# File

PE KE Work Review and Power
PE and KE Worksheet Solutions
Work
• Work is done on an object when a force causes
a displacement of the object.
• Work is done only when components of a force
are parallel to a displacement.
Indicate whether or not the following represent
examples of work.
a. A teacher applies a force to a wall and becomes
exhausted. NO
b. A weightlifter lifts a barbell above her head.
YES
c. A waiter carries a tray full of meals across a
dining room at a constant speed. NO
d. A rolling marble hits a note card and moves it
across a table. YES
A 100 N force is applied to move a 15 kg object a
horizontal distance of 5 meters at constant speed.
Calculate the work done.
W = (100N)(5m) = 500 J
Ben Pumpiniron applies an upward force to lift a
129-kg barbell to a height of 1.98 m at a constant
speed. Determine the work done.
W = (129)(9.81)(1.98) = 2505.67 J
A toy car is moving along with 0.40 joules of
kinetic energy. If its speed is doubled, then its new
kinetic energy will be _______.
a.
b.
c.
d.
e.
0.10 J
0.20 J
0.80 J
1.60 J
still 0.40 J
Which would ALWAYS be true of an object
possessing a kinetic energy of 0 joules?
a.
b.
c.
d.
e.
f.
g.
It is on the ground.
It is at rest.
It is moving on the ground
It is moving.
It is accelerating.
It is above the ground.
It is moving above ground level.
Which would ALWAYS be true of an object
possessing a potential energy of 0 joules?
a.
b.
c.
d.
e.
f.
g.
It is on the ground.
It is at rest.
It is moving.
It is accelerating.
It is at rest above ground level
It is above the ground.
It is moving above ground level.
The total mechanical energy of an object is the
______.
a. KE minus the PE of the object
b. PE minus the KE of the object
c. the initial KE plus the initial PE of the
object
d. final amount of KE and PE minus the initial
amount of KE and PE
Calculate the kinetic energy of a 5.2 kg object
moving at 2.4 m/s.
KE = &frac12;(5.2kg)(2.4m/s )2 =14.976 J
Calculate the total mechanical energy of a 5.2 kg
object moving at 2.4 m/s and positioned 5.8 m
above the ground.
KE = &frac12;(5.2)(2.4)2 = 14.976 J
PE = (5.2)(9.81)(5.8) = 295.8696 J
ME = PE + KE
ME = 310.8456 J
honors
Calculate the speed of a 5.2 kg object that
possesses 26.1 J of kinetic energy.
26.1 = &frac12; (5.2)v2
26.1 = 2.6v2
10.04 = v2
V = 3.17 m/s
Power
The quantity work has to do with a force causing a
displacement. Work has nothing to do with the amount
of time that this force acts to cause the displacement.
Sometimes, the work is done very quickly and other
times the work is done rather slowly. For example, a rock
climber takes an abnormally long time to elevate her
body up a few meters along the side of a cliff. On the
other hand, a trail hiker (who selects the easier path up
the mountain) might elevate her body a few meters in a
short amount of time. The two people might do the same
amount of work, yet the hiker does the work in
considerably less time than the rock climber. The
quantity that has to do with the rate at which a certain
amount of work is done is known as the power. The hiker
has a greater power rating than the rock climber.
The standard metric unit of power is the
Watt. As is implied by the equation for
power, a unit of power is equivalent to a
unit of work divided by a unit of time. Thus,
a Watt is equivalent to a Joule/second. For
historical reasons, the horsepower is
occasionally used to describe the power
delivered by a machine. One horsepower is
equivalent to approximately 750 Watts.
Power
• The rate of energy transfer
▫ Energy used or work done per second
Units = watts (W)
Horsepower (hp) is a unit used
1.00 hp = 746 W
A hair dryer uses 72,000 joules of
energy in 60 seconds. What is the
power of this hair dryer?
Water flows over a section of Niagara
Falls at the rate of 1.2 X 106 kg/s and falls
50.0 m. How much power is generated
by the falling water?
Two horses pull a cart. Each exerts a force of
250.0 N at a speed of 2.0 m/s for 10.0 min.
A. Calculate the power delivered by the horses.
B. How much work is done by the two horses?
Answers: 1.0 x 103 W and 6.0 x 105 J
An elevator lifts a 500-kg load up a
distance of 10 meters in 8 seconds.
A.Calculate the work done by the
elevator.
B.Calculate the elevator’s power.
W = (500)(9.81)(10) = 49,050 J
P = 49,050 J/8 seconds = 6131.25 watts