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Circular motion- friction exerted skates/ice and determine friction coefficient

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data

    A 65-kg speed skater with a velocity of 15.0m/s comes into a turn on the track of radius 25.0m.

    a) How much friction must be exerted between the skates and the ice to negotiate the turn?

    b) Determine the coefficient of friction between the skates and the surface of the ice.

    2. Relevant equations

    For part a) I used:
    Fc=Ff
    v2=μgR
    Ff=μmg

    For part b) I used:
    μ=Ff/Fn

    3. The attempt at a solution

    For part a) I rearranged the v2=μgR equation to solve for μ=0.918

    and then subbed that into the Ff=μmg equation and solved for Ff=585N

    For part b) I used μ=Ff/Fn and used m=65kg and g=9.8m/s2 to give μ=0.918

    If someone could verify that I did this question correctly that would be greatly appreciated! and that my significant figures are being respected! Thank you so much for your help and time :)
     
  2. jcsd
  3. May 5, 2012 #2

    tms

    User Avatar

    For part a you could have simply used [itex]F_f = F_c = m v^2 / r[/itex]. For part b you could have used the value for [itex]\mu[/itex] you calculated for part a.
     
  4. May 5, 2012 #3
    So are my answers wrong or are they correct but i just found a different way of finding them ? haha
     
  5. May 5, 2012 #4

    tms

    User Avatar

    I didn't check the numbers. As for the physics, see if what you said and what I said are equivalent.
     
  6. May 9, 2012 #5
    yup they do! thanks!
     
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