physics
posted by Anonymous .
an elevator weighing400kg is to be lifted up at a constant velocity of 3cms1 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.
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