Let input power = P Since 80 % of this power is being used, so effective power is \\dfrac{80P}{100} Mass of coal m = 10,000\\ kg Time t = 1 \ imes 60 \ imes 60\\ s Height h = 180\\ m Gravitational acceleration g = 10\\ m/s^2 (given)Now, Power = work / timeSo, \\dfrac{80P}{100} = \\dfrac{10,000 \ imes 9.8 \ imes 180}{60 \ imes 60} Simplifying we get: P = 6.25\\ kW