How to Solve Conservation of Energy Problems in 4 Lines

Quick answer: Conservation of energy says the total energy in a closed system never changes — it only moves between stores. To solve a problem with it, pick a start state and an end state, write every store of energy at each, set the two totals equal, and put anything lost to friction or air resistance on the end side.

I used to be bad at these for a silly reason: I treated energy as one more formula to hunt for instead of a way of thinking. I would reach for the motion equations, get stuck because I did not know the time, and lose four marks. Once I started writing "at the top" and "at the bottom" as two columns before touching a calculator, energy became the easiest marks on the paper.

What conservation of energy actually means

Energy is never created or destroyed; it is transferred between stores, and the total across a closed system stays fixed. The stores you meet at IGCSE, GCSE, CBSE, AP and WASSCE level are kinetic, gravitational potential, elastic potential, thermal, chemical, nuclear, electrostatic and magnetic. Energy moves between them in four ways: mechanically, electrically, by heating, and by radiation.

Two words matter. Closed means nothing crosses the boundary you drew. Dissipated means energy has spread into the thermal store of the surroundings and is no longer useful. Energy is never actually lost, only made useless, and examiners care about that difference.

The four-line method that picks up every mark

Write the same four lines every time and you will rarely lose a method mark.

1. Name state A and state B. Literally write "A: at the top, at rest" and "B: just before landing".
2. List every store holding energy at A, with numbers.
3. List every store at B, including anything dissipated.
4. Set total A equal to total B, then rearrange and solve.

The formulas that fill lines 2 and 3:

• Kinetic energy: KE = 1/2 m v^2
• Gravitational potential energy: GPE = m g h
• Elastic potential energy: EPE = 1/2 k x^2
• Work done by a force: W = F d
• Power: P = E / t

Then state your assumption, usually "assume air resistance is negligible". That sentence is sometimes worth its own mark and costs six seconds.

Worked example 1: the falling ball, and why mass cancels

A 2.0 kg ball is dropped from 5.0 m. Find its speed just before it lands. Take g = 9.81 N/kg.

A, at the top: GPE = m g h = 2.0 x 9.81 x 5.0 = 98.1 J, and KE = 0 because it starts at rest.

B, just before landing: GPE = 0 because h = 0, and KE = 1/2 x 2.0 x v^2 = 1.0 v^2.

Equate them: 98.1 = 1.0 v^2, so v = sqrt(98.1) = 9.9 m/s.

Now the algebra. m g h = 1/2 m v^2, so mass cancels and v = sqrt(2 g h). A 2 kg ball and a 200 kg ball land at the same speed once air resistance is ignored. If a question hands you a mass you never use, that is the point being tested.

It also works on paths no force diagram could handle: a pendulum, a bumpy hill, a skateboarder in a half-pipe. Know the vertical drop and you can find the speed, because GPE depends on height alone, not the route.

Worked example 2: when energy is dissipated

A 40 kg child slides down a slide with a vertical drop of 2.5 m and reaches the bottom at 5.0 m/s. The sloping surface is 6.0 m long. How much energy is dissipated, and what average friction force acts?

GPE lost = 40 x 9.81 x 2.5 = 981 J.

KE gained = 1/2 x 40 x 5.0^2 = 500 J.

Energy dissipated = 981 - 500 = 481 J.

Friction did 481 J of work over 6.0 m, and W = F d, so F = 481 / 6.0 = 80 N to two significant figures.

That last step catches people out and appears constantly: energy dissipated divided by distance gives an average resistive force with no force diagram at all. Do a few under timed conditions on a practice paper and it stops feeling like a trick.

Worked example 3: springs, and running the method backwards

A spring of spring constant k = 200 N/m is compressed 0.15 m and released, launching a 0.30 kg trolley along a frictionless track.

EPE stored = 1/2 k x^2 = 1/2 x 200 x 0.0225 = 2.25 J.

All of it becomes kinetic: 2.25 = 1/2 x 0.30 x v^2, so v^2 = 15 and v = 3.9 m/s.

If the trolley actually reached only 3.0 m/s, find the real KE, 1/2 x 0.30 x 9.0 = 1.35 J, and conclude 0.90 J was dissipated. Same method in reverse. When I get stuck it is almost always because I started writing before deciding which quantity was unknown.

Efficiency and power: the same idea, one extra step

Efficiency is useful output energy divided by total input energy, times 100 for a percentage. Power is energy transferred per second.

A pump lifts 300 kg of water through 12 m in 40 s. Useful output = 300 x 9.81 x 12 = 35 316 J, so useful power = 35 316 / 40 = 883 W. If the pump draws 1.2 kW, efficiency = 883 / 1200 = 0.74, or 74 per cent.

Two things get tested here. Efficiency can never exceed 100 per cent, so 130 per cent means you divided the wrong way round. And the wasted 26 per cent has not vanished; it sits in the thermal store of the motor, the pipes and the air. If rearranging trips you up, get one worked through step by step, then redo it on paper yourself.

Where conservation of energy is the wrong tool

Total energy is always conserved, but kinetic energy is not. In an inelastic collision the objects stick together and much of the KE becomes thermal energy and sound, so KE before does not equal KE after. Use momentum for collisions, energy for drops, slides, springs and pendulums.

The other trap is forgetting the system boundary. If a motor, a battery or a person's muscles are involved, energy is entering from a chemical store, and your closed system is not closed until you include it.

Test yourself

1. A 0.50 kg ball is thrown straight up at 12 m/s. Ignoring air resistance, how high does it go? Use g = 9.81 N/kg.
2. A 1200 kg car travelling at 20 m/s brakes to rest. How much energy is transferred to the thermal store of the brakes?
3. A lamp is supplied with 60 J and emits 9 J as light. What is its efficiency, and where did the rest go?

FAQ

Which value of g should I use?

Check the data sheet or the front of your paper first. Cambridge and AQA papers usually state 9.8 or 9.81 N/kg, CBSE questions commonly use 9.8, and WASSCE questions often specify 10. Use whatever the paper gives, and never mix two values in one calculation.

If energy is conserved, why does a bouncing ball stop bouncing?

Because energy is conserved but not always useful. Each bounce pushes some GPE into thermal energy in the ball, floor and air, plus sound. The total is unchanged, but the amount left in stores that could lift the ball again shrinks, so bounce height falls.

Do I lose marks for writing "energy is lost"?

Often, yes. Mark schemes want "transferred to the thermal store of the surroundings" or "dissipated as heat and sound". Naming the store and the destination earns the mark, so make that phrase automatic.

Can I use energy conservation when the path is curved?

Yes, and that is when it beats every other method. GPE depends only on vertical height change, so a ramp, a loop and a pendulum give identical answers for the same drop when friction is negligible. Try curved-track questions on a physics quiz or work through the physics topic hub.

In short: stop hunting for a formula and start writing two states. Total energy at A, total at B, anything dissipated on the B side, then solve. Mass often cancels, the route usually does not matter, and energy dissipated divided by distance quietly hands you the friction force.