A machine has an efficiency of 80%. If the input work is 200J, the output work is?

Answer Details

Efficiency measures how much of the energy put into a machine comes out as useful work, expressed as a percentage:

\[\eta = \frac{\text{work output}}{\text{work input}}\times 100\%.\]

Rearranging for the output and substituting the given values,

\[\text{work output} = \frac{\eta}{100}\times \text{work input} = \frac{80}{100}\times 200 = 160\ \text{J}.\]

The remaining \(200 - 160 = 40\ \text{J}\) is not destroyed; it is wasted mainly as heat and sound through friction in the moving parts, which is why no real machine reaches \(100\%\) efficiency.

The value \(250\ \text{J}\) comes from dividing by \(0.8\) instead of multiplying, and it should be rejected immediately on physical grounds: an output larger than the input would mean the machine creates energy, which violates the conservation of energy. Use that check every time. In any efficiency question the useful output must be smaller than the input, so the correct operation is always the one that reduces the number.