PhysicsICSE Class 10 · Different Types of Energy

Gravitational Potential Energy: Derivation of U = mgh and Solved Numericals (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Potential energy is the energy a body has because of its position or configuration. A mass m raised through a height h above a reference level gains gravitational potential energy U = mgh, equal to the work done against gravity in lifting it. It is measured in joules and depends on the reference level chosen.

Lifting mass m steadily through height h needs force mg, so the work mgh is stored as potential energy.

Illustration drawn for QuestionBuddy.

Understand Gravitational Potential Energy (U = mgh)

Potential energy (PE, $U$ ) is the energy stored in a body because of its position or its configuration (shape). It is called 'potential' because it can be turned into work later.

Gravitational PE: due to height above a reference level. Examples: water in an overhead tank or behind a dam, a raised hammer, a stone on a hilltop.
Elastic PE: due to a change of shape. Examples: a wound watch spring, a stretched bow or catapult, a compressed spring in a toy. (Only qualitative examples are needed.)

**Derivation of $U = m g h$ (asked in the syllabus).** Take a body of mass $m$ resting on the ground (the reference level). To lift it slowly, at steady speed, you must apply an upward force equal to its weight:

$F = m g .$

Lift it through a vertical height $h$ . The force and displacement are both upward, so the work done against gravity is

$W = F \times h = m g \times h = m g h .$

This work is not lost: it is stored in the body as gravitational potential energy. Hence

$U = m g h .$

If the body is allowed to fall back, it can do this same amount of work (for example, driving a nail or turning a turbine).

Reference level. $h$ is measured from a chosen zero level, usually the ground or the floor. The value of $U$ depends on this choice, but the change in PE between two positions does not:

$Δ U = m g (h_{2} - h_{1}) .$

If a body goes below the reference level, its PE is negative relative to that level.

Facts to remember.

$U$ is proportional to both $m$ and $h$ : double the height, double the PE.
Only the vertical height matters, not the path taken.
Unit: joule (J). Use $m$ in kg, $h$ in m and $g$ as given (10 or 9.8 m s⁻²).
For an extended body (a ladder, a tank of water), $h$ is the height of its centre of gravity.

Where it is tested. PE numericals in ICSE Class 10 are for translational motion only (the syllabus limit), usually 2-3 marks, and often lead into conservation of energy in Chapter 6.

Gravitational Potential Energy (U = mgh): formulas and key facts

Gravitational potential energy: $U = m g h$; h = vertical height above the reference level.
Change in potential energy: $Δ U = m g (h_{2} - h_{1})$; Independent of the reference level.
Work against gravity stored as PE: $W = F h = m g h = U$; For a slow, steady lift.
Height from PE: $h = \frac{U}{m g}$; Rearranged.

How to solve Gravitational Potential Energy (U = mgh) questions

Choose the reference level first, then use the vertical height from it.

1Note the reference level stated (ground, floor, table); if none, use the ground.
2Find the vertical height $h$ of the body (or of its C.G.) above that level.
3Convert mass to kg and height to m.
4Substitute in $U = m g h$ with the given $g$ ; write the unit J.
5For a change of position, use $Δ U = m g (h_{2} - h_{1})$ ; a negative answer means PE is lost.
6To find a height or mass, rearrange the formula before substituting.

Gravitational Potential Energy (U = mgh) in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Energy stored by climbing stairs

Suppose you carry a 15 kg bag of rice up to a third-floor flat 9 m above the ground.

How much gravitational PE does the bag gain? ( $g = 10 m s^{- 2}$ ) Where did this energy come from?

Try it first, then show the solutionSolution

$U = m g h = 15 \times 10 \times 9 = 1350$ J.
It came from the work your muscles did against gravity (from the chemical energy of food).

Answer: 1350 J, from the work you did (chemical energy of food).

Memory hook: The work you put in going up is stored as PE.

PE used to do work

Imagine a pile driver lifts a 200 kg hammer 5 m above the top of a post and then releases it.

Find the PE of the raised hammer relative to the post top, and say what it does with it. ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

$U = 200 \times 10 \times 5 = 10000$ J.
As it falls this becomes kinetic energy, which does work driving the post into the ground.

Answer: $1.0 \times 1 0^{4}$ J, which drives the post in.

Memory hook: Lift it high to hit it hard.

Tricks to remember Gravitational Potential Energy (U = mgh)

Short memory aids. Say each one aloud once, then use it on a question the same day.

M-G-H: 'My Great Height'
U = mgh: mass × g × height.
Use it when: Recalling the formula.
Work in, PE stored
The derivation in one line: force mg, displacement h, work mgh, stored as U.
Use it when: The 'derive U = mgh' question.
Zero is your choice
PE depends on where you put zero; the change in PE does not.
Use it when: Reference-level questions.
Shape or place
Spring, bow, catapult: shape (elastic). Tank, dam, hammer: place (gravitational).
Use it when: Examples of potential energy.

