WASSCE Physics · Work, Energy, and Power

Work, energy, power and conservation of energy

By the WASSCE Tutor team

Updated 10 November 2026 · 7 min read · Prepared against the WASSCE syllabus.

Energy is what makes things happen: it moves vehicles, lights homes, cooks food and keeps you alive. Physics gives it an exact meaning and a law that has never been broken: energy is never created or destroyed, only changed from one form into another.

Learning objectives

By the end of this lesson you should be able to:

  • define work, energy and power and state their units
  • calculate work done by a force, including from a force–distance graph
  • calculate kinetic and gravitational potential energy
  • apply the principle of conservation of energy

Work

Definition

Work is done when a force moves its point of application in the direction of the force:

W=F×sW = F \times s

where ss 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 angle θ to the direction of motion, only its component along the motion does work:

W=Fscos⁡θW = Fs\cos\theta
Common mistake

No work is done if there is no movement, or if the movement is perpendicular to the force. A man holding a heavy bag still does no work on it (though his muscles tire); a waiter carrying a tray horizontally at constant height does no work on the tray against gravity.

Worked example

A boy pulls a box 20 m along the floor with a rope inclined at 60° to the horizontal. The tension is 50 N. Find the work done by the boy.

W=Fscos⁡θ=50×20×cos⁡60°=500 JW = Fs\cos\theta = 50 \times 20 \times \cos 60° = 500\ \text{J}
W=Fscos⁡θ=50×20×cos⁡60°=500 J\begin{aligned}W &= Fs\cos\theta \\ &= 50 \times 20 \times \cos 60° \\ &= 500\ \text{J}\end{aligned}

Work from a graph

The area under a force–distance graph is the work done.

Force against extension for a spring: a straight line from the origin; the shaded triangle under it represents the work done, half F times x

Figure 1. Work done in stretching a spring is the area under the graph. WASSCE Tutor diagram.

For a spring stretched by xx with a final force FF: W=12Fx=12kx2W = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^2 (this is the elastic potential energy stored).

Energy

Definition

Energy is the capacity (ability) to do work. Its unit is also the joule (J).

Forms of energy include kinetic, gravitational potential, elastic potential, chemical, heat (thermal), electrical, light, sound and nuclear energy.

Kinetic energy

Definition

Kinetic energy is the energy a body has because of its motion:

Ek=12mv2E_k = \tfrac{1}{2}mv^2

Gravitational potential energy

Definition

Gravitational potential energy is the energy a body has because of its position (height) in a gravitational field:

Ep=mghE_p = mgh
Worked example

A 1200 kg car moves at 15 m s⁻¹. Find its kinetic energy. How high would it have to be lifted to have the same potential energy? (g=10g = 10 m s⁻²)

Ek=12×1200×152=135 000 JE_k = \tfrac{1}{2} \times 1200 \times 15^2 = 135\,000\ \text{J}
Ek=12×1200×152=135 000 J\begin{aligned}E_k &= \tfrac{1}{2} \times 1200 \times 15^2 \\ &= 135\,000\ \text{J}\end{aligned}
mgh=135 000⇒h=135 00012 000=11.25 mmgh = 135\,000 \Rightarrow h = \dfrac{135\,000}{12\,000} = 11.25\ \text{m}
mgh=135 000⇒h=135 00012 000=11.25 m\begin{gathered}mgh = 135\,000 \\[6pt] \begin{aligned}\Rightarrow h &= \dfrac{135\,000}{12\,000} \\ &= 11.25\ \text{m}\end{aligned}\end{gathered}

A crash at 54 km h⁻¹ is like falling from a three- or four-storey building.

The principle of conservation of energy

Definition

Principle of conservation of energy: energy can neither be created nor destroyed; it can only be changed from one form to another. The total energy of an isolated system remains constant.

For a body falling freely (no air resistance), potential energy turns into kinetic energy:

mgh=12mv2⇒v=2ghmgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}
mgh=12mv2⇒v=2gh\begin{gathered}mgh = \tfrac{1}{2}mv^2 \\[6pt] \Rightarrow v = \sqrt{2gh}\end{gathered}
Worked example

A 2 kg stone falls from rest from a height of 20 m. Find its kinetic energy and speed just before it hits the ground, and its kinetic energy halfway down. (g=10g = 10 m s⁻²)

At the top: Ep=2×10×20=400E_p = 2 \times 10 \times 20 = 400 J, Ek=0E_k = 0.

Just before landing: all 400 J is kinetic, so 12(2)v2=400⇒v=20\tfrac{1}{2}(2)v^2 = 400 \Rightarrow v = 20 m s⁻¹.

Halfway (10 m down): Ep=200E_p = 200 J, so Ek=400−200=200E_k = 400 - 200 = 200 J. The total stays 400 J throughout.

Energy transformations

Device or processEnergy change
Hydroelectric station (e.g. Bumbuna)gravitational PE of water → KE → electrical
Generator (gen-set)chemical (fuel) → heat → kinetic → electrical
Solar panellight → electrical
Electric fanelectrical → kinetic (+ some heat and sound)
Battery charging a phoneelectrical → chemical
Photosynthesislight → chemical
Burning firewood or charcoalchemical → heat and light

Power

Definition

Power is the rate of doing work or the rate of transferring energy:

P=Wt=EtP = \dfrac{W}{t} = \dfrac{E}{t}

Its unit is the watt (W): 1 W = 1 J s⁻¹. (1 horsepower ≈ 746 W.)

For a body moving at constant speed vv against a force FF:

P=Fst=FvP = \dfrac{Fs}{t} = Fv
Worked example

A 60 kg student runs up a flight of stairs 4 m high in 5 s. Find her power. (g=10g = 10 m s⁻²)

P=mght=60×10×45=480 WP = \dfrac{mgh}{t} = \dfrac{60 \times 10 \times 4}{5} = 480\ \text{W}
P=mght=60×10×45=480 W\begin{aligned}P &= \dfrac{mgh}{t} \\ &= \dfrac{60 \times 10 \times 4}{5} \\ &= 480\ \text{W}\end{aligned}

Key points

  • Work W=Fscos⁡θW = Fs\cos\theta (J); no movement in the direction of the force means no work.
  • Area under a force–distance graph = work done.
  • Ek=12mv2E_k = \tfrac{1}{2}mv^2; Ep=mghE_p = mgh.
  • Energy is conserved: it only changes form.
  • Power P=W/t=FvP = W/t = Fv (W).

WASSCE examination focus

WAEC tests this

Define work, energy and power with units, state the principle of conservation of energy, and solve problems on falling bodies, pendulums and inclines using mgh=12mv2mgh = \tfrac{1}{2}mv^2. Questions also ask for the energy changes in devices (a hydroelectric station, a generator, a solar panel) and for power in climbing stairs or lifting loads.

Quick check

  1. State the SI unit of power.
  2. A 0.5 kg ball moves at 8 m s⁻¹. Find its kinetic energy.
  3. A crane lifts a 500 kg load through 12 m in 20 s. Find its power output. (g=10g = 10 m s⁻²)

Answers: 1. The watt (W). 2. 12(0.5)(64)=16\tfrac{1}{2}(0.5)(64) = 16 J. 3. 500×10×1220=3000\dfrac{500 \times 10 \times 12}{20} = 3000 W.

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