A cubical vessel of height $1\mathrm{m}$ is full of water. What is the work done in pumping water out of the vessel?

Step 1: Given:

Height of the vessel, h$=1\mathrm{m}$

The centre of mass of the water in the cubical vessel lies at CM, i.e $\mathrm{h}\text{'}=\frac{\mathrm{h}}{2}=\frac{1}{2}\mathrm{m}$

The volume water, $\mathrm{V}={\mathrm{h}}^3=1^3=1{\mathrm{m}}^3$

∴ Mass of water,$\mathrm{m}=\mathrm{\rho V}=1000\times 1=1000\mathrm{kg}$

Step 2: Formula & Solution

Work done for pumping out water is equal to negative of change in its potential energy.

$$\begin{gathered} \mathrm{W}=-\mathrm{\Delta P}.\mathrm{E} \\ =-(\mathrm{P}.{\mathrm{E}}_{\mathrm{f}}-\mathrm{P}.{\mathrm{E}}_{\mathrm{i}}) \\ =-(0-\mathrm{mgh}\text{'}) \\ =\mathrm{mgh}\text{'} \\ =1000\times 10\times \frac{1}{2} \\ =5000\mathrm{J} \end{gathered}$$

Hence,the work done in pumping water out of the vessel is $5000\mathrm{J}$.