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Bab 6 Tenaga dan perpindahan tenaga
SOALAN-SOALAN
Q7.5
As a simple pendulum swings back and forth, the
forces acting on the suspended object are the
gravitational force, the tension in the supporting
cord, and air resistance. (a) Which of these forces,
if any, does no work on the pendulum? (b) Which
of these forces does negative woi k at all times
during its motion? (c) Describe the work done by
the gravitational force while the pendulum is
swinging.
Solution
(a) Tension
(b) Air resistance
(c) Positive in increasing velocity on the
downswing.Negative in decreasing velocity
on the upswing.
Q7.10
Can kinetic energy be negative? Explain.
Solution
Kinetic energy is always positive. Mass and
squared speed are both positive. A moving object
can always do positive work in striking another
object and causing it to move along the same
direction of motion.
Q7.12
52.10iie bullet has twice the mass of a second
bullet. If both are fired so that they have the same
speed, which has more kinetic energy? What is the
ratio of the kinetic energies of the two bullets?
Solution
Kinetic energy is proportional to mass. The first
bullet has twice as much kinetic energy.
Q7.14 (a) If the speed of a particle is doubled,
what happens to its kinetic energy? (b) What can
be said about the speed of a particle if the net
work done on it is zero?
Solution
(a) Kinetic energy is proportional to
squared speed. Doubling the speed makes an
object's kinetic energy four times larger.
(b) If the total work on an object is zero in some
process, its speed must be the same at the
final point as it was at the initial point.
MASALAH-MASALAH
1.
A block of mass 2.50 kg is pushed 2.20 m
along a frictionless horizontal table by a constant
16.0-N force directed 25.0 below the horizontal.
Determine the work done on the block by (a) the
applied force, (b) the normal force exerted by the
table, and (c) the gravitational force (d) Determine
the total work done on the block.
Solution
W  F r cos  16.0 N  2.20 m  cos 25.0
 31.9 J
(b), (c) The normal force and the weight are both
at 90° to the displacement in any time interval.
Both do 0 work.
(d)
W
 31.9 J 0  0  31.9 J
3.
Batman, whose mass is 80.0 kg, is dangling
on the free end of a 12.0-m rope, the other end of
which is fixed to a tree limb above. He is able to
get the rope in motion as only Batman knows how,
eventually getting it to swing enough that he can
reach a ledge when the rope makes a 60.0° angle
with the vertical. How much work was done by
the gravitational force on Batman in this
maneuver?
Solution
Method One.
Let  represent the instantaneous angle the rope
makes with the vertical as it is swinging up from
i  0 to  f  60 . In an incremental bit of motion
from angle  to   d , the definition of radian
measure implies that r   12 m  d . The angle 
between the incremental displacement and the
force of gravity is   90   . Then
cos  cos 90      sin  .
The work done by the gravitational force on
Batman is
1
(b)
 F  r 

 F r 
  cos 1 
 cos 1
FIG. P7.3
f
  60
i
 0
W   F cos dr 
 mg 12 m 

 6.00
16
2
  2.00 
2
 3.00
2
 1.00 
2
 36.9
mg   sin  12 m  d
60
 sin  d
0
  80 kg   9.8 m s 2  12 m   cos   0
60
  784 N 12 m   cos 60  1
13.
A particle is subject to a force Fx that varies
with position as in Figure P7.13. Find the work
done by the force on the particle as it moves (a)
from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x =
10.0 m, and (c) from x = 10.0 m to x = 15.0 m. (d)
What is the total work done by the force over the
distance x = 0 to x = 15.0 m?
 4.70 103 J
Method Two.
The force of gravity on Batman is


m g   80 kg 9.8 m s2  784 N down. Only his
vertical displacement contributes to the work
gravity does. His original y-coordinate below the
tree limb is –12 m. His final y-coordinate is
 12 m  cos60  6 m . His change in elevation is
6 m   12 m   6 m . The work done by gravity
is
W  F r cos   784 N  6 m  cos180
 4.70 kJ
7.
A force


