
volume to calculate the
difference in volumes of
the two cubes:
Tutorial Solutions
∆VAl − ∆VCu = 3 (α Al − α Cu )V0 ∆T
= 3 ( 2.4 − 1.7 ) × 10−5 ( C° )
−1
( 0.016m ) (97 − 21) °C
3
= 2.6 × 10−5 m3
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:
∆L = α L0 ∆T ⇒ T = T0 +
2. Insert the given data:
T = 22°C +
∆L
α L0
0.18 × 10−3 m
= 680 °C
1.7 × 10 −5 ( C° )−1  (1.6 × 10−2 m )


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
2. (b) Calculate the
initial volumes of the
saucepan and water:
 23 cm 
3
V0 = π r 2 h = π 
 ( 6.0 cm ) = 2493c m
 2 
3. Write the changes in
volume:
∆Vw = β wV0 ∆T
4. Subtract the change
on volume of the pan
from the water to
calculate the volume of
water spilled:
Vspill = ∆Vw − ∆VAl = ( β w − 3α Al )V0 ∆T
∆VAl = 3α AlV0 ∆T
= ( 21 − 3 × 2.4 ) × 10−5 ( C° )
−1
( 2493 cm ) (88 − 19 ) °C
3
Vspill = 24 cm3
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.
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Tutorial Solutions
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:
Q = mc∆T
= ( 0.055 kg ) 837 J/ ( kg ⋅ K )  (15C° ) = 0.69 kJ
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 :
∆T =
Q
mc
2. Set the heat equal to half the initial kinetic
energy:
∆T =
1
K
Q
= 2 =
mc mc
1
2
(
1
2
mv 2 )
mc
=
v2
4c
2
=
( 250 m/s )
= 120 K
4 128 J/ ( kg ⋅ K ) 
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:
2. Solve for the mass of silver:
QAg + Qw = 0
mAg cAg (T − TAg ) + mw cw (T − Tw ) = 0
mAg =
=
3. Divide by the mass of one pellet:
n=
mw cw (Tw − T )
cAg (T − TAg )
0.220 kg  4186 J/ ( kg ⋅ K )  (14 − 25 ) °C
 234 J/ ( kg ⋅ K )  ( 25 − 85 ) °C
mAg
mpellet
=
= 0.722 kg
0.722 kg
= 7.2 × 102 pellets
0.001 kg
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:
mCu =
=
mw cw (Tw − T )
cCu (T − TCu )
(0.220 kg)  4186 J/ ( kg ⋅ K )  (14 − 25 ) °C
387 J/ ( kg ⋅ K ) ( 25 − 85 ) °C
= 0.436 kg
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6. Divide by the mass of one
pellet:
Tutorial Solutions
n=
mCu
0.436 kg
=
= 4.4 × 102 pellets
mpellet 0.001 kg
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:
T=
mPb cPbTPb + mw cw Tw
mPb cPb + mw cw
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 ) 
T = 23.9 °C
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:
C=
Q
2200 J
=
= 0.18 kJ C°
∆T
12 C°
2. (b) Divide the heat capacity by
the mass:
c=
Q
C 0.183 kJ/ C°
= =
m ∆T m
0.190 kg
= 0.96 kJ ( kg ⋅ C° ) = 0.96 kJ ( kg ⋅ K )
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:
2. Solve for the change in
temperature:
U 0 + K0 = U + K
mgh + 0 = 0 + K
K = mgh = Q
∆T =
2
Q mgh gh ( 9.81 m/s ) ( 5.43 m )
=
=
=
mc mc
c
128 J/ ( kg ⋅ K )
= 0.416 K = 0.416 C°
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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 ∑ Q = 0
and solve for
QOb :
2. Solve for
the specific
heat:
0 = QOb + Qw + QAl
QOb = − ( Qw + QAl ) = − ( mw cw + mAl cAl ) ∆Tw
cOb =
Qob
mOb (T − TOb )
( mw cw + mAl cA1 )(Tw − T )
mOb (T − TOb )
{0.103 kg  4186 J/ ( kg ⋅ K ) + 0.155 kg 900 J/ ( kg ⋅ K )} ( 20 − 22 ) C°
=
0.0380 kg ( 22.0 − 100 ) C°
cOb =
cOb = 385 J (kg ⋅ C°) = 385 J (kg ⋅ K)
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:
T=
=
mh chTh + mw cw Tw
mh ch + mw cw
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 )
T = 24 °C
2. (b) Write the heat
capacities of the
lead and iron:
CPb = 1 kg × 128 J/ ( kg ⋅ K )  = 128 J/K
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.
CFe = 0.50 kg × 448 J/ ( kg ⋅ K ) = 224 J/K
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.
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Tutorial Solutions
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:
0 = Qcup + Qcof + Qcrm
2. Solve for
the equilibrium
temperature:
T=
= mcup ccup (T − Tcup ) + mcof cw (T − Tcof ) + mcrm cw (T − Tcrm )
0 = T  mcup ccup + ( mcof + mcrm ) cw  −  mcup ccupTcup + ( mcof Tcof + mcrmTcrm ) cw 
mcup ccupTcup + ( mcof Tcof + mcrmTcrm ) cw
mcup ccup + ( mcof + mcrm ) cw
( 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 ⋅ K )  

