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Elevator and unknown passenger mass

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    An elevator motor which is 60% efficient uses 80kW of power while lifting a 2000kg elevator 30m. If the elevator contains five passengers of the same mass and it takes 15 seconds to accomplish this task, then determine the mass of each passenger.


    2. Relevant equations
    d1=d0+v0t+1/2at2
    Fnet=ma
    W=FDcos[tex]\theta[/tex]
    [tex]\epsilon[/tex]=Eoutput/Einputx100%


    3. The attempt at a solution
    d1=d0+v0t+1/2at2
    30=0+0+1/2(152)a
    30=112.5a
    a=0.2666666

    Let x rep. each individual mass
    Fapplied-mg=ma
    F=(2000+5x)(0.26666)+(2000+5x)(9.8)
    F=(2000+5x)(10.06666)

    W=(2000+5x)(10.06666)(30)(1)
    W=(2000+5x)(301.9996)
    W=603999.2+1509.998x

    P=(603999.2+1509.998x)/15
    P=40266.61333+100.6665333x

    0.6=40266.61333+100.6665333x/80000
    0.6(80000)=40266.61333+100.6665333x
    7733.38667=100.6665333x
    x=76.822kg

    Is this correct?
     
    Last edited: Dec 7, 2008
  2. jcsd
  3. Dec 7, 2008 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    I think it is a little easier to figure than your approach.

    Your motor runs at 60% efficiency for 15 s. That means that it delivered 80K*(.6)*15 Joules.

    The motor lifted the car m*g*h.

    Doesn't that mean total M = 80K*(.6)*15/(30*9.8) ?
     
  4. Dec 7, 2008 #3
    Thank you for your reply.

    From what I understand, you're applying E=m*g*h, then rearranging to m=E/g*h right?
    Does this mean that time is proportional to mass in this situation because I thought the elevator motor's time is constant.

    I keep thinking that it is on two separate occasions, one where there are no passengers and one with five passengers.

    When calculating problems such as these, do you always use the output power/energy or is there a way to solve using input and the above efficiency equation.
     
  5. Dec 7, 2008 #4

    LowlyPion

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    Homework Helper

    I'm not sure what efficiency equation you would use, since the efficiency of the power conversion to mechanical work is apparently given.

    The presumption is that the motor only delivers 60% of its 80kw to mechanical power. The mechanical power available to work then is given by the time it took to accomplish the task.

    Power times time is work and work in this case would be increasing Potential Energy.
     
  6. Dec 7, 2008 #5
    Suppose it asks us to find the work done without the passengers. Would the 15 seconds still apply or will it change? Just need to clarify on what determines it an empty elevator or a loaded one.
     
  7. Dec 8, 2008 #6

    LowlyPion

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    Homework Helper

    In practice the motor would draw more current and create more torque within a range of loads and the added mass would be simply accounted for by just the increased power.

    But the statement of the problem only provides the single rate of delivering work. Hence you would expect that the time for an empty elevator would be less than a full one. I think the statement of the problem leaves something to be desired. I think the problem would have been better served to indicate that the power measured was for the loaded elevator.
     
  8. Dec 8, 2008 #7
    Yes! That's exactly what got me confused and made me think that there are two separate systems.

    Thank you very much for your clarification.
     
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