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Bab 4 Hukum-hukum Gerakan
SOALAN-SOALAN
Q5.4
In the motion picture It Happened One
Night (Columbia Pictures, 1934), Clark Gable is
standing inside a stationary bus in front of
Claudette Colbert, who is seated. The bus
suddenly starts moving forward and Clark falls into
Claudette's lap. Why did this happen?
Solution
When the bus starts moving, the mass of Claudette
is accelerated by the force of the back of the seat
on her body. Clark is standing, however, and the
only force on him is the friction between his shoes
and the floor of the bus. Thus, when the bus starts
moving, his feet start accelerating forward, but the
rest of his body experiences almost no accelerating
force (only that due to his being attached to his
accelerating feet!). As a consequence, his body
tends to stay almost at rest, according to Newton’s
first law, relative to the ground. Relative to
Claudette, however, he is moving toward her and
falls into her lap. (Both performers won Academy
Awards.)
Q5.7
A rubber ball is dropped onto the floor. What
force causes the ball to bounce?
Solution
The molecules of the floor resist the ball on impact
and push the ball back, upward. The actual force
acting is due to the forces between molecules that
allow the floor to keep its integrity and to prevent
the ball from passing through. Notice that for a
ball passing through a window, the molecular
forces weren’t strong enough.
the larger upward force that the floor exerts on
them together. Around the top of the weight’s
motion, the scale reads less than average. If the
iron is moving upward, the lifter can declare that
she has thrown it, just by letting go of it for a
moment, so our answer applies also to this case.
Q5.14
Identify the action-reaction pairs in the
following situation: a man takes a step; a snowball
hits a girl in the back; a baseball player catches a
ball; a gust of wind strikes a window
Solution
As a man takes a step, the action is the force his
foot exerts on the Earth; the reaction is the force of
the Earth on his foot. In the second case, the action
is the force exerted on the girl’s back by the
snowball; the reaction is the force exerted on the
snowball by the girl’s back. The third action is the
force of the glove on the ball; the reaction is the
force of the ball on the glove. The fourth action is
the force exerted on the window by the air
molecules; the reaction is the force on the air
molecules exerted by the window. We could in
each case interchange the terms ‘action’ and
‘reaction.’
Masalah-masalah
3.
A 3.00-kg object undergoes an acceleration
given by
Solution
As the barbell goes through the bottom of a cycle,
the lifter exerts an upward force on it, and the
scale reads

the resultant force acting on it and the magnitude
of the resultant force.
Solution
m  3.00 kg


ˆ 5.00ˆ
a  2.00i
j m s2
F  m a
Q5.10
A weightlifter stands on a bathroom
scale. He pumps a barbell up and down. What
happens to the reading on the bathroom scale as
this is done? What if he is strong enough to ctually
throw the barbell upward? How does the reading
on the scale vary now?

2
a  2.00ˆi  5.00ˆj m / s . Find
 6.00iˆ 15.0ˆj N
2
2
 F   6.00  15.0
N  16.2 N
5.
To model a spacecraft, a toy rocket engine
is securely fastened to a large puck, which can
glide with negligible friction over a horizontal
surface, taken as the xy plane. The 4.00-kg puck
has a velocity of
3.00ˆi m / s
at one instant.
Eight seconds later, its velocity is to be
8.00 ˆi  10.0ˆjm / s . Assuming the rocket
11.
Two forces F1 and F2 act on a 5.00-kg
object. If F1 = 20.0 N and F2 = 15.0 N, find the
accelerations in (a) and (b) of Figure P5.11.
engine exerts a constant horizontal force, find (a)
the components of the force and (b) its magnitude.
Solution
ˆ m s,
m  4.00 kg , vi  3.00i


ˆ 10.0ˆ
v8  8.00i
j m s , t 8.00 s
ˆ 10.0ˆ
j
v 5.00i

m s2
t
8.00
F  m a  2.50ˆ
i 5.00ˆ
j N
a

F

 2.50 2   5.00 2 
Figure P5.11
5.59 N
Solution
 F  F1  F2   20.0iˆ 15.0ˆj
(a)
7.
An electron of mass 9.11  10–31 kg has an
initial speed of 3.00  105 m/s. It travels in a
straight line, and its speed increases to 7.00  105
m/s in a distance of 5.00 cm. Assuming its
acceleration is constant, (a) determine the force
exerted on the electron and (b) compare this force
with the weight of the electron, which we
neglected.
F  m a:
 F  m a and
a
v2f  vi2
2x f
(b)

a  5.00 m s2 at  36.9

ˆ 13.0ˆ
F2  7.50i
j N
 F  F1  F2   27.5iˆ 13.0ˆj
 vi2
a

2x f
 F  9.111031

 
2

5
2
 3.00 105 m s 2
 7.00  10 m s
kg
2  0.050 0 m 
 3.64 1018 N .
(b)
or

v2f  vi2  2ax f or
.
2
f
The weight of the electron is



Fg  m g  9.11 1031 kg 9.80 m s2  8.93  1030 N
The accelerating force is
4.08  1011 tim es the w eightofthe electron.

