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Frictional Force/Kinematics Problem

  1. Apr 22, 2007 #1
    1. The problem statement, all variables and given/known data

    A 47,000 kg locomotive is traveling at 13 m/s when its engine and brakes both fail. How far will the locomotive roll before it comes to a stop?


    2. Relevant equations

    v_f^2 = v_i^2 + 2ad
    F=ma

    3. The attempt at a solution

    I've tried making a force diagram and working from there, but I'm not even sure if that's right. If I had to guess, I would solve for acceleration and use the equation above to get the desired distance, but I don't know how to do that! Please help!
     
    Last edited: Apr 22, 2007
  2. jcsd
  3. Apr 22, 2007 #2

    cristo

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    Are you not told the coefficient of friction between the train and the tracks?
     
  4. Apr 22, 2007 #3
    nope, not given that
     
  5. Apr 22, 2007 #4
    pleeeeeeeeeeasssssssssssssse help
     
  6. Apr 22, 2007 #5
    there has got to be some kind of resistant force some where, where is the friction? There's got to be friction! Any kind of drag! Please provide all the necessary info or we cannot help. State the question exactly how it was originally stated and show us any equations you think we should know about that has to do with the problem and show us that you tried to figure it out by posting some work. gravitational force is pushing your mass down on earth! provide a mew or something!
     
    Last edited: Apr 22, 2007
  7. Apr 22, 2007 #6
    ok i know there is some friction obviously because it brings it to a stop. but that is exactly how the problem is stated above. another equation to help would be f_k=u * N, where u is the coefficient of kinetic friction and N is the normal force. i know that N is just mg, but i still don't know the coefficient. i have a table of common coefficients of friction, but i don't know that we are supposed to use them. for example, i have steel on steel (dry) coefficient of kinetic friction of .60 and steel on steel (lubricated) one of .05. i tried both of these and they did not work.
     
  8. Apr 22, 2007 #7
    still no luck, any help is much appreciated...
     
  9. Apr 22, 2007 #8
    Do you have a time atleast?
     
  10. Apr 22, 2007 #9
    nope no time, all information given is presented in the first post, and the f_k = umg equation i guess, but i don't know if you need that
     
  11. Apr 22, 2007 #10
    Yup. the other posters are right.

    You have to have either:
    the coefficient of friction
    ----this will allow you to figure out the time and distance

    or

    the time it decellerated to a stop.

    As given, you do not have sufficient information to solve the problem.
     
  12. Apr 22, 2007 #11
    Random information about trains:

    Most "Deisel" engines aren't really powered by diesel engines.

    They are electric engines that use massive deisel generators for power.

    Not exactly pertinent, but interesting anyways.
     
  13. Apr 22, 2007 #12
    WOW i just got it...the key to the problem was approaching the friction as a rolling friction, which i should've got from the problem saying that the locomotive rolls. i looked up the coefficient of rolling friction between dry steel on steel, sure enough it was .002. This led to the right answer of 4311 m. thanks for trying to help, though, i really appreciate it.
     
  14. Apr 22, 2007 #13
    great observation!!
     
    Last edited: Apr 22, 2007
  15. Feb 14, 2010 #14
    Working it out the rest of the way for those who still don't fully understand:

    Initial Velocity, (Vi),=13 m/s
    Coefficient of Rolling Friction for Steel on Steel, (U),=0.002
    Distance Traveled, (d),= unknown
    Final Velocity, (Vf),=0

    Vf^2=Vi^2+2ad

    0=169+2ad

    Now, F=ma=Umg where again (U) is the coefficient of rolling friction, (a) is the acceleration, (g) is gravity, (m) is mass, and (F) is force.

    So, a=Ug

    a=(0.002)(-9.81)

    a=(-0.0196) m/s^2

    Returning to the previous formula,

    0=169+2ad

    0=169+2(-0.0196)d

    Solving for (d) gives

    d=4311.22 m

    which rounds to

    d=4311 m
     
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