LearnovaAll Physics notes
OCR Gateway PhysicsEnergy

Energy stores and transfers

Track energy as it moves between stores and pathways.

Start here

The key idea

Energy is conserved: it is transferred between stores rather than used up.

Equation to know

energy transferred = power x time

Energy Stores And Transfers
input energyusefulwasted

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

1

Energy stores

Energy is stored in different ways: kinetic (movement), gravitational potential (height in a gravitational field), elastic potential (stretched or compressed objects), thermal (internal), chemical (fuels, food, batteries), nuclear, magnetic and electrostatic.Energy is never created or destroyed, only transferred between stores.The total energy of a closed system stays constant — this is the principle of conservation of energy.

2

Energy transfers

Energy is shifted between stores by four pathways: mechanically (a force doing work), electrically (a current doing work), by heating, and by radiation (such as light or sound waves).For example, a falling object transfers energy mechanically from a gravitational potential store to a kinetic store.When a current flows through a resistor, energy is transferred electrically and then by heating to the surroundings.

3

Kinetic and gravitational energy

5 x m x v2, with energy in J, mass in kg and speed in m/s. 8 N/kg on Earth.5 x m x v2 = m x g x h if there is no air resistance.

4

Elastic potential energy

5 x k x e2, provided the limit of proportionality is not exceeded.Energy is in J, spring constant k in N/m and extension e in m.When the spring is released this store transfers to a kinetic store.

Key terms

Definitions to learn

Energy store

A way energy is held, such as kinetic, thermal or chemical.

Conservation of energy

Energy cannot be created or destroyed, only transferred.

Closed system

A system where no energy enters or leaves; total energy is constant.

Work done

Energy transferred when a force moves an object.

Dissipation

Energy spreading out to the surroundings, usually as wasted thermal energy.

Worked example

A 60 W lamp is switched on for 180 s. Calculate the energy transferred.

1

Use E = Pt.

2

Substitute E = 60 × 180.

3

Include the unit joules.

Final answer

10,800 J

Exam habit

Write the equation first, substitute values with units, calculate, then state the unit.Use 'transferred to' not 'used up' — energy is conserved.

Watch out

Do not confuse power in watts with energy in joules.

Examiner tips

How to score full marks

  • 1Use the words store and transfer carefully — never say energy is created, used up or lost.
  • 2For kinetic energy the speed is squared; doubling the speed gives four times the energy.
  • 3State units in your final answer; energy is always in joules (J) unless told otherwise.
Quick-fire quiz

Test yourself

Pick an answer — you'll see instantly if it's right.

A 50 W lamp runs for 60 s. How much energy is transferred?

Which statement about energy is correct?

A device has a power of 2 kW and runs for 3 minutes. What is the energy transferred in joules?

Which unit is used to measure energy?

A 400 W motor transfers 12,000 J. How long did it run?

Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1A 2 kW heater runs for 5 minutes. Calculate the energy transferred.
Mark scheme
  1. 1.Convert 2 kW to 2000 W and 5 minutes to 300 s.
  2. 2.Use E = Pt.
600,000 J
2A device transfers 36 kJ in 2 minutes. Calculate its power.
Mark scheme
  1. 1.Convert 36 kJ to 36,000 J and 2 minutes to 120 s.
  2. 2.Use P = E / t.
300 W
3Explain why a phone battery becomes warm while charging.
Mark scheme
  1. 1.Energy is transferred electrically.
  2. 2.Some energy is dissipated to the thermal store of the surroundings.
Some transferred energy is dissipated by heating, so the phone and surroundings become warmer.
4Name the energy store that increases as a car speeds up on a flat road.[1 mark]
Mark scheme
  1. 1.Speeding up means the car is moving faster.
  2. 2.Movement is linked to the kinetic store.
Kinetic energy store (1)
5Calculate the kinetic energy of a 1200 kg car travelling at 15 m/s.[2 marks]
Mark scheme
  1. 1.Use Ek = 0.5 x m x v2.
  2. 2.Substitute m = 1200 and v = 15.
  3. 3.Square the speed first, then multiply.
Ek = 0.5 × 1200 × 152 (1) = 0.5 × 1200 × 225 = 135000 J (1) (135 kJ)
6A 2.0 kg book is lifted 1.5 m onto a shelf. Calculate the gain in gravitational potential energy. Take g = 9.8 N/kg.[2 marks]
Mark scheme
  1. 1.Use Ep = m x g x h.
  2. 2.Substitute m = 2.0, g = 9.8, h = 1.5.
Ep = m x g x h (1) = 2.0 × 9.8 × 1.5 = 29.4 J (1)
7A spring with spring constant 200 N/m is stretched by 0.10 m. Calculate the elastic potential energy stored.[2 marks]
Mark scheme
  1. 1.Use Ee = 0.5 x k x e2.
  2. 2.Substitute k = 200 and e = 0.10.
  3. 3.Square the extension before multiplying.
Ee = 0.5 × 200 × 0.102 (1) = 0.5 × 200 × 0.01 = 1.0 J (1)
8A 0.50 kg ball is dropped from a height of 1.8 m. Assuming no air resistance, calculate its speed just before it hits the ground. Take g = 9.8 N/kg.[3 marks]
Mark scheme
  1. 1.At the top all energy is gravitational potential; at the bottom all is kinetic.
  2. 2.Set m x g x h = 0.5 x m x v2.
  3. 3.Mass cancels, so g x h = 0.5 x v2.
  4. 4.Rearrange to v = square root of (2 x g x h).
m x g x h = 0.5 x m x v2 (1); v2 = 2 x g x h = 2 × 9.8 × 1.8 = 35.28 (1); v = square root of 35.28 = 5.9 m/s (1)
9State the principle of conservation of energy and explain what it means for a closed system.[2 marks]
Mark scheme
  1. 1.Define conservation of energy.
  2. 2.Apply to a closed system.
Energy cannot be created or destroyed, only transferred between stores (1); in a closed system the total energy remains constant — any decrease in one store must be matched by an equal increase in another (1)
10A 3.0 kg toy car rolls down a ramp from a height of 0.50 m. Assuming all gravitational potential energy converts to kinetic energy, calculate the speed at the bottom. Take g = 9.8 N/kg.[3 marks]
Mark scheme
  1. 1.Calculate Ep = m x g x h.
  2. 2.Set Ep equal to Ek = 0.5 x m x v2.
  3. 3.Mass cancels; rearrange v = square root of (2 x g x h).
  4. 4.Substitute and evaluate.
Ep = 3.0 × 9.8 × 0.50 = 14.7 J (1); 14.7 = 0.5 × 3.0 x v2 so v2 = 14.7 / 1.5 = 9.8 (1); v = 3.1 m/s (1)
11The speed of the toy car in the previous question is measured at the bottom as 2.8 m/s, which is less than your calculated value. Explain this difference.[3 marks]
Mark scheme
  1. 1.Identify that kinetic energy is less than expected.
  2. 2.Link to energy being dissipated.
  3. 3.Name the pathway and destination store.
Some of the gravitational potential energy is dissipated by friction between the car's wheels and the ramp (1); this energy is transferred to the thermal store of the car, ramp and surroundings by heating (1); so less energy is available to be transferred to the kinetic store, giving a lower speed (1)
12Describe the energy store changes when a catapult is pulled back and then releases a stone. Name three stores involved.[3 marks]
Mark scheme
  1. 1.Identify energy in each stage.
  2. 2.Pulling back — elastic store.
  3. 3.Release — elastic to kinetic.
  4. 4.In flight — kinetic and gravitational potential.
Pulling the elastic back transfers chemical energy (from muscles) to the elastic potential store (1); when released, elastic potential energy is transferred to the kinetic store of the stone (1); as the stone rises, kinetic energy is transferred to the gravitational potential store (1)
13A skateboarder of mass 65 kg rolls along a flat surface at 6.0 m/s, then reaches the top of a ramp of height h before stopping. Calculate h, assuming no energy is dissipated. Take g = 9.8 N/kg.[3 marks]
Mark scheme
  1. 1.All kinetic energy converts to gravitational potential energy.
  2. 2.Set 0.5 x m x v2 = m x g x h.
  3. 3.Mass cancels; rearrange h = v2 / (2 x g).
  4. 4.Substitute v = 6.0 and g = 9.8.
0.5 x v2 = g x h (1); h = v2 / (2 x g) = 6.02 / (2 × 9.8) (1) = 36 / 19.6 = 1.84 m (accept 1.8 m) (1)
14Evaluate the claim: 'A heavier ball dropped from the same height will hit the ground faster than a lighter ball.' Use equations to support your answer, assuming no air resistance.[3 marks]
Mark scheme
  1. 1.Write the energy conservation equation: m x g x h = 0.5 x m x v2.
  2. 2.Cancel mass from both sides.
  3. 3.Show that speed depends only on g and h.
  4. 4.Conclude that mass has no effect.
Using m x g x h = 0.5 x m x v2, mass cancels to give v2 = 2 x g x h (1); this shows the final speed depends only on gravitational field strength and height, not on mass (1); therefore the claim is incorrect — both balls hit the ground at the same speed (assuming no air resistance) (1)
15A 500 g ball bearing is fired vertically upward from a spring gun. The spring has a constant of 800 N/m and was compressed by 0.12 m. Calculate the maximum height the ball reaches. Take g = 9.8 N/kg and assume no energy is dissipated.[3 marks]
Mark scheme
  1. 1.Calculate the elastic potential energy stored: Ee = 0.5 x k x e2.
  2. 2.Set Ee equal to gravitational potential energy at the top: m x g x h.
  3. 3.Rearrange for h = Ee / (m x g).
  4. 4.Convert 500 g to 0.50 kg.
Ee = 0.5 × 800 × 0.122 = 0.5 × 800 × 0.0144 = 5.76 J (1); m x g x h = 5.76, so h = 5.76 / (0.50 × 9.8) (1) = 5.76 / 4.9 = 1.18 m (accept 1.2 m) (1)
16Explain, using the ideas of energy stores and pathways, why no device can be 100% efficient. In your answer, refer to dissipation and the surroundings.[4 marks]
Mark scheme
  1. 1.Define what efficiency means in terms of energy stores.
  2. 2.Explain why some energy always escapes.
  3. 3.Name the pathway by which energy is wasted.
  4. 4.Refer to the surroundings.
Efficiency compares the useful energy transferred to the desired store with the total energy supplied (1); in any real device, some energy is always dissipated — usually by heating due to friction, air resistance or electrical resistance (1); this energy spreads out to the thermal store of the surroundings via the heating pathway (1); because this dissipated energy cannot be fully recovered and used, no device can transfer 100% of its input energy usefully, so efficiency is always less than 1 (1)
Official exam-board sources
Specification Past papers
Browse all Physics topics