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Bab 7 Tenaga keupayaan
SOALAN-SOALAN
Q3. One person drops a ball from the top of a building
while another person at the bottom observes its motion.
Will these two people agree on the value of the
gravitational potential energy of the ball-Earth system?
On the change in potential energy? On the kinetic
energy?
anyone gives a forward push to the ball anywhere along its
path, the demonstrator will have to duck.
MASALAH-MASALAH
5. A bead slides without friction around a loop-the-loop
(Fig. P8.5). The bead is released from a height h = 3.50R. (a)
What is its speed at point A? (b) How large is the normal
force on it if its mass is 5.00 g?
Solution
Both agree on the change in potential energy, and the
kinetic energy. They may disagree on the value of
gravitational potential energy, depending on their
choice of a zero point
Q9 You ride a bicycle. In what sense is your bicycle
solarpowered?
Solution
All the energy is supplied by foodstuffs that gained
their energy from the sun.
Q11 A bowling ball is suspended from the ceiling of a
lecture hall by a strong cord. The ball is drawn away
from its equilibrium position and released from rest at
the tip of the demonstrator's nose as in Figure Q8.1 1. If
the demonstrator remains stationary, explain why she is
not struck by the ball on its return swing. Would this
demonstrator be safe if the ball were given a push from
its starting position at her nose?
Figure P8.5
Solution
1
U i  K i  U f  K f : m gh  0  m g 2R   m v2
2
1 2
g 3.50R   2g R   v
2
v  3.00gR
v2
:
R
F  m
n m g  m
 v2

n  m   g  m
R


v2
R
 3.00gR

 R  g  2.00m g



n  2.00 5.00  103 kg 9.80 m s2

 0.098 0 N dow nw ard
Solution
The total energy of the ball-Earth system is conserved.
Since the system initially has gravitational energy mgh
and no kinetic energy, the ball will again have zero
kinetic energy when it returns to its original position.
Air resistance will cause the ball to come back to a point
slightly below its initial position. On the other hand, if
11. A block of mass 0.250 kg is placed on top of a light
vertical spring of force constant 5 000 N/m and pushed
downward, so that the spring is compressed by 0.100 m.
After the block is released from rest it travels upward and
then leaves the spring. To what maximum height above
the point of release does it rise?
Solution
From conservation of energy for the block-spring-Earth
system, U gt  U si,


 1
or  0.250 kg 9.80 m s2 h    5000 N m
 2
This gives a maximum height h  10.2 m
 0.100 m 2
13. Two objects are connected by a light string passing
over a light frictionless pulley as shown in Figure P8.13.
The object of mass 5.00-kg is released from rest. Using
the principle of conservation of energy, (a) determine
the speed of the 3.00-kg object just as the 5.00-kg object
hits the ground. (b) Find the maximum height to which
the 3.00-kg object rises.
21.
A 4.00-kg particle moves from the origin to
position C, having coordinates x = 5.00 m and y = 5.00 m.
One force on the particle is the gravitational force acting
in the negative y direction (Fig. P8.21). Using Equation
7.3, calculate the work done by the gravitational force in
going from O to C along (a) OAC, (b) OBC, (c) OC. Your
results should all be identical. Why?
Figure P8.21 Problem 21
Solution


Fg  m g   4.00 kg 9.80 m s2  39.2 N
(a)
Work along OAC  work along OA + work along
AC
 Fg  O A  cos90.0  Fg  A C  cos180
  39.2 N  5.00 m    39.2 N  5.00 m  1
 196 J
Figure P8.13 Problems 13 and 14.
.
Solution
Using conservation of energy for the system of the Earth
and the two objects
(a)
 5.00 kg g 4.00 m    3.00 kg g 4.00 m   12  5.00  3.00 v2
v  19.6  4.43 m s
(b)
Now we apply conservation of energy for the
system of the 3.00 kg object and the Earth during the
time interval between the instant when the string goes
slack and the instant at which the 3.00 kg object reaches
its highest position in its free fall.
1
 3.00 v2  m g y  3.00gy
2
y  1.00 m
ym ax  4.00 m  y  5.00 m
(b)
W along OBC  W along OB + W along BC
  39.2 N  5.00 m  cos180   39.2 N  5.00 m  cos90.0
 196 J
(c)
Work along OC  Fg  O C  cos135

