Work done by a force, kinetic and potential energy, the work-energy principle, conservation of mechanical energy, power as the rate of doing work, and work done against resistance.

What this dot point is asking

AQA wants you to calculate work done by a force, use kinetic and gravitational potential energy, apply the work-energy principle and conservation of mechanical energy, calculate power as the rate of doing work, and handle problems with resistance forces such as friction and air resistance.

Work and energy

Work is a scalar, so it adds and subtracts as ordinary numbers, but its sign matters: a force does positive work when its component along the displacement points the same way as the motion, and negative work when it opposes the motion. Gravity does positive work on a descending body and negative work on a rising one, while friction always does negative work because it acts against the direction of sliding. Gravitational potential energy $mgh$ is measured relative to a chosen reference level, and only differences in height matter, so you are free to set $h = 0$ wherever it is convenient, usually at the lowest point of the motion.

The work-energy principle

The power of this principle is that it bypasses the detailed motion entirely. You do not need the acceleration, the time taken, or the shape of the path: you only need the work each force does between the start and end points. For a body pulled up a rough slope, that means adding the work done by the applied force, subtracting the work done against friction, and subtracting the work done against gravity, then setting the total equal to the gain in kinetic energy. This is exactly the structure of most exam questions on the topic, and it is usually faster than resolving forces and integrating the acceleration.

Conservation of mechanical energy

When the only force doing work is gravity (a conservative force), the sum of kinetic and gravitational potential energy stays constant. As a body falls, potential energy converts into kinetic energy, so $\frac{1}{2}mv^2 + mgh$ takes the same value at every point of the motion. This gives a one-line route to the speed at any height without solving the equations of motion, which is why it is the preferred method for problems on smooth curves, pendulums and projectiles where the path is awkward. The moment a non-conservative force such as friction or air resistance does work, mechanical energy is no longer conserved, and you must switch to the full work-energy principle, including the work done against that force as a loss term.

Power

The relation $P = Fv$ is the key to the standard vehicle problem. An engine working at constant power produces a driving force $F = \frac{P}{v}$ that falls as the speed rises. The maximum speed on a given road is reached when this driving force has dropped to just balance the resistances and any gravity component, so the acceleration is zero. At lower speeds the larger driving force leaves a surplus over the resistances, and that surplus, divided by the mass, is the acceleration from Newton's second law. Working a problem therefore means computing the driving force at the stated speed, subtracting the resistance and weight components along the motion, and dividing by the mass.