Question

A simple machine does 30 J of work with an efficiency of 28%. How much energy was put into the machine?

Answer 1

Take into account that efficiency is given by the following expression:

$\text{ efficiency=(output work/input work)}\cdot100$

In this case, you have:

efficieny = 28%

output work = 30J

input work = ?

Replace the previous values of the parameters into the formula for efficieny, solve for input work and simplify:

28% = (30J/input work)*100%

input work = 30J*(100% / 28%)

input work = 107.14 J

Hence,about 107.14 J was put into the machine