F2x  15.0cos60.0  7.50 N
F2y  15.0sin 60.0  13.0 N
Therefore
v
F m

ˆ 3.00ˆ
a  4.00i
j m s2
Solution
(a)
ˆ 15.0ˆ
20.0i
j 5.00a
N
 
2
 5.50iˆ 2.60ˆj m
N  m a  5.00a
s2  6.08 m s2 at25.3
2
1
Fg
T1
1
2
T2
T3
23.
A 1.00-kg object is observed to accelerate
at 10.0 m/s2 in a direction 30.0° north of east (Fig.
P5.23). The force F2 acting on the object has a
magnitude of 5.00 N and is directed north.
Determine the magnitude and direction of the
force F1 acting on the object.
Figure P5.18 Problems 18 and 19
19.
A bag of cement of weight Fg hangs from
three wires as shown in Figure P5.18. Two of the
wires make angles  1 and  2 with the horizontal.
If the system is in equilibrium, show that the
tension in the left-hand wire is T1 = Fg cos  2
/sin(  1 +  2)
Solution
T3  Fg
(1)
T1 sin1  T2 sin2  Fg
(2)
T1 cos1  T2 cos 2
Eliminate T2 and solve for T1
(3)
T1 
Fg cos 2
 sin 1 cos 2  cos1 sin  2 

Figure P5.23
Fg cos 2
sin 1   2 
T3  Fg  325 N
Solution
 cos25.0 
T1  Fg 
 296 N
 sin 85.0 
Choose a coordinate system with î East and ĵ
North.
 cos1 
 cos60.0 
T2  T1 
 296 N 
 163 N
 cos25.0 
 cos 2 
 F  m a  1.00 kg 10.0 m

s2 at 30.0
 5.00 N  ˆj F1  10.0 N  30.0   5.00 N  ˆj  8.66 N  iˆ
 F1  8.66 N  East
25.
A block is given an initial velocity of 5.00
m/s up a frictionless 20.0° incline (Fig. P5.22). How
far up the incline does the block slide before
coming to rest?
Solution
After it leaves your hand, the block’s speed
changes only because of one component of its
weight:
 Fx  m ax
m g sin 20.0  m a

Figure P5.31
Solution
Forces acting on 2.00 kg block:

v2f  vi2  2a xf  xi .
T  m 1g  m 1a
Taking vf  0 , vi  5.00 m s, and
a   g sin  20.0 gives

Forces acting on 8.00 kg block:

0   5.00  2 9.80 sin  20.0 xf  0
2
Fx  T  m 2a
or
xf 
25.0
 3.73 m
2 9.80 sin  20.0
.
(1)
(a)
(2)
Eliminate T and solve for a:
a
Fx  m 1g
m1  m2
a  0 forFx  m 1g  19.6 N .
(b)
Eliminate a and solve for T:
T
m1
 Fx  m 2g
m1  m2
T  0 forFx  m 2g  78.4 N .
31.
In the system shown in Figure P5.31, a
horizontal force Fx acts on the 8.00-kg object. The
horizontal surface is frictionless. (a) For what
values of Fx does the 2.00-kg object accelerate
upward? (b) For what values of Fx is the tension in
the cord zero? (c) Plot the acceleration of the 8.00kg object versus Fx. Include values of Fx from –100
N to +100 N.
Newton’s second law is:
 Fy  m ay


S   72.0 kg 9.80 m s2   72.0 kg ay
33.
A 72.0-kg man stands on a spring scale in
an elevator. Starting from rest, the elevator
ascends, attaining its maximum speed of 1.20 m/s
in 0.800 s. It travels with this constant speed for
the next 5.00 s. The elevator then undergoes a
uniform acceleration in the negative y direction for
1.50 s and comes to rest. What does the spring
scale register (a) before the elevator starts to
move? (b) during the first 0.800 s? (c) while the
elevator is traveling at constant speed? (d) during
the time it is slowing down?
Solution
First, we will compute the needed accelerations:
(1)
Before it starts to move:
ay  0
(2)
During the first 0.800 s:
ay 
(3)
While moving at constant velocity:
 4
During the last 1.50 s:
v yf  v yi

