Potential energy

A stone perched on a cliff edge, water held back behind a dam, a stretched catapult before release — none of these are moving, yet each is "waiting" to do work the instant it is let go. This stored, positional energy is called potential energy, and it is the second core, formula-based idea in the CBSE Class 9 chapter "Work and Energy," studied right after kinetic energy. The syllabus asks for one quantitative formula — gravitational potential energy, $E_p = mgh$ — derived as the work done in raising a body against gravity, plus the idea (by name only, no formula) that a stretched spring or bent bow also stores energy because of its changed shape. This article builds that one formula carefully, works through it with real numbers, and connects it to where the stored energy goes next.

The core insight

A body stores gravitational potential energy $E_p = mgh$ purely because of its height above a chosen reference level, and this energy equals the work someone did against gravity to place it there.

The core idea

Potential energy is the energy a body possesses by virtue of its position or its configuration (shape or arrangement), not its motion. A body can be completely at rest and still hold energy — because it is positioned somewhere it could fall, or shaped in a way that could spring back — and that stored energy could later do work on something else.

Two flavours, one formula at this level. Class 9 studies gravitational potential energy quantitatively; elastic potential energy (a compressed spring, a drawn bow) is recognised only conceptually, as energy stored due to a change in configuration rather than height. Its formula, $U = \tfrac12 kx^2$ , belongs to Class 11 and should never be used here.

Gravitational potential energy is the energy a body has due to its height above a reference level, usually the ground:

E_p = mgh

where $m$ is the mass of the body in kilograms, $g$ is the acceleration due to gravity ( $9.8\ \text{m/s}^2$ on Earth), and $h$ is the height above the chosen reference level, in metres. Like every form of energy, $E_p$ is measured in joules (J), and one joule is the energy stored by raising roughly a 100 g mass by about 1 m. The formula hides an important idea worth stating plainly: potential energy is always measured relative to something. Lift a book from the floor to a shelf, and its potential energy relative to the floor increases; but relative to a table beside the shelf, the increase could be smaller or even negative. There is no absolute zero of potential energy — only a chosen reference level, and the physics of interest is almost always the change in $E_p$ , which does not depend on which reference you picked.

Figure

A body being lifted slowly through height h against gravity by an applied force equal to its weight, with the work done stored as gravitational potential energy

Derivation AGravitational potential energy as work done against gravity

The goal is to show, using only the definition of work and the fact that a slowly lifted body needs an applied force equal to its own weight, that the energy stored by raising mass $m$ through height $h$ is exactly $E_p = mgh$ , with no calculus and no assumption beyond Newton's laws at constant, slow speed.

Step 1. Identify the force needed. To raise a body of mass $m$ steadily (no acceleration, so no extra force is "wasted" speeding it up), the applied force must exactly balance the downward pull of gravity, the weight $W_{\text{wt}} = mg$ . So the minimum lifting force is $F = mg$ , directed upward.

Step 2. Apply the definition of work done by a constant force acting along the direction of displacement: $W = F \times s$ , where $s$ is the distance moved in the direction of the force. Here the body moves straight up through height $h$ , so $s = h$ .

Step 3. Substitute: $W = F \times s = mg \times h = mgh$ . This is the work you personally did on the body while lifting it — energy you supplied by your own muscles or a motor, transferred into the body.

Step 4. Recognise where that work went. The body's speed did not change (it was lifted slowly and is now at rest again at height $h$ ), so the work did not become kinetic energy. It must be stored in the body, available to be released as work later — for instance, if the body is allowed to fall and drives a hammer or turns a wheel on the way down. That stored quantity is defined as the gravitational potential energy, so $E_p = W = mgh$ . Check: the formula has the right units, $\text{kg} \times \text{m/s}^2 \times \text{m} = \text{kg}\,\cdot\,\text{m}^2/\text{s}^2 = \text{J}$ , matching the joule as the unit of work and energy, and $E_p$ correctly increases with any of $m$ , $g$ , or $h$ — a heavier body, a stronger gravity, or a greater height all mean more stored energy, exactly as intuition demands.