F  6 ˆi – 2 ˆj N
.
acts on a
particle that undergoes a displacement


r  3 ˆi  ˆj m .
Find (a) the work done
by the force on the particle and (b) the angle
between F and  r.
Figure P7.13 Problems 13 and 28
Solution
W   Fxdx and W equals the area under the ForceDisplacement curve
(a)
For the region 0  x  5.00 m ,
 3.00 N  5.00 m 
W 
 7.50 J
2
(b)
For the region 5.00  x  10.0 ,
W   3.00 N  5.00 m   15.0 J
(c)
Solution
(a)
W  F  r  Fx x  Fy y
  6.00  3.00  N  m   2.00 1.00  N  m
 16.0 J
(d)
For the region 10.0  x  15.0 ,
 3.00 N  5.00 m 
W 
 7.50 J
2
For the region 0  x  15.0
W   7.50  7.50  15.0 J 30.0 J
19. If it takes 4.00 J of work to stretch a
Hooke's-law spring 10.0 cm from its unstressed
length, determine the extra work required to
2

stretch it an additional 10.0 cm.
Solution
1
2
4.00 J k 0.100 m 
2
 k  800 N m and to stretch the spring to 0.200
m requires
1
2
W   800 0.200  4.00 J 12.0 J
2
27. A 2 100-kg pile driver is used to drive a
steel I-beam into the ground. The pile driver falls
5.00 m before coming into contact with the top of
the beam, and it drives the beam 12.0 cm farther
into the ground before coming to rest. Using
energy considerations, calculate the average force
the beam exerts on the pile driver while the pile
driver is brought to rest.
Solution
Consider the work done on the pile driver from
the time it starts from rest until it comes to rest at
the end of the fall. Let d  5.00 m represent the
distance over which the driver falls freely, and
h  0.12 m the distance it moves the piling.
W
 K W gravity  W beam 
1 2 1 2
m vf  m vi
2
2
so
 m g  h d cos0   F  d cos180  0 0 .
Thus,
F

 mg  h  d 
d
 2 100 kg   9.80 m s 2   5.12 m 
force, (e) the change in kinetic energy of the box,
and (f) the final speed of the box.
Solution
 Fy  m ay :
n 392 N  0
n  392 N
fk  kn   0.300 392 N   118 N
(a)
W F  Frcos  130 5.00 cos0  650 J
(b)
Eint  fkx  118 5.00  588 J
(c)
W n  nrcos   392 5.00 cos90  0
(d) W g  m grcos   392 5.00 cos 90  0
(e)
(f)
K  K f  K i   W other  Eint
1 2
m vf  0  650 J 588 J 0  0  62.0 J
2
2K f
2 62.0 J
vf 

 1.76 m s
m
40.0 kg
33.
A crate of mass 10.0 kg is pulled up a
rough incline with an initial speed of 1.50 m/s.
The
pulling force is 100 N parallel to the incline, which
makes an angle of 20.0° with the horizontal. The
coefficient of kinetic friction is 0.400, and the crate
is pulled 5.00 m. (a) How much work is done by
the gravitational force on the crate? (b) Determine
the increase in internal energy of the crate-incline
system due to friction. (c) How much work is done
by the 100-N force on the crate? (d) What is the
change in kinetic energy of the crate? (e) What is
the speed of the crate after being pulled 5.00 m?
0.120 m
 8.78 10 N
5
The force on the pile driver is upw ard .
31.
A 40.0-kg box initially at rest is pushed
5.00 m along a rough, horizontal floor with a
constant applied horizontal force of 130 N. If the
coefficient of friction between box and floor is
0.300, find (a) the work done by the applied force,
(b) the increase in internal energy in the box-floor
system due to friction, (c) the work done by the
normal force, (d) the work done by the
gravitational
Solution
(a) W g  m g cos 90.0   


W g  10.0 kg 9.80 m s2  5.00 m  cos110  168 J
(b) fk  kn  km g cos
Eint  fk  km g cos
Eint   5.00 m  0.400 10.0 9.80 cos20.0  184 J
(c) W F  F  100 5.00  500 J
(d)
K   Wother  Eint  WF  Wg  Eint
 148 J
3
(e)
vf 
1 2 1 2
m vf  m vi
2
2
2 K 
2
148
2
 vi2 
  1.50  5.65 m s
m
10.0
K 
Solution
(a)
fuel needed


35. A sled of mass m is given a kick on a frozen
pond. The kick imparts to it an initial speed of
2.00 m/s. The coefficient of kinetic friction
between sled and ice is 0.100. Use energy
considerations to find the distance the sled moves
before it stops.
k  0.100
1
K i  fkx  W other  K f : m vi2  fkx  0
2
1 2
m vi  km gx
2
 2.00 m s  2.04 m
vi2

2kg 2 0.100 9.80
2
37.
A 700-N Marine in basic training climbs a
10.0-m vertical rope at a constant speed in 8.00 s.
What is his power output?
Solution
Pow er 
usefulenergy pergallon
1
2
 900 kg 24.6 m s
 0.150 1.34  108
(b)
73.8
(c)
power

2
1
2 m vf  0
eff.  energy contentoffuel
2

J gal
 1.35  102 gal
8
 1 gal  55.0 mi   1.00 h   1.34  10 J 


  0.150 



 38.0 mi  1.00 h   3 600 s   1 gal

 8.08 kW
Solution
vi  2.00 m s
x 
2
2
1
1
2 m vf  2 m vi
W
t
m gh  700 N  10.0 m 
P

 875 W
t
8.00 s
49.
A 4.00-kg particle moves along the x axis.
Its position varies with time according to x = t +
2.0t3, where x is in meters and t is in seconds. Find
(a) the kinetic energy at any time t, (b) the
acceleration of the particle and the force acting on
it at time t, (c) the power being delivered to the
particle at time t, and (d) the work done on the
particle in the interval t = 0 to t = 2.00 s.
Solution
(a)
x  t 2.00t3
Therefore,
dx
v
 1 6.00t2
dt
1
1
K  m v2   4.00 1 6.00t2
2
2

(b)
45.
A compact car of mass 900 kg has an
overall motor efficiency of 15.0%. (That is, 15% of
the energy supplied by the fuel is delivered to the
wheels of the car.) (a) If burning one gallon of
gasoline supplies 1.34  108 J of energy, find the
amount of gasoline used in accelerating the car
from rest to 55.0 mi/h. Here you may ignore the
effects of air resistance and rolling friction. (b)
How many such accelerations will one gallon
provide? (c) The mileage claimed for the car is 38.0
mi/gal at 55 mi/h. What power is delivered to the
wheels (to overcome frictional effects) when the
car is driven at this speed?
a
dv

dt
12.0t
F  m a  4.0012.0t 


2

 2.00  24.0t  72.0t  J
2.00

 48.0t
N
  48.0t 288t  W
0
2.00
P dt
4
m s2
(c) P  Fv   48.0t 1 6.00t2 
(d) W 
2

0
3
 48.0t 288t  dt
3
1250 J
61. A 200-g block is pressed against a spring of
force constant 1.40 kN/m until the block
compresses the spring 10.0 cm. The spring rests at
the bottom of a ramp inclined at 60.0° to the
horizontal. Using energy considerations,
determine how far up the incline the block moves
4
before it stops (a) if there is no friction between the
block and the ramp and (b) if the coefficient of
kinetic friction is 0.400.
Solution
(a)
W  K :
W sW g  0
1 2
kxi  0  mg x cos  90  60   0
2
1
2
1.40 103 N m   0.100    0.200  9.80  sin 60.0  x
2
0

1
1.20 N cm  5.00 cm  0.050 0 m 
2


  0.100 kg  9.80 m s 2  0.050 0 m  sin10.0
1
 0.100 kg  v 2 0.150 J  8.51103 J   0.050 0 kg  v 2
2
0.141
v
 1.68 m s
0.050 0


x  4.12 m
(b)
W
 K  Eint :
W s  W g  Eint  0
1 2
kxi  mg x cos150   k mg cos 60x  0
2
1
2
1.40  103 N m   0.100    0.200  9.80  sin 60.0  x
2
  0.200  9.80  0.400  cos 60.0  x  0


x  3.35 m
63. The ball launcher in a pinball machine has a
spring that has a force constant of 1.20 N/cm (Fig.
P7.63). The surface on which the ball moves is
inclined 10.0° with respect to the horizontal. If the
spring is initially compressed 5.00 cm, find the
launching speed of a 100-g ball when the plunger
is released. Friction and the mass of the plunger
are negligible.
Figure P7.63
Solution
Ki W s  W
g
Kf
1 2 1 2 1 2
1
m vi  kxi  kx f  m gx cos  m v2f
2
2
2
2
1 2
1 2
0  kxi  0  m gxicos100  m vf
2
2
5