Power

Power as the rate of doing work, average power = work ÷ time, and the key working relationship P = Fv for a driving force — used to find engine power at constant speed and, by combining P = Fv with Newton's second law, the instantaneous acceleration of a vehicle on a level road or a hill.

Mechanics · Work, energy & power

Power

Why it works

Power is the rate at which work is done, measured in watts (

1\text{ W} = 1\text{ J s}^{-1}

). Over a period of time,

\text{average power} = \frac{\text{work done}}{\text{time taken}}.

The relationship $P = Fv$ . For a force

F

driving an object along its direction of motion at speed

v

, the work done per second is

F

times the distance covered per second, i.e.

F \times v

. So the instantaneous power of the driving force is

P = Fv.

This is the workhorse formula for vehicle problems. Rearranged, the driving (tractive) force at a given speed is

F = \dfrac{P}{v}

— note it is larger at low speed and falls as the vehicle speeds up.

Constant speed. When a vehicle moves at constant speed, its acceleration is zero, so by Newton's second law the driving force exactly balances the forces opposing motion. On a level road that means

F = \text{resistance}

; going up a slope of angle

\theta

it means

F = \text{resistance} + mg\sin\theta,

because the engine must also overcome the component of weight down the slope (take

g = 10

). The power is then

P = Fv

.

Instantaneous acceleration. When the vehicle is not at its top speed, use

P = Fv

to get the driving force at that instant, then apply Newton's second law to the resultant force:

F = \frac{P}{v}, \qquad F - (\text{resistance} + mg\sin\theta) = ma.

The maximum speed is the special case

a = 0

: the driving force has dropped until it just balances the resistance (and any weight component), so

v_{\max} = \dfrac{P}{F}

with

F

equal to that balancing force.