Download Power and Efficiency

Power and Efficiency
And a review of the work-energy theorem
Whiteboards 
 A bungee jumper jumps from a bridge.
Explain the energy changes that occur
during the jump, what work is being done,
and what force(s) is/are doing the work as
the jumper completes his jump
 A 1.5 kg block slides along a rough table
and is brought to rest after travelling a
distance of 2.4 m. The frictional force is
assumed constant at 3.2 N.
 What work is done by friction?
W  F  d  (3.2 N )(2.4m)  7.7 J
 What is the coefficient of dynamic friction
between the block and the table?
𝐹𝑓 = 𝜇 · 𝐹𝑁  3.2 = 𝜇 · (1.5)(9.81)  𝜇 = 0.22
 What was the initial velocity of the block?
W  K
1
2
2
 7.68 J  (m)(v f  vi )
2
2
 7.68  (0.5)(1.5)(0  vi )
vi  3.2 m s
Power
 The rate at which work is performed
 In other words…the rate at which energy
is transferred
W E
P

t
t
Power
 Power can also be defined in terms of how
quickly an object moves as a result of a
force applied:
W F d
P

 F v
t
t
Efficiency
 The power supplied to a motor is not
always the same as the power that is
actually used by the motor.
 The ratio of Power available to the Power
used is a measure of the efficiency
 The larger the percentage of available
power that is used, the greater the
efficiency
Pused
% Eff 
x100
Pavailable
Sample:
The motor of an elevator develops
power at a rate of 2500 W.
At what constant speed can a 120 kg
load be raised?
P  F  v  (mg )  v
P
2500W
v

 2.12 m s
mg (120kg)(9.81 m s 2 )
It was measured that the load was
actually lifted at a speed of 1.5 m/s.
What is the efficiency of the motor for
this elevator?
% Eff 
Pused
100
Pavailable
% Eff 
(120)(9.81)(1.5)
100  70.6%  71%
2500