Define power as the rate of doing work or transferring energy, and apply $P = W / t = Fv$ to mechanical systems, including efficiency calculations

What this dot point is asking

QCAA wants you to use power as the rate quantity that links work and time, and to use efficiency as the ratio of useful energy output to total energy input. Both ideas appear in IA1 stimulus and EA Paper 1, often as small calculations attached to a larger motion or energy problem.

Power

Power is the rate at which work is done (or energy is transferred):

SI unit: watt (W = J s$^{-1}$). For a force pushing an object at constant velocity:

where $F$ is the component of force in the direction of motion. This form is useful when the force is constant and the speed is steady (cruising car, elevator at constant speed, conveyor belt).

Constant-power problems

When a vehicle delivers constant power, the driving force decreases as speed increases:

This is why a car has plenty of acceleration at low speeds but accelerates only slightly near top speed: the engine power is the same but $v$ is large.

At top speed (steady velocity), driving force equals total resistive force, so $P = F_{\text{resistive}} v_{\max}$.

Efficiency

Efficiency is the dimensionless ratio of useful output to total input:

Expressed as a fraction or a percentage. Always less than $1$ in real machines; the rest is energy lost as heat, sound, friction, or electrical resistance.

Worked example

A car of mass $1200$ kg travels at a steady $25$ m s$^{-1}$ against a total resistive force of $600$ N. Find the engine power output.

At steady speed, driving force equals resistive force: $F = 600$ N.

$P = F v = (600)(25) = 15\,000$ W $= 15$ kW.

If the engine consumes $50$ kW of fuel energy, efficiency is $\eta = 15 / 50 = 30\%$, typical for a petrol engine.

Common traps

Treating $P = Fv$ as instantaneous when it is not. $P = Fv$ gives instantaneous power for a constant force; it gives average power for an average force.

Mixing up watts and joules. Watts are power (J s$^{-1}$). Joules are energy. A $100$ W bulb running for $1$ s consumes $100$ J of electrical energy.

Computing efficiency from forces, not energies. Efficiency compares energy or power, not raw forces. The lifting force and pulling force in a pulley system can differ by a factor of two while still delivering the same work.

Forgetting that efficiency is dimensionless. Both numerator and denominator must use the same unit (joules with joules, watts with watts).

Connection to the rest of Unit 2

Power links to the work and mechanical energy dot point ($P = W/t$ converts a kinematic work calculation into a rate) and to motion graphs (constant-power motion gives a non-linear $v$-$t$ curve because acceleration depends on $v$). In Unit 3, the same power-efficiency framework reappears for transformers and AC transmission.

Try it: For a horizontal-motion power calculation, use the SUVAT calculator to get the steady velocity, then apply $P = Fv$ by hand.

In one sentence

Power is the rate of doing work or transferring energy, defined as $P = W/t$ and equal to $Fv$ for a constant force, and efficiency is the dimensionless ratio of useful energy output to total energy input ($\eta = P_{\text{out}}/P_{\text{in}}$), with the difference appearing as friction, heat, sound or resistive losses.