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Statuc and kinetic frictional forces

  1. Oct 10, 2009 #1
    A 103kg baseball player slides into second base. Then coefficient of kinetic friction between the player and the ground is 0.607 a.) what is the magntiude of tfictional force? b.) if the player comes to rest after 2.22sec, what is his intial speed?

    a.) Fs max=Ms*Fn
    Ms=.607
    Fn=103
    0.607*103=62.521
    but it's wrong
    b.) I thought could use V=Vo+At
    Vo=?
    V=O
    t=2.22
    A=)Fk/m

    so 0=Vo+(Fk/m)*2.22
    Is this the right set up?
     
  2. jcsd
  3. Oct 10, 2009 #2
    Remember that frictional force is equal to the normal force times the coefficient, not the mass.

    Your method for part B seems fine, you just need to adjust your acceleration after recalculating the force of friction.
     
  4. Oct 10, 2009 #3
    0=Vo+(Fk/m)*2.22
    for part b
    and I set it up where
    v0= -(fK/m)/2.22
    Fk is .607
    and m is 103
    so -(.607/103)/2.22=-.002655
    but its wrong what am I doing wrong. thank you
     
  5. Oct 11, 2009 #4

    rl.bhat

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    Homework Helper

    Normal force Fn = m*g
     
  6. Oct 12, 2009 #5
    So i redid this problem by doing this:
    Since fk=mk*Fn and Fn=m*g so (mk*m*g)/m but the m gets cancelled out so...
    Vo=-(Mk*g)/2.22
    =(.607*9.8)/2.22
    =-2.6795
    But it's still wrong what am i doing wrong?
     
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