Download Work/Energy Review KEY

Lyzinski Physics
Work and Energy Review Packet
SOLUTIONS!!!
1. How much work is required to slow a 45 kg snowboarder down from 30 m/s to 15
m/s? How much power was exerted (in Watts) if the snowboarder was slowed down
by this amount in 2.5 seconds.
1
1
1
1
2
2
mv2  mv1  (45kg)(15 ms ) 2  (45kg)(30 ms ) 2  15,187.5 J
2
2
2
2
P  W / t  (15,187.5 J ) /( 2.5 sec)  6,075W
W  KE 
2. A 1200 kg car is traveling at 6.67 m/s (about 15 mph) in a school zone. If its brakes
can supply a force of 6674 Newtons, determine the stopping distance.
W  KE 
W  Fd
1
1
1
1
2
2
mv2  mv1  (1200kg)(0 ms ) 2  (1200kg)(6.67 ms ) 2  26,693.34 J
2
2
2
2
W  26,693.34 J
 d 

 4m
F
 6674 N
3. If the same car in (from the previous question above) with the same brakes was
traveling at 13.3 m/s (twice as fast), determine the stopping distance.
W  KE 
W  Fd
1
1
1
1
2
2
mv2  mv1  (1200kg)(0 ms ) 2  (1200kg)(13.3 ms ) 2  106,134 J
2
2
2
2
W  106,134 J
 d 

 15.9m
F
 6674 N
4. A horse pulls a 20 kg carriage by pulling with a horizontal force of 500 Newtons over
a distance of 200 meters. How much work is done?
W  Fd  (500N )(200m)  100,000J
5. A motorcycle engine supplies a force of 1000 Newtons to a 100 kg bike. Determine
the final velocity of the motorcycle if it starts at 15 m/s for 20 meters.
a.
b.
c.
d.
Determine the final velocity of the bike.
What is the work done by the engine?
What force is required to make the bike come to a stop in 20 meters?
What is the change in kinetic energy of the motorcycle while coming to rest?
W  KE 
1
1
2
2
mv2  mv1
2
2

Fd 
1
1
2
2
mv2  mv1
2
2
(1000 N )(20m) 
1
1
2
(100kg)v 2  * (100kg)(15 ms ) 2
2
2
v 2  25 ms
W  Fd  (1000 N )(20m)  20,000 J
1
mv2  12 mv1
(100)(0) 2  12 (100)(25) 2
W  Fd  mv2  mv1

F
2
 1,562.5 N
d
20d
2
2
KE  12 mv2  12 mv1  12 (100)(0) 2  12 (100)(25) 2  31,250 J
1
2
2
1
2
1
2
2
2
2
6. What variables affect the period of a mass-spring system that is oscillating up and
down? What 2 things (surprisingly) don’t affect the period?
k & m affect the mass-spring period. Surprisingly, g & x do not!!!
7. A pendulum has a period of 2 seconds on the Earth. What is its length?
2
l
T 
T  2
 l  g
  .993m
g
 2 
8. A 2 kg mass is hung on a 15 cm long spring. It lowers and comes to a rest (stays at
rest) when the spring has reached a length of 32 cm. (a) find the spring constant and
(b) determine the period of the same spring is a 4 kg mass is placed on it and allowed
to oscillate (remove the 2 kg mass).
F  kx
T  2

k
(2kg)(9.8 sm2 )
F mg
N


 115.3
x
x
(.32m  .15m)
m
m
4kg
 2
 1.17 sec
N
k
115.3
m
9. A 40 kg boy riding on a 60 kg go-kart has a kinetic energy of 3125 Joules. Determine
the velocity of the go-kart and boy.
KE 
1 2
mv
2

