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Speed and spring on a track

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A spring with constant 72500.0 N/m is compressed 54.6 cm and used to launch a 109.0 kg physics student. The student starts decending down a 13.4 m high track that is frictionless until the student starts up the incline again. The student's coefficient of kinetic friction on the 28.5° incline is 0.129. What is the student's speed just after losing contact with the spring?

    2. Relevant equations

    spring = 0.5kx2

    3. The attempt at a solution
    alright for this question i equated the mgh+0.5mv2 = 0.5kx2
    and solved for v .... is this the wrong thinking for this question pls. help

    Attached Files:

  2. jcsd
  3. May 20, 2009 #2


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    Is this a V shaped track? It goes down and then up again?

    As to your approach, can you think of why they gave you the coefficient of friction?
  4. May 20, 2009 #3
    its more of a u-shaped curve...and the coefficient of friction is given for the incline which isnt frictionless..but theyre asking for speed right as it leaves the spring so i didnt think the coefficient mattered
  5. May 20, 2009 #4


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    The down half is frictionless right? So there is no need to account for it there. But there is on the other side.

    Like the last problem I just posted about (the truck speed ramp), the KE budget can be given by

    KE_bottom = PE_g_up + W_frict_up

    But it is also at that point derived from

    PE_spring + PE_gravity_down_the_ramp = KE_bottom
  6. May 20, 2009 #5
    if we use this equation KE_bottom = PE_g_up + W_frict_up to solve for the speed we dont know the height that the boy is travelling up...and also maybe a stupid question but why would we need the the other side of the hill if the speed is too be found as soon as the student leaves the spring
  7. May 20, 2009 #6


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    One word: Friction.

    Friction makes the trip non-conservative.

    If it was all frictionless, then all you would need to know is what his downward speed was at the moment of leaving the spring. The height from the bottom wouldn't matter.

    Edit: Apparently they only want the speed off the spring after compression .56m. In what I suppose must just be the first part of a several part question. In which case for that won't that just be 1/2*mv2 = 1/2*k*x2 + m*g*sin28.5*.56m ?
    Last edited: May 20, 2009
  8. May 29, 2009 #7
    i have a quick question.. i did this question a different way using only 0.5*k*x^2=0.5m*v^2.... i didnt use m*g*sin28.5*.56m but i still got the right answer and i was wondering how was that possible??i got 14.1 from my way and 14.3 from the way you did not a big difference but still puzzling to me
  9. May 29, 2009 #8


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    Now that I see the image, now that it's approved, there was the assumption on my part at the time that the spring was pointing down hill. The additional term would take care of the additional assist from gravity over the launching of the boy. Since he was launched horizontally, then that term is not needed.
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