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Power
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Unit: Momentum & Energy
Power
Unit: Momentum & Energy
NGSS Standards: N/A
MA Curriculum Frameworks (2006): 2.4
Skills:
 calculate power
Language Objectives:
 Understand and correctly use the term “power.”
 Accurately describe and apply the concepts described in this section using
appropriate academic language.
 Set up and solve word problems involving power.
Notes:
power: a measure of the rate at which energy is applied or work is done. Power
is calculated by dividing work (or energy) by time.
P
W E k U


t
t
t
Power is a scalar quantity and is measured in Watts (W).
1 W  1 sJ  1 Nsm  1 kgsm
3
2
Note that utility companies measure energy in kilowatt-hours. This is because
W
P  , which means energy = W = Pt.
t
Because 1 kW = 1000 W and 1 h = 3600 s, this means
1 kWh = (1000 W)(3600 s) = 3 600 000 J
Use this space for summary and/or additional notes.
Copyright © 2010–2017 Mr. Bigler.
Physics
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Mr. Bigler
Power
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Unit: Momentum & Energy
Solving Power Problems
Most power problems require you to calculate the amount of work done or
change in energy, which you should recall is:
W  Fd
if the force and displacement are in the same direction
W  Fd cos
if the force and displacement are in different directions
and the angle between them is  .
E k  12 m(v 2  v o2 )
if the change in energy was caused by a change in
velocity
Ug  mg h
if the change in energy was caused by a change in
height
Once you have the work or energy, you plug it in for either W, E k or U , use
the appropriate parts of the formula:
P
W E k U


t
t
t
and solve for the missing variable.
Sample Problems
Q: What is the power output of an engine that pulls with a force of 500. N over
a distance of 100. m in 25 s?
A: W  Fd  (500)(100)  50 000 J
P
W 50 000

 2 000 W
t
25
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Physics
Mr. Bigler
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Power
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Unit: Momentum & Energy
Q: A 60. W light bulb is powered by a generator that is powered by a falling
1.0 kg mass on a rope. Assuming the generator is 100% efficient (i.e., no
energy is lost when the generator converts its motion to electricity), how far
must the mass fall in order to power the bulb at full brightness for
1.0 minute?
A:
U g
mg h
t
t
(1)(9.8) h
60 
60
3600  9.8 h
P
h 

3600
 367 m
9.8
Note that 367 m is approximately the height of the top floor of the Empire
State Building. Think about that the next time you decide that it doesn’t
matter if you leave the lights on when you’re not in a room!
Use this space for summary and/or additional notes.
Physics
Mr. Bigler