- How to describe energy changes using stores and transfer pathways.
- Why energy is conserved in a closed system.
- How to calculate work done, gravitational potential energy and kinetic energy.
- Why energy is often dissipated into less useful stores.
In GCSE Physics, we do not usually say that energy is “used up”. Instead, we say energy is transferred between different energy stores.
Energy
Energy is a quantity measured in joules (J). It can be stored in different ways and transferred when a system changes.
A system is the object or group of objects you are focusing on. For example, if a ball is falling, the system might be “the ball and the Earth”.
Common energy stores include:
- kinetic store — energy due to movement
- gravitational potential store — energy due to height in a gravitational field
- elastic potential store — energy in stretched or squashed objects
- chemical store — energy in fuels, food and batteries
- thermal store — energy due to the temperature of an object
- nuclear store, magnetic store and electrostatic store
Energy changes are best described by saying which store decreases, which store increases, and how the transfer happens.
This schematic shows the key GCSE idea: energy moves between stores by transfer pathways, but in a closed system the total amount of energy stays the same.

Describing energy changes in a falling ball
A ball falls from rest towards the ground. Describe the energy changes.
- Choose the system: the ball and the Earth, because the gravitational potential store depends on the ball’s position in Earth’s gravitational field.
- As the ball falls, its height decreases, so energy is transferred away from the gravitational potential store.
- As the ball speeds up, energy is transferred to the kinetic store. If air resistance is included, some energy is also transferred to the thermal stores of the ball and surroundings.
A transfer pathway is the way energy is moved from one store to another.
For this part of the Edexcel 1PH0 specification, the main pathways you need are:
- work done by forces — a force moves something through a distance
- electrical work — energy is transferred by moving charges in electrical equipment
- heating — energy is transferred because of a temperature difference
How to write energy-change answers
A strong answer usually has this pattern: energy is transferred from [store] to [store] by [pathway].
For example, when a battery-powered motor lifts a toy, energy is transferred from the battery’s chemical store by electrical work to the motor, then by work done by forces to the toy’s gravitational potential store.
Closed system
A closed system is a system where no energy is transferred into it or out of it.
In a closed system, there is no net change to the total energy. Energy can move between stores, but the total amount stays constant.
This is called conservation of energy.
Using conservation of energy
A toy car has 120 J in its elastic potential store before it is released. After release, 85 J is in its kinetic store. The rest has been transferred to thermal stores and sound. Calculate the dissipated energy.
- Treat the energy before and after as equal, because the total energy is conserved.
- Subtract the useful kinetic energy from the starting energy:
120 J−85 J=35 J120\ \text{J} - 85\ \text{J} = 35\ \text{J}120 J−85 J=35 J.
- So 35 J has been transferred to thermal stores and sound.
You may be asked to draw or interpret diagrams showing energy transfers.
Two common diagram styles are:
- store-transfer diagrams — boxes or labels show stores; arrows show transfer pathways
- Sankey diagrams — arrow widths represent the amount of energy transferred
In a Sankey diagram, the total input energy must equal the total output energy. The useful energy usually continues forwards, while dissipated energy is often shown going off to the side or downwards.

Sankey diagram check
The output arrows should add up to the input arrow. If the input is 100 J, and 70 J is useful, then 30 J must be dissipated.
When a force causes an object to move, the force does work. This means energy is transferred.
Work done
Work done is the energy transferred by a force moving an object through a distance. Work done is measured in joules (J).
You must recall and use:
E=F×dE = F \times dE=F×d
where:
- EEE = work done, or energy transferred, in joules (J)
- FFF = force in newtons (N)
- ddd = distance moved in the direction of the force in metres (m)
To measure work done by a force, measure the force with a newton meter or force sensor, measure the distance moved in the direction of the force with a ruler or tape measure, then multiply them.

