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Homework Help: Work done by motor pulling up elevator

  1. May 10, 2010 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    W=F * D D= displacment (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

    3. 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 dont know what any thing will help thank you
  2. jcsd
  3. May 10, 2010 #2


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    Gold Member

    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? :tongue2:
  4. May 10, 2010 #3
    sorry don't follow you on the work done per unit time but i do understand the 0 acceleration
  5. May 10, 2010 #4
    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?
  6. May 10, 2010 #5


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    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?
  7. May 10, 2010 #6
    i got it thank you
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