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CHAPTER 4
FORCE AND MOTION
Multiple Choice Questions:
1.
(d).
2.
(c).
3.
(c).
4.
(d).
5.
(d).
6.
(b).
7.
(c).
8.
(d).
9.
(c).
10.
(c).
11.
(c).
12.
(a).
13.
(d).
14.
(c).
15.
(c).
16.
(b).
17.
(a).
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.5/6/2017
60
College Physics Seventh Edition: Instructor Solutions Manual
Conceptual Questions:
1.
(a) No . If an object remains at rest, the net force is zero. There could still be forces acting on it as long as the net
force is zero.
(b) No . The object could still be moving with constant velocity.
2.
According to Newton’s first law, your tendency is to remain at rest or move with constant velocity. However, the
plane is accelerating to a velocity faster than yours, so you are “behind” and feel “pushed” into the seat. The seat
actually supplies a forward force to accelerate you to the same velocity as the plane.
3.
4.
No same mass same inertia .
(a) The bubble moves forward in the direction of velocity or acceleration, because the inertia of the liquid will
resist the forward acceleration. So the bubble of negligible mass or inertia moves forward relative to the liquid.
Then it moves backward opposite the velocity (or in the direction of acceleration) for the same reason.
(b) The principle is based on the liquid inertia .
5.
(a) Balloon moves forward. The air has more inertia and tends to stay at the rear of the car. (b) Balloon moves
backward. The air has more inertia and tends to stay at the front of the car.
6.
According to Newton’s first law, or the law of inertia, the dishes at rest tend to remain at rest. The quick pull of the
tablecloth requires a force that exceeds the maximum static friction (discussed in Section 4.6) so the cloth can move
relative to the dishes.
7.
(a) Gradually increase the downward pull of the lower string. For balance, the tension in the upper string must be
equal to the pull plus the weight of the object, so it will break first. (b) Pull the lower string with a sudden jerk. By
Newton’s third law, the object will tend to remain at rest, so the tension in the upper string will not increase as much
as the tension in the lower string, hence breaking the lower one.
8.
The scale will read more than 600 N. In general, the scale reads the downward force (N) exerted on it from the
object placed on it, in this case, the student. If the student is not accelerating then the upward force on the student IS
equal in magnitude to the student’s weight. In reaction to that the student pushes down with the same size force,
hence the scale reads the student’s push N which in that case equals w. However for the student to accelerate upward
the scale force on the student N must exceed w. In turn, the student‘s reaction force down on the scale is greater in
magnitude than his weight. Hence the scale reads more 600 N as long as the student is in contact with it and
accelerating upward.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 4
9.
Force and Motion
61
The weight is zero in space, because there is no gravity. The mass is still 70 kg , because mass is a measure of
inertia, and it does not change.
10.
If the mass of an object changes, its acceleration will not be constant. For example, a pickup truck in a snowstorm
will experience less acceleration (with the same force) as it accumulates extra mass and a launched rocket has greater
acceleration (with the same force) as time goes on and it burns fuel, reducing its mass.
11.
No both mass and gravity decrease . This is not a violation of Newton’s laws. The thrust overcomes the
gravitational force and accelerates the rocket. However, when fuel is burned, the mass of the rocket decreases. Also,
farther up in space, there is less gravity. So the constant thrust has less gravity to overcome and less rocket to
accelerate. This, in turn, increases the velocity and acceleration of the rocket.
12.
“Soft hands” result in longer contact time between the ball and the hands. The increase in contact time decreases the
magnitude of the acceleration. From Newton’s second law, this, in turn, decreases the force required to stop the ball
and its reaction force, the force on the hands.
13.
The forces act on different objects (one on horse, one on cart) and therefore cannot cancel.
14.
Yes the forces on different objects (one on ball, one on bat) cannot cancel. .
15.
(a)
(b)
N
N
F
F
mg
mg
The force F is the combination of the pushing by the seat back and the friction force by the seat surface.
16.
Friction on block
The force on the block by the wall Fwall on block and the force on the wall by the
block Fblock on wall are an action-reaction pair. The friction on the block and on
Fperson on block
the wall are also an action-reaction pair.
Fwall on block
Fblock on wall
Friction on wall
17.
When the arms are quickly raised, an upward force is required to do this.
Weight
This force, ultimately, is provided by the normal force of the scale on the
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College Physics Seventh Edition: Instructor Solutions Manual
person. From Newton’s 3rd law, the person will push down on the scale more than the weight of the person, so the
scale reading increases. Conversely, a downward force is required to lower the arms, and the end result is a decrease
in the scale reading.
18.
(a) There is no friction in this case.
(b) The direction of the friction force is opposite the direction of velocity .
(c) It is sideways or perpendicular to the direction of velocity.
(d) It is forward in direction of velocity .
19.
This is because kinetic friction (sliding) is less than static friction (rolling) . A greater friction force can decrease
the stopping distance.
20.
The down force will increase the normal force on the car to more than the weight of the car. Since friction force is
directly proportional to the normal force, this will increase friction or “grip.” If a car is simply made more massive
to get more normal force, it will be more difficult to accelerate according to Newton’s second law. So the wings can
make the car grip better without significantly increasing the weight of the car—net result is more acceleration.
21.
(a) No, there is no inconsistency. Here the friction force OPPOSES slipping.
(b) Wind can increase or decrease air friction depending on wind directions. If wind is in the direction of motion,
friction decreases, and vice versa.
22
The treads are designed to displace water, so cars with regular tires can drive in the rain. However, the wide and
smooth drag racing tires increase friction because they use a softer compound, therefore there is a large coefficient of
friction. This does not contradict the fact that friction is independent of the surface area. A common misconception is
that the larger area of the racing tires contributes to larger friction. Soft compound tires are required to be wider for
the sidewall to support the weight of the car. A narrow, soft tire would not be strong enough, nor would it last very
long. Wear in a tire is related to surface area.
23.
Run at a speed you can estimate (say 2 m/s), then slide and measure the distance you move before stopping. First use
kinematics: v2 = v02 + 2a(x – x0) gives the acceleration. Now use Newton’s second law: Fnet,x = µkmg = ma gives µk =
a/g.
Exercises:
1.
m =  V,
Or

