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Kinetic friction, mass on track when will it stop.

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A block of mass m is initially held in place on the left side of a track at a distance h above the bottom of the track. The track is completely frictionless, with the exception of a rough horizontal section of track of length h which has a coefficient of kinetic friction μ = 0.15. If the block is released from rest, how many times does the block completely cross the rough section of track before it stops?




    2. Relevant equations




    3. The attempt at a solution

    So I have worked out the height to be 0.85 m, but am stuck on finding an expression for when the ball will stop I have what I think it is

    x=number of times without stopping= v^2/2gh

    does this see right, and I worked out v = 11.43 but when I plug it in I get 7.84 times... in the answer booklet it says seven but im not sure if thats because you cant round up because it doesnt make sense so they round down to 7??
     
  2. jcsd
  3. Mar 25, 2012 #2
    Hey,


    I hope this makes it clear enough :-)


     
  4. Mar 25, 2012 #3
    Wondering how you get the value of h=0.85 m since only a coefficient of kinetic friction μ = 0.15 is given. And not mentioning anything about a ball.
     
  5. Mar 25, 2012 #4
    by the work energy theorem, I worked out how much energy was lost through the distance of friction patch... the height of the track is the same 'height' or distance as the friction patch.
     
  6. Mar 25, 2012 #5
    Yeah thanks, I was assuming that just wanted to check... I did not really pick up on the COMPLETELY part of the question.
     
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