Power

Work is done by a force on an object when the object moves in a direction parallel to the direction of the force.  Power is a measure of how quickly this work is done, it is the rate at which work is done.

P = ∆W/∆t = average power

Example:

You can load a scientific instrument to measure wind speed into your backpack and hike up a mountain, or you can load it into your car and drive up the mountain.  Either way, you do the same amount of work on the instrument.  This work is equal to mgh, where mg is the weight of the instrument and h is the difference in altitude.  The power, the rate at which you do this work, however, is different.  The hike takes 4 hours, while the drive takes 30 minutes.

Units of power:

The SI unit of power is Joule/s = Watt.  (J/s = W.)  Often used units are kilowatt (kW), megawatt (MW), and milliwatt (mW).
1 kW = 1000 W.
1MW = 1000000 W = 106 W.
One horsepower (hp) = 745.7 W.

Unit of energy:

Power * time = energy transferred

When you pay your electric bill, you pay for each kilowatt-hour (kWh) used.

Kilowatt-hour is a unit of energy or work, not power.

1 kWh = (1 kW)(1 hour) = (1000 J/s)(3600 s) = 3600000 J = 3.6*106 J.

Translational Power:

Power = ∆W/∆t.  But the work ∆W = F*∆x.  We may therefore write P = F*∆x/∆t = F*v.

In one dimension, the power or rate at which work is done by a force on an object is equal to the magnitude of the force, multiplied by the velocity of the object.


Problem:

A certain automobile engine delivers 30 hp (2.24*104 W) to its wheels when moving at a constant speed of 27 m/s (~60 mi/h).  What is the resistive force acting on the automobile at that speed?

Solution:
The car is moving with constant velocity, so the net force on the car is zero. The magnitude of the work done by the frictional force in a given time interval is equal to the work done by the engine in the same time interval.
P = Fv = 2.24*104 W.
The work done per second by the resistive force is -2.24*104 J.  The magnitude of the resistive force f is f = (2.24*104 J)/(27 m/s) = 830 N.

Problem:

If a person who normally requires an average of 12,000 kJ (3000 kcal) of food energy per day consumes 13,000 kJ per day, he will steadily gain weight.  How much bicycling per day is required to work off this extra 1000 kJ?  The rate stored energy converted into other forms, ( i. e. the power) is approximately 400 W for cycling at a moderate speed.

Solution:
Power = energy/time.  t = (106 J)/(400 W) = (2500 s)*(1 min /60 s) = 42 min.

Problem:

The average power consumption when rapidly climbing stairs at ~116 stairs per minute is 685 W.
(a)  How long can a person rapidly climb stairs (116/min) on the 3.9*105 J of energy in a 10 g pat of butter?
(b)  How many flights is this if each flight has 16 stairs?

Solution:
(a)  Power = energy/time.  t = (3.9*105 J)/(685 W) = (569 s)*(1 min /60 s) = 9.5 min.
(b)  (116 stairs/min)*9.5 m = (1100 stairs)*(1 flight/16 stairs) = ~70 flights.


Link:  The physics classroom:  Work, Energy and Power