=
= 70.5 °C
( 0.116 kg ) 1090 J/ ( kg ⋅ K )  + ( 0.225 kg + 0.0122 kg ) 4186 J/ ( kg ⋅ K ) 
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:
 ∆T
Q = kA 
 L

−3
2  8.5 °C 
t = 34.3 W/ ( m ⋅ K ) (1.2 ×10 m ) 
 (1.0 s ) = 2.3 J

 0.15 m 
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:
Qtotal QAl Qst
 ∆T
=
+
= kAl A 
t
t
t
 L
2. Solve for the
length of the bars:
L=
=

 ∆T
 + kst A 

 L

 ∆T 
 = A
 ( kAl + kst )

 L 
A ∆T (kAl + kst )
( Qtotal t )
1
4
2
π ( 0.0350 m ) (118 − 20 ) °C  ( 217 + 16.3) W/ ( m ⋅ K )
= 0.800 m
( 27.5 J/s )
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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; QAl t = 25.6 J s and
Qst t = 1.9 J s .
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:
2. Solve for
the diameter
of the
copper rod:
Qtotal QCu QPb
 ∆T 
 ∆T 
=
+
= kCu ACu 
 + kPb APb 

t
t
t
L


 L 
 π 2 
 π 2    ∆T 
 π∆T 
2
2
=  kCu  d Cu
 + kPb  d Pb   
 =  kCu d Cu + kPb d Pb  


4
  L 
 4L 
 4
d Cu =
 Qtotal   4 L 
2
 t   π∆T  − kPb d Pb 




kCu
( 33.2 J/s )
=
4 ( 0.650 m )
− 34.3 W/ ( m ⋅ K )  ( 0.0276 m )
π (112 − 21) °C 
395 W/ ( m ⋅ K )
2
= 2.64 cm
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:
T2 − T1
t
L
(1.41 J )( 0.525 m )
QL
T1 = T2 −
= 106 °C −
= 98°C
2
k At
395W/ ( m ⋅ K )  ( 0.015 m ) (1 s ec )
Q=kA
Insight: This problem can also be solved by finding the temperature difference across the lead:
(1.41 J )( 0.525 m )
QL
T2 = T1 +
= 2 °C +
= 98°C
2
k At
34.3W/ ( m ⋅ K )  ( 0.015 m ) (1 sec )
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Tutorial Solutions
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.
∆T
∆TAl
= kPb A Pb
LAl
LPb
2. (b) Set
the rate of
heat flow
through the
rods equal:
kAl ⋅ A
3. Solve for
the length of
the
aluminum
rod:
 k   ∆T
LAl =  Al   Al
 kPb   ∆TPb
 217 W/ ( m ⋅ K )   80.0 °C − 50.0 °C 


