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-   -   Coeffecient of Kinetic Friction (https://www.askmehelpdesk.com/showthread.php?t=516331)

  • Oct 13, 2010, 10:12 AM
    heyyalliee
    Coeffecient of Kinetic Friction
    Given:After hitting a long fly ball that goes over the right fielder's head and lands in the outfield, the batter decides to keep going past second base and try for third base. The 58 kg player begins sliding 3.40 m from the base with a speed of 4.35 ms.

    Okay the first questions wanted to know how much work was done on player by friction and I got -552J which is right.
    It's second part I'm having a hard time with...
    If the player comes to rest at third base, What was the coefficient of kinetic friction between the player and the ground?
    I am not sure how to find that. Any tips would be awesome.
  • Oct 13, 2010, 11:08 AM
    Unknown008

    I don't know how you got the energy, but now that you have it, you can work out the force that the friction exerts.





    And

    R is the normal reaction of the batter and is at the same time his weight. Find mu.

    Or,

    To find a, use the kimatics equation:



    Then use F = ma to get the force.

    Since

    We have:



  • Oct 14, 2010, 10:31 AM
    heyyalliee

    So how do I know what the normal reaction time of the batter is?

    Sorry I am still a little confused. Since I have work I solved for F. F=W/d
    F= .552/ 3.40? I assumed d stood for distance. Am I on the right track?
  • Oct 14, 2010, 10:37 AM
    Unknown008

    Yes, the normal force of the batter is his weight :)
  • Oct 14, 2010, 02:05 PM
    heyyalliee

    Okay don't think I am an idiot but I'm not getting the right answer.

    I find Force=-552/3.40
    I get -162.35. If R is his weight 58. I divide the force I have by 58 and I get -2.79 or -2.8. It's saying that's not the answer. Haha I think I have completely messed up this problem.
  • Oct 15, 2010, 06:43 AM
    Unknown008

    Well, remember that 58 kg is his mass and not his weight? :rolleyes:

    Weight = Mass x Acceleration due to gravity
  • Oct 15, 2010, 07:38 AM
    heyyalliee

    Awesome thanks for you help! :)
  • Oct 15, 2010, 07:41 AM
    Unknown008

    You're welcome :)

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