Gravitational Potential Energy (U = mgh): solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: PE of a raised body · 1 mark · MCQBoard level

A body of mass 2 kg is at a height of 5 m above the ground. Its gravitational potential energy is ( $g = 10 m s^{- 2}$ ):
(a) 10 J
(b) 50 J
(c) 100 J
(d) 25 J

Try it first, then show the solutionSolution

$U = m g h = 2 \times 10 \times 5$ .
$= 100$ J.

Answer: (c) 100 J

Example 2: Derivation of U = mgh · 3 marks · Short answerBoard level

A body of mass $m$ is raised slowly from a height $h_{1}$ to a height $h_{2}$ above the ground. Derive an expression for the gain in its gravitational potential energy, and hence show that a body at a height $h$ above the ground has $U = m g h$ .

Try it first, then show the solutionSolution

To raise the body at steady speed, an upward force equal to its weight is needed: $F = m g$ . (1)
The displacement $(h_{2} - h_{1})$ is in the direction of the force, so the work done against gravity is $W = m g (h_{2} - h_{1})$ ; this work is stored in the body as gravitational PE: $Δ U = m g (h_{2} - h_{1})$ . (1)
Taking the ground as the reference level, a body lifted from the ground ( $h_{1} = 0$ ) to a height $h$ has $U = m g (h - 0) = m g h$ . (1)

Answer: $Δ U = m g (h_{2} - h_{1})$ ; with $h_{1} = 0$ , $U = m g h$

Example 3: Three reference levels · 3 marks · Short answerHarder board level

A 3 kg flower pot stands on a balcony 6 m above the ground. Near the building is a well 4 m deep. Find the PE of the pot relative to (a) the ground, (b) the bottom of the well, (c) the balcony floor. ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

(a) $h = 6$ m: $U = 3 \times 10 \times 6 = 180$ J. (1)
(b) $h = 6 + 4 = 10$ m: $U = 3 \times 10 \times 10 = 300$ J. (1)
(c) $h = 0$ : $U = 0$ . (1)

Answer: (a) 180 J (b) 300 J (c) 0 J

Example 4: Change in PE on a mountain · 3 marks · Short answerHarder board level

A climber of mass 60 kg climbs from a camp at 1200 m to a ridge at 1500 m, then comes down 100 m to rest. Find the change in her PE (a) during the climb, (b) during the descent. ( $g = 9.8 m s^{- 2}$ )

Try it first, then show the solutionSolution

(a) $Δ U = m g (h_{2} - h_{1}) = 60 \times 9.8 \times 300 = 1.764 \times 1 0^{5}$ J (gain). (1½)
(b) $Δ U = 60 \times 9.8 \times (- 100) = - 5.88 \times 1 0^{4}$ J (loss). (1½)

Answer: (a) Gain of $1.764 \times 1 0^{5}$ J (b) Loss of $5.88 \times 1 0^{4}$ J

Example 5: PE of water in a roof tank · 3 marks · Short answerHOTS / competency

A roof tank 2 m tall stands with its base 15 m above a pump. When full it holds 2000 kg of water. ( $g = 10 m s^{- 2}$ )
(a) At what height does the water's weight act?
(b) Find the PE of the water relative to the pump.
(c) A student uses $h = 15$ m. By how much is the answer wrong?

Try it first, then show the solutionSolution

(a) At the C.G. of the water, the middle of the 2 m column: $15 + 1 = 16$ m above the pump. (1)
(b) $U = 2000 \times 10 \times 16 = 3.2 \times 1 0^{5}$ J. (1)
(c) $2000 \times 10 \times 15 = 3.0 \times 1 0^{5}$ J, too small by $2 \times 1 0^{4}$ J. (1)

Answer: (a) 16 m (b) $3.2 \times 1 0^{5}$ J (c) $2 \times 1 0^{4}$ J too small

Example 6: Case study: stacking bricks · 4 marks · Case-basedHOTS / competency

Five identical bricks, each of mass 2 kg and 0.1 m thick, lie flat side by side on the ground. A mason stacks them one on top of another. ( $g = 10 m s^{- 2}$ )
(a) Through what height is the second brick's C.G. raised? And the fifth's?
(b) Find the total work done (the gain in PE).
(c) Check your answer using the C.G. of all five bricks together.

Try it first, then show the solutionSolution

(a) The bottom brick stays; the second rises 0.1 m, the third 0.2 m, the fourth 0.3 m, the fifth 0.4 m. (1)
(b) $W = 2 \times 10 \times (0.1 + 0.2 + 0.3 + 0.4) = 20 \times 1.0 = 20$ J. (1)
(c) Lying flat, the C.G. of all bricks is at 0.05 m; stacked, at the middle of a 0.5 m pile, 0.25 m. (1)
Gain $= 10 \times 10 \times (0.25 - 0.05) = 20$ J, the same. (1)

Answer: (a) 0.1 m; 0.4 m (b) 20 J (c) 20 J, using the C.G. rising from 0.05 m to 0.25 m

Where students lose marks in Gravitational Potential Energy (U = mgh)

The slips examiners see most often, and what to write instead.

