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How efficient is a machine, and how can wasteful heating be reduced?

Efficiency of forces: calculating efficiency for a machine, why machines waste energy by heating, and reducing wasteful transfers by lubrication and streamlining.

A focused answer to Edexcel GCSE Physics 8.15 (and 8.10 to 8.11), covering the efficiency equation for a machine doing work, why mechanical processes waste energy as heat, and how lubrication and streamlining reduce wasteful energy transfers, with worked calculations.

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. The efficiency equation
  3. Why machines waste energy
  4. Reducing wasteful transfers
  5. How Edexcel examines this
  6. Try this

What this dot point is asking

Edexcel statement 8.15 (with 8.10 and 8.11) wants you to recall and use the efficiency equation for a machine, to explain that mechanical processes become wasteful when they raise the temperature (because of friction), and to describe how to reduce wasteful energy transfers, for example by lubrication.

The efficiency equation

Efficiency measures how much of the input energy a machine transfers usefully rather than wasting. It is a ratio, so it has no unit, and it is always less than 11 (or 100%100\%) for a real machine. The equation is on the Edexcel equation sheet, but you must be able to use it and rearrange it (for example to find the useful or total energy).

Why machines waste energy

The wasted energy in a machine almost always ends up as heating. The moving parts rub, friction does work against the motion, and the energy is transferred to thermal energy, warming the machine and the surroundings. The more friction, the more energy is wasted and the lower the efficiency.

Reducing wasteful transfers

Lubrication is the main method for moving parts: a slippery layer keeps surfaces from rubbing directly, reducing the frictional force and the energy wasted as heat. Streamlining helps where an object moves through air or water. Reducing the wasted energy directly raises the efficiency, because efficiency is the useful output over the total input.

How Edexcel examines this

This dot point is examined on both tiers, mostly as an efficiency calculation worth two or three marks, often followed by an explanation of why a machine is never fully efficient or how to improve it. The mark scheme for the calculation rewards the correct ratio (useful over total), the value and the percentage, so set it out clearly. A frequent style gives the total input and useful output and asks for the percentage efficiency, or gives the efficiency and one energy and asks you to rearrange for the other. The explanation question rewards linking the wasted energy to friction (and air resistance) transferring energy to the thermal store of the surroundings, raising the temperature, and a valid improvement such as lubrication or streamlining. Examiners penalise inverting the ratio and claiming any real machine can reach 100%100\% efficiency. Because this statement overlaps with the Topic 3 efficiency and the reducing-transfer ideas, the same techniques apply, and a power form of the equation may appear if the data are in watts.

Try this

Q1. A machine does 60 J60\,\text{J} of useful work from a total input of 240 J240\,\text{J}. Calculate its efficiency as a percentage. [2 marks]

  • Cue. efficiency =60240=0.25=25%= \dfrac{60}{240} = 0.25 = 25\%.

Q2. State one way to make a machine with moving parts more efficient. [1 mark]

  • Cue. Lubricate the moving parts (to reduce friction).

Exam-style practice questions

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

Edexcel 20203 marksA machine is supplied with 1200 J1200\,\text{J} of energy and does 900 J900\,\text{J} of useful work. Calculate the efficiency of the machine as a percentage.
Show worked answer →

Use efficiency =useful energy transferredtotal energy supplied= \dfrac{\text{useful energy transferred}}{\text{total energy supplied}} with useful =900 J= 900\,\text{J} and total =1200 J= 1200\,\text{J} (1 mark). Substitute: efficiency =9001200=0.75= \dfrac{900}{1200} = 0.75 (1 mark), and as a percentage 0.75×100=75%0.75 \times 100 = 75\% (1 mark). Markers reward the correct ratio, the value and the percentage. Putting the total on top or using the wasted energy is the usual error.

Edexcel 20223 marksExplain why a machine with moving parts is never 100% efficient, and describe one way to make it more efficient.
Show worked answer →

A machine with moving parts is never 100% efficient because friction between the moving parts (and air resistance) always transfers some energy to the thermal store of the surroundings, raising the temperature, so not all the input energy ends up as useful output (2 marks). It can be made more efficient by lubricating the moving parts to reduce friction (or by streamlining to reduce air resistance), so less energy is wasted as heat (1 mark). Markers reward linking the wasted energy to friction heating the surroundings and a valid way to reduce friction such as lubrication.

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