Worked example

A crane lifts a 500 kg concrete block from the ground to the top of a building, 25 m high, then the block is allowed to fall freely. Find (a) the potential energy gained on the way up, and (b) the potential energy remaining once the block has fallen 10 m. Take $g = 9.8\ \text{m/s}^2$ .

Step 1. List the knowns: $m = 500\ \text{kg}$ , initial height $h_1 = 25\ \text{m}$ , $g = 9.8\ \text{m/s}^2$ .

Step 2. Part (a) — apply $E_p = mgh$ at the top: $E_{p,1} = 500 \times 9.8 \times 25$ . Working it out, $500 \times 9.8 = 4900$ , and $4900 \times 25 = 122{,}500\ \text{J}$ , so $E_{p,1} = 1.225\times10^5\ \text{J}$ (122.5 kJ) is gained as the block is raised.

Step 3. Part (b) — after falling 10 m, the block's height above the ground is now $h_2 = 25 - 10 = 15\ \text{m}$ . Apply $E_p = mgh$ again with this new, smaller height: $E_{p,2} = 500 \times 9.8 \times 15 = 4900 \times 15 = 73{,}500\ \text{J}$ .

So the potential energy remaining is $E_{p,2} = 7.35\times10^4\ \text{J}$ (73.5 kJ).

Competitive stretch

Where did the missing energy go? The block lost $122{,}500 - 73{,}500 = 49{,}000\ \text{J}$ of potential energy as it fell those first 10 m, but energy is never destroyed — it converted into kinetic energy, the energy of the block's growing speed. This is the seed of the law of conservation of mechanical energy (total energy = $E_p + E_k$ stays constant for a freely falling body, ignoring air resistance): the potential energy lost exactly equals the kinetic energy gained at every instant of the fall, a relationship developed fully once kinetic energy and free fall are combined.

Common traps

Writing $E_p = mh$ and forgetting to include $g$ — potential energy needs all three factors, mass, gravity, and height.
Using an inconsistent reference level: potential energy is always measured relative to a chosen level (often the ground), and switching that reference mid-problem changes the numerical answer even though nothing physical has changed.
Assuming a stationary object can never have energy. A motionless, raised object still stores gravitational potential energy simply because of where it is — energy and motion are not the same thing.
Treating "potential energy" as automatically meaning the spring formula $U = \tfrac12 kx^2$ . That formula is not on this syllabus; only $E_p = mgh$ is required, and elastic potential energy should be discussed in words only.

Ask the tutor: Stuck on picking the right reference height when a body moves between two nonzero heights, or want to see how lost potential energy turns into kinetic energy step by step? Ask the AI tutor to walk through it with you.

Why is potential energy always relative to a reference level, but kinetic energy is not?

Kinetic energy, $E_k = \tfrac12 mv^2$ , depends only on speed, which is a real, measurable property of the body right now — there is no ambiguity about how fast something is moving. Potential energy, however, is defined through height, and "height" only makes sense once you pick a level to measure from — the floor, the ground, sea level, or anything else. Because the underlying physics only ever needs the change in potential energy (how much was gained or lost), the choice of reference never affects real predictions like speeds or energy transferred, even though the raw number for $E_p$ itself does change with the reference chosen.

If I lift a book and then put it back down at the same height, has it gained any potential energy overall?

No — if the final height equals the initial height, the net change in gravitational potential energy is zero, since $E_p = mgh$ depends only on the height, not on the path taken to get there or how many times it went up and down along the way. Energy was certainly transferred during the lifting (you did positive work) and during the lowering (gravity did positive work on the book, or your hand did negative work slowing its descent), but these cancel out for the round trip. This "path independence" is a hallmark of gravitational potential energy and is why it is called a conservative form of energy.

Does a book lying flat on a table have zero potential energy?

It depends entirely on your reference level, and both answers can be correct at once. Relative to the tabletop itself, yes, $h = 0$ so $E_p = 0$ . But relative to the floor, the book is still at some height above the ground (the height of the table), so it has a positive, nonzero $E_p = mgh$ using that floor-based height. Neither answer is "more true" than the other — always state the reference level you are using, and in a numerical problem, use whichever level the question specifies or implies (usually the ground or the lowest point mentioned).

Potential energy

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads