The key idea
The topic that connects everything: a falling object is a motion question, an energy question and often an efficiency question in one. Use the energy-store and transfer terms required by your syllabus, and explain the physical change.
Understand it · 01
Choose the system and follow the energy
An energy explanation needs both a store and a physical process. A raised object and Earth have gravitational potential energy; as the object falls, energy can transfer to its kinetic store. If air resistance matters, some also increases the internal energy of the object and surrounding air. Naming only a list of energy words misses the connections the explanation is meant to show.
Choose a system boundary that makes the transfers clear. Energy entering the system plus its initial energy equals energy leaving plus its final energy. Dissipated energy is spread into less useful stores, often the thermal energy of the surroundings; it has not vanished. The useful output depends on the device's purpose: heating is useful in a kettle but often unwanted in a motor used to lift a load.
Understand it · 02
Match each distance to its equation
For a constant force parallel to motion, work done is force multiplied by distance moved in that direction. The force can transfer energy into or out of the object's kinetic store. A braking force acts against motion and reduces kinetic energy. In a simple braking calculation, the positive magnitude of work done against resistance equals the kinetic energy decrease.
Gravitational potential energy change near Earth's surface is mgΔh, where Δh is the vertical height change, not the length of a slope. Two routes reaching the same height give the same gravitational energy increase for the same mass, although friction can make the total input work different. An ideal ramp trades a smaller force for a longer travel distance; it does not create free energy. Use the question's g and keep the mass in kilograms.
Understand it · 03
Speed changes energy quadratically; time changes power
Kinetic energy is one half of mass multiplied by speed squared. Doubling speed quadruples kinetic energy for an unchanged mass. This helps explain why braking distance can grow strongly with speed: a much larger energy transfer is needed before the vehicle stops. A simple constant-braking-force model predicts braking distance proportional to speed squared, although actual braking also depends on tyres, road conditions and other factors.
Power describes how quickly energy is transferred. Two motors may both lift a load through the same height and do the same useful work, but the motor that completes the lift sooner has greater average useful power. Keep input and output power separate when efficiency is involved. A watt means one joule per second; it is not a unit of energy, so a time is needed to convert a power into an energy transfer.
Understand it · 04
Compare efficiency and energy resources fairly
Efficiency is useful output divided by total input, using either energies or powers consistently. Multiply by 100 only when a percentage is required. A real device commonly transfers some input into unwanted forms, so its efficiency is below 100%. An ideal model can have 100% efficiency, but a correctly defined device cannot deliver more useful energy than its total energy input.
When comparing energy resources, separate efficiency from availability, reliability and environmental effects. A renewable resource is replenished on a human timescale; that does not mean its output is constant or that construction has no impact. Solar and wind output vary with conditions, while a stored energy source can help match supply to demand. Explain a particular advantage and its consequence instead of calling a resource simply good, clean or bad.
Key definitions
- Principle of conservation of energy
- Energy cannot be created or destroyed; it is only transferred between stores. The total energy is constant.
- Work done
- The energy transferred when a force moves an object: W = F × d, where d is the distance moved in the direction of the force.
- Power
- The work done, or energy transferred, per unit time.
- Efficiency
- The fraction of the input energy that is usefully transferred: (useful output ÷ total input) × 100%.
Equations, units & conditions
Kinetic energy
KE = ½ × m × v²
Units: J
Gravitational PE
ΔGPE = m × g × Δh
Units: J
Work done
W = F × d
Units: J
Power
P = W ÷ t = E ÷ t
Units: W
Efficiency
efficiency = (useful energy out ÷ total energy in) × 100%
Worked examples, step by step
Example 1
A 0.50 kg ball is dropped from a height of 20 m. Ignoring air resistance, calculate its speed just before it hits the ground. (g = 10 m/s²)
Follow the method
- Energy conservation: KE gained = GPE lost, so ½mv² = mgh.
- The mass cancels: v² = 2gh = 2 × 10 × 20 = 400.
- v = √400.
v = 20 m/s
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 5 suggested marks
1. Useful power and motor efficiency
A motor lifts a 12 kg load vertically through 2.5 m in 6.0 s. During the lift it receives 500 J electrically. Use g = 10 N/kg. Calculate useful energy output, useful average power and efficiency.
Show solution and self-check: question 1
- Useful energy = mgΔh = 12 × 10 × 2.5 = 300 J.
- Useful average power = 300/6.0 = 50 W.
- Efficiency = 300/500 × 100% = 60%.
Useful output = 300 J; useful average power = 50 W; efficiency = 60%.
Self-check · one mark per point
- Uses vertical height in mgΔh.
- Finds 300 J.
- Divides useful energy by time to obtain 50 W.
- Uses useful output divided by total input.
- Finds 60%.
Stretch · 4 suggested marks
2. Use energy to find stopping distance
An 800 kg car travels at 15 m/s. Its engine stops driving and a constant resultant braking force of 3000 N acts until it stops. Calculate its initial kinetic energy and braking distance. Ignore any change in height.
Show solution and self-check: question 2
- Initial kinetic energy = ½ × 800 × 15² = 90 000 J.
- The braking force removes this kinetic energy: Fd = 90 000 J.
- Distance = 90 000/3000 = 30 m.
Initial kinetic energy = 90 000 J; braking distance = 30 m.
Self-check · one mark per point
- Squares the speed in ½mv².
- Finds 90 000 J.
- Equates the work done against braking to the kinetic energy decrease.
- Finds 30 m.
Common mistakes & how to avoid them
- Forgetting to square the velocity in kinetic energy — or squaring after multiplying by the mass. Substitute v², then multiply.
- Using the slope length instead of the vertical height in GPE. The h in mgh is always the vertical height gained.
- Writing 'energy is lost'. Energy is never lost — it is transferred to less useful stores, usually the internal (thermal) energy of the surroundings. That wording is worth the mark.
- Giving efficiency greater than 100%, then not noticing. If your efficiency exceeds 100%, the useful and total values are swapped.
Test yourself
10 questions · instant marking
Question 1 of 10
A 2 kg object moves at 3 m/s. Its kinetic energy is:
Question 2 of 10
A 4 kg mass is lifted 5 m (g = 10 m/s²). The gain in gravitational PE is:
Question 3 of 10
A motor transfers 3000 J in 60 s. Its power is:
Question 4 of 10
A lamp receives 200 J and usefully transfers 30 J as light. Its efficiency is:
Question 5 of 10
Energy that is 'wasted' by a machine is usually transferred to:
Question 6 of 10
A force of 50 N pushes a box 4 m in the direction of the force. Work done is:
Question 7 of 10
A ball is dropped and air resistance is negligible. Just before landing, its kinetic energy equals:
Question 8 of 10
Which energy resource does NOT originate from the Sun?
Question 9 of 10
Doubling the speed of an object multiplies its kinetic energy by:
Question 10 of 10
A crane lifts a 100 kg load 6 m in 12 s (g = 10 m/s²). Its useful output power is:
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
What are the energy stores in IGCSE physics?
Common energy stores include kinetic, gravitational potential, elastic, chemical, internal, nuclear and electrostatic stores; magnetic energy is also a useful physics concept. Follow your specification's required list. Describe transfers separately, such as mechanical work, electrical work, heating or radiation, rather than treating every transfer process as a store.
A store describes energy associated with a system's state, while a transfer describes a process moving energy between states or systems. For example, lifting a load transfers energy mechanically into the gravitational potential store of the load–Earth system. The word lifting is not an extra store, and the load does not acquire a store called work.
Choose a clear start and end when answering an unfamiliar scenario. A battery-powered fan draws on a chemical store, transfers energy electrically to the motor, and increases motion while also warming its surroundings. Naming the unwanted thermal transfer helps complete the account. Use the terminology taught for your particular board without losing the physical explanation behind the labels.
How do you calculate efficiency?
Divide useful energy output by total energy input, or useful power output by total power input. Multiply by 100 to express the result as a percentage. Keep numerator and denominator in matching units and identify what output is useful for the device's stated purpose.
A common mistake is dividing wasted output by useful output, or subtracting the two and calling the difference efficiency. First make an energy account: total input equals useful output plus other transfers for a complete interval with no unaccounted stored-energy change. Then select the ratio the definition actually requires.
A value above 100% usually signals reversed quantities, mismatched units or an incomplete account of the input. Real devices commonly dissipate part of their input, while an ideal model may be assigned 100% efficiency. Whether heating is useful depends on the device: it is the purpose of a heater but can be an unwanted transfer in a lifting motor.
Find useful energy from the other transfers
A motor receives 900 J during a lift and transfers 225 J to unwanted heating and sound. Assuming no remaining stored-energy change, calculate its efficiency for lifting.
- Useful lifting output = 900 − 225 = 675 J.
- Efficiency = 675/900 × 100% = 75%.
Efficiency = 75%.
What is the difference between work and power?
Work is an energy transfer, measured in joules. Power is the rate of doing work or transferring energy, measured in watts, where one watt equals one joule per second. Two devices can do the same work but have different average powers if they take different times.
