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
First, I assume you meant constant acceleration of 3m/s
Net force=ma + mg
Net force=force applied - friction
= (ma+mg+friction )* velocity
ok well i see that the answer is 66 kw but i don't see how
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
yes and thankyou and i do believe i said constant in questions thansks for help
You said a constant force, not a constant speed