Download ENERGY AND ITS CONSERVATION (K) KEY

ENERGY AND ITS CONSERVATION (K) KEY
1.
Natasha weighs 530 N. What is her kinetic
energy as she swims at a constant speed,
covering a distance of 72 m in 1.0 min?

 0.346 m/s
3.
x
v
t
 72 m  1 min 



 1.0 min  60 s 
Fg  mg
Fg
m
g
1
KE  mv2
2
1 Fg v2

2 g
1 (530 N)(1.2 m/s)2

2
9.8 N/kg
 39 J
PE  PEf  PEi
 mghf  mghi  mgh
 (78.0 kg)(9.8 N/kg)(1.20 m)
 917 J
b. What is the change in the kinetic energy
of Zeke and the mat when they slide to
1.20 m below the crest?
KEi  PEi  KEf  PEf
KE  KEf  KEi  PEi  PEf
A 6.00-g block, initially at rest, is pulled
to the right along a frictionless horizontal
surface by a constant horizontal force of
1.20102 N for a distance of 3.00 cm.
 (PEf  PEi)  PE
917 J
c. How fast are Zeke and the mat moving
when they are 1.20 m below the crest?
KE  KEf  KEi
1
KEi  mvi2  0
2
KEf  KE
1
mvf2  KE
2
2 KE
vf 
m
a. What is the work done by the force?
W  Fx
 (1.20102 N)(3.00102 m)
 3.60104 J
b. What is the change in the kinetic energy
of the block?
KE  W
 3.60104 J
c. What is the speed of the block after the
force is removed?
KE  KEf  KEi
1
KEi  mvi2  0
2
KEf  KE
1
mvf2  KE
2
2 KE
vf 
m
Zeke slides down a snow hill on a rubber
mat. Zeke’s mass is 76.0 kg and the mass of
the mat is 2.00 kg. Zeke starts from rest at
the crest of the hill. Frictional forces may be
disregarded.
a. What is the change in the gravitational
potential energy of Zeke and the mat when they
slide to 1.20 m below the crest?
m  mZeke  mmat  78.0 kg
 1.2 m/s
2.
2(3.60  104 J)
6.00  103 kg

2(917 J)
78.0 kg
 4.85 m/s
4.
As shown below, a 450-kg roller-coaster car starts
from rest at point A at a height of 47 m, rolls down the
track, reaching point B at a speed of 25 m/s, and then
rolls up a second hill where it reaches a height of 23
m before coming to rest (at point C). What are the
gravitational potential energy and kinetic energy of
the car when it is at points A, B, and C?
5.
An arrow with a mass of 320 g is shot straight up into
the air and reaches a height of 150 m before stopping
and falling back to the ground.
a. How much work is done by gravity as
the arrow reaches its maximum height?
Wg  mgh
 (0.32 kg)(9.8 N/kg)(150 m)
 470 J
b. If the arrow’s kinetic energy is zero
at its maximum height, calculate the
initial velocity of this arrow as it is
shot from the ground.
KEi  Wg  KEf, and KEf  0 J
6.
Rohit can consistently throw a 0.200-kg ball at a
speed of 12.0 m/s. On one such throw, Rohit throws
the ball straight upward and it passes the top of a
flagpole when it is 6.00 m above the ball’s initial
position.
a. What is the ball’s gravitational potential energy
when it passes the top of the flagpole? (Assume
the ball’s initial gravitational energy is 0 J.)
PEf  mghflagpole
 (0.200 kg)(9.8 N/kg)(6.00 m)
 11.8 J
b. What is the ball’s kinetic energy as it passes the
top of the flagpole?
KEi  PEi  KEf  PEf
PEi  0
KEf  KEi  PEf
1
 mvi2  PEf
2
1
 (0.200 kg)(12.0 m/s)2  11.8 J
2
 2.6 J
Point A:
PEA  2.1105 J
KEA  0
Point B:
PEB  0
KEB  1.4105 J
Point C:
PEC  1.0105 J
KEC  0
c. What is the ball’s velocity as it first passes the top
of the flagpole?
1
KEf  mvf2
2
2KEf
vf 
m

2(2.6 J)
0.200 kg
 5.1 m/s, upward
d. What is the maximum height to which the ball will
rise?
KEi  PEi  KEf  PEf
PEi  0; KEf  0 since vf  0 at maximum
height
PEf  KEi
1
mghmax  mvi2
2
vi2
hmax 
2g
(12.0 m/s)2

2(9.8 N/kg)
 7.35 m
7.
In a hardware store, paint cans, which weigh 46.0 N
each, are transported from storage to the back of the
paint department by placing them on a ramp that is
inclined at an angle of 24.0 above the horizontal. The
cans slide down the ramp at a constant speed of 3.40
m/s and then slide onto a table made of the same
material as the ramp. How far does each can slide on
the table’s horizontal surface before coming to rest?
On the ramp, using a coordinate system
in which the positive x-axis is oriented
down the ramp:
y-direction:
Fnet, y  may  0
FN  Fgy  0
FN  Fgy  mg cos 
Ff  kFN  k mg cos 
x-direction:
Fnet, x  max  0
Fgx  mg sin 
Fgx  Ff  0
mg sin   k mg cos   0
k 
sin 
 tan 
cos 
On the table:
y-direction:
FN  mg
Ff  kFN  k mg
W  KE
8.
Meena releases her 10.5-kg toboggan from rest on a
hill. The toboggan glides down the frictionless slope
of the hill, and at the bottom of the slope it moves
along a rough
horizontal surface, which exerts a constant
frictional force on the toboggan. When the toboggan
is released from a height of 15.0 m, it travels 6.0 m
along the horizontal surface before coming to rest.
From what height should the toboggan be released so
that it stops after traveling 10.0 m on the horizontal
surface?
W   Ff x   PEi  mgh
mgh
Ff
x is proportional to h, so to increase x from 6.00 m
to 10.0 m, h must increase from
15.0 m to
x
 10.0 m 
(15.0 m)  

 6.00 m 
h  25.0 m