mAl
(2.7 g/cm3)(10 cm3)
=
= 1.4.
mwater
(1.0 g/cm3)(20 cm3)
mAl = 1.4mwater .
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 4
2.
Force and Motion
63
In the horizontal direction, Fnet = +30.0 N + (35.0 N) = 5.0 N.
Fnet = ma,

a=
Fnet
5.0 N
=
= 1.0 m/s2 .
m
5.0 kg
The negative sign indicates that the acceleration is along the x-axis.
3.
In the horizontal direction, Fnet = +30.0 N  (35.0 N) cos 40 = 3.19 N.
So the acceleration is
4.
Fnet = ma,
5.
(a)

m=
3.19 N
= 0.64 m/s2 .
5.0 kg
Fnet
4.0 N
=
= 0.40 kg .
a
10 m/s2
2.0 kg:
F = w = mg = (2.0 kg)(9.80 m/s2) = 20 N .
6.0 kg:
F = (6.0 kg)(9.80 m/s2) = 59 N .
(b) The acceleration is the same for both, that is g = 9.80 m/s2 .
6.
(a) The answer is (3) the upward force is the same in both situations . In both situations, there is no acceleration in
the vertical direction so the net force in the vertical direction is zero so the normal force is equal to the weight.
(b) Fnet = N  w = 0,
7.

N = w = 0.50 lb .
F 1 = (5.5 N)[(cos 30) ^x  (sin 30) ŷ ] = (4.76 N) ^x + (2.75 N) ŷ ,
F 2 = (3.5 N)[(cos 37) ^x + (sin 37) ŷ ] = (2.80 N) ^x + (2.11 N) ŷ .
 F = F 1 + F 2 + F 3 = 0,
8.

F 3 = ( F 1 + F 2) = (–7.6 N) x̂ + (0.64 N) ŷ .
(a) The answer is (3) either (1) or (2) is possible because “at rest” and “constant velocity” both have zero
acceleration.
(b) F 1 = (3.6 N)[(cos 74) ^x  (sin 74) ŷ ] = (0.99 N) ^x + (3.46 N) ŷ .
F 2 = (3.6 N)[(cos 34) ^x + (sin 34) ŷ ] = (2.98 N) ^x + (2.01 N) ŷ .
If a = 0, then  F = 0 from Newton’s first law.
 F = F 1 + F 2 = (1.99 N) ^x + (1.45 N) ŷ  0.
So the answer is yes , there must be a third force to make  F = 0.
 F = F 1 + F 2 + F 3 = 0,
F3 =

F 3 = ( F 1 + F 2) = (1.99 N) ^x + (1.45 N) ŷ .
(1.99 N)2 + (1.45 N)2 = 2.5 N .
 1.45 N 
 = tan1 
 = 36 above the +x axis .
 1.99 N 
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64
9.
College Physics Seventh Edition: Instructor Solutions Manual
(a) The answer is (3) the tension is the same in both situations . In both, there is no acceleration in the vertical
direction, so the net force in that direction is zero, so the tension is equal to the weight.
(b) Fnet = T  w = 0,
10.

N = w = 25 lb .
(a) F 1 = (5.0 N)[(cos 37) ^x + (sin 37) ŷ ] = (3.99 N) ^x + (3.01 N) ŷ ,
F 2 = (2.5 N) ^x,
F1 = 5.0 N
y
F 3 = (3.5 N)[(cos 45) ^x + (sin 45) ŷ ] = (2.47 N) ^x + (2.47 N) ^y,
37
F 4 = (1.5 N) ŷ .
x
45
 F = F 1 + F 2 + F 3 + F 4 = (4.02 N) ^x + (0.96 N) ŷ  0. No .
(b)  F = F 1 + F 2 + F 3 + F 4 + F 5 = 0,
11.
F3 = 3.5 N
so
F 5 = ( F 1 + F 2 + F 3 + F 4) = (4.0 N) ^x + (0.96 N) ŷ ,
or
F5 =
(4.0 N)2 + (0.96 N)2 = 4.1 N ,
F2 = 2.0 N
F4 = 1.5 N
 0.96 N 
 = tan1 
 = 13 above the x axis .
  4.0 N 
(a) For the net force to be zero, the unknown (3 rd force) must be in
F3
(2) the second quadrant . The x-component of the unknown force must be
F2x
F1
due west and the y-component of the unknown force must be due north.
(b) From  F = F 1 + F 2 + F 3 = 0,
F2y
F 3 = 0  ( F 1 + F 2) =  F 1  F 2
= [(150 lb)  (30.0 lb)] ^x  (40.0 lb) ŷ = (180 N) ^x + (40.0 N) ŷ ,
or
F3 =
(180 N)2 + (40.0 N)2 = 184 N ,

a=
40.0 N 
 = tan1 
 = 12.5 above the x-axis .
 180 N 
Fnet
6.0 N
2
=
= 4.0 m/s in the direction of the net force .
m
1.5 kg
12.
Fnet = ma,
13.
(a) Using Newton’s second law and taking the ratio of the accelerations, we get
a2 F / m2 m1
1.5 kg


, which gives a2  (3.0 m/s2 )
 1.8 m/s2
a1 F / m1 m2
2.5 kg
(b) F = ma = (1.5 kg)(3.0 m/s2) = 4.5 N
14.
Fnet = ma = (2.0  105 kg)(3.5 m/s2) = 7.0  105 N .
15.
(a) The mass is still (3) 6.0 kg because mass is a measure of inertia, and it does not change.
(b) w = mgMoon =
(6.0 kg)(9.80 m/s2)
= 9.8 N .
6
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 4
16.
Force and Motion
65
First use kinematics to find the acceleration.
v2 = v02 + 2a(x – x0)

