Work, energy and conservation of energyEdexcel International A Level Physics: Revision notes

Work done is the energy transferred when a force moves its point of application. For a force along the line of motion:

$W = Fs$

where $s$ is the distance moved in the direction of the force. The unit is the joule (J): 1 J is the work done when a force of 1 N moves its point of application 1 m in the direction of the force.

If the force acts at an angle $\theta$ to the direction of motion, only the component along the motion does work:

$W = Fs\cos\theta$

A rope pulling a sledge with $50\ \text{N}$ at $30^\circ$ above the horizontal for $8.0\ \text{m}$ does $50 \times 8.0 \times \cos 30^\circ = 346\ \text{J}$.

The kinetic energy of a body of mass $m$ moving at speed $v$ is:

$E_k = \frac{1}{2}mv^2$

It comes from work done to accelerate the body. For a body starting from rest, $W = Fs = ma \times s$ and $v^2 = 2as$, so $W = \frac{1}{2}mv^2$. Kinetic energy depends on the square of the speed, so doubling the speed quadruples the kinetic energy. It is a scalar and is never negative.

Near the Earth's surface, where the gravitational field strength $g$ is constant, the change in gravitational potential energy when a mass $m$ is raised through a vertical height $\Delta h$ is:

$\Delta E_{grav} = mg\Delta h$

Only the difference in height matters, so you may choose any zero level. $g = 9.81\ \text{N kg}^{-1}$ (or m s⁻²). Raising $0.15\ \text{kg}$ by $9.0\ \text{m}$ gains $0.15 \times 9.81 \times 9.0 = 13.2\ \text{J}$.

The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred from one form to another, so the total energy of an isolated system is constant.

When a body falls without resistive forces, the loss in gravitational potential energy equals the gain in kinetic energy: $mg\Delta h = \frac{1}{2}mv^2$, so $v = \sqrt{2g\Delta h}$, independent of mass. When resistive forces act, work is done against them: $\text{energy dissipated} = \text{work done against friction} = Fs$. This appears as thermal energy in the surroundings, so energy is dissipated, not lost.

A $30\ \text{kg}$ child slides $4.0\ \text{m}$ down a slide from rest, falling through $2.5\ \text{m}$, and reaches $5.0\ \text{m s}^{-1}$.

Energy released: $mg\Delta h = 30 \times 9.81 \times 2.5 = 736\ \text{J}$.

Kinetic energy gained: $\frac{1}{2} \times 30 \times 5.0^2 = 375\ \text{J}$.

Energy dissipated: $736 - 375 = 361\ \text{J}$.

Average friction: $F = W/s = 361/4.0 \approx 90\ \text{N}$.