 LPb = 
 (14 cm ) = 89 cm
 34.3 W/ ( m ⋅ K )   50.0 °C − 20.0°C 

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:
2. Substitute for the heat flows:
Q Q23
=
t
t
 T23 − 21°C 
 105 °C − 21°C 
kA 
 = kA 

85 cm


 23 cm 
 23 cm 
T23 = 21°C + 
 ( 84 °C ) = 21°C + 23 °C = 44 °C
 85 cm 
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.
3. Now solve for T23:
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
4
P1 eσ A (T1 − Ts ) ( 273.15 + 95 ) K  − ( 273.15 + 21) K 
=
=
= 26
P2 eσ A (T24 − Ts4 ) ( 273.15 + 25 ) K  4 − ( 273.15 + 21) K  4

 

4
4
Insight: Since the cooler object’s temperature is close to the room temperature, its net power
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Tutorial Solutions
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.
L
2L
3L
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 A = 2 L × L :
 ∆T
P = kA 
 L′

2  ∆T
 = k ( 2L ) 

 3L
 2
 = k L ∆T
 3
2. (a) Set Pa equal to the heat
flow through the length 2L
and A = 3L × L :
9
 ∆T  3
Pa = k ( 3L2 ) 
P
 = k L ∆T =
L
2
2
4


3. (b) Set Pb equal to the heat
flow through the length L and
A = 3L × 2 L :
 ∆T 
Pb = k ( 6 L2 ) 
 = 6 k L ∆T = 9 P
 L 
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.
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Tutorial Solutions
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:
F = ma = evB sin 90°
a=
−19
evB (1.6 × 10 C ) ( 9.5 m/s )(1.6 T )
=
= 1.5 ×109 m/s 2
m
1.673 × 10−27 kg
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:
v=
F
8.9 × 10−15 N
=
= 4.6 ×105 m/s
e B sin 90° (1.6 × 10−19 C ) ( 0.12 T )
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:
F = qvB sin 0° = 0
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:
a=
−19
−5
evB sin 90° (1.60 × 10 C ) ( 355 m/s ) ( 4.05 × 10 T )
=
m
1.673 × 10−27 kg
= 1.38 ×106 m/s 2
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 :
 F 

 qvB 
θ = sin −1 
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
2. Insert the numerical values for F =
4.8 µN:
θ = sin −1 

 = 81°
C ) (16 m/s )( 0.95 T ) 
4.8 × 10−6 N
 ( 0.32 × 10−6


3.0 × 10−6 N
3. (b) Insert the numerical values for F θ = sin −1 
 = 38°
 ( 0.32 × 10−6 C ) (16 m/s )( 0.95 T ) 
= 3.0 µN:


