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Coefficents of Friction

  1. Dec 18, 2008 #1
    1. The problem statement, all variables and given/known data
    A young skier (25kg) pushes off with ski poles to give herself an initial velocity of 3.5 m/s down a hill with 5o slope with a coefficient of friction of 0.20.
    Find the time until skier comes to stop and her displacement.
    m = 25kg
    vi = 3.5m/s
    [tex]\theta[/tex] = 5o
    [tex]\mu[/tex] = 0.20

    2. Relevant equations
    All kinematic-based equations
    Newton's Laws (specifically [tex]\stackrel{\rightarrow}{F}[/tex] = m([tex]\Delta[/tex][tex]\stackrel{\rightarrow}{v}/[/tex][tex]\Delta[/tex]t)
    FF = [tex]\mu[/tex]FN

    3. The attempt at a solution
    My idea for this was to find the FNET and sub that into Newton's Second Law, solving for [tex]\Delta[/tex]t (assuming that [tex]\stackrel{\rightarrow}{v}[/tex]f is zero)
    My solution for [tex]\Delta[/tex]t was 1.3 seconds, correct to two significant digits (to me, that seems quite off)

    For the second half of the question, finding the distance, I used [tex]\Delta[/tex][tex]\stackrel{\rightarrow}{d}[/tex] = [tex]\stackrel{\rightarrow}{v}[/tex]i + [tex]\frac{1}{2}[/tex][tex]\stackrel{\rightarrow}{a}[/tex]([tex]\Delta[/tex]t)2 using [tex]\stackrel{\rightarrow}{a}[/tex] as the component force of gravity acting parallel to the hill and I got 23m, correct to 2 significant digits.

    I was then told I completed the question completely wrong. Now I'm lost. What was I supposed to do?
  2. jcsd
  3. Dec 18, 2008 #2


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    Homework Helper

    Welcome to PF.

    Maybe consider it is a Kinetic Energy to work from friction problem?

    Work would be the (μ*mg*cos5° - mg*sin5°) times the distance and that would equal the KE.
  4. Dec 18, 2008 #3
    I haven't learned kinetic energy yet...

    BTW, this is a correspondence coarse for Grade 12 Physics in Ontario..
  5. Dec 19, 2008 #4
    You know the frictional force, so find the net acceleration on the skier due to friction and gravity.
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