EASA / CAR Part-66 · Module 2 · 2.2 Mechanics · Cat A
Objective: calculate work as F × d, potential energy as mgh and power as work ÷ time, and see clearly that lifting the same load faster changes the power but not the work.
Lifting a load stores potential energy equal to the work done against gravity.
Work is done when a force moves through a distance: W = F × d, measured in joules. Lifting a load means working against its weight, so the work done is mgh — and that work is not lost, it is stored in the load as potential energy. Let the load back down and you get it all back, converted into kinetic energy as it falls. No work is done if nothing moves: straining against a load that will not budge is tiring, but in the physics sense it is zero work.
Power is the rate of doing work, W/t, measured in watts. This is the distinction the two hoists are built to show: both lift the identical load to the identical height, so both do exactly the same work and store the same energy. The fast hoist simply does it in half the time, so it needs twice the power. Power sizes the motor; work sizes the energy bill.
Because the load rises at a steady speed, the power is also F × speed, which is the form you meet in engine and hydraulic ratings. Energy itself is never created or destroyed, only converted — here electrical energy becomes potential energy, with some always lost as heat in a real machine, which is what efficiency measures.