Physics 9702 · Paper 2

Gravitational Potential Energy

Lift an object and you do work against gravity. That work is stored: let go, and it comes back as motion.
The stored energy is (GPE).

The formula and its derivation

\Delta E_p = mg\,\Delta h

Symbols

$\Delta E_p$ = change in gravitational potential energy (J)
$m$ = mass (kg)
$g$ = acceleration of free fall (m s⁻²)
$\Delta h$ = change in vertical height (m)

Worked example

The derivation the exam asks for

A block of mass m is raised vertically at constant speed through a height Δh. Derive an expression for its gain in gravitational potential energy. (9702/21/O/N/24 Q3(b))

At constant speed the lifting force just balances the weight: $F = mg$ , where g is the acceleration of free fall.
Work done by the lifting force: $W = F \times \Delta h = mg\,\Delta h$ .
That work is stored as potential energy: $\Delta E_p = mg\,\Delta h$ .

Answer

\Delta E_p = mg\,\Delta h

Both marks come from naming the force (weight, mg) and from linking the energy to work done. Lifting at steady speed does not need a force bigger than the weight.

$\Delta h$ is the vertical height change, never the slope or path length. The two facts the formula needs are the weight and the vertical rise (9702/11/M/J/25 Q17).
The formula assumes g is constant, so it works near the Earth's surface but not out at satellite heights.
If a question gives the weight in newtons, it is already mg: use $\Delta E_p = W \times \Delta h$ directly.

Worked example

Smallest case: climbing a ladder

A 65 kg student climbs 2.0 m vertically. Find the gain in GPE.

$\Delta E_p = mg\Delta h = 65 \times 9.81 \times 2.0 = 1.3 \times 10^{3} \text{ J}$ .

Answer

\Delta E_p = 1.3 \times 10^{3} \text{ J}

Height care: the two standard traps

Rebounds use the net height. A ball falls from 2.4 m and bounces back up to 1.8 m. For the overall GPE change, the height that matters is $2.4 - 1.8 = 0.6 \text{ m}$ (9702/12/O/N/25 Q18).
Ramps do not change the answer. The minimum work to raise a 50 kg barrel onto a lorry 1.6 m high is $mgh = 50 \times 9.81 \times 1.6 = 780 \text{ J}$ , whether it goes straight up or along a 3.4 m plank. The plank length is a distractor (9702/12/O/N/24 Q18).
Watch the mass unit: 28 g raised 4.6 m is $0.028 \times 9.81 \times 4.6 = 1.3 \text{ J}$ , not 1300 J (9702/12/M/J/24 Q17).

GPE cares about one thing only: the vertical height change. Path, slope and speed do not enter the formula.

PreviousWork Done by a Gas NextKinetic Energy

Gravitational Potential Energy

Lift an object and you do work against gravity. That work is stored: let go, and it comes back as motion.
The stored energy is (GPE).

The formula and its derivation

\Delta E_p = mg\,\Delta h

Symbols

$\Delta E_p$ = change in gravitational potential energy (J)
$m$ = mass (kg)
$g$ = acceleration of free fall (m s⁻²)
$\Delta h$ = change in vertical height (m)

Worked example

The derivation the exam asks for

A block of mass m is raised vertically at constant speed through a height Δh. Derive an expression for its gain in gravitational potential energy. (9702/21/O/N/24 Q3(b))

At constant speed the lifting force just balances the weight: $F = mg$ , where g is the acceleration of free fall.
Work done by the lifting force: $W = F \times \Delta h = mg\,\Delta h$ .
That work is stored as potential energy: $\Delta E_p = mg\,\Delta h$ .

Answer

\Delta E_p = mg\,\Delta h

Both marks come from naming the force (weight, mg) and from linking the energy to work done. Lifting at steady speed does not need a force bigger than the weight.

$\Delta h$ is the vertical height change, never the slope or path length. The two facts the formula needs are the weight and the vertical rise (9702/11/M/J/25 Q17).
The formula assumes g is constant, so it works near the Earth's surface but not out at satellite heights.
If a question gives the weight in newtons, it is already mg: use $\Delta E_p = W \times \Delta h$ directly.

Worked example

Smallest case: climbing a ladder

A 65 kg student climbs 2.0 m vertically. Find the gain in GPE.

$\Delta E_p = mg\Delta h = 65 \times 9.81 \times 2.0 = 1.3 \times 10^{3} \text{ J}$ .

Answer

\Delta E_p = 1.3 \times 10^{3} \text{ J}

Height care: the two standard traps

Rebounds use the net height. A ball falls from 2.4 m and bounces back up to 1.8 m. For the overall GPE change, the height that matters is $2.4 - 1.8 = 0.6 \text{ m}$ (9702/12/O/N/25 Q18).
Ramps do not change the answer. The minimum work to raise a 50 kg barrel onto a lorry 1.6 m high is $mgh = 50 \times 9.81 \times 1.6 = 780 \text{ J}$ , whether it goes straight up or along a 3.4 m plank. The plank length is a distractor (9702/12/O/N/24 Q18).
Watch the mass unit: 28 g raised 4.6 m is $0.028 \times 9.81 \times 4.6 = 1.3 \text{ J}$ , not 1300 J (9702/12/M/J/24 Q17).

GPE cares about one thing only: the vertical height change. Path, slope and speed do not enter the formula.