(100 m/s)2 = 0 + 2a(0.90 m)
a = 5560 m/s2
Newton’s second law gives
F = ma = (0.050 kg)(5560 m/s2) = 280 N
17.
(a) This label is correct (1) on the Earth . 1 lb is equivalent to 454 g, or 454 g has a weight of 1 lb.
(b) The gravity on the Moon is only 1/6 of that on the Earth. So it takes 6 times the mass to have the same weight on
the Moon. Since 1 lb on the Earth is 454 g, 2.0 lb on the Moon would require
2.0  6  (454 g) = 5.4 kg. So the label should be 5.4 kg (2.0 lb).
18.
(a) The total upward force = 18(400 N) = 7.20  103 N = weight of car.
w
7.20  10 N
= 735 kg .
=
2
g
9.80 m/s
3
m=
(b) w = (7.20  103 N) 
19.
(a) Fnet = ma,
So

1 lb
= 1.62  103 lb .
4.45 N
a=
Fnet
.
m
a2
F2
m
1
1
 1 =

= 1/4.
=
a1
F1
m2
2
2
Thus the answer is (4) one fourth as great .
(b)
20.
a2
2
2

= 4.
=
a1
1
1
So
a2 = 4a1 = 4(1.0 m/s2) = 4.0 m/s2 .
Newton’s second law gives
F
F
1
1



 3.0 m/s2
1 1
1
1
m1  m 2 F F



a1 a2 a1 a2
4.0 m/s2 12 m/s2
Note that we do not need to know the force F. An alternative way is to first find the masses using F.
a
m1 = F/a1 = (50 N)/(4.0 m/s2) = 12.5 kg; m2 = F/a2 = (50 N)/(12 m/s2) = 4.17 kg
a = F/(m1 + m2) = (50 N)/(12.5 kg + 4.17 kg) = 3.0m/s2
21.
Apply Newton’s second law:
Fscale – mg = ma
900 N – 800 N = [(800 N)/(9.80 m/s2)]a
22.

The resistive force is opposite the forward force. a =
a = 1.23 m/s2
Fnet
15.0 N  8.0 N
2
=
= 7.0m/s .
m
1.0 kg
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
66
College Physics Seventh Edition: Instructor Solutions Manual
Fnet
300 N  120 N
=
= 2.40m/s2 .
m
75 kg
23.
The friction force is opposite the horizontal force. a =
24.
(a) There is only (2) one force acting on the crate, the normal force due to the floor.
(b) In the vertical direction: Fy = N = ma = (75.0 kg)(9.80 m/s2) = 735 N .
It would experience the same force if it were at rest on the surface of the Earth.
25.
(a) Since the object slides upward, there is no acceleration in the horizontal direction, so Fx = 0.
Fx = N  F cos 60 = 0,

N = F cos 60 = (60 N) cos 60 = 30 N .
(b) Fy = F sin 60  w = = F sin 60  mg = may,
ay =
so
F sin 60
(60 N) sin 60
g=
 9.80 m/s2 = 4.60 m/s2 .
m
10.0 kg
The negative sign means that the object is slowing down while it is sliding upward.
26.
First find the acceleration from kinematics.
vo = 90 km/h = 25 m/s,
So
a=
v = 0,
t = 5.5 s.
v  vo
0  25 m/s
=
= 4.55 m/s2.
t
5.5 s
Fnet = ma = (60 kg)(4.55 m/s2) =  2.7  102 N .
The negative sign indicates that the force is opposite the motion or the velocity.
27.
(a) The ball tends to remain at rest as the RV accelerates forward, so the ball hangs backward .
(b) Call T the tension in the string and  the angle the string makes with the vertical. Balancing vertical forces gives
T cos  = mg. Applying Newton’s second law horizontally gives T sin  = ma. Combining these two equations gives
a = g tan  = (9.80 m/s2) tan 3.0° = 0.51 m/s 2
28.
(a) Applying Newton’s second law to the two blocks as a system gives
10 N = (5.0 kg)a

a = 2.0 m/s2
(b) Applying Newton’s second law to block B gives
FA = mBa = (2.0 kg)(2.00 m/s2) = 4.0 N to the right
(c) By Newton’s third law, the force is 4.0 N to the left
29.
Applying Newton’s second law along the x- and y-axes gives
Fx = max = -ma cos 30° = –(2.0 kg)(1.5 m/s2) cos 30° = –2.6 N
Fy = may = ma sin 30° = (2.0 kg)(1.5 m/s2) sin 30° = 1.5 N
F = –2.6 N xˆ + 1.5 N yˆ
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 4
30.
(a)
Force and Motion
67
Fp
mg
(b) First find the man’s acceleration using kinematics:
x = ½ at2
7.0 m = ½ a (2.5 s)2

a = 2.24 m/s2, downward
Now apply Newton’s second law to the man:
mg – Fp = ma
Fp = m(g – a) = (90 kg)(9.80 m/s2 – 2.24 m/s2) = 680 N
(c) Using Newton’s second law to find the person’s acceleration, we have
mg – Fp = ma
(75 kg)(9.80 m/s2) – (0.80)(680 N) = (75 kg)a
a = 2.54 m/s2
Using kinematics to find the time gives
x = ½ at2
7.0 m = ½ (2.54 m/s2)t2
31.

t = 2.3 s
(a) There are (2) two forces acting on the book, the gravitational force (weight, w) and the normal force from the
surface, N.
(b) The reaction to w is an upward force on the Earth by the book, and the reaction force to N is a downward force
on the horizontal surface by the book.
32.
The force on the female by the male is F = ma = (45 kg)(2.0 m/s2) = 90 N.
The force on the male by the female is also 90 N according to Newton’s third law.
So
33.
amale =
90 N
2
= 1.4 m/s  opposite to her’s .
65 kg
(a) (3) The force the blocks exerted forward on him is the reaction to the sprinter’s pushing force.
(b) Fnet = ma,
34.
F
200 N
net
 a = m = 65.0 kg = 3.08 m/s2 .
(a) For Jane: T = ma = (50 kg)(0.92 m/s2) = 46 N.
The force on John is also 46 N (Newton’s 3rd law).
aJohn =
46 N
= 0.77 m/s2 toward Jane .
60 kg
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
68
College Physics Seventh Edition: Instructor Solutions Manual
(b) xJohn = 12 aJohn t 2,
So
Also
xJane
aJane
0.92
=
=
= 1.2.
xJohn
aJohn
0.767
xJohn + xJane = 10 m,
Therefore
35.
xJane = 12 aJane t 2.
or
xJohn + 1.2 xJohn = 10 m.
xJohn = 4.5 m from John’s original position .
(a) The answer is (4) the pull of the rope on the girl . The pull of the rope on the girl is the reaction to the pull of
the girl on the rope.
(b) Fy = T  w = T  mg = ma,
so
36.
T = m(a + g) = (25.0 kg)(0.75 m/s2 + 9.80 m/s2) = 264 N .
(a) The scale reading is equal to the normal force on the person.
Choose upward as positive.
Fy = N  w = ma = 0,