t
 1.50 m s 2
ay  0
1.20 m s  0
0.800 s
(a)
When ay  0 , S  706 N .
(b)
When ay  1.50 m s2 , S  814 N .
(c)
When ay  0 , S  706 N .
(d)
When ay  0.800 m s2 , S  648 N .
41.
A 3.00-kg block starts from rest at the top
of a 30.0° incline and slides a distance of 2.00 m
down the incline in 1.50 s. Find (a) the magnitude
of the acceleration of the block, (b) the coefficient
of kinetic friction between block and plane, (c) the
friction force acting on the block, and (d) the speed
of the block after it has slid 2.00 m.
Solution
m  3.00 kg ,   30.0 , x  2.00 m , t 1.50 s
1
(a)
x  at2 :
2
1
2
2.00 m  a 1.50 s
2
4.00
a
 1.78 m s2
2
1.50
v yf  v yi
0  1.20 m s

t
1.50 s
 0.800 m s 2
ay 
S  706 N   72.0 kg ay .
 F  n  f m g  m a :
A long x: 0  f m g sin 30.0  m a
f  m  g sin 30.0  a
A long y: n  0  m g cos30.0  0
n  m g cos30.0
f m  g sin 30.0  a

,
n
m g cos30.0
a
k  tan 30.0 
 0.368
gcos30.0
(b)
k 
(c)
f  m  g sin 30.0  a ,
f 3.00 9.80sin30.0  1.78  9.37 N


(d)
v2f  vi2  2a xf  xi
where
xf  xi  2.00 m
v2f  0  2 1.78 2.00  7.11 m
vf  7.11 m
2
2
s2
s2  2.67 m s
45.
Two blocks connected by a rope of
negligible mass are being dragged by a horizontal
force F (Fig. P5.45). Suppose that F = 68.0 N, m1 =
12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic
friction between each block and the surface is
0.100. (a) Draw a free-body diagram for each
block. (b) Determine the tension T and the
magnitude of the acceleration of the system.
over a frictionless pulley (Fig. P5.51), Pat pulls on
the loose end of the rope with such a force that the
spring scale reads 250 N. Pat's true weight is
320 N, and the chair weighs 160 N. (a) Draw freebody diagrams for Pat and the chair considered as
separate systems, and another diagram for Pat and
the chair considered as one system. (b) Show that
the acceleration of the system is upward and find
its magnitude. (c) Find the force Pat exerts on the
chair.
Figure P5.45
Solution
(a) See Figure
T
m1
m2
F
Figure P5.51
n1
m1
f1 =  nk
n2
T
T
f2 =  nk
1
m1g = 118 N
(b)
m2
F
Solution
(a)
2
m2g = 176 N
68.0  T  m 2g  m 2a (Block #2)
T  m 1g  m 1a (Block #1)
Adding,
68.0    m 1  m 2  g   m 1  m 2  a
a
(b)
First consider Pat and the chair as the
system. Note that two ropes support the system,
and T  250 N in each rope. Applying  F  m a
68.0
  g  1.29 m s2
m

m
 1 2
T  m 1a  m 1g  27.2 N
51.
An inventive child named Pat wants to
reach an apple in a tree without climbing the tree.
Sitting in a chair connected to a rope that passes
2T  480  m a , where m 
480
 49.0 kg .
9.80
Solving for a gives
500  480
 0.408 m s2 .
49.0
 F  m a on Pat:
a
(c)
 F  n  T  320  m a, where
320
 32.7 kg
9.80
n  m a 320  T  32.7 0.408  320  250  83.3 N
m
55.
An object of mass M is held in place by an
applied force F and a pulley system as shown in
Figure P5.55. The pulleys are massless and
frictionless. Find (a) the tension in each section of
rope, T1, T2, T3, T4, and T5 and (b) the magnitude of
F. Suggestion: Draw a free-body diagram for
each pulley.
61.
What horizontal force must be applied to
the cart shown in Figure P5.61 in order that the
blocks remain stationary relative to the cart?
Assume all surfaces, wheels, and pulley are
frictionless. (Hint: Note that the force exerted by
the string accelerates m1.)
Solution
(a)
First, we note that F  T1 . Next, we focus
on the mass M and write T5  M g . Next, we focus
on the bottom pulley and write T5  T2  T3 .
Finally, we focus on the top pulley and write
T4  T1  T2  T3 .
Since the pulleys are not starting to rotate and are
frictionless, T1  T3 , and T2  T3 . From this
Mg
.
2
Mg
3M g
Then T1  T2  T3 
, and T4 
, and
2
2
information, we have T5  2T2 , so T2 
T5  M g .
(b)
Since F  T1 , we have F 
Mg
.
2
Figure P5.61 Problems 61 and 63
Solution
F  ma
For m 1 : T  m 1a
For m 2 : T  m 2g  0
Eliminating T, a 
m 2g
. For all 3 blocks:
m1
F   M  m 1  m 2 a
M
 m g
 m 1  m 2  2 
 m1 