Objective: To define and apply the concepts of
power and mechanical efficiency.
Power
The rate at which work is done.
The power is the work done divided
by the time it takes to do the work.
W
P
t
Units:
J/s = Watt
1 horsepower =746 Watts
1000 Watts = 1 kW
Kilowatt-hours (kwh) measure energy.
J

1kwh  1000 3600 s   3.6 106 J
s

SRP charges about 10 cents for 1kwh. How much
would it cost to leave a 60W light bulb on for a full
day?
60W  0.060kW
(0.060kW )( 24hours )
 1.44kWh
10 cents
1.44kwh 
 14.4 cents
kwh
Leaving a 60W light bulb on for 24 hours will
cost about 15 cents.
This equation can be written in
another form:
W
P
t
W  F  x  Fx cos
Fx cos 
P
t
P  Fv cos
Which has more power?
A 60kg man takes 5 seconds to run up a set
of stairs at a constant velocity. What was
the man’s power?
Work is transferred to
gravitational potential Energy
W Ug
P

t
t
mgh

t
Mass = 60kg
15m
(60)(10)(15)

5
J
 1800 (Watts)
s
A man pushes a 12kg crate horizontally with a
constant velocity of 2m/s. The coefficient of friction
between the crate and the floor is 0.25. What
power does the man have as he pushes the crate?
f  F
120N
f  (.25)(120)
f  30 N
P  Fv cos
P  (F )( 2)(1)
P  (30)( 2)(1)
f
F
P  60W
-120N
A crane hoist lifts a load of 1000kg a vertical
distance of 25m in 9.0s at a constant velocity.
What is the useful power output of the crane?
W U g mgh
P 

t
t
t
(1000)(10)( 25)
4
P
 2.8 10 W
9
If the hoist motor of the previous crane is
rated at 70hp, what percentage of this power
goes to useful work?
746w
4
70hp 
 5.2 10 W
1hp
2.8 10 W
%
4
5.2 10 W
4
 54%
Efficiency – A measure of what you get out
for what you put in.
work output Wout


energy input Ein
The Efficiency of a typical car is about 25%!
The motor of an electric drill with an
efficiency of 80% has a power of input of
600 W. How much useful work is done by
the drill in a time of 30 seconds?
J
600W  600  (30 s )  18000 J
s
work output Wout


energy input Ein
Wout
0.80 
18000
Wout  14400 J