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Skier's coefficient

  1. Dec 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A skier traveling 12.0 m/s reaches the foot of a steady upward 18° incline and glides 14.2 m up along this slope before coming to rest. What was the average coefficient of friction?



    2. Relevant equations
    Wnet=delta KE

    coefficient=Ff/Fn
    and a ton of other Forces equations


    3. The attempt at a solution

    This problem was amazingly hard..so many steps but here's what i did
    KEi=Wnet+PEf
    72=coefficient(9.33+3.03)(14.2)+43
    Coefficient=0.755

    Is that correct?
     
  2. jcsd
  3. Dec 13, 2007 #2
    First,

    [tex]KE_0=PE+W_f[/tex]

    Make sure this equation makes sense to you intuitively. The object starts with only kinetic energy and this energy is transferred to other forms of energy--potential energy and 'wasted' energy of friction. Since the skier is stopped, he/she has zero kinetic energy because at that exact moment he/she is not moving (v=0).

    With:

    [tex]W_f=F_fD[/tex]

    [tex]F_f=\mu_kF_n[/tex]

    [tex]KE_0=\frac{mv^2}{2}[/tex]

    [tex]PE=mgh[/tex]

    Take care noting the difference between D, the distance travelled along the slope, and h, the height above the beginning of the slope. Use trigonometry to relate D,h,and the angle of the slope. Plugging in and with those equations you are left with one unknown. Do you know what it is? If you do, how did you go about calculating it? If you show your work without plugging in numbers, we can help much more easily :smile:

    If you're still stuck, write out your form of my first equation plugging in everything that is known and we'll go from there.
     
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