an elevator weighing400kg is to be lifted up at a constant velocity of 3cms-1 what would be the minimum power of motor to be used

Fe = m*g = 400kg * 9.8N/kg = 3920 N. =

Force of elevator.

P = F * V = 3920 * 0,03 = 117.6 Joules/s
= 117.6 Watts.