A go-kart engine is 25% efficient. One lap needs 500 J of useful work from it.
How much energy must its fuel supply?
Think in shares: 25% efficient means each 100 J of fuel energy delivers 25 J of useful work.
So the lap's 500 J of useful work needs far more than 500 J of fuel energy.
Count it: 500 ÷ 0.25 = 2000. The fuel must supply 2000 J.
That move is the efficiency equation solved for the input.
efficiency = useful energy out ÷ total energy in rearranges for either unknown.
To find the useful output: useful energy out = efficiency × total energy in.
To find the input: total energy in = useful energy out ÷ efficiency.
Write the efficiency as a decimal before using it: 25% is 0.25.
Q: A go-kart engine is 25% efficient. Calculate the fuel energy it needs to deliver 500 J of useful work.
Write down the values:
efficiency = 25% = 0.25, useful energy out = 500 J, total energy in = ?
Write down the arrangement whose unknown stands alone:
total energy in = useful energy out ÷ efficiency
Substitute in the values, and calculate:
total energy in = 500 ÷ 0.25 = 2000
The fuel must supply 2000 J.
Q: A dirt bike engine is 20% efficient and takes in 1500 J from its fuel. Calculate its useful work output.
Write down the values:
efficiency = 20% = 0.20, total energy in = 1500 J, useful energy out = ?
Write down the arrangement whose unknown stands alone:
useful energy out = efficiency × total energy in
Substitute in the values, and calculate:
useful energy out = 0.20 × 1500 = 300
The engine delivers 300 J of useful work.
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