1.0 × 10−7 N
4. (c) Insert the numerical values for F θ = sin −1 
 = 1.2°
 ( 0.32 × 10−6 C ) (16 m/s )( 0.95 T ) 
= 0.10 µN:
Insight: The magnetic force is the largest when the velocity and the magnetic field are
r
r
perpendicular, and it is zero when v and B 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:
Fnew qvnew B sin θ new vnew sin θ new ( 6.3 m/s ) sin 25°
=
=
=
= 0.099
Fold
qvold B sin θ old
vold sin θ old
( 27 m/s ) sin 90°
2. Now solve for Fnew :
Fnew = 0.099 Fold = ( 0.099 ) ( 2.2 × 10−4 N ) = 2.2 ×10−5 N
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 ŷ direction, so we conclude that the magnetic field also points in the
ŷ direction. When the ion travels in the xy plane and along the line y = x,
r
it moves at an angle of 45° with respect to B. When it moves in the x̂
direction, it experiences the maximum magnetic force.
Solution: Letting Fmax = q vB :
F = Fmax sin 45° = ( 6.2 × 10 −16 N ) sin 45° = 4.4 × 10 −16 N
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 x̂ direction, so we
conclude that the magnetic field also points along either the + xˆ or − xˆ 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 ( −zˆ direction) and the fingers of your left hand in the ŷ direction to find that the
magnetic field must point in the − xˆ direction. Then use the force experienced by the electron when
r
it moves in the + yˆ direction to find the magnitude of B.
r
Solution: Find B :
B=
( 2.0 ×10−13 N ) sin 90° = 1.4 T ⇒ Br = −1.4 T xˆ
F sin θ
=
(
)
ev
(1.6 ×10−19 C )(9.1×105 m/s )
Insight: If instead the electron were to travel in the + zˆ direction, it would experience a force
r
F = ( 2.0 × 10−13 N ) yˆ .
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9.
Tutorial Solutions
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.
v1 F q1 B sin θ q2
q
1
=
=
= 2 =
v2 F q2 B sin θ q1 4q2
4
2. (b) Find the ratio:
Insight: Suppose both charges were allowed to travel in circles as described in section 22-3.
r
mv1 q1 B v1q2
v q
1
Assuming both charges have the same mass, we find that 1 =
=
= 1 2 = ;
r2 mv2 q2 B v2 q1 4v1 4q2 16
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:
r=
−31
5
mv ( 9.11× 10 kg )( 6.27 × 10 m/s )
=
= 7.9 ×10−6 m = 7.9 µ m
eB
(1.6 ×10−19 C ) ( 0.45 T )
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.
mv (1.673 × 10 kg )( 6.27 × 10 m/s )
=
= 1.5 × 10−2 m = 1.5 cm
eB
(1.6 ×10−19 C ) ( 0.45 T )
−27
Solution: Apply equation
directly:
r=
5
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:
2. Solve for B:
v=
2 e ∆V
=
m
2 (1.60 × 10−19 C ) ( 410 V )
( 9.11×10
−31
kg )
= 1.2 × 107 m/s
−31
7
mv ( 9.11× 10 kg )(1.2 × 10 m/s )
B=
=
= 4.0 ×10 −4 T = 0.40 mT
er
(1.60 ×10−19 C ) ( 0.17 m )
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).
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Tutorial Solutions
13. Picture the Problem: The magnetic force on a charged particle
travelling at constant speed causes it to move in a circle.
Strategy: The magnetic force provides the centripetal force
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.
Solution: 1.
(a) Solve for
v:
2. (b)
Divide the
circumference by
the speed:
v=
−6
qrB (12.5 ×10 C ) ( 26.8 m )(1.01 T )
=
m
2.80 ×10−5 kg
= 12.1 m/s
t=
2π r 2π m 2π ( 26.8 m )
=
=
= 13.9 s
v
qB
(12.1 m/s )
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
travelling at constant speed causes it to move in a circle.
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.
−19
erB (1.60 × 10 C ) ( 0.520 m )( 0.180 T )
1.0 u
=
×
= 1.5 u
6
v
1.67 ×10−27 kg
( 6.0 ×10 m/s )
r
Insight: Another way to answer part (a) is to point the fingers of your hand along v and curl them
r
r
into the page in the direction of B. For which hand does the thumb (and therefore F ) 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.
2. (b) Solve for m:
m=
15. Picture the Problem: The magnetic force on a proton