N = w = mg = (75.0 kg)(9.80 m/s2) = 735 N .
N
(b) a is still zero. So N = 735 N .
(c) Fy = N  w = ma,
a
N = w + ma = mg + ma = m(g + a) = (75.0 kg)(9.80 m/s2 + 2.00 m/s2) = 885 N .
37.
w
Fy = w  N = ma,
From Exercise 36:
N = w  ma = mg  ma = m(g  a) = (75.0 kg)(9.80 m/s2  2.00 m/s2) = 585 N .
38.
(a) The normal force is (1) less than the weight of the object.
The component of the weight perpendicular to the inclined plane is w cos .
In the y-direction, Fy = N  w cos  = may.
Since the object does not accelerate in the y-direction, ay = 0.
So
N = w cos  < w (for any   0).
(b) w = mg = (10 kg)(9.80 m/s2) = 98 N . N = (98 N) cos 30 = 85 N .
39.
(a) The measured weight is the normal force on the object, N. Choose downward as positive.
Fy = w  N = ma,

N
N = w – ma.
a
So N can take any value, depending on the value of a.
For example, if a = 0, then N = w; if a > 0, N < w; if a < 0, then N > w.
w
The answer is (4) all of the preceding .
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 4
69
Force and Motion
wN
(500 kg)(9.80 m/s )  4000 N
=
= 1.8 m/s2.
m
500 kg
2
(b) a =
So it is a = 1.8 m/s2 downward .
40.
y
(a) Fx = F cos 30 = (25 N) cos 30 = 21.65 N = max,
so
ax =
N
F
21.65 N
2
= 0.72m/s .
30 kg
30
x
(b) Fy = N + F sin 30  w = may = 0,
so
41.
N = w  F sin 30 = (30 kg)(9.80 m/s2)  (25 N) sin 30 = 2.8  102 N .
w = mg
(a) Fx = F cos = (30 N) cos 37 = 23.96 N = max,
so
y
N
23.96 N
2
ax =
= 0.96m / s .
25 kg
F

(b) Fy = N  F sin  w = may = 0,
x
N = w + F sin 37 = (25 kg)(9.80 m/s2) + (30 N) sin 37 = 2.6  102 N .
w = mg
42.
(a) Since the car and the truck accelerate together, they have the same acceleration, a.
For the truck:
Fx = 3200 N  T = (3000 kg)a.
(1)
For the car:
Fx = T = (1500 kg)a.
(2)
car
Equation (1) + Equation (2) results:
T
T
3200 N
1500 kg
3200 N = (3000 kg + 1500 kg)a.
So
truck
3000 kg
a = 0.711 m/s2 .
Alternate method: Consider the car and truck as a system of mass.
3000 kg + 1500 kg = 4500 kg.
F = 3200 N = (4500 kg)a,
so
a = 0.711 m/s2.
y
N
(b) From Equation (1) in (a), T = (1500 kg)(0.711 m/s2) = 1067 N .
F
43.
The acceleration is zero. The direction of the force F must be along and up the incline.
Fx = mg sin  F = max = 0,
so
x
30
F = mg sin  = (25.0 kg)(9.80 m/s2 ) sin 30 = 123 N up the incline .
mg
y
N
44.
(a) There are (2) two forces on the skier, the weight and the normal force.
(b) The x-component of the weight is the side opposite to the angle  shown,
so sine is used.
Fx = mg sin = max,

x
mg
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70
College Physics Seventh Edition: Instructor Solutions Manual
so
ax = g sin = (9.80 m/s2) sin 37 = 5.9 m/s2 .
(c) From v2 = vo2 + 2ax, v =
45.
(5.0 m/s)2 + 2(5.9 m/s2)(35 m) = 21 m/s .
From Exercise 44(b), ax = g sin = (9.80 m/s2) sin 30 = 4.9 m/s2 down the slope.
If up the slope is chosen as positive, then a = 4.9 m/s2. Taking xo = 0.
v = v + 2a(x  xo),
2
46.
2
o
x=
 0 2 –  25m / s 2

2 –4.9m / s2

= 64 m .
The free-body diagrams of the three objects are
For m1: F1 = T1  m1 g = m1 a,
(1)
For m3: F3 = T2  T1 = m3 a,
a
m3
(2)
For m2: F2 = m2 g  T2 = m2 a,
T1
T2
T1
(3)
T2
a
Equation (1) + Equation (2) + Equation (3) gives
(m2  m1)g = (m1 + m2 + m3)a,
47.
so
m1
(m2  m1)g
a=
.
m1 + m2 + m3
m2
a
m1 g
(a) a =
(0.50 kg  0.25 kg)(9.80 m/s2)
= 2.5 m/s2 to right .
0.25 kg + 0.50 kg + 0.25 kg
(b) a =
(0.15 kg  0.35 kg)(9.80 m/s2)
= 2.0 m/s2. So it is 2.0 m/s2 to left .
0.35 kg + 0.15 kg + 0.50 kg
m2 g
(a) The angle the rope makes with the horizontal, , depends on both the tree
separation and the sag.
Fy = 2T sin   mg = 0,

T=
mg
.
2 sin 
T


T

T
So the tension depends on (3) both the tree separation and sag .
 0.20 m 
(b)  = tan1 
 = 2.29.
 5.0 m 
48.
T=
mg
(5.0 kg)(9.80 m/s2)
= 6.1  102 N .
2 sin 2.29
(b)
(a) There are two rings. Fy = T + T  w = 0.
So
T=
w
mg
(50 kg)(9.80 m/s2)
=
=
= 2.7  102 N .
2
2
2
T

