I believe the answer is 35%. Since the formula for energy = useful work / total work input, you would then plug 35 j into the useful work and 100 under the total work input. You divide that and you get .35. Then, you multiply .35 by 100, and you get 35%.
70% efficient Take: Useful work out --------------------- X 100 = efficiency Work in
The machine efficiency is 35 percent (35/100).
The efficiency is (output energy)/(input energy) = 340/400 = 85%
Efficiency
input force x input distance > output force x output distance -Novanet
The mechanical efficiency of this machine is 30 percent.
The idea is to divide the useful work by the input energy.
70% efficient Take: Useful work out --------------------- X 100 = efficiency Work in
50
The machine efficiency is 35 percent (35/100).
80%
The efficiency is (output energy)/(input energy) = 340/400 = 85%
If a machine has 100 percent efficiency, the output work = the input work. That's actually basically what the efficiency of a machine is - output work / input work * 100.
Efficiency
input force x input distance > output force x output distance -Novanet
The relationship between input work and useful output work can be represented by efficiency. Efficiency is the ratio of useful output work to input work. A higher efficiency indicates that a larger proportion of the input work is converted into useful output work, while a lower efficiency suggests that more of the input work is wasted or converted into non-useful forms.
efficiency