Work, energy and power (SACE Stage 2 Physics)

What this dot point is asking

You need to calculate work done by a force (including when the force is at an angle to the motion), use the work-energy theorem to connect net work and change in kinetic energy, and define and calculate power.

Work done by a force

Work is done only when a force has a component along the displacement.

The $\cos\theta$ factor is important:

• Force along the motion ($\theta = 0$): $W = Fs$ (maximum positive work).
• Force perpendicular to the motion ($\theta = 90^\circ$): $W = 0$. A centripetal force, or the normal force on a sliding box, does no work.
• Force opposing the motion ($\theta = 180^\circ$): $W = -Fs$ (negative work, removes energy, e.g. friction).

The work-energy theorem

The total (net) work done on an object equals its change in kinetic energy.

Kinetic energy itself is $E_k = \tfrac{1}{2}mv^2$. The theorem follows directly from $F=ma$ combined with $v^2 = u^2 + 2as$. The key word is net: you must add the work done by every force, including the negative work done by friction or drag.

Power

Power is how quickly work is done, the rate of energy transfer.

A more powerful engine does the same work in less time, or more work in the same time. At a steady speed against a constant resistive force, the driving power equals $Fv$ because the driving force just balances the resistance.

Gravitational potential energy

Lifting a mass through a height $h$ stores gravitational potential energy $E_p = mgh$. The work you do against gravity equals the energy stored. In energy problems, total mechanical energy ($E_k + E_p$) is conserved when only gravity does work; friction or drag does negative work and converts mechanical energy to heat. This conservation lets you find the speed at the bottom of a ramp or the height a thrown ball reaches without using forces at all.

How SACE assesses this

SACE Stage 2 work-energy questions frequently combine several ideas in one item: compute the work done by a force (watching for the $\cos\theta$ factor), apply the work-energy theorem to find a final speed, and then find the average power. The most common error is forgetting that the work-energy theorem uses the net work, so any friction or drag (which does negative work) must be subtracted before equating to $\Delta E_k$. Set out each force's work separately, sum them for the net work, then solve for the unknown. For power, decide whether the question wants average power ($P = W/t$) or instantaneous power ($P = Fv$), and quote the correct form before substituting.