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Coeffiecient of Kinetic Friction problem? (Shuffleboard and Distance traveled)

  1. Oct 14, 2008 #1
    Coeffiecient of Kinetic Friction problem??? (Shuffleboard and Distance traveled)

    1. The problem statement, all variables and given/known data

    A shuffleboard disk is accelerated to a speed of 5.8m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.31, how far does the disk go before it comes to a stop? The courts are 15.8m long.

    Known info:
    vi = 5.8m/s
    vf = 0m/s
    [tex]\mu[/tex]k = 0.31

    2. Relevant equations

    I have no idea!

    3. The attempt at a solution

    So, I originally tried vf2 = vi2+ 2a[tex]\Delta[/tex]d, but that would make no sense because I used 15.8 as the distance but then the problem would be solved....
  2. jcsd
  3. Oct 15, 2008 #2


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    Science Advisor
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    Welcome to PF!

    Hi glindawantsme! Welcome to PF! :smile:

    (have a delta: ∆ :smile:)

    Yes, that is the right equation …

    So what is vf? what is vi? and what is a?

    Put them all in, and find ∆d. :smile:
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