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 s 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?

Question

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

and

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

You said a constant force, not a constant speed