Post a New Question

physics

posted by .

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.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question