Derivation of P = Fv

The power delivered to a moving object is described by the equation:

$P = F v$

Where:
- P = power in watts (W)
- F = force in newtons (N)
- v = velocity in metres per second (m s^-1)

This equation is only relevant where a constant force moves a body at constant velocity
Power is delivered to the object by a force in order to produce an acceleration
The force must be applied in the same direction as the velocity

Derivation

Power is the rate of doing work

$P = \frac{W}{t}$

Work is done when a force moves an object over a distance

$W = F s$

At a constant velocity, the distance moved by the object can be described as:

$d = v t$

Substituting this into the work done equation gives:

$W = F v t$

And substituting this into the power equation gives:

$P = \frac{F v t}{t}$

$P = F v$

Worked example

A lorry moves up a road that is inclined at 14.5° to the horizontal.

The lorry has mass 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

$P = F v$

Step 3: Calculate the component of the applied force which overcomes the weight

5-1-5-worked-example-p-fv-2-cie-new

$F = m g \sin θ$

$F = 3500 \times 9.81 \times \sin (14.5)$

$F = 8596.8 N$

Step 4: Substitute the known values into the power equation to calculate

$P = 8596.8 \times 9.4$

$P = 81 000 W (2 s . f .)$

Examiner Tip

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.

Derivation of P = Fv (CIE AS Physics)

Revision Note