NEETPhysicsWork, Power and Energy
An escalator in a shopping mall moves passengers to a floor 10 m above. A steady stream of passengers steps onto the escalator at a rate of 30 passengers per minute . If the average mass of a passenger is 60 kg and 20 % of the input electrical energy is lost to friction, what is the minimum input electrical power required by the escalator motor? (Take g = 10 m s ⁻² )
Options
- A3.75 kW
- B2.40 kW
- C3.00 kW
- D225 kW
Correct answer
A. 3.75 kW
Step-by-step solution
Average mass flow rate of passengers, dm dt = 30 60 60 = 30 kg s ⁻¹ Useful mechanical power output required, P_ out = ( dm dt )gh = 30 10 10 = 3000 W = 3.0 kW Since 20 % of the input energy is lost to friction, the efficiency of the motor is 80 % . Input electrical power, P_ in = P_ out 0.80 = 3.0 0.80 = 3.75 kW Answer: 3.75 kW