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,
[tex]\eta=\dfrac{1000\times 0.02}{200\times 1.5}\\\\\eta=0.067[/tex]
Thus, the efficiency rating of the jack is 0.067.