3125 J 
1
(40kg  60kg)v 2
2

v  7.9
m
s
10.A bike experiences a force as shown by the graph below. The bike and its rider have
a mass of 110 kg and an initial velocity of 6 m/s. Determine the final velocity of the
bike and rider at 120 meters.
Work  area under curve  5500 J  (2100 J )  3400 J
1
1
2
2
W  KE  mv2  mv1
2
2
1
1
2
3400 J  (110kg)(v2 )  (110kg)(6 ms ) 2
2
2
m
v2  9.9
s
F
100
(N)
100
40
120
70
- 60
11.A 20 kg box is lifted up a set of steps to a vertical height of 12 meters. How much
work is done? How much gravitational potential energy did the box gain?
Work  PE g  mgh2  mgh1  (20kg)(9.8 sm2 )(12m)  2,352 J
PE g gained  2,352 J
12.A ball is thrown up at 6 m/s from the top of a 20 meter cliff. It passes by the edge of
the cliff on the way down and hits the ground. With what speed did the ball hit the
ground? (Do not use the equations of motion to solve this problem. Do it another
way, more suitable to this chapter )
Work  KE  PE g
1
1
2
2
mv2  mv1  mgh2  mgh1
2
2
1 2 1 2
0  v2  v1  0  gh1
2
2
1 2 1 m 2
0  v2  (6 s )  0  (9.8 sm2 )(20m)
2
2
W 

v2  20.69
m
[downward]
s
d (m)
13.A 20 gram pellet is placed up against a spring with force constant 6000 N/m. The
spring is initially compressed 2 cm. The pellet is fired from rest from the compressed
spring and an initial height of 2 meters from the ground. At what speed will the pellet
hit the ground?
Work  KE  PE g  PE s
1
1
1
1 2
2
2
2
mv2  mv1  mgh2  mgh1  kx2  kx1
2
2
2
2
1
1
2
2
0  mv2  0  0  mgh1  0  kx1
2
2
1
1
2
0  (.02kg)v 2  (.02kg)(9.8 sm2 )(2m)  (6000 Nm )(.02m) 2
2
2
W

v 2  12.62
m
s
14.A ball is rolling along level ground at 10 m/s. It begins to climb a hill with an angle
of 30o. How far up the incline will the ball roll?
Work  KE  PE g

1
2
0  0  mv1  mgh2  0
2
sin  
h2
d

d 
1
1
2
2
mv2  mv1  mgh2  mgh1
2
2
1
1 2
2
mv1  mgh2

v1  gh2
2
2
W 


h2 
2
(10 ms ) 2
v1

 5.102m
2 g 2(9.8 sm2 )
h2
5.102m

 10.2m
sin  sin(30)
15.A 6 kg block is pressed against a spring with force constant 600 N/m at the top of a
hill 4 meters high. The spring is initially compressed 40 cm. (a) What is the speed of
the block at the bottom of the hill after sliding down the frictionless surface? (b) How
far will the block slide if it encounters a rough level surface with coefficient of
friction 0.4 at the bottom of the hill?
Work  KE  PE g  PE S
1
1
1
1 2
2
2
2
mv2  mv1  mgh2  mgh1  kx2  kx1
2
2
2
2
1
1
2
2
0  mv 2  0  0  mgh1 0  kx1
2
2
1
1
2
0  (6kg)v 2  (6kg)(9.8 sm2 )(4m)  (600 Nm )(.4m) 2
2
2
W

v 2  9.716
m
s
At the bottom of the hill, friction slows down the block.
F  Ff  FN  mg
1
1
2
2
mv2  mv1
2
2
2
2
m 2
(9.716 s )
mv1
v
d 
 1 
 12.04m
2 mg 2 g 2(.4)(9.8 sm2 )
Work  KE

Fd 

1
2
mgd   mv1 2
16.A 40 kg little girl slides down a 10 meter slide from rest at an angle of 40o. She
reaches the bottom with a speed of 4 m/s. What was the force of friction along the
incline?
Work  KE  PE g
1
1
1
1 2
2
2
2
mv2  mv1  mgh2  mgh1  kx2  kx1
2
2
2
2
1
2
Fd  mv2  0  0  mgh1
2
1
1
2
mv2  mgh1
(40kg)(4 ms ) 2  (40kg)(9.8 sm2 )(10 sin 40 o )
F 2
 2
 220 N
d
10
Fd 
220 N [up the inclined plane]
17.A 6 kg block moving to the right at 10 m/s collides head on with a 4 kg block moving
to the left at 8 m/s. After they hit, the 4 kg block is moving to the right at 10 m/s.
Determine the final velocity of the 6 kg block after the collision (using the
conservation of momentum).
m1v1  m2 v 2  m1v1 ' m2 v 2 '
(6kg)(10 ms )  (4kg)(8 ms )  (6kg)v1 '(4kg)(10 ms )