 1
  39.2 N  5.00  2 m  
  196 J

2
The results should all be the same, since gravitational
forces are conservative.
23.

A force acting on a particle moving in the xy plane
is given by


2
F  2y iˆ  x ˆj N , where x and y are in
meters. The particle moves from the origin to a final
position having coordinates x = 5.00 m and y = 5.00 m, as in
Figure P8.21. Calculate the work done by F along (a) OAC,
(b) OBC, (c) OC. (d) Is F conservative or nonconservative?
Explain
Solution
(a)
5.00 m

W OA 
0


5.00 m
dxˆ
i 2yˆ
i x2ˆ
j

U i  K i  Em ech  U f  K f : m 2gh  fh 
2ydx
0
f  n  m 1g
and since along this path, y  0 W O A  0
5.00 m

W AC 
0


5.00 m
dyˆ
j 2yˆ
i x2ˆ
j

x dy
0
v2 
W O AC  0  125  125 J
v
(b)
5.00 m

W OB 
0


dyˆ
j 2yˆ
i x2ˆ
j
5.00 m

5.00 m

W BC 
0


dxˆ
i 2yˆ
i x2ˆ
j
2
x dy


2ydx
0
 

W O C   dxˆ
i dyˆ
j  2yˆ
i x2ˆ
j   2ydx  x2dy
5.00 m
Since x  y along OC, W O C 

8.00 kg
W OB  0
W O BC  0  50.0  50.0 J


2 9.80 m s 2 1.50 m  5.00 kg  0.400  3.00 kg 
5.00 m
W BC  50.0 J
since y  5.00 m ,
m1  m2
 3.74 m s
0
since along this path, x  0 ,
(c)
1
 m 1  m 2  v2
2
2 m 2  m 1   hg
m 2gh  m 1gh 
2
For x  5.00 m , W AC  125 J
and
1
1
m 1v2  m 2v2
2
2

0
 2x  x  dx 
2
66.7 J
33.
A 5.00-kg block is set into motion up an inclined
plane with an initial speed of 8.00 m/s (Fig. P8.33). The
block comes to rest after traveling 3.00 m along the plane,
which is inclined at an angle of 30.0° to the horizontal. For
this motion determine (a) the change in the block's kinetic
energy, (b) the change in the potential energy of the blockEarth system, and (c) the friction force exerted on the block
(assumed to be constant). (d) What is the coefficient of
kinetic friction?
(d)
F is nonconservative since the work done is path
dependent.
31.
The coefficient of friction between the 3.00-kg
block and the surface in Figure P8.31 is 0.400. The
system starts from rest. What is the speed of the 5.00-kg
ball when it has fallen 1.50 m?
Figure P8.33
Solution
(a)
(b)
(c)
86.5 J


1
1
m v2f  vi2   m vi2  160 J
2
2
U  m g 3.00 m  sin30.0  73.5 J
K 
The mechanical energy converted due to friction is
86.5 J
 28.8 N
3.00 m
(d)
f  kn  km g cos30.0  28.8 N
28.8 N
k 
 0.679
 5.00 kg 9.80 m s2 cos30.0
f
Figure P8.31
Solution

41.

A single conservative force acts on a 5.00-kg
particle. The equation Fx = (2x + 4) N describes the
force, where x is in meters. As the particle moves along
the x axis from x = 1.00 m to x = 5.00 m, calculate (a) the
work done by this force, (b) the change in the potential
energy of the system, and (c) the kinetic energy of the
particle at
x = 5.00 m if its speed is 3.00 m/s at
x = 1.00 m.
Figure P8.52 Problems 52 and 53.
Solution
(a)
W   Fx dx 
5.00 m

1
5.00 m
 2 x2

 2 x  4  dx    4 x 
 2
1
 25.0  20.0  1.00  4.00  40.0 J
(b) K  U  0 U  K  W  40.0 J
m v12
m v12
(c) K  K f 
, K f  K 
 62.5 J
2
2
43.
The potential energy of a system of two
particles separated by a distance r is given by U(r) = A/r,
where A is a constant. Find the radial force Fr that each
particle exerts on the other.
Solution
A
U  r 
r
U
d  A
A
Fr  
     2 . The positive value


r
dr r
r
indicates a force of repulsion.
53.
The particle (Fig. P8.52) is released from rest at
A, and the surface of the bowl is rough. The speed of
the particle at B is 1.50 m/s. (a) What is its kinetic
energy at B? (b) How much mechanical energy is
transformed into internal energy as the particle moves
from A to B ? (c) Is it possible to determine the
coefficient of friction from these results in any simple
manner? Explain.
Solution
1
1
2
(a) K B  m vB2   0.200 kg1.50 m s  0.225 J
2
2
(b)
Em ech  K  U  K B  K A  U B  U A
 K B  m g  hB  hA 


