An escalator at a metro station is designed to transport 40 passengers per minute from the ground floor to a vertical height of 12 m . The average mass of a passenger is 60 kg . If 20 % of the electrical input power is lost against friction in the mechanical parts, what is the minimum electrical power required by the motor to operate the escalator? (Take g = 10 m s ⁻² )

Step-by-step solution

Total mass of passengers transported per minute is M = 40 60 = 2400 kg . The useful work done by the escalator per minute in lifting the passengers is W = Mgh = 2400 10 12 = 288000 J . The useful power output of the motor is P_ out = W t = 288000 60 = 4800 W = 4.8 kW . Since 20 % of the electrical input power is lost against friction, the efficiency of the motor is 80 % . Let the electrical input power be P_ in . Then 0.8 P_ in = P_ out . P_ in = 4.8 0.8 = 6.0 kW . Answer: 6.00 kW