Mistake: Using grams or centimetres
The joule needs kg and m.
Wrong working
$U = 500 \times 10 \times 160$ for 500 g at 160 cm.
Right working
$U = 0.5 \times 10 \times 1.6 = 8$ J.
Do this: Convert before substituting.
Usually costs: The answer mark.
Mistake: Wrong reference level
The height must be measured from the level named in the question.
Wrong working
PE relative to the floor of a room 8 m above ground, using $h = 9.5$ m for a shelf 1.5 m above that floor.
Right working
Use $h = 1.5$ m relative to the floor.
Do this: Underline 'relative to …' in the question.
Usually costs: The answer mark.
Mistake: Incomplete derivation
The derivation needs the force ( $m g$ ), the work ( $F h$ ) and the statement that work is stored as PE.
Wrong working
$U = m g h$ because PE depends on mass and height.
Right working
Force $= m g$ ; work $= m g \times h$ ; this work is stored as $U = m g h$ .
Do this: Write the three steps in order.
Usually costs: Up to 2 of 3 derivation marks.
Mistake: Mixing up PE and work done by gravity
When a body rises, its PE increases but gravity does negative work.
Wrong working
Work done by gravity as a 2 kg body rises 3 m: +60 J.
Right working
Gravity does −60 J; the body gains 60 J of PE.
Do this: Say whose work: yours (positive) or gravity's (negative).
Usually costs: Sign marks.

Exam tips

The derivation of $U = m g h$ is named in the 2027 syllabus: practise writing it in three numbered steps.
Always state the value of $g$ you use and the reference level.
PE numericals are for translational motion only; you will not be asked about rotational energy calculations.
Give both gravitational and elastic examples when asked for 'examples of potential energy'.

Practice questions on Gravitational Potential Energy (U = mgh)

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of these has potential energy because of its configuration rather than its height?
(a) a wound watch spring
(b) water in an overhead tank
(c) a raised hammer
(d) a stone on a hilltop
Hint
Configuration means shape.
Answer
(a)
Q2 · 1 mark · Very short answerBoard level
A ball of mass 500 g rests on a shelf 1.6 m above the floor. Find its PE relative to the floor. ( $g = 10 m s^{- 2}$ )
Hint
Convert 500 g to kg.
Answer
$0.5 \times 10 \times 1.6 = 8$ J.
Q3 · 2 marks · Short answerHarder board level
At what height above the ground will a 4 kg body have a PE of 196 J? ( $g = 9.8 m s^{- 2}$ )
Hint
$h = \frac{U}{m g}$ .
Answer
$h = \frac{196}{4 \times 9.8} = 5$ m.
Q4 · 2 marks · Short answerHarder board level
Body A (2 kg) is 10 m above the ground and body B (5 kg) is 3 m above the ground. Which has more PE, and what is the ratio $U_{A} : U_{B}$ ? ( $g = 10 m s^{- 2}$ )
Hint
Compare the products $mh$ .
Answer
$U_{A} = 200$ J, $U_{B} = 150$ J; A has more; $4 : 3$ .
Q5 · 2 marks · Short answerHarder board level
A body has a PE of 600 J at a certain height. What is its PE if (a) the height is doubled, (b) the mass is halved and the height tripled?
Hint
$U \propto mh$ .
Answer
(a) 1200 J (b) $600 \times \frac{1}{2} \times 3 = 900$ J.
Q6 · 2 marks · Short answerHarder board level
A 1 kg stone lies at the bottom of a pit 2 m deep. Taking the ground as the reference level, find its PE. What does the sign mean? ( $g = 10 m s^{- 2}$ )
Hint
$h$ is below the zero level.
Answer
$U = 1 \times 10 \times (- 2) = - 20$ J. Negative means work must be done (20 J) to bring it up to ground level.
Q7 · 3 marks · Short answerHOTS / competency
A uniform ladder 4 m long and of mass 10 kg lies flat on the ground. How much work is needed to stand it upright on one end? ( $g = 10 m s^{- 2}$ )
Hint
Track the height of its centre of gravity.
Answer
The C.G. rises from ground level to 2 m: $W = 10 \times 10 \times 2 = 200$ J.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.PE of 1 kg at 1 m ( $g = 10$ )?Show answer
10 J.
2.What happens to PE if the height is halved?Show answer
It halves.
3.Give one example of elastic PE.Show answer
A wound watch spring (or a stretched bow).
4.Does the change in PE depend on the reference level?Show answer
No.
5.Height of a 5 kg body with PE 250 J ( $g = 10$ )?Show answer
5 m.

Gravitational Potential Energy (U = mgh): frequently asked questions

What is gravitational potential energy?

It is the energy a body has because of its height above a reference level. It equals the work done against gravity in lifting it: U = mgh.

Why does potential energy depend on the reference level?

Because h is measured from a chosen zero level. Changing that level changes h, and so U; but the change in PE between two positions is the same whatever level is chosen.

Is the derivation of U = mgh in the ICSE Class 10 syllabus?

Yes. The 2027 syllabus says 'derivation included' for U = mgh, so learn it in steps: force mg, work mgh, stored as PE.