Work, Energy, and Power

Work is done when a force causes displacement. The work-energy theorem connects net work to change in kinetic energy.

Work

W = F \cdot d \cdot \cos\theta

F = force (N), d = displacement (m), θ = angle between force and displacement.

When force and displacement are in the same direction ( $\theta = 0$ ):

$W = F \cdot d$

Kinetic Energy

E_k = \dfrac{1}{2}mv^2

m = mass (kg), v = speed (m/s), Ek in Joules (J).

Work-Energy Theorem

$W_{\text{net}} = \Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Potential Energy (Gravitational)

Gravitational Potential Energy

E_p = mgh

m = mass (kg), g ≈ 9.81 m/s², h = height above reference (m).

Power

P = \dfrac{W}{t} = F \cdot v

Power is the rate of doing work. Unit: Watt (W = J/s).

A 2 kg ball falls 5 m. Find its speed at the bottom (from rest).

Energy conservation: $mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9 \text{ m/s}$

Real world: Pushing a stalled car

A mechanic pushes a stalled 900 kg car with a constant force of 300 N over 12 m (θ = 0°).

Work done: $W = F \times d = 300 \times 12 = 3{,}600 \text{ J}$

By the work-energy theorem, this equals the car's gain in KE: $v = \sqrt{\dfrac{2W}{m}} = \sqrt{\dfrac{7200}{900}} \approx 2.8 \text{ m/s}$

Real world: Cyclist braking to a stop

A cyclist and bike together have mass 80 kg and are travelling at 10 m/s. The brakes bring them to rest.

Kinetic energy to remove: $\Delta E_k = \tfrac{1}{2} \times 80 \times 10^2 = 4{,}000 \text{ J}$

The brakes do $-4{,}000 \text{ J}$ of work (opposing motion). All this energy converts to heat in the brake pads.

Real world: Sprinter accelerating off the blocks

A 70 kg sprinter accelerates from rest to 9 m/s over a distance of 20 m.

Net work done: $W = \Delta E_k = \tfrac{1}{2} \times 70 \times 9^2 = 2{,}835 \text{ J}$

Average net force applied: $F = \dfrac{W}{d} = \dfrac{2835}{20} \approx 142 \text{ N}$

Tip

Where do you see this in real life? Every time a car accelerates, a cyclist brakes, or an athlete sprints, the work-energy theorem is at work — force acting over a distance changes the kinetic energy of the object.

Practice Problems

A force of 50 N moves an object 8 m. How much work is done?

Show answer

W = 50 × 8 = 400 J

A 3 kg object moving at 6 m/s. Find its kinetic energy.

Show answer

Ek = ½ × 3 × 36 = 54 J