Power (OCR AS Physics): Revision Note
Exam code: H156
Power
Power is defined as:
The rate of energy transfer
Since work done is equal to the energy transferred, power can also be defined as:
Work done per unit time
Power can be calculated using the following equations:
Where:
P = power in watts (W)
E = energy in joules (J)
W = work done in joules (J) or newton metres (N m)
t = time in seconds (s)
Power is also used in electricity, with labels on lightbulbs which indicate their power, such as 60 W or 100 W
These indicate the amount of energy transferred by an electrical current rather than by a force doing work
The Watt
The watt (W) is the unit for power (and radiant flux)
It is defined as 1 W = 1 J s-1
In SI base units, the watt is kg m2 s–3
One watt is defined as:
A transfer of energy of 1 J in 1 s
Worked Example
A car engine exerts the following force for 1.0 km in 200 s.

What is the average power developed by the engine?
Answer:
Step 1: List the known quantities and convert to SI units
Force, F = 500 N
Displacement, s = 1.0×103 m
Time, t = 200 s
Step 2: Substitute the work done equation into the power equation
Step 3: Substitute the known values to calculate
Examiner Tips and Tricks
Think of power as “energy per second”. Thinking of it this way will help you to remember the relationship between power and energy: “Watt is the unit of power?”
Power & Force
If an object is moving at constant velocity with a constant force, the power can be calculated by:
Where:
= force (N)
= velocity (m s–1)
This equation is only relevant where a constant force moves a body at constant velocity
Power is required in order to produce an acceleration
The force must be in the direction of the velocity
Otherwise, the component of the force in the direction of the velocity must be used instead
Derivation of P = Fv
Power is the rate of doing work
Work is done when a force moves an object over a distance
At a constant velocity, the distance moved by the object can be described as:
Substituting this into the work done equation gives:
And substituting this into the power equation gives:
Worked Example
A lorry moves up a road that is inclined at 14.5° to the horizontal.

The lorry has a mass of 3500 kg and is travelling at a constant speed of 9.4 m s-1. The force due to air resistance is negligible.
Calculate the useful power from the engine to move the lorry up the road.
Answer:
Step 1: List the known quantities
Mass, m = 3500 kg
Speed, v = 9.4 m s-1
Angle of incline = 14.5°
Step 2: State the equation for power in motion
Step 3: Calculate the component of the applied force which overcomes the weight

Step 4: Substitute the known values into the power equation to calculate
Examiner Tips and Tricks
The force represented in exam questions will often be a drag force. Whilst this is in the opposite direction to its velocity, remember the force needed to calculate the power is equal to (or above) this drag force to overcome it, therefore, you equate it to that value.
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