K
EY
Name:________________________
Energy In-Class Problems (1 of 2)
1. A ball having a weight of 1.5 N is dropped from a height of 4 meters. (neglect air friction)
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What net force is acting on the ball after being dropped? ___________
1.5 N
What is the potential energy just before being dropped? _________
6J
(reference ground)
What is the kinetic energy just before hitting the ground? ___________
6J
In what direction is the kinetic energy just before hitting the ground? ______
N/A
What mechanical energy is lost just before it hits? _____________
None
2. In a similar situation to problem 1 above, it is found that the average frictional force on
the1.5 N falling ball is 0.2 N.
 What net force is acting on the ball after being dropped? ___________
1.3N
 What is the kinetic energy just before hitting the ground? ___________
5.2 J
 What mechanical energy is lost? _____________
0.8 J
3. In problem 1 above, from what height must the ball be dropped to have twice the kinetic
energy as from 4 meters? ____________.
Explain.
8m
PEgrav ~ height
4. In problem 1 above, from what height must the ball be dropped to have twice the velocity
as from the 4 meters? _____________
Explain.
16 m
KE ~ v2, so doubling the velocity will quadruple the kinetic energy.
5. The speed for an object to escape the Earth’s gravity is called the escape velocity. This
speed is calculated from energy considerations. An object must have enough kinetic energy to
overcome the gravitational potential energy. For a 30 kg satellite in a circular orbit around the
Earth at a distance of 7 x 106 m from the center, we find the orbital speed to be 7.56 x 103 m/s.
From that location calculate the gravitational PE, the kinetic energy, and the escape velocity.
Gravitational PE = _______________J
- 1.7 x 109
Orbital KE = __________________J
8.5 x 108
Escape velocity = _______________
m/s
1.1 x 104
PEgrav = -Gm1m2/r
= - (6.67x10-8N∙m2/kg2)(5.98x1024kg)(30kg)/(7x106m)
= - 1.7 x 109 J
KE = ½mv2 = ½(30kg)(7.56 x 103m/s)2
= 8.5 x 108 J
NOTE:
For a satellite in orbit, KE is ½ of
the PE (but positive). For escape
velocity, KE must be equal and
opposite of PE, or double the
orbital KE. Since KE ~ v2, to
double the KE you must increase
the velocity by √2, or about 1.414.
6. The pulley system shown to the right has 6 pulleys and thus it requires
only 100 N to lift a 600 N weight. Using energy considerations and assuming
that the unit is 100% efficient (no friction), how far must the rope be pulled to
raise the weight a distance of 20cm?
1.2
Distance = __________________
m.
Wo = W i
F2d2 = F1d1
(600 N)(0.20 m) = (100 N)( d1 )
d1 = 1.2 m
7. A simple machine often studied in IPC classes is the lever. One use of a lever is to
multiply the force of a person. A physics student uses a lever with a length of 3m on one
side of the fulcrum and a length of 1m on the other side. The student finds that it requires a
force of 110 N pushing through a distance of 30cm to raise a 300 N weight a distance of 10
cm. Find the efficiency of this simple machine.
e = ______________
90.9%
e = (100%)W o / W i
= (100%)(300N)(0.10m) / (110N)(0.30m)
= 90.9%
8. Which ones of the following forces are conservative and non-conservative. Place an N or
C in the space to designate the force as non-conservative or conservative.
C Gravity
_____
C Bow (for arrow)
_____
C Spring
_____
N Friction
_____
C Electrostatic
_____
N Air friction
_____
9. If a woman lifts a 500N weight a distance of 1.5 meters up in 0.5 seconds, what is her
power output?
1500 W
P = ______
Power = Work / time
= (500N)(1.5m) / (0.5s)
= 1500 W