Question

What would be the power of an engine required to lift 9 metric tons of coal per hour from a mine 200 m deep (g = 9.8 m $s^{-2}$).

Answer 1

$1$ metric ton $=1000kg$

So, mass $m=90$ metric ton $=90\times {10}^{3}kg$

Here, the work done $W=$ potential enegy $=mgh$

The required power $P=\frac{W}{t}=\frac{mgh}{t}=\frac{(90\times {10}^3)\times 9.8\times 200}{3600}=49000watt$