Question

A motor is used to lift water from well of depth $10m$ and to fill a water tank of volume $30{{m}^{3\,}}\,10\,\text{mins}$. The tank is at height of $20m$ above ground. If energy is wasted, the power of the motor is in (kW).

Answer 1

Hint: Work done by external force is equal to the change in potential energy of the body. Total power input to the system is equal to the sum of power leakage and the power used for doing the work.
Total energy of the system is the sum of potential energy of the system and the kinetic energy of the system. For an ideal motor total input power is equal to total power used to do work.

Complete step by step solution:
If a force lifts a body of mass $m$ to a height $h$ vertically up, then the work done by the force is given by
$W=mgh$
Where,
$g$ is the acceleration due to gravity.
$h$ is the height through which the body is lifted.
$m$ is the mass of the body.
The density of the water is $\rho =1000kg/{{m}^{3}}$
If $\rho $ is the density of the body and $V$ is the volume of the body,
Then the mass of the body is given by,
$m=\rho V$
In 10 minutes, total volume of the water lifted is $30{{m}^{3}}$
Mass of the water lifted in 10 minutes is, $m=\rho V$
$m=10\times 1000kg=10000kg$
Total height of lift is, $h=10m+20m=30m$
Total work done by the motor in 10 minutes is,
$\begin{align}
  & W=mgh \\
 & =10000\times 9.8\times 30J \\
 & =2.94\times {{10}^{6}}J
\end{align}$
Total energy wasted $=20%$
Then total energy used to do work is $80%$
If power of the motor is $P$
$P\times \frac{80}{100}$$=\frac{Work}{time}$
$\begin{align}
  & =\frac{2.94\times {{10}^{6}}}{10\times 60}W \\
 & =4.90\times {{10}^{3}}W \\
 & =4.90kW
\end{align}$
$\begin{align}
  & 0.8P=4.90kW \\
 & P=\frac{4.90}{0.8}kW \\
 & =6.13kW
\end{align}$

Therefore, the power of the pump is $6.13kW$.

Note: Using the work energy theorem the work done is a change in total energy of the system. If net force acting on the system is zero then total energy of the system remains conserved.