Question

State the prove law of conservation of energy in case of a freely falling body. A pump is required to lift $600kg$ of water per minute from a well of $25m$ deep and to eject it with a speed of $50m{s}^{-1}$. Calculate the power required to perform the above task. $(g=10m{s}^{-2})$

Answer 1

Suppose a body of mass $m$ is falling under accelerating due to gravity $g$ from height $h$.
Energy of a body before free fall (Potential energy i.e. $P.E.$)$=mgh$ .
Initially velocity $u=0$. Therefore kinetic energy $(K.E)=0$
Total energy $=P.E.+K.E.=mgh$
Now suppose it falls upto height $x$ from top.
Potential energy $=mg(h-x)$
since the body is now at height $h-x$ from ground
velocity acquired $v^2=u^2+2gx=2gx$
Therefore $K.E.=\frac{m{v}^2}{2}=mgx$

$P.E.+K.E.=mg(h-x)+mgx=mgh$.
Therefore energy is conserved.
Here $m=600kg,h=25m,v=50m/s,g=10$$m/s^2$, $t=60s$.
Total work done $=mgh+$$\frac{m{v}^2}{2}=600(10\times 25+\frac{{50}^2}{2})=600\times 1500$
Power required $=\frac{\text{Total work done}}{time}=\frac{600\times 1500}{60}=15000watts$$=15kW$