Compare two motors lifting the same load through the same vertical height. Ignoring differences in efficiency, each supplies the same useful energy increase, but the faster lift has greater useful power. The greater power does not mean the final gravitational potential energy is greater when the load and height are unchanged.
Use average power when dividing the total work by the full time interval. A motor's instantaneous power can vary during its operation. Also distinguish useful output power from electrical input power if energy is dissipated. A label in watts alone is not an energy consumption; multiply by an appropriate operating time to calculate energy transferred.
Same work at different rates
Two machines each do 600 J of useful work. Machine A takes 3.0 s and machine B takes 12 s. Compare their average useful powers.
- Power A = 600/3.0 = 200 W.
- Power B = 600/12 = 50 W.
A supplies 200 W and B supplies 50 W of average useful power. Their useful work is the same.
Why does mgh use vertical height rather than the length of a ramp?
Gravitational potential energy changes with vertical separation in the approximately uniform gravitational field used by mgh. A longer sloping route to the same height produces the same gravitational energy increase. The ramp changes the force and travel distance required, while friction can add extra input work.
An ideal shallow ramp lets you raise a load using a smaller force over a greater distance. The reduced force does not reduce the final gravitational energy increase, because the product of the force and the distance along the ideal ramp supplies the same required energy. There is no energy saving simply from making the route longer.
Read the diagram for the vertical rise rather than choosing its most prominent labelled length. If the ramp is rough, input work also increases internal energy through friction, so input work exceeds mgh. If the load speeds up as well, include its kinetic-energy increase. The mgh term still represents only the change associated with height.
Separate route length and height
A 2.0 kg load moves up a 5.0 m ramp to a point 3.0 m higher. Use g = 10 N/kg. Find its gravitational energy increase, and the ideal constant pulling force along the ramp if speed is constant and friction is negligible.
- GPE increase = 2.0 × 10 × 3.0 = 60 J.
- For the ideal constant-speed lift, F × 5.0 = 60, so F = 12 N.
GPE increases by 60 J. The ideal pulling force along the ramp is 12 N.
Why does doubling speed quadruple kinetic energy?
Kinetic energy is proportional to speed squared for a fixed mass: Eₖ = ½mv². Replacing v with 2v multiplies v² by four. That means an object travelling twice as fast requires four times as much energy removal to stop, under the same mass assumption.
Separate a proportional comparison from an absolute calculation. If you only need the factor of change and mass stays fixed, the one-half and mass cancel when you take the ratio. Tripling the speed gives nine times the kinetic energy, while doubling mass at unchanged speed only doubles the energy.
A constant braking-force model links this to distance through Fd = kinetic-energy decrease. With unchanged force, four times the energy requires four times the braking distance. This does not mean every real vehicle's full stopping distance exactly quadruples: thinking distance, changing braking conditions, road surface and other effects can also matter.
Check the square numerically
Calculate the kinetic energy of a 2.0 kg object at 3.0 m/s and at 6.0 m/s.
- At 3.0 m/s: KE = ½ × 2.0 × 3.0² = 9.0 J.
- At 6.0 m/s: KE = ½ × 2.0 × 6.0² = 36 J.
- The ratio is 36/9.0 = 4.
The energy rises from 9.0 J to 36 J when speed doubles.
Where does wasted energy go if energy is conserved?
Wasted energy is transferred into forms or places that are less useful for the intended task, commonly the internal energy of the surroundings. It has not been destroyed. Conservation counts all transfers, whereas usefulness describes whether a transfer contributes to the device's purpose.
When a moving object slows through friction, its kinetic energy can become internal energy in the surfaces and surrounding material. The motion disappears, but the energy account includes the resulting warming, sound and deformation. Calling that energy lost is acceptable only if you clearly mean lost from a particular store or system, not from existence.
Energy spread through a large environment is difficult to collect back into the original useful form. That practical limitation explains why conservation does not imply perfect efficiency. Define your system boundary: heat leaving a motor is output from the motor but input to its surroundings. Both descriptions refer to the same transfer and should not be counted twice.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Describe energy stores (kinetic, gravitational potential, chemical, elastic, internal, nuclear) and transfers between them
- Apply the principle of conservation of energy
- Calculate kinetic energy, gravitational potential energy, work done and power
- Define and calculate efficiency
- Evaluate energy resources: fossil fuels, nuclear, solar, wind, hydroelectric, geothermal
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
- Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
- Pearson Edexcel International GCSE Physics 4PH1 — specification
Reference this page
Dr Desouky Physics Academy. IGCSE Physics: Energy, work and power. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/energy-work-and-power
Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.