If the maximum load carried by an elevator is 1400 kg ( 600 kg -Passengers + 800 kg -elevator ) , which is moving up with a uniform speed of 3 m s - 1 and the frictional force acting on it is 2000 N , then the maximum power used by the motor is _____________ kW . g = 10 m s - 2

Step-by-step solution

The maximum force acting is, F m a x = 14000 + 2000   N =   16000   N The formula for power is given by, P = F m a x v . It is given, v = 3   m   s - 1 . ⇒ P   =   16000   N   ×   3   m   s - 1   =   48000   W = 48   kW ⇒ W   =   48   kW