Chapter 9, Problem 19A

What is the efficiency of a machine that requires 100 J of energy to do 35 J of work?

To determine

Efficiency of machine.

Answer to Problem 19A

Efficiency is $\eta =35\%$

Explanation of Solution

Given:

Energy required or input energy is 100 J.

Energy produced or output energy is 35 J

Formula Used:

The efficiency is given as,

  $\eta \%=\frac{E_O}{E_I}\times 100$

Where,

  $E_I$ is energy input.

  $E_O$ is energy output.

Calculation:

For the given values, the efficieny of the machine will be,

  $\begin{array}{l}\eta =\frac{35}{100}\times 100\\ \Rightarrow \eta =35\%\end{array}$

Conclusion:

Therefore, the efficiency of the machine is $\eta =35\%$ .