3.3.3 - Power and efficiency

Power as a rate of doing work

Two machines can do the same amount of work but not be equally powerful. The one that does the work in the shorter time has the greater power.

Average power over an interval is:

Here $P$ is power in watts, $W$ is work done or energy transferred in joules, and $t$ is time in seconds.

The unit of power is the watt, symbol $\mathrm{W}$.

Be careful with notation. In an equation, $W$ can mean work done. After a numerical value, $\mathrm{W}$ is the unit watt. Context tells you which one is being used.

Worked example: a small motor transfers $18\,\mathrm{kJ}$ of energy in $12\,\mathrm{s}$. Calculate its average power.

Convert the energy first:

$18\,\mathrm{kJ} = 18 000\,\mathrm{J}$

Then use the definition of power:

So the motor's average power is:

$1.5 \times 10^{3}\,\mathrm{W}$, or $1.5\,\mathrm{kW}$.

The result is sensible because transferring thousands of joules every few seconds should give a kilowatt-scale power, not a few watts.

Power is not the same as total energy. A powerful machine transfers energy quickly; it does not automatically transfer more total energy unless it operates for long enough.

Deriving power from force and velocity

The relationship $P = Fv$ connects power to force and speed. OCR expects you to derive it from first principles, so do not treat it as a detached formula.

Start with work done by a constant force acting in the direction of motion:

$W = Fs$

where $F$ is force and $s$ is displacement in the direction of the force.

Substitute this into the power definition:

$P = W/t = Fs/t$

For straight motion in one direction, the average speed over this interval is:

$v = s/t$

For instantaneous power, use a small displacement and take the limit as $\Delta t\to0$: $P=\lim_{\Delta t\to0}F\Delta s/\Delta t=Fv$.

This form applies when $F$ is the force, or component of force, in the direction of motion. If a vehicle moves at constant speed, the driving force at the wheels can still be doing work against resistive forces even though the resultant force is zero.

[DIAGRAM: asset_name: 3.3.3 - Power and efficiency - diagram 1; asset_slug: 3_03_3_power_and_efficiency__diagram_01; recommended_method: drawn_physics; caption: Force acting along the motion links work, speed and power.; description: A trolley moves in the direction of an applied force, linking work, displacement, time and power.]

Worked example: a cyclist moves at a steady speed of $8.0\,\mathrm{m}\,\mathrm{s}^{-1}$. The useful forward driving force at the wheel is $95\,\mathrm{N}$. Calculate the useful mechanical power output.

Use:

$P = Fv$

Substitute:

$P = 95 \times 8.0 = 760\,\mathrm{W}$

The useful output power is $760\,\mathrm{W}$.

Notice what the equation is saying physically. At a higher speed, the same force moves through a larger distance each second, so it does more work each second.

The common trap is to say "constant speed means no power is needed". Constant speed means zero resultant force. It does not mean zero driving force or zero energy transfer when resistive forces are present.

Power At Speed

For many moving systems, the useful mechanical power is found from the force needed to maintain the motion and the speed of the object:

$P = Fv$

At constant speed on a level road or track, the driving force is equal in magnitude to the total resistive force. This is because the resultant force is zero, not because the resistive forces have disappeared.

Worked example: a vehicle moves at constant speed. At this speed the total resistive force is $620\,\mathrm{N}$, and the vehicle travels at $24\,\mathrm{m}\,\mathrm{s}^{-1}$. Calculate the useful power needed to maintain this speed.

At constant speed:

Use:

So the useful power is about:

$1.5 \times 10^{4}\,\mathrm{W}$, or $15\,\mathrm{kW}$.

A frequent OCR-style extension is to give a model for the resistive force. For example, if the resistive force is described by:

$F = kv^{2}$

then using $P = Fv$ gives:

$P = (kv^{2})v = kv^{3}$

That means the power needed rises very quickly with speed.

Worked example: the resistive force on a small racing car is modelled by $F = 0.90v^{2}$, where $F$ is in newtons and $v$ is in $\mathrm{m} \mathrm{s}^{-1}$. The maximum useful mechanical power is $75\,\mathrm{kW}$. Estimate the maximum speed.

At maximum steady speed:

$P = Fv = (0.90v^{2})v = 0.90v^{3}$

Convert the power:

$75\,\mathrm{kW} = 75 000\,\mathrm{W}$

Now solve:

$75 000 = 0.90v^{3}$

$v^{3} = 75 000 / 0.90 = 83 333$

$v = 43.7\,\mathrm{m}\,\mathrm{s}^{-1}$

To two significant figures, the maximum speed is $44\,\mathrm{m}\,\mathrm{s}^{-1}$.