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 . We
must first find the speed v of the proton by using the
definition of kinetic energy, then find the radius r.
1
Solution: 1. Solve for K = mv 2 ⇒ v =
2
v:
2. Find r:
mv m 2 K
r=
=
=
eB eB m
2K
m
2mK
=
eB
2 (1.673 × 10−27 kg )( 4.9 × 10−16 J )
(1.6 ×10
−19
C ) ( 0.26 T )
= 3.1 cm
Insight: The magnetic field must be increased in order to decrease the radius of the proton’s path.
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Tutorial Solutions
16. Picture the Problem: The magnetic force on an alpha particle
travelling at constant speed causes it to move in a circle.
Strategy: The magnetic force provides the centripetal force
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.
Solution: 1. (a) Find
T 1  2π r  π r π  mv  π m
t= = 
= 
half of the orbit
=
=
v  qB  qB
2 2 v  v
period, then substitute
for v:
2. Insert numerical
values:
t=
π ( 6.64 × 10−27 kg )
(1.60 ×10
−19
C ) ( 0.155 T )
= 4.21× 10−7 s = 421 ns
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:
t = 4.21×10−7 s = 421 ns
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 relectron rproton for each of two cases: (a) the particles have equal momenta,
and (b) the particles have equal kinetic energies.
Solution: 1. (a) Calculate re rp when
pe = pp :
2. (b) Calculate re rp when K e = K p :
re me ve qe B pe e B
=
=
= 1.00
rp mp vp qp B pp e B
re me ve qe B me 2 K e me e B
=
=
=
rp mp vp qp B mp 2 K p mp e B
=
me K e
=
mp K p
me
mp
9.11× 10−31 kg
= 0.0233
1.673 × 10−27 kg
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:
F = I L B sin 90° = ( 0.695 A )( 2.15 m )( 0.720 T )(1) = 1.08 N
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.
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Tutorial Solutions
19. 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.
F = I L B sin θ = ( 2.8 A )( 2.25 m )( 0.88 T ) sin 36.0° = 3.3 N
Solution: Find F:
Insight: The maximum force occurs at θ = 90°, at which angle the force amounts to 2.5 N of force
per meter of wire.
20. Picture the Problem: A current-carrying wire experiences a force due to the presence of a
magnetic field.
Strategy: Solve for the angle θ that would produce the given magnetic force on the wire.
sin θ =
Solution: Solve for θ:
F
I LB
θ = sin −1
F
1.6 N
= sin −1
= 63°
I LB
( 3.0 A )(1.2 m )( 0.50 T )
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:
 0° or 
Ftop = Fbottom = I L B sin 
= 0
180° 
2. The left and right wires
are perpendicular to the
field:
Fleft = Fright = I L B sin 90° = ( 9.5 A )( 0.46 m )( 0.34 T ) = 1.5 N
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
r
the weight of the wire to find the magnitude of B. Then
r
use the Right-Hand Rule to determine the direction that B
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:
2. Solve for B:
F = mg = I L B sin 90° = I L B
B=
m g ( 0.17 kg ) ( 9.81 m/s
=
IL
(11 A )( 0.45 m )
2
) = 0.34 T
3. Let upward in the figure be the positive y-direction, and the positive x-direction be to the right.
r
The RHR stipulates that B must point into the page, or in the −zˆ direction. Therefore,
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Tutorial Solutions
r
B = ( − 0.34 T ) zˆ .
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).
Solution: 1. (a) Write
equation in terms of
force per unit length:
F
= I B sin θ
L
2. Solve for B:
B=
2. (b) Solve for θ :
θ = sin −1
F L
0.033 N/m
=
= 0.041 T = 41 mT
I sin θ ( 6.2 A ) sin 7.5°
F L
0.015 N/m
= sin −1
= 3.4°
IB
( 6.2 A )( 0.041 T )
Insight: At small angles the sin θ 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:
2. Solve for I:
F = mg = I L B sin 90° = I L B
I=
mg ( 0.75 kg ) ( 9.81 m/s
=
LB
( 2.7 m )( 0.84 T )
2
) = 3.2 A
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 x̂ point east, ŷ point
north, and ẑ point upward. Converting the magnetic field from gauss to
