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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

  • physics -

    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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