Question

A pump is used to lift 500 kg of water from a depth of 80 m in 10 s. Calculate:
(A) The work done by the pump,
(B) The power at which the pump works,
(C) The power rating of the pump if its efficiency is 40%

Answer 1

Hint
Efficiency signifies a peak level of performance that uses the least amount of inputs to achieve the highest amount of output. Efficiency requires reducing the number of unnecessary resources used to produce a given output including personal time and energy.

Complete step by step answer
1) According to the work done formula
Work done (W)= mgh
Here m = mass of the water = 500 kg,
g = gravitational force,
h = depth = 80m
put the values in the equation we get,
 $ W = 500 \times 10 \times 80 = 4 \times {10^5} $ J
2) Now, the power $ P = \dfrac{{work done}}{{time}} $
We know the work done
So, $ P = \dfrac{{4 \times {{10}^5}}}{{10}} = 4 \times {10^4} $
3) Now efficiency is equal to useful power which the pump work upon power input
Now efficiency of the pump is 40%
So, $ \dfrac{{40}}{{100}} = \dfrac{{4 \times {{10}^4}}}{P} $
 $ \therefore P = \dfrac{{4 \times {{10}^4} \times 100}}{{40}} $
So, $P = 105$.

Note
There is a relation between work done and power, for example, the work done against gravity is equal to the change in the potential energy of the body and the work done against all resistive forces is equal to the change in the total energy. Power is the rate at which work is done (measured in watts (W)), in other words the work done per second.