Post a New Question


posted by .

A 1.0 e 3 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of 4.0 e 3 N retards the elevator's motion upward. Waht minum power in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant force of 3.00 m/s?

I have no idea how to do this

I believe I'm suppose to use the formula for power

P= W/delta t = Fd/delta t = mgd/delta t


P = Fv

I don't know how to do this problem using any of these formulas

please tell me which formula to use and how to do this thanks

  • Physics -

    First, I assume you meant constant acceleration of 3m/s

    Net force=ma + mg
    Net force=force applied - friction

    forceapplied-friction=ma+ mg
    power required=forceapplied*velocity
    = (ma+mg+friction )* velocity

  • Physics -

    ok well i see that the answer is 66 kw but i don't see how

  • Physics -

    Do you mean a constant SPEED of 3.00 m/s?

    At maximum load, the motor must provide a maximum force
    (1000 kg)*g + 800 kg*g + 4000 N = 21,640 N.
    Maximum power requirement = Force x velocity = 65 kw

  • Physics -

    yes and thankyou and i do believe i said constant in questions thansks for help

  • Physics -

    You said a constant force, not a constant speed

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question