If a person does 100 Joules of work on a hammer, and the hammer does 92.5 Joules, what is the efficiency?

Efficiency in a mechanical context is calculated by comparing the useful work output to the total work input. In this scenario, work input is the total amount of work done on the hammer, which is 100 Joules, and the useful work output is the amount of work the hammer does, which is 92.5 Joules.

To find efficiency, you can use the formula:

[ \text{Efficiency} = \left( \frac{\text{Useful Work Output}}{\text{Total Work Input}} \right) \times 100% ]

Substituting the values:

[ \text{Efficiency} = \left( \frac{92.5 , \text{Joules}}{100 , \text{Joules}} \right) \times 100% ]

Calculating this gives:

[ \text{Efficiency} = 0.925 \times 100% = 92.5% ]

Thus, the efficiency of the hammer is 92.5%, indicating that 92.5 Joules of the input energy is effectively converted into useful work, while the remaining energy may be lost to factors such as friction or sound.

This understanding is fundamental when