v1 '  2
m
s
18. Was the collision in the previous problem above elastic or not? How much energy
was lost if any?
If the collision was elastic, then energy would have been conserved (ie. E Before  E After )
1
1
1
1
2
2
m1 v1  m 2 v 2  m1 (v1 ' ) 2  m 2 (v 2 ' ) 2
2
2
2
2
1
1
1
1
(6kg)(10 ms ) 2  (4kg)(8 ms ) 2  (6kg)(2 ms ) 2  (4kg)(10 ms ) 2
2
2
2
2
300 J  128 J  12 J  200 J (?????)
428 J  212 J !!!!
NO!!! Energy was not conserved, and therefore the collision was NOTelastic.
428J - 212J  216Jof energy were lost!!!!!
19.A 40 gram bullet is fired into a 4 kg block with a speed of 350 m/s. The 4 kg block is
hanging from a 3 meter string. How high will the block rise (vertically) after the
bullet embeds itself into the block (sticks)? What will the angle be between the string
and the vertical at its highest height?
STICK
mv  mv  Mv
m
s

1
1
2
2
W  KE  PE g  mv2  mv1  mgh2  mgh1
2
2
1
2
0  0  mv1  mgh2  0
2
2
(3.465 ms ) 2
v1
1
1 2
2
mv1  mgh2

v1  gh2
h2 

 .6126m
2
2
2g
2(9.8 sm2 )
(.040kg)(350 ms )  (4kg)(0 ms )  (4.040kg)v
3m  .6126m  2.387 m

cos  

2.387 m
3m
v  3.465

 2.387 m 
o
  37.3
 3m 
  cos 1 
20.A 30 gram bullet is fired into a 2 kg block with a speed of 380 m/s. It passes through
the block and exits with a speed of 120 m/s. The block slides along a level
frictionless surface and collides with a spring of force constant of 760 N/m. How
much will the spring compress?
Part #1 : conservation of momentum (b  bullet, B  block)
mb vb  mB v B  mb vb ' mB v B '
m
s

Part #2 : conservation of energy
1
1
1
1 2
2
2
2
W  KE  PE S  mv2  mv1  kx2  kx1
2
2
2
2
1
1
2
2
0  0  mv1  kx2  0
2
2
(.030kg)(380 ms )  (2kg)(0 ms )  (.030kg)(120 ms )  (2kg)(v B ' )
1
1
2
2
mv1  kx2
2
2

mv1  kx2
2
2


v B '  3.9
2
(2kg)(3.9 ms ) 2
mv1
x

 .2m  20cm
k
760 Nm
21.A student performs the spring lab in class. She measures the distance from the floor
to the bottom of the unloaded spring to be 60 cm. She hangs the 1 kg mass on the
spring and allows it to come to a rest. The bottom of the spring is now 23.5 cm above
the floor. (This is the time to determine the “k” value of the spring). She then pulls
the mass down so that the bottom of the spring is 8 cm above the floor. She releases
the mass from rest and it rises to a height of 40 cm. Determine the change in potential
energy of the spring and the change in the gravitational potential energy. What is the
percentage of error if you use PEsp as the theoretical value?
m
F (1kg)(9.8 s 2 )
N
Part #1 : F  kx

k 
 26.85
x (.60  .235)
m

Part #2 : conservation of energy
W  KE  PE g  PE S
KE  0
PE g  mgh2  mgh1  mg (h2  h1 )  (1kg)(9.8 sm2 )(.4  .08)  3.136 J
PE S 


