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Take into account that efficiency is given by the following expression:
[tex]\text{ efficiency=(output work/input work)}\cdot100[/tex]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