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Homework Help: Coefficient of Friction. Please Help?

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A skier traveling 12.6 m/s reaches the foot of a steady upward 15.2° incline and glides 12.0 m up along this slope before coming to rest. What was the average coefficient of friction?

    2. Relevant equations



    3. The attempt at a solution

    D = 12m
    Velocity initial = 12.3 m/s
    V final = 0 m/s
    Angle = 15.2° incline

    Use the equation Vf^2 = Vi^2 + 2*a*d
    Find a

    0 = 12.3^2 + 2*a*12
    2*a*12 = -12.3^2

    Divide both sides by 2*12
    a = 6.30375 m/s^2


    Force parallel = mass * g * sin θ
    Force normal = mass*g* cos θ
    Friction Force = µ * mass*g* cos θ

    Going up an incline 2 forces decrease your velocity
    Force parallel and Friction, so we add these 2 forces to find the total force decreasing your velocity. Since these forces decrease your velocity they are negative.
    a = 6.30375 m/s^2


    ∑ Forces = mass * acceleration
    (-mass * g * sin θ) + -(µ * mass*g* cos θ) = mass * 6.30375
    Notice mass cancels


    (- g * sin 15.2°) + (-µ * g* cos 15.2° ) = 6.30375
    6.30375 = -2.57 + -9.457 µ
    6.30375 = -12.027 µ
    µ = 0.524

    But the answer is incorrect. Can someone please tell me where I messed up?
     
  2. jcsd
  3. Mar 11, 2010 #2
    By how much? Your initial velocity suddenly went from 12.6 m/s to 12.3m/s.
     
  4. Mar 11, 2010 #3
    I just recalculated my mistake, and I got [tex]\mu[/tex] = 0.549.

    Does that seem right to you?
     
  5. Mar 11, 2010 #4
    (- g * sin 15.2°) + (-µ * g* cos 15.2° ) = 6.30375
    6.30375 = -2.57 + -9.457 µ
    6.30375 = -12.027 µ
    µ = 0.524

    But the answer is incorrect. Can someone please tell me where I messed up?[/QUOTE]

    In your original calculation you have put the net acceleration +ve which isn't the case as forces are along the acceleration.
     
  6. Mar 11, 2010 #5
    So, I'm slightly confused. Do you mean it should be:

    (- g * sin 15.2°) + (-µ * g* cos 15.2° ) = 6.615
    6.615 = -2.57 + 9.457 µ
    6.615 = 6.887 µ
    µ = 0.961

    I'm drawing a blank. =/
     
  7. Mar 11, 2010 #6
    No I meant the opposite. It should be
    -g*sin15.2 - µ*g*cos15.2 = -6.615.
    I had only quoted you.
    One more thing, why are you always adding the the terms on left when there is a unknown
    µ with one of the terms. Like here(this is yours):
    6.615 = -2.57 + 9.457 µ
    6.615 = 6.887 µ​
    [/CENTER]

    I noticed it in the first post too.
     
  8. Mar 11, 2010 #7
    So I redid it:

    -g*sin15.2 - µ*g*cos15.2 = -6.615.
    -2.572 - 9.457 µ = -6.615
    -9.457 µ = -4.043
    µ = 0.4275

    Is that correct?
     
  9. Mar 11, 2010 #8
    Yes looks good to me.
     
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