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Coefficient of Kinetic Friction, Normal Force.

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    "A box initially moving at 3.00m/s on a rough, level floor comes to a stop after sliding 6.00m. Find the coefficient of kinetic friction between the box and the floor."


    2. Relevant equations
    Fk=[tex]\mu[/tex]kFN
    V2-V02 / 2x = a


    3. The attempt at a solution

    The first thing I did was write down all of the known variables:
    x = 6.00m
    V0x = 3.00m/s (initial)
    Vx = 0m/s (final)

    The next thing I did was draw a free body diagram, showing the weight (mg) pointing down, the fk pointing to the left (in the negative x direction), and Normal force (N) pointing up.

    I found the acceleration by plugging given values into the second relevant equation:
    02 - (3.00m/s)2 / 2 (6.00m) = -0.75m/s2

    Then I found [tex]\mu[/tex]k by dividing the found acceleration (-0.75m/s2) by g (-9.8m/22) to come up with 0.0765 for [tex]\mu[/tex]k.

    According to the solutions page, the answers are correct, but I have no idea why.
    Why does a/g = [tex]\mu[/tex]k?

    Also, if you are not given the mass of an object, how can you find the normal force of the object if N=mg?

    Any and all feedback is appreciated. Thank you :)
     
  2. jcsd
  3. Oct 20, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    Welcome to PF, Laura.
    Your answer looks good to me. I got .0765 but I used g=9.81.
    The equation of motion is Ff = umg = ma so dividing by m gives you
    u = a/g. Yes, you can't find the normal force mg but you can still do the problem! Think of it as finding the normal force per kg of mass - you are just doing each kg separately.
     
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