Work done by motor pulling up elevator

AI Thread Summary
The discussion revolves around calculating the average power required by a motor to lift a freight elevator with a mass of 1490 kg over a distance of 58 meters in 2.1 minutes, considering a counterweight of 985 kg. Participants suggest using the concept of constant velocity, implying zero acceleration, to simplify calculations. They emphasize the importance of calculating the work done to raise the elevator and the impact of the counterweight on the system's energy dynamics. The conversation highlights the need to determine the potential energy difference and the energy required per second for the lift. Ultimately, understanding these principles is crucial for solving the problem accurately.
JUSTaROCK
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Homework Statement


A fully loaded freight elevator has a cab with a total mass of 1490 kg, which is required to travel upward 58
m in 2.1 minutes, starting and ending at rest. The elevator's counterweight has a mass of only 985 kg, so the
elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?

Homework Equations



F=ma
W=F * D D= displacement (distance traveled)
P = W/t P= power, t = time
v=u + at u = initial velocity, v = final velocity
v^2 = u^2 +2aD
D= ut + 1/2(a)t^2

The Attempt at a Solution



I have tried to solve for a by using the formula D= ut + 1/2(a)t^2 then taking that a and multiplying it with the mass then multiplying that by distance and then dividing by the time needed to be done but i can't get it right and i feel like i should do something with the counterweight but i don't know what any thing will help thank you
 
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JUSTaROCK said:
I have tried to solve for a by using the formula D= ut + 1/2(a)t^2 then taking that a and multiplying it with the mass then multiplying that by distance and then dividing by the time needed to be done but i can't get it right and i feel like i should do something with the counterweight but i don't know what any thing will help thank you

I don't recommend using kinematics on this one. I think you're supposed to assume that the elevator travels up slowly, at a constant velocity. (Constant velocity corresponds to 0 acceleration :wink: .)

How much work is required to raise the elevator all the way up? All that work is done in how much time? So how much work is done per unit time? What does that mean? :-p
 
sorry don't follow you on the work done per unit time but i do understand the 0 acceleration
 
Are you considering average power or instantaneous power in your calculation?
How much work is actually being done? How is the counterweight affecting the system?
 
JUSTaROCK said:
sorry don't follow you on the work done per unit time

What is the potential energy difference between the situations where the elevator is at the top and bottom (considering the counterweight too, which is helping the situation)? How much energy per second would be required to get it all that change in energy done in 2.1 minutes?
 
i got it thank you
 

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