1
1 2 1
1
N
2
2
2
kx2  kx1  k ( x 2  x1 )  (26.85 ) (.6m  .4m) 2  (.6m  .08m) 2 )  3.093 J
2
2
2
2
m
% error 
PE g  (PE S )
PE S
 100 
3.136 J  [(3.093 J )]
 100  .0139  100  1.39% error
 3.093 J
22.A boy pulls on a 100 kg crate with a force of 800 N at an angle of 60o above the
horizontal. A frictional force of 200 N acts on the crate at the same time. If he pulls
the crate through a distance of 15 m, determine (a) the work done by the boy, (b)
determine the work done by the frictional force, (c) determine the net work done on
the box.
Wboy  F pull,|| d  ( F cos  )d  [(800 N ) cos 60 o ](15m)  6,000 N
W friction  F f d  (200 N )(15m)  3,000 J
Wnet  6,000 J  (3000 J )  3000 J
OR
Wnet  Fnet d  (400 N  200 N )(15m)  3000 J
23.A 6 kg cat is hoisted up into the air a distance of 20 meters. How much work is done
on the cat? How much power is exerted (in W, kW, and hp) if the work is done in 3
seconds?
W  PE g  mgh2  mgh1  mg (h2  h1 )  (6kg)(9.8 sm2 )(20m  0m)  1,176 J
P
W 1,176 J

 392W  .392kW  .53hp
t
3 sec
24.A 200 kg jet ski experiences a force as shown on the Force vs. position graph below.
Its initial velocity at x = 0 is 15 m/s.
F (N)
F
(a) How much work is done from 0 to 8 sec?
(b) How fast is the car moving at 8 sec?
(c) How much work is done from 15 to 18 m?
(d) How fast is the ski moving at 18 m?
(e) What is the ski’s change in kinetic energy
from 0 to 15 meters?
1000
d (m)
0
6
8
15
-600
Work  area under curve
1
1
1
1
W0-8  bh  bh  (6m)(1000 N )  (2m)(600 N )  3000 J1000
 600 J  2400 J
2
2
2
2
1
1
1
1
2
2
2
W0-8  mv2  mv1

2400 J  (200kg)v 2  (200kg)(15 ms ) 2

2
2
2
2
0
6
___________________________
-600
W15-18  0 J
F
v 2  15.8
8
m
s
15
___________________________
W0-18  3000 J  600 J  4200 J  0 J  1800 J
1
1
1
1
2
2
2
mv2  mv1

 1800 J  (200kg)v 2  (200kg)(15 ms ) 2
2
2
2
2
___________________________
W0-15  3000 J  600 J  4200 J  1800 J

W0-15  KE 015  1800 J
W0-18 
18

v 2  14.4
m
s
18
25. An 800 kg block is pushed up a 5 m long ramp with an incline angle of 20 o. Find:
a) the work needed to push the box up if the ramp is frictionless.
b) the work needed to push the box up if the coefficient of friction is 0.20.
c) the horsepower needed to do the work in part b above, assuming that the block is
pushed up the ramp in 10 seconds.
W  PE g  mgh2  mgh1  mg (h2  h1 )  (800kg)(9.8 sm2 )(5 sin 20 o  0)  13,407 J

Fnet  F||  F f  F||  FN  mg sin   mg cos 
 (800kg)(9.8 sm2 )(sin 20 o )  (.20)(800kg)(9.8 sm2 )(cos 20 o )  4,154.876 N
W  Fnet d  (4,154.876 N )(5m)  20,774 J

W 20,774 J
P

 2,077W  2.8hp
t
10 sec
26.TOUGH!!! A HORIZONTAL force of “F” (not parallel to the incline, but truly
horizontal) pushes a 200 kg block up a 20o incline that has a coefficient of kinetic
friction of 0.2. Find the force necessary to give the block a speed of 4 m/s after
moving 3 meters (starting from rest). Then, find the work done by the horizontal
force “F”.
Since the force “F” has both a parallel AND a perpendicular component, the normal
force is NOT mgcos, but rather is mgcos + Fperp.
W pushing force  F|| d  F sin(20 o )d

need to find the value of " F".
--------------v1  0
v2  4
d  3
v 22  v12  2ad

4 2  0 2  2a(3)

a  2.667 sm2
--------------Ff   mg cos   F cos 20  (.2)200 * 9.8 * cos 20  F cos 20
--------------Fnet  ma
F||  W||  F f  ma

F sin 20  200 * 9.8 * sin(20)   (.2)200 * 9.8 * cos 20  F cos 20  200(2.667)
F  10,203.157 N
--------------W pushing force  F|| d  F sin 20d  (10,203.157)(sin 20)(3)   10,469.1J