Calculating work done
A student pushes a box with a force of 80 N. The box moves 3.5 m in the direction of the force. Calculate the work done.
- Identify the values: F=80 NF = 80\ \text{N}F=80 N and d=3.5 md = 3.5\ \text{m}d=3.5 m.
- Substitute into the equation:
E=80 N×3.5 mE = 80\ \text{N} \times 3.5\ \text{m}E=80 N×3.5 m.
- Calculate the energy transferred:
E=280 JE = 280\ \text{J}E=280 J.
Using the wrong distance
In E=F×dE = F \times dE=F×d, the distance must be the distance moved in the direction of the force, not just any distance mentioned in the question.
An object has more energy in its gravitational potential store when it is higher up in a gravitational field.
You must recall and use:
ΔGPE=m×g×Δh\Delta GPE = m \times g \times \Delta hΔGPE=m×g×Δh
where:
- ΔGPE\Delta GPEΔGPE = change in gravitational potential energy in joules (J)
- mmm = mass in kilograms (kg)
- ggg = gravitational field strength in newtons per kilogram (N/kg)
- Δh\Delta hΔh = change in vertical height in metres (m)
On Earth, ggg is often taken as 10 N/kg at GCSE, but use the value given in the question.
Calculating gravitational potential energy gained
A 2.0 kg bag is lifted vertically by 1.5 m. Take g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg. Calculate the increase in gravitational potential energy.
- Identify the values: m=2.0 kgm = 2.0\ \text{kg}m=2.0 kg, g=10 N/kgg = 10\ \text{N/kg}g=10 N/kg and Δh=1.5 m\Delta h = 1.5\ \text{m}Δh=1.5 m.
- Substitute into the equation:
ΔGPE=2.0 kg×10 N/kg×1.5 m\Delta GPE = 2.0\ \text{kg} \times 10\ \text{N/kg} \times 1.5\ \text{m}ΔGPE=2.0 kg×10 N/kg×1.5 m.
- Calculate the change:
ΔGPE=30 J\Delta GPE = 30\ \text{J}ΔGPE=30 J.
So 30 J is transferred to the bag’s gravitational potential store.
Using slope distance instead of height
For gravitational potential energy, use the vertical height change, not the distance travelled along a ramp or slope.
A moving object has energy in its kinetic store. The faster it moves, the more kinetic energy it has.
You must recall and use:
KE=12×m×v2KE = \frac{1}{2} \times m \times v^2KE=21×m×v2
where:
- KEKEKE = kinetic energy in joules (J)
- mmm = mass in kilograms (kg)
- vvv = speed in metres per second (m/s)
Calculating kinetic energy
A trolley of mass 0.50 kg moves at 4.0 m/s. Calculate its kinetic energy.
- Identify the values: m=0.50 kgm = 0.50\ \text{kg}m=0.50 kg and v=4.0 m/sv = 4.0\ \text{m/s}v=4.0 m/s.
- Square the speed:
v2=(4.0 m/s)2=16 (m/s)2v^2 = (4.0\ \text{m/s})^2 = 16\ \text{(m/s)}^2v2=(4.0 m/s)2=16 (m/s)2.
- Substitute and calculate:
KE=12×0.50 kg×16 (m/s)2=4.0 JKE = \frac{1}{2} \times 0.50\ \text{kg} \times 16\ \text{(m/s)}^2 = 4.0\ \text{J}KE=21×0.50 kg×16 (m/s)2=4.0 J.
Speed matters a lot
Because the speed is squared, doubling the speed makes the kinetic energy four times bigger, if the mass stays the same.
In real energy transfers, some energy is almost always transferred to less useful stores. This is called dissipation.
Dissipated energy
Dissipated energy is energy transferred to less useful stores, usually thermal stores of the surroundings, so it becomes spread out and harder to use.
For example, when a car brakes, energy is transferred from the car’s kinetic store to the thermal stores of the brakes, tyres and surroundings. The energy has not disappeared, but it is now less useful.
Calculating dissipated energy
An electric motor receives 600 J of electrical energy. It transfers 450 J usefully to the kinetic store of a load. Calculate the energy dissipated.
- Apply conservation of energy: total input energy equals useful output energy plus dissipated energy.
- Rearrange by subtracting the useful energy from the input energy:
600 J−450 J=150 J600\ \text{J} - 450\ \text{J} = 150\ \text{J}600 J−450 J=150 J.
- So 150 J is dissipated, mainly to thermal stores of the motor and surroundings.
Energy is conserved, but usefulness can decrease
Energy is never destroyed, but it can become spread out in thermal stores of the surroundings, making it less useful for doing work.
In the exam
- For description questions, name the starting store, the ending store, and the transfer pathway.
- For calculations, write the equation first, then substitute values with SI units: kg, m, m/s, N and J.
- Watch the key words: use distance in the direction of the force for work done, and vertical height for gravitational potential energy.
- For diagrams, check conservation: total input energy must equal useful energy plus dissipated energy.
Check yourself
- A cyclist speeds up on a flat road. Which energy store increases, and where might the energy have come from?
- What is the difference between energy being “dissipated” and energy being “destroyed”?
- Which equation would you use for a falling object’s kinetic store just before it hits the ground?