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Finding the Work Done by Friction w/out Friction

  1. Dec 10, 2014 #1
    I feel like I'm missing some obvious or simple piece but I can't seem to figure out the first part. If I can get the first part, the second part is easy.

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

    An 80-kg baseball player running at 6.5 m/s goes into a slide 3.0 m from second base. He comes to rest just as he touches the bag. What is the work done by friction? What is the coefficient of kinetic friction between the player and the ground?

    2. Relevant equations
    W = F * d
    W = ΔK
    F = μN

    3. The attempt at a solution
    This seemed straight forward. I calculated the kinetic energy to get the overall work:
    W = Kf - Ki = 1/2mv_f^2 - 1/2mv_i^2 = -169 J
    This should be the same work as the friction but I've been told the correct answer should be -1700 J. Am I wrong about something here?
     
    Last edited: Dec 10, 2014
  2. jcsd
  3. Dec 10, 2014 #2

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    Check your arithmetic.
     
  4. Dec 10, 2014 #3
    m = 80kg
    v_i = 6.5m/s
    v_f = 0

    1/2(80)(0)^2 - 1/2(80)(6.5)^2 =
    0 - 1/2(80)(42.25) =
    0 - 1/2(338) =
    0 - 169 = -169

    Same answer.
     
  5. Dec 10, 2014 #4

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    Check it again --- it's glaring at me with big red eyes.
     
  6. Dec 10, 2014 #5
    Wow! How did I miss that twice!

    1/2(80)(0)^2 - 1/2(80)(6.5)^2 =
    0 - 1/2(80)(42.25) =
    0 - 1/2(3380) =
    0 - 1690 = -1690 ≈ -1700
     
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