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Homework Help: Find coefficient of kinetic friction?

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data

    60kg block slides along the top of 100kg block. 60kg block has an acceleration = 3.0m/s^2 while a horizontal force of 320N is applied to it. There is no friction between the 100kg block and the frictionless surface but there is friction between the two block. Find the coefficient of kinetic friction between the blocks

    2. Relevant equations

    coefficient of kinetic friction = fk / Fn

    3. The attempt at a solution

    Well since the normal force is easy to find, Fn=ma (right?) and the fk I was thinking could be found by with mg...but when I divide ma/mg I get an answer that's just a little higher than the right answer (.306m/s^2 vs. .24m/s^2)

    I feel like I might just be missing a simple step somewhere or confusing something, any insight on this would be much appreciated.
  2. jcsd
  3. Oct 8, 2009 #2
    Are you summing your forces properly? What are all of the forces that lead to a 3.0 m/s^2 acceleration? You're right about the normal force, but you seem to be disregarding that whole 320 N that's also being applied. Unless I'm missing something in your attempted solution.
  4. Oct 8, 2009 #3
    Actually that 320N was confusing me, I'm not really sure what I'm supposed to do with it or how it comes into play finding the coefficient of kinetic friction.
  5. Oct 8, 2009 #4
    draw out your FBD and see where that 320 goes. It will have a significant effect on your acceleration.
  6. Oct 8, 2009 #5
    Well I have drawn out a FBD and I can see that the 320N give the block a Fnet in the +x direction (since it does have acceleration), but I'm just not sure how I'm supposed to use that 320N in any equation.
  7. Oct 9, 2009 #6
    Well if you are given or know the answer (.24), you can reverse the problem from there to find the force that is needed to get that coefficient of friction.
    ~~(I know that in a normal assignment you won't know the answer, but this is good experience to see what is going wrong on this problem.)~~

    Fk = (.24)(Fn) = (.24)(9.8)(60) = (.24)(588) = 141.12N

    Now that we know the force that is needed for the proper coefficient, we can then reverse that force to find the acceleration needed to achieve that.

    a = F/m = 141.12/60 = 2.352m/s2

    We know that that acceleration is lower than the current horizontal acceleration, therefore I assumed that the 320N force is heading in the opposite direction as the already given 3.0m/s2 horizontal acceleration.
    So i calculated the acceleration caused by the 320N.

    a = F/m = 320/60 = 5.3333m/s2
    5.3333m/s2 - 3.0m/s2 = 2.3333m/s2

    2.3333m/s2 is very close to what we found would be the proper acceleration needed for that coefficient of friction.

    When working the problem back using 2.3333m/s2 as my horizontal acceleration, i get a coefficient of kinetic friction that is 0.2380... which can then be rounded to .24.

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