Work, Energy and Power: Complete A Level H2 Physics Cheatsheet

Work done by a force

Work is a scalar: W = Fs cosθ, where θ is the angle between the force and the direction of displacement. Only the force component along the displacement actually does work.

If a force acts exactly perpendicular to the displacement (θ = 90°), it does zero work, even if it's a large force. This applies to the normal contact force on level ground, and to the centripetal force in circular motion.

Trap: A force can act on an object, and even feel like hard work to apply, without doing any physics 'work' on it. Holding a heavy bag stationary does zero work (no displacement). Carrying it at constant height across a level floor also does zero work by your vertical support force, since that force is perpendicular to the horizontal displacement.

The work-energy theorem and conservation of energy

Work-energy theorem: the net (resultant) work done on an object equals its change in kinetic energy, W(net) = ΔKE.

Conservation of energy: the total energy of an isolated system stays constant. Energy changes form, e.g. gravitational potential energy converting to kinetic energy, but total energy is never created or destroyed. Where friction or air resistance acts, mechanical energy is converted into thermal energy (heat) and sound, not 'lost'.

Trick: For problems combining height loss/gain, changing speed, and resistive forces, set up a single energy equation, initial energy = final energy + energy converted to heat (work done against friction), rather than tracking forces with F = ma at every stage. It's usually much faster and avoids needing to know the object's acceleration at all.

Trap: Avoid describing energy as 'lost'. In an exam answer, state specifically what it converts into, almost always thermal energy (and sometimes sound) due to a resistive force, since total energy is always conserved.

Power and efficiency

Power is the rate of doing work (or transferring energy): P = W/t. For a constant force acting in the direction of motion, this simplifies to the instantaneous power P = Fv.

Efficiency = useful energy (or power) output ÷ total energy (or power) input × 100%. It is always less than 100% for a real process, since some energy is always dissipated, usually as heat, in a way that isn't useful for the task at hand.

Trick: For an object moving at **constant velocity**, the resultant force on it is zero (equilibrium, from Chapter 4). This means the driving force must exactly balance every resistive force (and any component of weight along an incline), letting you find the driving force first from equilibrium, then use P = Fv to get the power needed.

Worked Example

A block of mass 5.0 kg is pulled at a constant speed of 2.0 m/s up a rough incline of angle 20° to the horizontal, by a force acting parallel to the incline. A constant frictional force of 6.0 N acts on the block. Take g = 9.81 m/s².

Why this question is a good test of the topic: it doesn't stand alone, part (b) only works by applying Chapter 4's equilibrium condition first, before this chapter's P = Fv can be used. Work, Energy and Power questions rarely appear in isolation from Forces on the actual exam.

Frequently Asked Questions