(b) In the vertical direction: Fy = T sin + T sin  mg = 0.
So
49.
T=
mg
(55 kg)(9.80 m/s2)
=
= 3.8  102 N .
2 sin
2 sin 45
mg
We first draw the free-body diagram of the lead weight. T is the tension in the string. The acceleration in the y
direction is zero, ay = 0.
In the horizontal direction:
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Chapter 4
Fx = T sin  = max
Force and Motion
71
(1)
In the vertical direction:
Fy = T cos   mg = may = 0,
T
(2)

mg
T=
.
cos 
Equation (2) gives
Substituting the above result for T into Equation (1),
T sin 
mg sin 
ax =
=
= g tan  = (9.80 m/s2) tan 15.0 = 2.63 m/s2 .
m
m cos 
50.
y
x
mg
Isolate point A (or B). The vertical component of the tension T in the topmost cord supports half of the 10-kg weight.
Therefore
T sin 45° = (5 kg)(9.80 m/s2) = 69.3 N
The horizontal forces at A balance, giving
TAB = T cos 45° = (69.3 N) cos 45° = 49 N
51.
The vertical component of the tension T in the lower cords attached at A and B supports the 10-kg weight, so

2T cos 30° = mg = (10 kg)(9.80 m/s2)
T = 57 N
The tension in the vertical cord attached to the weight supports the full weight, so its tension is mg = 98 N . From
Exercise 50, the other two tensions are 49 N and 69 N .
52.
First find the acceleration from kinematics.
Given:
vo = 25.0 m/s,
v2 = vo2 + 2a(x  xo) ,
v = 0,

a=
x = 100 m.
Find:
a.
(Take xo = 0.)
v2  vo2
02  (25.0 m/s)2
=
= 3.125 m/s2.
2x
2(100 m)
F = ma = (2.00  105 kg)(3.125 m/s2) =  6.25  105 N .
53.
First find the acceleration from kinematics.
Given:
x = 0.750 m,
v2 = vo2 + 2a(x  xo)
Then
54.
vo = 0,

a=
v = 300 m/s.
Find:
a.
(Take xo = 0.)
v2  vo2
(300 m/s)2  (0)2
4
2
=
= 6.00  10 m/s .
2x
2(0.750 m)
F = ma = (0.0250 kg)(6.00  104 m/s2) = 1.50  103 N .
First find the mass of the first block.
m1 =
F
40 N
=
= 16 kg.
a
2.5 m/s2
Then
40 N
a  = 16 kg + 4.0 kg = 2.0 m/s2 .
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72
55.
College Physics Seventh Edition: Instructor Solutions Manual
(a) Since the two objects accelerate together, they have the same acceleration, a. Also the tension on m1 (up) is the
same as the tension on m2 (up).
For m1:
F1 = T  m1 g = m1 a.
(1)
For m2:
F2 = m2 g – T = m2 a.
(2)
Equation (1) + Equation (2) gives
T
T
m1
(m2  m1)g = (m1 + m2)a,
m2
a
(m2  m1)g
(0.80 kg  0.55 kg)(9.80 m/s )
=
= 1.8 m/s2 .
m1 + m2
0.55 kg + 0.80 kg
a
2
so
a=
2
m1 g
m2 g
2
(b) From Equation (1), T = m1 (a + g) = (0.55 kg)(1.8 m/s + 9.80 m/s ) = 6.4 N .
(m2  m1)g
(0.250 kg  0.200 kg)(9.80 m/s2)
=
= 1.1 m/s2 up .
m1 + m2
0.250 kg + 0.200 kg
56.
From the result of Exercise 55, a =
57.
(a) First find the acceleration from dynamics. (Take xo = 0.)
From the result of Exercise 55(a),
a=
x = xo + vo t + 12 at 2 = 12 at 2,

Now
(m2  m1)g
(0.255 kg  0.215 kg)(9.80 m/s2)
=
= 0.834 m/s2.
m1 + m2
0.215 kg + 0.255 kg
t=
2x
2(1.10 m)
=
= 1.62 s .
a
0.834 m/s2
(b) When m2 hits the floor, m1 has traveled up a distance of 1.10 m and has a velocity of v, which is obtained from:
v2 = v2o + 2a(x  xo) = 0 + 2(0.834 m/s2)(1.10 m) = 1.835 m2/s2,
so
v = 1.35 m/s.
Now m1 will move up as a “free fall” with an initial velocity of 1.35 m/s.
The height from that point is v2 = 0 = v2o – 2g(y  yo),
therefore y =
v2o
(1.35 m/s)2
=
= 0.0936 m.
2g
2(9.80 m/s2)
Thus m1 will ascend from the floor by 1.10 m + 0.0936 m = 1.19 m before stopping.
F
58.
(a) According to Newton’s 3rd law, the tensions at the ends of the string are the same
because the string is massless. Since m2 is accelerating up, T > w2. Since F is responsible
for accelerating both m1 and m2 upward, F > T. Therefore the answer is
m1
w1 = m1 g
(1) T > w2 and T < F .
T
(b) For m1: F1 = F  T  m1 g = m1 a
(1)
For m2: F2 = T  m2 g = m2 a
(2)
Equation (1) + Equation (2) gives
T
m2
F  m1 g  m2 g = m1 a + m2 a = (m1 + m2)a.
F = (m1 + m2)(a + g) = (50.0 kg + 100 kg)(1.50 m/s2 + 9.80 m/s2) = 1.70  103 N .
w2 = m2 g
(c) From Equation (2), T = m2 (a + g) = (100 kg)(1.50 m/s2 + 9.80 m/s2) = 1.13  103 N .
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Chapter 4
59.
(a) Chose the +x-axis to the left. Both blocks are accelerating with an
acceleration of 1.50 m/s2.
73
Force and Motion
5.00 kg
10.0 kg
F1
Left block:
Fx1 = N  F1 = m1 a
(1)
Right block:
Fx2 = F2  N = m2 a
(2)
N
N
F2
F2  F1 = (m1 + m2)a.
Equation (1) + Equation (2) gives
F2 = F1 + (m1 + m2)a = 75.0 N + (5.00 kg + 10.0 kg)(1.50 m/s2) = 97.5 N .
(b) From Equation (1) in (a), N = F1 + m1 a = 75.0 N + (5.00 kg) )(1.50 m/s2) = 82.5 N .
60.
The acceleration is zero if both are at rest.
For m1: Fx1 = T  m1 g sin = m1ax = 0,
For m2: Fy2 = m2 g  T = m2 ay = 0,
So


