Energy, work & power

GPE, KE and efficiency

Efficiency Core 4 min read

Gravitational potential energy (GPE)

The gravitational potential energy of an object is the energy stored due to its height above a reference level. \[ \Delta \text{GPE} = mg\Delta h \]

Symbol	Quantity	Unit
\( \Delta \text{GPE} \)	Change in gravitational potential energy	joule, J
\( m \)	Mass	kilogram, kg
\( g \)	Gravitational field strength	N/kg (= 10 N/kg near Earth's surface)
\( \Delta h \)	Change in height	metre, m

Kinetic energy (KE)

The kinetic energy of a moving object is the energy stored due to its motion. \[ \text{KE} = \tfrac{1}{2}mv^2 \]

Symbol	Quantity	Unit
\( \text{KE} \)	Kinetic energy	joule, J
\( m \)	Mass	kilogram, kg
\( v \)	Speed	metre per second, m/s

Doubling the speed quadruples the KE (since KE ∝ v²). This is why speed is so dangerous in road accidents.

GPE and KE conversions

For a freely falling object (no air resistance), energy transfers between GPE and KE with none wasted: \[ \text{Loss in GPE} = \text{Gain in KE} \] \[ mg\Delta h = \tfrac{1}{2}mv^2 \] Mass cancels, giving: \( v = \sqrt{2g\Delta h} \) — the final speed of a freely falling object depends only on \( g \) and the height fallen, not on the mass.

Example 1 — find speed from height (free fall)

A ball is dropped from a height of 5.0 m. Find the speed just before it hits the ground. (g = 10 N/kg, ignore air resistance)

Loss in GPE = gain in KE:

\( mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5.0} = \sqrt{100} = 10\,\text{m/s} \)

Example 2 — find height from speed (projectile)

A ball is thrown upward at 14 m/s. Find the maximum height reached. (g = 10 N/kg, ignore air resistance)

All KE converts to GPE at the top:

\( \tfrac{1}{2}mv^2 = mgh \Rightarrow h = \dfrac{v^2}{2g} = \dfrac{14^2}{2 \times 10} = \dfrac{196}{20} = 9.8\,\text{m} \)

Example 3 — energy with air resistance

A 2.0 kg rock falls 10 m and reaches a speed of 12 m/s (not 14 m/s as predicted by free fall). How much energy was transferred to thermal energy by air resistance?

Initial GPE lost: \( mgh = 2.0 \times 10 \times 10 = 200\,\text{J} \)

Final KE gained: \( \tfrac{1}{2} \times 2.0 \times 12^2 = 144\,\text{J} \)

Thermal energy = \( 200 - 144 = 56\,\text{J} \)

Efficiency

Efficiency is the fraction of the input energy that is usefully transferred. \[ \text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \] As a percentage: \[ \text{efficiency (\%)} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\% \] Extended (Supplement) — can also be expressed as: \[ \text{efficiency} = \frac{\text{useful power output}}{\text{total power input}} \]

Efficiency has no unit (or is expressed as a percentage).
Efficiency is always ≤ 1 (or ≤ 100%) — you can never get more useful energy out than you put in.
A value > 1 (or > 100%) is impossible and indicates an error in calculation.
The wasted energy = total input − useful output.

Example 4 — calculate efficiency

A motor takes in 500 J of electrical energy and produces 350 J of useful kinetic energy. Calculate the efficiency.

\( \text{efficiency} = \dfrac{350}{500} = 0.70 \quad (= 70\%) \)

Wasted energy = 500 − 350 = 150 J (transferred as thermal energy).

Example 5 — efficiency using power (Extended)

A petrol engine has an input power of 80 kW and a useful output power of 20 kW. Find the efficiency.

\( \text{efficiency} = \dfrac{20}{80} = 0.25 \quad (= 25\%) \)

Example 6 — find useful output energy from efficiency

A light bulb is 12% efficient and uses 50 J of electrical energy. How much light energy does it emit?

\( \text{useful output} = 0.12 \times 50 = 6.0\,\text{J} \)

Wasted as thermal = 50 − 6 = 44 J.

Do not confuse efficiency with percentage change. Efficiency = useful output ÷ total input — always divide by the total input, never the useful output. And remember: no real machine is 100% efficient because some energy is always transferred to thermal energy by friction or resistance.

For 6-mark "describe and explain" efficiency questions: (1) state the formula; (2) identify useful and wasted energy stores; (3) explain where wasted energy goes (friction → thermal; air resistance → thermal; electrical resistance → thermal). Relate Sankey diagram arrow widths to energy amounts.