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Why didn't this work? (Pun intended)

  1. Jul 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mineshaft by a cable connected to a winch. The shaft is inclined at 30o above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 12.0 s and then continues at constant speed. (a) What power must the winch motor provide when the car is moving at constant speed? (b) What maximum power must the winch motor provide? (c) What total energy has transferred out of the winch motor by work by the time the car moves off the end of the track, which is of length 1250 m?
    2. Relevant equations
    P=W/t
    W=ΔK+ΔU
    v=vo+at
    x=xo+vot+½at2
    Sinθ=O/H

    3. The attempt at a solution
    I'm okay with (a), but I'm working on part (b). I assume that the max power the winch must provide can be found by considering the total change in mechanical energy of the system over a time t.

    My result is,

    Pmax=½mv[(v/t)+gsinθ]

    This gives me half of the max power. Does the total change in mechanical energy of the system not govern the max power output of the winch?

    Does the net force on the cart govern the max power the winch must provide?
     
  2. jcsd
  3. Jul 22, 2015 #2

    TSny

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    Gold Member

    Question (b) is asking for the maximum instantaneous power. Do you know how to express the instantaneous power in terms of force F and velocity v?
     
  4. Jul 22, 2015 #3
    Yeah,

    P=F⋅v

    right?..

    I used P=Fv, then I used Newtons second law to show that,

    Pmax=mv[(v/t)+gsinθ]

    Which is the correct answer.. I'm wondering why this approach has given me the right answer but my approach gave me half of the correct answer.
     
  5. Jul 22, 2015 #4

    TSny

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    Homework Helper
    Gold Member

    Total change in energy over time gives average power for the interval of time rather than an instantaneous power.

    At the start, t = 0, the power is zero since P = Fv and v = 0 at t = 0.
    At time tf = 12.0s just before the acceleration goes to zero the instantaneous power is P = Fvf where vf is the velocity at time tf. P = Fvf is the maximum instantaneous power.

    Due to the fact that the acceleration is constant between t = 0 and t = tf, the average power is just the average of the initial and final power:

    Pavg = (1/2)(0 + Fvf) = (1/2)Fvf = (1/2)Pmax.
     
  6. Jul 22, 2015 #5
    Ahhh, okay. I understand.

    Thank you for your help. Much appreciated
     
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