Question

A force of 250 N is required to lift a 75 kg mass through a pulley system. In order to lift the mass through 3m, the rope has to be pulled through d = 12m. The percent efficiency of system is
A.50
B.75
C.33
D.90

Answer 1

The percent efficiency is the ratio of the useful work done by the pulley to the work put into the pulley by the effort multiplied by 100.

Step by step answer: A body of mass m is to be lifted through a pulley system. We have to pull the rope from one side by d =12 m in order to lift the mass by 3 m to the other side. The height h through which it is to be lifted is 3m. Input work is defined as the amount of effort given by us and output work is the amount of work done by the pulley in lifting the mass.
Amount of work done W by a force F = $F \times d$
(Input work) $W = 250 \times 12 = 3000$ J
Amount of work done against the gravity = mgh [since the body is being lifted in the upward direction i.e., opposite to the direction of weight]
Output Work =$75 \times 10 \times 3 = 2250$ J
Percent efficiency = $\left( {\dfrac{{Outputwork}}{{inputwork}}} \right) \times 100\%$
$\left( {\dfrac{{2250}}{{3000}}} \right) \times 100\%$ = $75\%$
Therefore, option B is correct.

Note: The mass has to be pulled up by 3 m so that we have to pull the rope downwards through 12 m. Thus, 3m is the distance used in calculating output work and 12m in calculating input work.