 0.225 J  0.200 kg 9.80 m s2  0  0.300 m

 0.225 J 0.588 J 0.363 J
(c)
It’s possible to find an effective coefficient of
friction, but not the actual value of  since n and f vary
with position.
57.
A 10.0-kg block is released from point A in Figure
P8.57. The track is frictionless except for the portion
between points B and C , which has a length of 6.00 m. The
block travels down the track, hits a spring of force constant
2 250 N/m, and compresses the spring 0.300 m from its
equilibrium position before coming to rest momentarily.
Determine the coefficient of kinetic friction between the
block and the rough surface between B and C.
Figure P8.57
Solution
Em ech   fx
What speed do you predict for the block at the top of the
track? (c) Does the block actually reach the top of the track,
or does it fall off before reaching the top?
E f  Ei   f dBC
1 2
kx  m gh    m gdBC
2
m gh  12 kx2

 0.328
m gdBC
59.
A 20.0-kg block is connected to a 30.0-kg block
by a string that passes over a light frictionless pulley.
The 30.0-kg block is connected to a spring that has
negligible mass and a force constant of 250 N/m, as
shown in Figure P8.59. The spring is unstretched when
the system is as shown in the figure, and the incline is
frictionless. The 20.0-kg block is pulled 20.0 cm down
the incline (so that the 30.0-kg block is 40.0 cm above
the floor) and released from rest. Find the speed of each
block when the 30.0-kg block is 20.0 cm above the floor
(that is, when the spring is unstretched).
Figure P8.61
Solution
(a)
Initial compression of spring:
1
1
2
2
1 2 1 2 2  450 N m   x  2  0.500 kg12.0 m s
kx  m v
2
2
x  0.400 m
(b)
Speed of block at top of track:
Em ech   fx
 K  U i   K  U  f


Figure P8.59

0   30.0 kg  9.80 m s 2  0.200 m  

1
 250 N m  0.200 m 2
2

1
 50.0 kg  v2   20.0 kg  9.80 m s2  0.200 m  sin 40.0
2
58.8 J 5.00 J  25.0 kg v2  25.2 J

1 2 
1 2

 mghT  2 mvT    mghB  2 mvB    f  R 

 

1
 0.500 kg  9.80 m s2  2.00 m    0.500 kg  vT2
2
1
2
  0.500 kg 12.0 m s     7.00 N  1.00 m 
2
0.250vT2  4.21
v  1.24 m s
61.
A block of mass 0.500 kg is pushed against a
horizontal spring of negligible mass until the spring is
compressed a distance x (Fig. P8.61). The force constant
of the spring is 450 N/m. When it is released, the block
travels along a frictionless, horizontal surface to point
B, the bottom of a vertical circular track of radius R =
1.00 m, and continues to move up the track. The speed
of the block at the bottom of the track is vB = 12.0 m/s,
and the block experiences an average friction force of
7.00 N while sliding up the track. (a) What is x? (b)

 vT  4.10 m s
(c)
Does block fall off at or before top of track? Block
falls if ac  g
vT2  4.10

 16.8 m s2
R
1.00
Therefore ac  g and the block stays on the track .
2
ac 
71.
A ball whirls around in a vertical circle at the end
of a string. If the total energy of the ball-Earth system
remains constant, show that the tension in the string at the
bottom is greater than the tension at the top by six times
the weight of the ball
Solution
Applying Newton’s second law at the bottom (b) and top
(t) of the circle gives
Tb  m g 
m vb2
m vt2
and Tt  m g  
R
R
Adding these gives
Tb  Tt  2m g 

m vb2  vt2

R
Also, energy must be conserved and U  K  0
So,

m vb2  vt2
  0 2m gR   0 and m v  v   4m g
2
b
2
t
2
R
Substituting into the above equation gives Tb  Tt  6m g