Posted by HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! on Thursday, March 12, 2009 at 6:01pm.
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
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

Physics  bobpursley, Thursday, March 12, 2009 at 6:17pm
First, I assume you meant constant acceleration of 3m/s
Net force=ma + mg
Net force=force applied  friction
forceappliedfriction=ma+ mg
power required=forceapplied*velocity
= (ma+mg+friction )* velocity

Physics  HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!, Thursday, March 12, 2009 at 6:26pm
ok well i see that the answer is 66 kw but i don't see how

Physics  drwls, Thursday, March 12, 2009 at 8:02pm
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  HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!, Thursday, March 12, 2009 at 8:57pm
yes and thankyou and i do believe i said constant in questions thansks for help

Physics  drwls, Friday, March 13, 2009 at 12:15am
You said a constant force, not a constant speed
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