Question

calculate the power of engine to lift 90 metric tonnes of coal per hour from a coal mine of depth 200m

Answer 1

Here m= 90,000 kg ; and d = 200 m ; so force F= mg =90,0000 N
Now power $P=\frac{W}{t}=\frac{Fd}{t}=\frac{90,0000\times 200}{1}=18\times {10}^{7}J/s$