Work, Energy and Power

Forces and acceleration answer how a body moves. Energy answers how much, and often does so in one line where $F = ma$ would need three. The rule of thumb: when a question involves speeds at two points and a distance between them, energy is usually the shorter route.

Take $g = 9.81\ \text{m s}^{-2}$. Work and energy are both measured in joules (J), and power in watts (W).

Work done by a force

$$W = Fd\cos\theta$$

where $\theta$ is the angle between the force and the displacement.

A box is dragged 8 m by a 50 N force at $30°$ to the horizontal.

$$W = 50(8)\cos 30° = 346\ \text{J}$$

Kinetic energy

$$\text{KE} = \tfrac12 mv^2$$

A car of mass 1200 kg at $20\ \text{m s}^{-1}$ has $\tfrac12(1200)(400) = 240\,000$ J $= 240$ kJ.

Gravitational potential energy

$$\text{PE} = mgh$$

with $h$ measured from whatever level you choose as zero. Raising 5 kg through 3 m stores $5(9.81)(3) = 147$ J.

The work–energy principle

$$\text{work done by the resultant force} = \text{change in kinetic energy}$$

$$W = \tfrac12 mv^2 - \tfrac12 mu^2$$

A 2 kg body speeds up from $3$ to $7\ \text{m s}^{-1}$.

$$W = \tfrac12(2)(49) - \tfrac12(2)(9) = 49 - 9 = 40\ \text{J}$$

Conservation of mechanical energy

When no resistive force does work,

$$\text{KE} + \text{PE} = \text{constant}$$

A ball is dropped from 20 m. With PE zero at the ground,

$$mgh = \tfrac12 mv^2 \quad\Longrightarrow\quad v = \sqrt{2gh} = \sqrt{2(9.81)(20)} = 19.8\ \text{m s}^{-1}$$

The mass cancels — every body falls the same way in this model.

When friction is present, the energy it removes must be accounted for:

$$\text{PE lost} = \text{KE gained} + \text{work done against friction}$$

A 60 kg child slides down 5 m of vertical drop, with friction doing 800 J of work.

$$\tfrac12(60)v^2 = 60(9.81)(5) - 800 = 2943 - 800 = 2143$$

$$v^2 = 71.4 \quad\Longrightarrow\quad v = 8.45\ \text{m s}^{-1}$$

Power

Power is the rate of doing work:

$$P = \frac{W}{t} \qquad\text{and, for a force moving with the body,}\qquad P = Fv$$

A car travels at a steady $30\ \text{m s}^{-1}$ against a resistance of 600 N.

At constant velocity the driving force equals the resistance, so

$$P = 600(30) = 18\,000\ \text{W} = 18\ \text{kW}$$

A pump raises 200 kg of water through 12 m every minute.

$$P = \frac{mgh}{t} = \frac{200(9.81)(12)}{60} = 392\ \text{W}$$

An engine works at 24 kW. At $20\ \text{m s}^{-1}$ against a 500 N resistance, a 1200 kg car accelerates at:

$$F = \frac{P}{v} = \frac{24\,000}{20} = 1200\ \text{N}$$

$$a = \frac{1200 - 500}{1200} = 0.58\ \text{m s}^{-2}$$