T = m1 g sin.
m2 g = T = m1 g sin.
m2 = m1 sin = (2.0 kg) sin 37 = 1.2 kg .
If both are moving at constant velocity, the answer is the same 1.2 kg
because the acceleration is still zero and the forces must still balance out.
61.
(a) For m1:
Fx1 = T  m1 g sin = m1ax.
(1)
For m2:
Fy2 = m2 g  T = m2 ay.
(2)
ax = ay = a in magnitude.
a=
Equation (1) + Equation (2) gives (m2  m1 sin)g = (m1 + m2)a.
m2  m1 sin
2.50 kg  (3.00 kg) sin 37
 (9.80 m/s2) = 1.24 m/s2
g=
m1 + m2
2.50 kg + 3.00 kg
=.1.2 m/s2 (m1 up and m2 down)
2
2
(b) From Equation (2) in (a), T = m2 (g  a) = (2.50 kg)(9.80 m/s  1.24 m/s ) = 21 N .
62.
(a) The acceleration will (3) increase but more than double . There is constant friction involved in this exercise, so
the net force more than doubles. For a 20-kg object to accelerate at 1.0 m/s2, the net force is (20 kg)(1.0 m/s2) = 20
N. That means there is a 100-N friction force (120 N – 20 N = 100 N). When the force doubles from 120 N to 240
N, the net force is then 240 N – 100 N = 140 N. So the acceleration is (160 N)/(20 kg) = 7.0 m/s 2, more than double.
(b) As calculated in (a), it is 7.0 m/s2 .
y
N
63.
Fy = N  mg = may = 0,

F
N = mg.
fsmax = s N = s mg = 0.500(50.0 kg)(9.80 m/s2) = 245 N.
x
f
mg
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74
College Physics Seventh Edition: Instructor Solutions Manual
fk = k mg = 0.400(50.0 kg)(9.80 m/s2) = 196 N.
(a) Since 250 N > fsmax = 245 N, the object will accelerate or start moving. Once it starts moving, we have kinetic
friction.
So the net force is Fx = 250 N  196 N = 54 N.
Therefore a =
Fx
54 N
=
= 1.1 m/s2 .
m
50.0 kg
(b) Since 235 N < fsmax = 245 N, the object will not move. So the acceleration is 0 .
64.
(a) Fy = N  mg = may = 0,

N = mg.
y
Fx = F  fs = max = 0 (on the verge of moving),
so
fsmax = s N = F,
or
F
N = mg
275 N
F
s = mg = (35.0 kg)(9.80 m/s2) = 0.802 .
x
f
(b) Similarly,
195 N
k = (35.0 kg)(9.80 m/s2) = 0.569 .
mg
F = fsmax = s N = s mg = 0.69(40 kg)(9.80 m/s2) = 2.7  102 N .
65.
See the diagram in Exercise 64.
66.
(a) First find the minimum angle of inclination required for an object to slide.
Fy = N  mg cos = may = 0,
On the verge of sliding, ax = 0.
Therefore
N = mg cos.

So
Fx = mg sin  fs = 0.
mg sin  = s N = s (mg cos).
Therefore
s =
sin
= tan ,
cos
Thus the minimum angle is  = tan1 0.65 = 33 > 20 . So it will not move .
67.
(a) First find acceleration from dynamics and use k from Table 4.1.
Fx = fk =  k mg = ma,
So
vo = 90 km/h = 25 m/s,
v = v + 2a(x  xo),
2
o

v = 0.
(b) a = 0.60(9.80 m/s ) = 5.88 m/s .
Use the result of Exercise 67(a),
a=
N = mg
(Take xo = 0.)
v2  v2o
0  (25 m/s)2
x=
=
= 38 m .
2a
2(8.33 m/s2)
2
68.
y
a =  k g = 0.85(9.80 m/s2) = 8.33 m/s2,
and
2
a
k =  g .

0  (5.0 m/s)2
= 0.625 m/s2,
2(20 m)
2
So
0  (25 m/s)2
x=
= 53 m .
2(5.88 m/s2)
x
fk
mg
a
k =  g .
so
0.625 m/s2
k =  9.80 m/s2 = 0.064 .
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Chapter 4
69.
75
Force and Motion
First get the maximum acceleration that is possible without causing the crate to slide.

µsmg = mamax
amax = µg
Now get the minimum distance to stop with this acceleration.
v2 = v02 + 2a(x – x0)

0 = (13.9 m/s)2 + 2(–0.30)(9.80 m/s2)(x – x0)
70.
x – x0 = 33 m
First use kinematics to find the acceleration.
v2 = v02 + 2a(x – x0)

0 = (2.5 m/s)2 + 2a(1.5 m)
a = 2.08 m/s2
Now use Newton’s second law to find the coefficient of kinetic friction.
µkmg = ma
µk = a/g = (2.08 m/s2)/(9.80 m/s2) = 0.21
71.
Use Newton’s second law to find the acceleration.
µkmg = ma

a = µkg = (0.60)(9.80 m/s2) = 5.88 m/s2
Now use kinematics to get the distance.
v2 = v02 + 2a(x – x0)
0 = (3.0 m/s)2 + 2(–5.88 m/s2)(x – x0)

x – x0 = 0.77 m
72.
(a) The answer is (2) pulling at the same angle . When it is pushed, the normal force equals the sum of the weight
and the vertical component of the pushing force.
That is, N = w + F sin.
N
When it is pulled, N = w  F sin.
F
Since the friction force is directly proportional to the normal

