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Skier going down slope

  1. Sep 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A skier with a mass of 53 kg starts from rest and skis down an icy (frictionless) slope that has a length of 59 m at an angle of 32° with respect to the horizontal. At the bottom of the slope, the path levels out and becomes horizontal, the snow becomes less icy, and the skier begins to slow down, coming to rest in a distance of 155 m along the horizontal path.

    (a) What is the speed of the skier at the bottom of the slope?
    (b) What is the coefficient of kinetic friction between the skier and the horizontal surface?

    I just need help with (b)

    2. Relevant equations

    Vf2 - Vi2 = 2ay
    fk = uk*N

    3. The attempt at a solution

    Don't need help with (a) but I'll list what I did anyway...

    (a) What is the speed of the skier at the bottom of the slope?

    Got the height first.

    -->sin32 = y/59

    Vi = 0m/s
    -->Vf2 = 2ay
    -->Vf2 = 2*-9.8m/s2*-31.265m
    -->Vf = 24.755 m/s


    Vi = 24.755m/s
    Vf = 0m/s
    x = 155m

    -->Vf2-Vi2 = 2*a*x
    -->0 - 24.7552 = 2*a*155m
    -->a = -1.9768m/s2

    -->F - Fk = ma
    -->mg - ukN = ma
    -->mg - ukmg = ma
    -->g - ukg = a
    -->uk = (g-a) / g
    -->uk = (-9.8m/s2 - (-1.9768m/s2)) / (-9.8m/s2)
    -->uk = 0.7983

    Somehow this is wrong??? Any help is much appreciated!
  2. jcsd
  3. Sep 30, 2010 #2
    The speed you calculated in part (a) is incorrect. First the vertical component of acceleration is not -9.8 (the skier isn't in free fall, there are forces pushing up on him as well as pulling down). Second, you only considered the vertical speed of the skier and completely neglected the horizontal speed.

    The speed you want is the magnitude of the velocity that is parallel to the ski slope, since once he reaches the bottom all that velocity will be redirected horizontally.

    Start by drawing a free body diagram for the skier. Then find the components of each force that is parallel to the ski slope. Use those components in Newtons 2nd law to find the acceleration.

    It would be easier to use energy, but I'm assuming you haven't learned that in class yet.
  4. Sep 30, 2010 #3
    Actually, it is correct as I checked the answer. I'm guessing they just want the y-component. We haven't learned energy yet but I went ahead and looked it up and got the same answer using the potential energy to get the KE at the bottom.

    Only (b) is incorrect.
  5. Sep 30, 2010 #4
    lol, you're right, the speed is correct. I didn't even bother to calculate the speed or look at part b, once I saw your equation. You got lucky there actually because the reasoning is still incorrect, It just happens to work out to the same as the energy equation, but like i said it appears to be blind luck.

    Anyway, in part B, only horizontal forces are slowing him down. The only horizontal force acting on the skier is friction. The skiers weight has no horizontal component.
  6. Sep 30, 2010 #5
    Ahhh, I see now. Thanks! I was compensating for the vertical forces as well.

    For those who are reading this thread in the future, the coefficient of kinetic friction between the skier and the horizontal surface is just:

    -->Fk = ma
    -->ukN = ma
    -->ukg = a
    -->uk = a/g
    -->uk = -1.9768m/s2 / -9.8m/s2 = 0.2017
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