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How are work, power, kinetic and potential energy calculated?

Work done, power, kinetic and gravitational potential energy, and the conservation of energy in mechanical situations.

A focused answer to the WJEC GCSE Science Double Award Unit 6 topic on work and energy, covering work done, power, kinetic and gravitational potential energy, and the conservation of energy in mechanical situations.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Work and power
  3. Kinetic and potential energy
  4. Conservation of energy
  5. Power and everyday devices
  6. Springs and elastic energy
  7. Where energy goes against friction
  8. Try this

What this dot point is asking

WJEC Double Award Unit 6 wants you to calculate work done, power, kinetic energy and gravitational potential energy, and apply the conservation of energy.

Work and power

Work and energy use the same unit (the joule) because doing work transfers energy. A more powerful device transfers the same energy in less time.

Kinetic and potential energy

Note the v2v^2 in kinetic energy: doubling the speed quadruples the kinetic energy, which matters for stopping distances.

Conservation of energy

Energy conservation lets you work out, for example, the speed of a falling object by setting the potential energy lost equal to the kinetic energy gained.

Power and everyday devices

Power tells you how quickly a device transfers energy, and it is why appliances are labelled in watts or kilowatts. A 2000 W kettle transfers energy twice as fast as a 1000 W one, so it boils water more quickly but uses energy at a faster rate. Power can also be worked out for a person or machine: a person who does 600 J of work running upstairs in 3 s has a power of 6003=200W\dfrac{600}{3} = 200\,\text{W}. Recognising power as the rate of energy transfer, and using P=WtP = \dfrac{W}{t}, is a common calculation in this topic.

Springs and elastic energy

When a force stretches or squashes a spring, it stores elastic potential energy. While the spring is not stretched too far (within its limit), the extension is proportional to the force (Hooke's law), so a force-extension graph is a straight line through the origin. The stored elastic energy is released when the spring returns to shape, which is how a catapult or a spring-loaded toy works. Knowing that a stretched spring stores elastic potential energy, and that extension is proportional to force up to a limit, links forces to energy stores.

Where energy goes against friction

When a force does work against friction or air resistance, the energy is transferred to the surroundings as heat. This is why brakes get hot when a car stops, and why rubbing your hands warms them. In real machines, some of the input energy is always wasted as heat in this way, which is why no machine is 100% efficient. Recognising that work done against friction becomes heat links this topic to energy efficiency.

Try this

Q1. State the equation for work done. [1 mark]

  • Cue. Work = force x distance.

Q2. A 3 kg object is lifted 2 m (g = 10 N/kg). Calculate the gain in gravitational potential energy. [2 marks]

  • Cue. Ep=mgh=3×10×2=60JE_p = mgh = 3 \times 10 \times 2 = 60\,\text{J}.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC style3 marksA crane lifts a 400 kg load to a height of 15 m. Calculate the gain in gravitational potential energy (g = 10 N/kg).
Show worked answer →

A Unit 6 calculation. Use Ep=mghE_p = mgh (1 mark). Substitute m=400m = 400, g=10g = 10 and h=15h = 15: Ep=400×10×15E_p = 400 \times 10 \times 15 (1 mark) =60000J= 60\,000\,\text{J} (1 mark). Markers reward the equation, the substitution and the unit joules. A common error is to leave out g.

WJEC style3 marksA motor does 9000 J of work in 30 s. Calculate its power.
Show worked answer →

A Unit 6 calculation. Use P=WtP = \dfrac{W}{t} (1 mark). Substitute: P=900030P = \dfrac{9000}{30} (1 mark) =300W= 300\,\text{W} (1 mark). Markers reward the equation, the substitution and the unit watts. A common error is to multiply work and time.

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