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Finding Coefficient of Friction

  1. Nov 12, 2007 #1
    Finding Coefficient of Friction [solved]

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

    A girl is ice skating at 10 m/s. She stops skating and glides to a stop in 100m. Find the coefficient of friction of the ice.

    2. Relevant equations

    uk = Fk / Fn
    F=ma

    3. The attempt at a solution

    I found the acceleration to be -1/2 m/s^2 since

    Vf^2 = Vi^2 + 2a * displacement
    so
    -100 / 200 = a.

    Since it didn't give me her weight I don't know how to find gravitational force on her. If the surface were frictionless her acceleration would be zero.

    Since her acceleration isn't zero and it takes her 20 seconds to come to a stop, I know that there must be a uk and that it also must be very small. Without her weight, however, I'm lost. Can someone point me in the right direction?
     
    Last edited: Nov 12, 2007
  2. jcsd
  3. Nov 12, 2007 #2
    you know the gravitational acceleration.

    Ff=(mu)Fn

    Fn=Fg

    An=Ag

    take out the masses, since they are constant
     
  4. Nov 12, 2007 #3
    Substitute the formulas for Fk and Fn into the formula for [tex]\mu[/tex]
     
  5. Nov 12, 2007 #4
    Okay, so

    Ff=(mu)9.81 m/s^2

    and the acceleration of the skater is -1/2 m/s^2.

    I'm not sure what the An = Ag means. Normal acceleration equals acceleration due to gravity?

    Does this mean that if force is mass * acceleration and the masses are the same I can just do Ff=(mu)*Fn (letting the mass be 1)

    (1)(-1/2 m/s^2) = (mu)(1)(9.81 m/s^2)

    SO then (mu) = 0.051?
     
    Last edited: Nov 12, 2007
  6. Nov 12, 2007 #5
    As I said, write out the full formula. You will see that the masses cancel, leaving just the accelerations.

    Your calculations appear to be correct to me.
     
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