Calculate the work done by a force, kinetic energy and gravitational potential energy; apply the work-energy principle and conservation of mechanical energy; and calculate power.

What this dot point is asking

Energy methods give a powerful shortcut: they relate the start and end states of a motion without ever solving for the acceleration. The SQA wants you to calculate the work done by a force, the kinetic and gravitational potential energy of a body, and the power of an engine, and to use the work-energy principle and conservation of mechanical energy to solve problems.

Work, kinetic energy and potential energy

Work is done when a force moves its point of application. Only the component of force along the displacement does work, which is why the angle appears.

A force perpendicular to the motion (such as the normal reaction, or the tension in a conical pendulum) does no work, because $\cos 90^\circ = 0$. Friction does negative work, removing energy from the body. All quantities are in joules (J).

The variable-force integral $W = \int F\,dx$ is what you use whenever the force changes along the path, most often for the work done in stretching a spring. Because a spring's tension grows as $kx$, the work done in stretching it from its natural length to extension $e$ is $\int_0^e kx\,dx = \tfrac{1}{2}ke^2$, the elastic potential energy stored. Gravitational potential energy is always measured relative to a chosen reference level, so only changes in $E_p$ are physically meaningful; pick the lowest point of the motion as the zero to keep the arithmetic clean.

The work-energy principle

The work-energy principle ties work directly to kinetic energy: do net positive work on a body and it speeds up; do net negative work and it slows down.

Conservation of energy and power

When the only forces doing work are conservative (gravity, an ideal spring), no mechanical energy is lost, so kinetic and potential energy simply interchange.

When friction or a driving force is present, mechanical energy is not conserved, so use the broader balance: work input equals the gain in kinetic energy plus the gain in potential energy plus the energy lost to resistance. The relation $P = Fv$ is the key to vehicle problems: at top speed the acceleration is zero, so the driving force $\dfrac{P}{v}$ exactly balances the resistance.

Try this

Q1. A $3$ kg mass falls freely through $5$ m from rest. Use energy to find its speed ($g = 9.8$). [3 marks]

• Cue. $mgh = \tfrac{1}{2}mv^2$, so $v^2 = 2gh = 2(9.8)(5) = 98$, giving $v \approx 9.9$ m s$^{-1}$.

Q2. An engine does $30000$ J of work in $10$ s. Find its average power. [2 marks]

• Cue. $P = \dfrac{W}{t} = \dfrac{30000}{10} = 3000$ W $= 3$ kW.