Question

What should be the power of an engine used to lift $90$ metric tones of coal per hour from a minimum of depth $200m$? $g=10m/s^2$.

Answer 1

We know that $1$ metric tonnes $=1000kg$

Work done , $W=mgh=90\times 1000\times 10\times 200=18\times {10}^{7}J$

So, Power of an engine , $P=\frac{W}{t}=\frac{18\times {10}^7}{3600}=5\times {10}^{4}W$