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Coeffecient of train and the tracks?

  1. Dec 17, 2014 #1
    1. The problem statement, all variables and given/known data
    What is the coefficient of friction between the train and the track, given that the train has 20,000 kg and is going at velocity of 26m/s. Note that when the train conductor saw a pedestrian he started braking and also the distance between the pedestrian to the train is 500m
    Other given information:
    X= 500m
    A= -.676 ( because he was stopping the train)
    braking force is -13520 N

    2. Relevant equations
    μkN=Fk

    N means normal force in this case



    3. The attempt at a solution
    So at first I tried using the simple friction formula and tried isolating the coefficient of friction so that i would have ----------- μ=Nf---------- but i have everything except friction force. So i then thought from that I thought i can just divide a and m to get the force, which in turn is the friction force. Overall i am not sure if this is correct because i thought you can just divide the breaking force and the mass for the friction force. It's either one or the other, although i am not completely sure :))))
     
  2. jcsd
  3. Dec 17, 2014 #2
    In this problem, it's more about the Work done to the train to get it to come to a halt. First they give you the train's mass and velocity.
    (Remember: Kinetic Energy http://www.sciweavers.org/upload/Tex2Img_1418868535/render.png [Broken] )

    If we look at the the energy of the system, it goes to zero. As you've said before, there is a frictional force which acts on the system. This (frictional) Work must then equal the Kinetic Energy lost when braking the train.

    Also, Work is equal to the force applied over a distance http://www.sciweavers.org/upload/Tex2Img_1418868266/render.png [Broken] .

    I hope this helps point you in the right direction :)
     
    Last edited by a moderator: May 7, 2017
  4. Dec 17, 2014 #3

    haruspex

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    The braking force you (correctly) calculated is the friction force.
     
  5. Dec 17, 2014 #4
    Thank you for everybody that answered, I get it now :)
     
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