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Calculating the Coefficient of Friction

  1. Nov 2, 2009 #1
    1. A 50 kg Skater pushes off horizontally for a distance of 0.4m with a force of 150N. If she travels 80m before stopping, what is the coefficient of friction?

    I am completely lost on this problem, can someone please assist me?

    Thanks.
     
  2. jcsd
  3. Nov 2, 2009 #2
    First, you should find skater's velocity after pushing off. This velocity will be initial one in second path.

    In the second path, you should find acceleration using v^2 - v0^2 = 2as, then plug a into Newton's 2nd law eq to find c.f.
     
  4. Nov 3, 2009 #3
    I'm not quite sure how to find the velocity after the push off. And would our total distance be 79.6m (80-0.4)?
     
  5. Nov 3, 2009 #4
    to find the velocity after the push of;

    you have your net Force acting on your skater and you have the mass of the skater; so you can use the newton's law F=m.a to find the acceleration during the push of period; Now you have to find the velocity...

    you can use X=1/2at^2 to find it you know the acceleration and the displacement during the push of so you can calculate the t....
     
  6. Nov 3, 2009 #5
    was your last post unfinished?
     
  7. Nov 3, 2009 #6
    Can you find skater's in 1st path when you have F, s?
     
  8. Nov 3, 2009 #7
    so far i have a=3m/s2 t=7.285s v= 21.855m/s

    i am still stuck on how to get the friction coefficient from this information.
     
  9. Nov 3, 2009 #8
    When you have initial speed in 2nd path and its distance, you can find a in this path (a < 0). Now from N's 2nd law, you have a, m => F => c.f

    is this termin speed in 1st path? In 1st path, i have: v^2 = 2as = 2 * 3 * 0.4 = 2.4
     
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