Work, energy and power

5.1 Energy conservation

Work done

$W = Fs$

Work done is force × displacement in the direction of the force. If the force is at angle $\theta$ to the displacement, $W = Fs\cos\theta$. The unit is the joule. A force at right angles to the motion does no work, for example the centripetal force on an object in a circle.

Conservation of energy

Work done is how energy is transferred by a force.

Efficiency

$\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} = \frac{\text{useful power output}}{\text{total power input}}$

Efficiency is often given as a percentage, and it can never exceed 100%. The rest of the energy usually ends up as thermal energy in the surroundings.

Power

Power is the rate of doing work, or the work done per unit time:

$P = \frac{W}{t}$

The unit is the watt (1 W = 1 J/s).

Deriving P = Fv

A constant force $F$ moves an object through a displacement $s$ in time $t$. Then $P = W/t = Fs/t$, and $s/t = v$, so:

$P = Fv$

If the car is accelerating, the driving force is greater than the resistance. Use $F_{\text{driving}} = P/v$, then $F_{\text{driving}} - F_{\text{resistance}} = ma$.

5.2 Gravitational potential energy and kinetic energy

Gravitational potential energy

To lift a mass $m$ at steady speed through a height $\Delta h$, the force needed equals the weight, $mg$. The work done is force × distance:

$\Delta E_P = mg\Delta h$

This holds only where the field is uniform, which means small height changes near the Earth's surface. Chapter 13 deals with larger distances.

Kinetic energy

Take a mass $m$ accelerated from rest by a force $F$ through a distance $s$, reaching speed $v$. The work done is $W = Fs = (ma)s$. From $v^2 = u^2 + 2as$ with $u = 0$, $as = \tfrac{1}{2}v^2$, so:

$E_K = \tfrac{1}{2}mv^2$

Kinetic energy depends on the square of the speed, so doubling the speed makes it four times as large.

Using energy to solve problems

When there is no friction or drag, the loss of gravitational potential energy equals the gain of kinetic energy, so $mg\Delta h = \tfrac{1}{2}mv^2$ and $v = \sqrt{2g\Delta h}$. The mass cancels.