Question

A power station conveyor lifts 500 tons of coal per hour to a height of 80 feet. What average

horsepower is required?

Answer 1

$$P=\frac{W}{t}=\frac{mgh}{t}$$ $$={\rm\frac{500*10^3\: kg*9.8\: N/kg*80*0.305\: m}{3600\: s}}$$

$$=3.32*10^4\:\rm W=44.5\: hp$$