N
F

force, it is easier to pull than to push.
(b) At steady speed, the acceleration is zero.
Pull:
fk
fk
w
Fx = F cos   fk = ma = 0.
w
F cos   fk = F cos   k N = F cos   k (w  F sin) = F cos   k (mg  F sin) = 0.
Solving,
Push:
Solving,
73.
f=
k mg
(0.750)(50.0 kg)(9.80 m/s2)
=
= 296 N .
cos  + k sin 
cos 30 + (0.750) sin 30
= F cos   fk = F cos   k N = F cos   k (w + F sin) = F cos   k (mg + F sin) = 0.
f=
k mg
(0.750)(50.0 kg)(9.80 m/s2)
=
= 748 N .
cos   k sin 
cos 30  (0.750) sin 30
Yes , the coefficient of kinetic friction can be found.
Fy = N  mg cos = may = 0,

N = mg cos.
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76
College Physics Seventh Edition: Instructor Solutions Manual
For constant velocity,
Or
74.
ax = 0.
Fx = mg sin  fk = 0.
So
mg sin = k N = k (mg cos).
Therefore
sin
= tan .
cos
(a) Use the result of Exercise 66(a). When the object starts sliding down the plane,
s =
sin
= tan ,
cos
so
 = tan1 0.58 = 30 (s from Table 4.1).
(b) Replace s with k (or see Exercise 73),
75.
k =
(a) Use the result from Exercise 66(a).
 = tan10.40 = 22 .
s = tan, independent of m, size, etc. So it is still 30 .
(b) s = tan 30 = 0.58 .
76.
(a) When the system is on the verge of (but not quite) moving, the acceleration is still zero.
For m1: Fy = N  m1 g = 0,

N = m1 g.
Fx = T  fsmax = T  s N = T  s m1 g = 0,
For m2: F = m2 g  T = 0,
So

y
so
N
T = s m1 g.
T
T = m2 g = s m1 g.
m1
T
x
m2 = s m1 = 0.60(10 kg) = 6.0 kg .
fsmax
m2
(b) Once the system starts moving, the kinetic friction force is used.
For m1: Fx = T  fk = T  k N = T  k m1 g = m1 a.
(1)
For m2: F = m2 g  T = m2 a.
(2)
(1) + (2) gives
So
m1g
m2g
m2 g  k m1 g = (m1 + m2)a.
y
(m2  k m1)g
[6.0 kg  0.40(10 kg)](9.80 m/s2)
2
a=
=
= 1.2m / s .
m1 + m2
10 kg + 6.0 kg
N
x
F
77.
(a) If there were no friction, the minimum pushing force is equal to
F = mg sin  = (35.0 kg)(9.80 m/s2) sin 20 = 117 N.
fk

mg
Since the person has to use a 150 N force, there is friction.
So the answer is no , the incline is not frictionless.
(b) fk = 150 N  117 N = 33 N .
y
78.
(a) See the free-body diagram.
(b) In the horizontal direction, the object does not have any acceleration.
Fx = N  F cos  = 0,
so
N
x
60
N = F cos  = (60 N) cos 60  = 30 N .
fk
(c) The velocity is constant, so the vertical acceleration is also zero.
F
mg
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 4
77
Force and Motion
Fy = F sin   fk  mg = 0,
so
79.
fk = = F sin   mg = (60 N) sin 60  (3.0 kg)(9.80 m/s2) = 23 N
(a) First assume that m1 has the tendency to move up the incline.
For m1: Fy1 = N1  m1 g cos = 0,
N1 = m1 g cos.

Fx1 = T  m1 g sin  fsmax = T  m1 g sin  s N1
= T  m1 g sin  s m1 g cos = 0.
So
T = m1 g sin + s m1 g cos = m1 g(sin + s cos).
For m2: Fx2 = m2 g  T = 0,
m2 g = T = m1 g(sin + s cos).

