Question

A person applies a force of 200 N over 1.5 m to a jack. The jack exerts a 1000-N force on a car a distance of 0.02 m. What is the efficiency rating of the jack? HINT: Use work formula first to solve for WORKout and again for WORKin...then use efficiency formula.

Answer 1

Answer:

The efficiency rating of the jack is 0.067.

Explanation:

We have, a person applies a force of 200 N over 1.5 m to a jack. The jack exerts a 1000-N force on a car a distance of 0.02 m.

It is required to find the the efficiency rating of the jack. It is equal to the ratio of output work to the ratio of input work. So,

$\eta=\dfrac{1000\times 0.02}{200\times 1.5}\\\\\eta=0.067$

Thus, the efficiency rating of the jack is 0.067.