Potential energy problems

A crane lifts a 500 kg steel beam to the top of a building. The beam does not move once it is up there, yet it now holds something it did not have on the ground — the capacity to fall, to smash, to do damage if the cable snapped. That stored capacity is gravitational potential energy, and the whole of this topic is one formula, $U = mgh$ , used in three directions: straight substitution to find $U$ , and two rearrangements to find the mass or the height when the energy is already known. The physics is not hard. What separates a correct answer from a wrong one is discipline — writing down what is given, choosing the right rearrangement, keeping units in kilograms, metres and joules, and checking the final number makes sense. This article drills exactly that discipline with fully worked problems, so that a rearrangement question feels no harder than a plug-in question.

The core insight

Every gravitational potential energy problem is $U = mgh$ solved for whichever one of $U$ , $m$ or $h$ is missing, with $g = 9.8\ \text{m/s}^2$ and all quantities in SI units.

The core idea

One formula, three uses. The gravitational potential energy of an object relative to a reference level is

U = mgh

where: $U$ is potential energy in joules (J), $m$ is mass in kilograms (kg), $h$ is height above the reference level in metres (m), and $g = 9.8\ \text{m/s}^2$ is the acceleration due to gravity. Every problem in this topic gives two of $\{U, m, h\}$ and asks for the third — nothing more exotic than that.

The two rearrangements. If $U$ is already known, mass or height can be isolated:

m = \frac{U}{gh} \qquad \text{and} \qquad h = \frac{U}{mg}

Both come from dividing both sides of $U = mgh$ by whichever quantity is not wanted. There is no need to memorise these separately — re-derive whichever one you need on the spot from $U = mgh$ , so a moment of forgetfulness never costs the whole problem.

A fixed method. For any of the three question types: (1) list what is given and what is asked, (2) pick the correct form of the equation, (3) substitute with height in metres and mass in kilograms, and (4) check the unit and the size of the answer. Skipping step 1 is the single biggest cause of picking the wrong rearrangement under exam pressure.

Comparing two objects. Some questions describe two objects at different heights and different masses and ask which has more potential energy. Mass and height cannot be compared separately — a lighter object high up can have more or less $U$ than a heavier object lower down, depending on the actual numbers. The only safe method is to compute $U = mgh$ for each object in full and then compare the two results.

Figure

Bar chart comparing two boxes showing mass height and the resulting potential energy for each

Derivation AGravitational potential energy from work done against gravity

The goal is to show that the formula $U = mgh$ , which so far has only been used as a rule to substitute into, is not an arbitrary label but a direct consequence of the definition of work — the energy stored in a raised object equals the work someone had to do to lift it there at a steady speed, and tracing that work through in symbols is what produces the $mgh$ that every problem in this topic substitutes into.

Step 1. To lift an object of mass $m$ straight up without speeding it up or slowing it down, an external force must exactly balance its weight. The weight of the object is $F = mg$ , so the lifting force needed is also $F = mg$ , applied over the height $h$ through which the object rises.

Step 2. Work done is force multiplied by distance moved in the direction of the force: $W = F \times h$ . Substituting the lifting force from Step 1, $W = mg \times h = mgh$ .

Step 3. This work does not disappear — it is stored in the object as gravitational potential energy, because the object could now do that same amount of work on something else while falling back down (for example, driving a pile into the ground, or turning a wheel). By the work-energy link, the potential energy gained equals the work done in lifting: $U = W = mgh$ .

Step 4. Sanity check: if $h = 0$ (the object stays at the reference level), $U = 0$ , since no lifting work was done — matching the everyday sense that an object on the ground has no stored height-energy relative to the ground.

Worked example

Question: A 4 kg box is raised to a height of 5 m and a 9 kg box is raised to a height of 2 m, both above the same floor. Which box has more gravitational potential energy, and by how much? (Take $g = 9.8\ \text{m/s}^2$ .)

Step 1. List the given values. Box A: $m_A = 4\ \text{kg}$ , $h_A = 5\ \text{m}$ . Box B: $m_B = 9\ \text{kg}$ , $h_B = 2\ \text{m}$ . Asked: compare $U_A$ and $U_B$ .

Step 2. Compute $U_A = m_A g h_A = 4 \times 9.8 \times 5 = 196\ \text{J}$ .

Step 3. Compute $U_B = m_B g h_B = 9 \times 9.8 \times 2 = 176.4\ \text{J}$ .

Step 4. Compare: $196\ \text{J} > 176.4\ \text{J}$ , so Box A, the lighter box placed higher, has more potential energy, exceeding Box B by $196 - 176.4 = 19.6\ \text{J}$ . Box A has more potential energy, by 19.6 J, even though Box B is more than twice as massive — confirming that mass and height must both be combined through the formula before comparing.

Competitive stretch

Suppose Box B in the example above is instead raised to a height $h_B'$ such that both boxes end up with exactly equal potential energy. Find $h_B'$ .

Step 1. Set $U_B' = U_A = 196\ \text{J}$ , with $m_B = 9\ \text{kg}$ still fixed.

Step 2. Rearrange $U = mgh$ for height: $h_B' = \dfrac{U_B'}{m_B g} = \dfrac{196}{9 \times 9.8} = \dfrac{196}{88.2}$ .

Step 3. $h_B' = 2.22\ \text{m}$ (to 3 significant figures). Box B would need to be raised to about 2.22 m for the two boxes to store equal potential energy.

Common traps

Dropping $g$ out of a rearranged formula, writing $m = U/h$ or $h = U/m$ instead of dividing by $gh$ or $mg$ .
Substituting height in centimetres or mass in grams directly, without first converting to metres and kilograms — this throws the answer off by a factor of 100 or 1000.
Comparing two objects' energy by mass alone or height alone, instead of calculating $U = mgh$ separately for each and comparing the actual joule values.
Using an inconsistent value of $g$ partway through a multi-part calculation, or confusing mass with weight when substituting into $U = mgh$ .

Ask the tutor: Stuck on a rearrangement or want more mixed practice problems with $U = mgh$ ? Ask the AI tutor to generate and walk through a few with you.

Do I always have to use g equals 9.8, or can I round it to 10?

For ICSE Class 8 problems in this topic, always use $g = 9.8\ \text{m/s}^2$ unless the question explicitly states a different value. Rounding to 10 changes every answer slightly and can make a "clean" answer like 5 kg come out as 4.9 kg instead, which examiners mark as an error. Keep $g = 9.8$ fixed throughout a calculation — mixing 9.8 in one step and 10 in another is a common and easily avoided mistake.

Why does a heavier object not always have more potential energy than a lighter one?

Because potential energy depends on the product of mass and height, not on either quantity alone. A heavy object close to the ground and a light object high up can have similar, greater, or smaller potential energy than each other depending on the exact numbers — you cannot judge by eye. That is exactly why comparison questions require computing $U = mgh$ fully for each object rather than reasoning from mass or height in isolation, as shown in the worked example above.

I keep picking the wrong rearranged formula under time pressure — any trick?

Do not memorise $m = U/(gh)$ and $h = U/(mg)$ as separate facts to recall; instead, always start from $U = mgh$ and divide both sides by whatever you do not want to isolate. If you are solving for $m$ , divide by $gh$ ; if solving for $h$ , divide by $mg$ . Practising this re-derivation a few times makes it faster than trying to recall two separate formulas, and it never fails even if your memory does.

Potential energy problems

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads