A mixer-grinder has an efficiency of 65 percent. If the output work is 50 joules, what is the amount of input work required?

Question

Bryan Maynard

Physics

6 May, 08:07

A mixer-grinder has an efficiency of 65 percent. If the output work is 50 joules, what is the amount of input work required?

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Answers (1)

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Nyasia Curry · Answer 1

We are to look for the input work that would yield 50 J when it is only at 65% efficiency. To show this mathematically: Input work x (65/100) = 50 J. So to find the input work, we are to divide 50 J by (65/100). This is shown below:

Input work = 50 / (65/100) = 76.92 J

Therefore, the input work needed is 76.92 J.