IB Physics HLTopic 1 — Motion, Forces & EnergyPaper 1 & 2Work, Energy & Power~9 min read
Efficiency Formula
No machine gives back everything you put into it. Feed a motor electrical energy and only some emerges as useful lifting or turning — the rest leaks away, usually as heat. Efficiency is the score that measures how good a device is at this: what fraction of the energy going in actually does the job you wanted. A high-efficiency device wastes little; a low-efficiency one throws most of it away. It’s a simple ratio, but it ties together everything you’ve learned about energy, work and power.
📘 What you need to know
Efficiency measures how successfully a device transfers energy into the useful form
It’s the ratio of useful output to total input
η = Eout / Ein = Pout / Pin
Efficiency can be a ratio (0 to 1) or a percentage (0% to 100%) — multiply by 100 to convert
Efficiency has no units — it’s a ratio of two energies (or two powers)
It can never exceed 100% (or 1) — you can’t get out more than you put in
The energy (or power) you’re not using is wasted, usually as heat
The efficiency equation
Efficiency, given the Greek letter η (“eta”), is the ratio of the useful energy (or power) coming out of a device to the total energy (or power) going in:
Efficiency
η = Eout / Ein = Pout / Pin
You can work with energies or powers — both give the same efficiency, since power is just energy per second. In words:
Efficiency in words
η = useful output ÷ total input
Efficiency is the useful (green) slice of the total energy input. The bigger the green part compared with the whole bar, the higher the efficiency; the grey part is wasted, usually as heat.
WE 1
A device is supplied with 800 J of energy and produces 600 J of useful output. Calculate its efficiency as a percentage.
Step 1 — write the efficiency equation
η = Eout ÷ EinStep 2 — substituteη = 600 ÷ 800 = 0.75Step 3 — convert to a percentage (× 100)Efficiency = 75%The other 200 J (25%) is wasted — dissipated to the surroundings, usually as heat.
Ratio or percentage?
Efficiency comes in two flavours that mean the same thing. As a ratio it’s a number between 0 and 1; as a percentage it’s between 0% and 100%. To go from ratio to percentage, multiply by 100:
Ratio 0.75
× 100 →
Percentage 75%
÷ 100 ←
back to ratio
Read the question carefully: if it asks for efficiency “as a ratio” or “as a decimal”, give 0.75; if it asks for a percentage, give 75%. Either way, efficiency has no units — it’s one energy divided by another, so the units cancel.
A quick sanity check that catches most mistakes: efficiency can never be more than 100% (or 1). If your answer comes out above that, you’ve almost certainly put the input and output the wrong way round in the fraction. Useful output goes on top; total input goes on the bottom — always the smaller number over the bigger one.
Comparing efficiencies
Because efficiency is a ratio, two devices doing the same job can be judged side by side. A modern, efficient device delivers most of its input as useful output; an old or poorly designed one wastes most of it. The same total input can produce very different useful outputs:
Same energy input, very different results. The 80%-efficient device turns most of it into useful output (green); the 25%-efficient one wastes most as heat (grey). Efficiency is the green fraction of the whole bar.
WE 2
A pump is 40% efficient. Its useful output power is 120 W. Calculate the input power it needs.
Step 1 — use the power form and rearrange for input
η = Pout ÷ Pin, so Pin = Pout ÷ η
Step 2 — use the ratio form of efficiency (40% = 0.40)Pin = 120 ÷ 0.40Pin = 300 WConvert the percentage to a ratio (0.40) before dividing — a very common slip.
WE 3
A 60% efficient electric motor lifts a 15 kg load through 4.0 m. How much electrical energy must be supplied to the motor? (Take g = 9.81 m s−2.)
Step 1 — the useful output is the GPE gained
Eout = mgΔh
Eout = 15 × 9.81 × 4.0 = 589 JStep 2 — rearrange η = Eout ÷ Ein for the input
Ein = Eout ÷ η
Step 3 — substitute (60% = 0.60)Ein = 589 ÷ 0.60Ein = 980 J (2 s.f.)The motor needs 980 J to deliver 589 J of useful lifting — the extra ~390 J is wasted as heat and sound.
🛠️ Solving an efficiency problem
Identify the useful output — often the GPE gained, KE gained, or the stated useful power.
Identify the total input — the electrical energy or power supplied.
Put useful output on top, total input on the bottom: η = output ÷ input.
Rearrange first if you’re solving for the input or the output.
Convert percentages to ratios (divide by 100) before calculating, and multiply back by 100 at the end if a percentage is wanted.
💡 Top tips
Useful over total. The smaller number always goes on top — efficiency can’t exceed 1 (or 100%).
Energies or powers. Both give the same efficiency — just don’t mix an energy with a power in the same fraction.
Percentage to ratio. Turn a percentage into a decimal (40% → 0.40) before dividing.
No units. Efficiency is a pure ratio, so it never carries joules, watts or anything else.
Quick recap: Efficiency measures how much of the input energy becomes useful output: η = Eout / Ein = Pout / Pin. It’s a unitless ratio (0 to 1) or percentage (0% to 100%), can never exceed 100%, and the rest of the energy is wasted, usually as heat.
⚠ Common mistakes
Putting input over output — useful output always goes on top
Forgetting to convert a percentage to a ratio before dividing
Giving efficiency units — it’s a pure ratio with none
Mixing an energy with a power in the same fraction — keep them consistent
Accepting an answer over 100% instead of spotting the fraction is upside down
Efficiency rounds off the heart of the energy section — you can now track energy in, useful energy out, and everything wasted along the way. One last piece completes the picture: how much energy a given amount of fuel actually holds. That’s energy density, which is where we go next.
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