m2 = m1 (sin + s cos) = (2.0 kg)(sin 37 + 0.30 cos 37) = 1.7 kg.
Therefore
Next assume that m1 has the tendency to move down the incline.
The static friction force will now be pointing up the incline, so the term s m1
g cos becomes negative. Repeating the calculation: m2 = m1 (sin  s
cos) = (2.0 kg)(sin 37  0.30 cos 37) = 0.72 kg.
Thus m2 can be anywhere between 0.72 kg and 1.7 kg .
(b) When both are moving at constant velocity, the acceleration is still zero.
However, s is replaced by k.
Assume that m1 has the tendency to move up the incline.
m2 = m1 (sin + k cos) = (2.0 kg)(sin 37 + 0.20 cos 37) = 1.5 kg.
Assume that m1 has the tendency to move down the incline.
m2 = m1 (sin  k cos) = (2.0 kg)(sin 37  0.20 cos 37) = 0.88 kg.
Therefore m2 can be anywhere between 0.88 kg and 1.5 kg .
80.
For m1: F1 = T1  m1 g = 0,
(1)
For m3: F3 = T2  T1  fsmax = 0,
(2)
For m2: F2 = m2 g  T2 = 0,
(3)
Equation (1) + Equation (2) + Equation (3) gives
a
m3
T1
T2
T1
T2
a
fs
m1
m2
(m2  m1)g  fs = 0,
a
or
(m2  m1)g  s N3 = (m2  m1)g  s m3 g = 0,
so
s =
m1 g
m2 g
m2  m1
0.50 kg  0.25 kg
=
= 0.33 .
m3
0.75 kg
Alternate method (more conceptual).
Since a = 0,
T1 = m1 g
and
So for m3,
m2 g  m1 g  fs = 0,
Therefore
s =
T2 = m2 g.
or
m2 g  m1 g  s m3 g = 0.
m2  m1
.
m3
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78
81.
College Physics Seventh Edition: Instructor Solutions Manual
Use the diagram in Exercise 80 but with fk replacing fs.
For m1: F1 = T1  m1 g = m1 a,
(1)
For m3: F3 = T2  T1  fk = m3 a,
(2), where fk = k N3 = k m3 g.
For m2: F2 = m2 g  T2 = m2 a,
(3)
(m2  m1  k m3)g = (m1 + m2 + m3)a,
Equation (1) + Equation (2) + Equation (3) gives
so
a=
(m2  m1  k m3)g
.
m1 + m2 + m3
(a) For constant speed,
(b) a =
82.
a = 0.
So
m3 =
m2  m1
k
=
0.250 kg  0.150 kg
= 0.179 kg .
0.560
0.250 kg  0.150 kg  (0.560)(0.100 kg)
(9.80 m/s2) = 0.862 m/s2 .
0.150 kg + 0.250 kg + 0.100 kg
(a) See the free-body diagrams. There are a few action-reaction pairs. The normal force on A by B (NA) has a
reaction force on B by A. The static friction force on A by B (fAs) has a
M
M
NA
reaction force on B by A. Since A does not touch the walkway, (1) and (3)
are not right. The force in (2) is in the vertical direction, so it is not
M
fAs
M
correct. Thus the answer is
M
M
(4) the force of static friction acting on A due to the top surface of B . A
NB
NA
fAs
M
mA g
fBs
is moved to the right by the static friction force on A by B. This force
M
exists on the top surface of B.
(b) For A:
FxA = fAs = mA a = (2.00 kg)(2.50 m/s2) = 5.00 N .
For B:
FxB = fBs  fAs = mB a,
So
83.
M
mB g
N
fBs = fAs + mB a = 5.00 N + (5.00 kg)(2.50 m/s2) = 17.5 N .
M
F
(a) Consider the two-object system as one object with a mass of
M
M = 5.00 kg + 10.0 kg = 15.0 kg.
fBk
See the free-body diagram of this combined system.
M
Mg
In the vertical direction: Fy = N  Mg = may = 0,
so
N = Mg = (15.0 kg)(9.80 m/s2) = 147 N.
M
In the horizontal direction: Fx = F  fBk = F  k N = Max,
so
F  k N
200 N  (0.800)(147 N)
ax =
=
= 5.5 m/s2 .
M
15.0 kg
(b) Now we draw the free-body diagrams of each object.
M
M
NA
M
F
M
NB
NA
M
fAs
fAs
M
mA g
There are a few action-reaction pairs on the two objects.
M
fBk
The normal force on A by B (NA) has a reaction force on B
by A. The static friction force on A by B (fAs) has a reaction
M
mB g
force on B by A.
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Chapter 4
Force and Motion
79
For A: FxA = F  fAs = mA ax,
fAs = F  mA ax = 200 N  (5.00 kg)(5.49 m/s2) = 173 N .
84.
(a) As the boat rises, the cables will go flatter (the angle  decreases). The sum of the vertical components of the
tensions in the cable must support the weight of the boat. The weight of the boat remains the same. As  decreases,
the tension must (1) increase .
(b) Fy = 2T sin   mg = ma = 0.
At 45, T =
mg
(500 kg)(9.80 m/s2)
=
= 3.46  103 N .
2 sin 
2 sin 45
At 30, T =
(500 kg)(9.80 m/s2)
= 4.90  103 N .
2 sin 30
As we see, the decrease in angle increases the tension.
85.
(a) Convert units: 65.2 mi/h = 29.14 m/s
Use kinematics to find the acceleration.
v2 = v02 + 2a(x – x0)
0 = (29.14 m/s)2 + 2a(51.5 m)

a = 8.24 m/s2
(b) Newton’s second law gives
µkmg = ma
µk = a/g = (8.24 m/s2)/(9.80 m/s2) = 0.824
(c) Kinematics gives
v2 = v02 + 2a(x – x0)
0 = v02 + 2(–8.24 m/s2)(57.3 m)
86.

v0 = 30.7 m/s = 68.8 mi/h
(a) It exerts more force when it is dropped and is therefore more likely to dent the floor because in that case the
force of the floor on the ball is greater than the ball’s weight in order to accelerate it upward and stop it.
(b) Balancing forces gives
Ffloor – mg = 0
Ffloor = mg = w
(c) Get the speed of the ball just as it reaches the floor, using kinematics.
v2 = v02 + 2a(x – x0) = 0 + 2(9.80 m/s2)(2.00 m)

v = 6.26 m/s
Now get the average acceleration while the ball is slowing down.
v2 = v02 + 2a(x – x0)
0 = (6.26 m/s)2 + 2a(0.0115 m)

a = 1700 m/s2
Apply Newton’s second law while it is in contact with the floor.
Ffloor – mg = ma
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80
College Physics Seventh Edition: Instructor Solutions Manual
Ffloor = m(a + g) = (w/g)(a + g) = w(a/g + 1) = w[(1700 m/s2)/(9.80 m/s2) + 1]
Ffloor = 170w
87.
(a) Convert units: 122 mi/h = 54.5 m/s, 47 mi/h = 21.0 m/s
First use kinematics to find the acceleration.
v = v0 + at
–21.0 m/s = 54.5 m/s + a(0.025 s)

a = 3018 m/s2
Now apply Newton’s second law:
Fpuck = ma = (0.170 kg)(3018 m/s2) = 513 N
The force is in the same direction as the rebounded puck .
(b) By Newton’s third law, the goalie and the puck feel the same force. Using Newton’s second law gives
agoalie = (mpuck/mgoalie)apuck = [(0.170 kg)/(85 kg)](3018 m/s2) = 6.04 m/s2
v = v0 + at = 0 + (6.04 m/s2)(0.025 s) = 0.151 m/s
88.
(a)
x
y
fk
N
mg

Since the block is moving downward on the incline but slowing down (because it comes to rest), its acceleration
must be up the incline; call this the +x-direction, and call the +y-direction perpendicular to this pointing upward
from the surface of the incline. (It would also be possible to call the +x-axis pointing down the plane.)
(b) Forces along the y-axis balance.
F
y
 N  mg cos   0
Acceleration is along the x-axis.
F
x
 f k  mg sin   ma
(c) Using kinematics to find the acceleration gives
v2 = v02 + 2a(x – x0)
0 = (1.60 m/s)2 + 2a(1.10 m)

a = 1.164 m/s2 up the incline
fk = µkN = µkmg cos 
Putting this into the result from Newton’s second law gives
µkmg cos –- mg sin  = ma
Solving for µk gives
1.164 m/s2
 sin 30
2
a / g  sin 
k 
 9.80 m/s
 0.714
cos 
cos 30
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.