tesla, we find that 0.59 G ×1.0 × 10−4 T/G = 5.9 × 10−5 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:
Tutorial Solutions
F = I L B sin θ
= (110 A )( 250 m ) ( 5.9 ×10−5 T ) sin 90° = 1.6 N
3. (b) According to the RHR, the magnetic force points toward the east.
4. Calculate F:
(
)
F = I L B sin θ = (110 A )( 250 m ) 5.9 × 10−5 T sin 72° = 1.5 N
Insight: These are the forces if the wires carry direct current. In reality, the electrical current in highvoltage 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
r
the current I 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:
∑F
∑F
2. Solve each expression from step 1 for T: T =
3. Rearrange to solve for θ :
y
= 0 = T cos θ − mg
x
= 0 = T sin θ − I LB
mg
I LB
=
cos θ sin θ
tan θ =
I LB
mg
θ = tan −1
I LB
mg
Insight: The angle θ can be increased by increasing I or B and decreased by increasing m.
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Tutorial Solutions
Solutions to Tutorial 10
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: Φ = BA cos θ
2
= ( 0.055 T ) π ( 0.031 m ) cos16°
Φ = 1.6 × 10−4 Wb
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:
Φ = BA cos θ = ( 0.0250 T )( 0.325 m )( 0.120 m ) cos 0° = 9.75 ×10−4 Wb .
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:
B=
Φ
4.8 × 10−5 Tm 2
=
= 17 mT
A cos θ ( 0.055 m )( 0.068 m ) cos 42°
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.
Solution: Calculate the
magnetic flux:
Φ = BA cos θ = B⊥ A = ( 4.2 × 10−5 T ) ( 22 m )(18 m ) = 1.7 × 10−2 Wb
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.
2
 1.2 m 
Solution: Calculate the magnetic flux: Φ = BA cos θ = (1.7 T ) π 
 cos 0° = 1.9 Wb
 2 
Insight: Note that the length of the solenoid does not affect the flux through the solenoid.
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Tutorial Solutions
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°.
Solution:
Calculate the
flux:
Φ = BA cos θ = ( 5.9 × 10−5 T ) (1.30 m )( 0.82 m ) cos ( 90° − 72° ) = 6.0 ×10−5 Wb
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.
2
0.25 T ) π ( 0.15 m )
Solution: Calculate the ε = N ∆Φ = N BA − 0 = 53 (
= 7.8 V
t
t
0.12 s
∆
∆
emf:
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:
2. (b Calculate
the emf at
t = 0.15 s:
ε = − N ∆Φ = − 10 Wb − 0 =
3. (c)
Calculate the
emf at
t = 0.50 s:
ε = − −5 Wb − 10 Wb =
∆t
ε=
0.1 s
− 0.1 kV
0
0.6 s − 0.2 s
0.04 kV
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:
∆Φ −5 Wb − 10 Wb
=
= − 37.5 Wb/s
∆t
0.6 s − 0.2 s
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EE1427 Engineering Science – Dr. Daniel Nankoo
Tutorial Solutions
ε = − N ∆Φ = −1( − 37.5 Wb/s ) =
4. Calculate the induced emf :
0.04 kV
∆t
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 :
2. Solve for the
change in
magnetic field:
ε
= IR = ( 0.32 A )(110 Ω ) = 35.2 V
∆Φ
A∆B
=N
∆t
∆t
ε
∆B
35.2 V
=
=
= 4.8 × 102 T/s
∆t
ΝΑ 1( 7.4 ×10− 2 m 2 )
ε
=N
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
:
ε
=N
BA − ( − BA )
∆Φ
2 BA
=N
=N
∆t
∆t
∆t
= 120
2 ( 0.20 T ) ( 0.050 m 2 )
0.34 s
= 7.1 V
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:
C = 2π r
C
r=
= 0.1783 m
2π
2.
Calculat
e the
flux
through
the
circle:
Φ circle = BA = Bπ r 2 = ( 0.105 T ) π ( 0.1783 m )
= 0.0105 Wb
2
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Tutorial Solutions
3.
Calculate
the side
of the
square:
P = 4s
P 1.12 m
s= =
= 0.28 m
4
4
4.
Calculate
the flux
through
the
square:
Φ square = BA = Bs 2 = ( 0.105 T )( 0.28 m ) = 0.00823 Wb
5.
Calculate
the emf :
ε
2
=N
Φ circle − Φ square
∆t
=1
0.0105 − 0.00823 Wb
= 5.3 × 10−4 V
4.25 s
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
r = 12 d = 12 ( 0.12 m ) = 0.060 m.
Solution: Calculate the
number of coils:
ε
=N
∆Φ
∆t
ε
∆Φ
∆t
N=
−1
=
ε
BA
∆t
−1
= 6.0 V
( 0.20 T ) π ( 0.060 m )
1.5 s
2 −1
= 4.0 × 103
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.
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