Using 15 percent as machine energy efficiency, what is the actual work output if the total work input is 7500 kilojoules?
A. 500 kilojoules
B. 750 kilojoules
C. 1,125 kilojoules
D. 7,500 kilojoules

The actual work output is 1,125 kilojoules.

Explanation:

Energy efficiency is defined as the efficient use of energy. An appliance, process, or facility is energy efficient when it consumes less than the average amount of energy to perform an activity. Then, energy efficiency is the ratio between the amount of energy used in an activity and the amount expected to be carried out.

Efficiency is calculated as:

efficiency = output / input

Where output is the amount of mechanical work (in watts) or energy consumed by the process (in joules), and input (input) is the amount of work or energy that is used as input to carry out the process.

In this case:

Replacing:

0.15=output/7500 kilojoules

Solving:

Output=0.15* 7500 kilojoules

output=1,125 kilojoules

The actual work output is 1,125 kilojoules.

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