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?

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

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

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

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

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

You said a constant force, not a constant speed

To solve this problem, you need to use the formula for power:

P = Fv

where P is power, F is force, and v is velocity. In this case, you are given the force (F) to be constant at 3.00 m/s, and you need to find the minimum power required (P).

First, find the net force acting on the elevator. The net force can be calculated by subtracting the frictional force from the weight of the elevator and its maximum load.

Net force = (Weight of the elevator + Maximum Load) - Frictional force
Net force = (mass of elevator + mass of maximum load) * acceleration due to gravity - Frictional force

mass of elevator + mass of maximum load = 1.0e3 kg + 800.0 kg = 1800.0 kg
acceleration due to gravity, g = 9.8 m/s^2

Net force = (1800.0 kg) * (9.8 m/s^2) - (4.0e3 N)
Net force = 17,640 N - 4000 N
Net force = 13,640 N

Next, multiply the net force by the given velocity to find the power required:

P = Net force * velocity
P = (13,640 N) * (3.00 m/s)
P = 40,920 N.m/s

Finally, convert the power from watts to kilowatts by dividing it by 1000:

P = 40,920 N.m/s / 1000
P = 40.92 kW

Therefore, the minimum power that the motor must deliver to lift the fully loaded elevator at a constant force of 3.00 m/s is 40.92 kilowatts.