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Skier's coefficient of velocity

  1. Dec 13, 2007 #1
    [SOLVED] Skier's coefficient of velocity

    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

    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
    Wnet=delta KE
    Fnetd=Delta KE
    Fa-(coefficient Fn+ Fp)d=1/2 mv2
    Fa- (coefficient Cos 18 g + Sin 18 g)d=1/2mv2

    Finally i got my answers which came up to 18.7 is that right?

    and also i couldn't figure out how to get Fa so i didn't even put in anything for that in my calculations , so i know it's wrong. Thanks for your help
  2. jcsd
  3. Dec 13, 2007 #2


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    What is Fa meant to be? Applied force? If so, then this is zero, since the skier starts with constant velocity. What does Fp stand for?

    You're going about it in more or less the right way, in that the net decelerating force multiplied by the distance travelled is equal to the change in energy; but note that the change in energy is not equal to the initial kinetic energy. There is final potential energy too! However, that is quite straightforward to work out. Apart from that, you're more or less there. Try again, putting this in.
  4. Dec 13, 2007 #3
    Fp is force parallel
  5. Dec 13, 2007 #4
    wait so i added final potential energy to KE in the equation? Fnetd=KE+PEf ??
  6. Dec 13, 2007 #5
    wait wat is my final height? how do i get height? is it 14.2?
  7. Dec 13, 2007 #6


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    No, the initial kinetic energy is equal to the work done by the decelerating force plus the final potential energy.

    No, that is the distance up the slope. You need the vertical distance. You have an angle and a side of a right angled triangle; how do you find another side?
  8. Dec 13, 2007 #7
    oh duh..the height is Sin(18)14.2 right? so..4.39..i can never get these problems unless i draw them out ..i drew it out and saw it haha
  9. Dec 13, 2007 #8
    okay i worked it all out and got 0.165 correcto? (oh god...oh god..plz) :)
  10. Dec 13, 2007 #9
    i just entered it and it wasn't correct..what did i do wrong


    average coefficient? is that the same?
  11. Dec 13, 2007 #10
    Okay, I'm a little confused by your equations but here's a shot at what they mean (or what they should mean).
    -1/2mv[tex]^{2}_{i}[/tex] +mgy[tex]^{}_{f}[/tex] = W[tex]^{}_{friction}[/tex]
    We should know that W[tex]_{friction}[/tex] = [tex]\Delta[/tex]SNcos180.
    From this the coefficient of friction is easily found.
    I'm not really sure what you did with your second and third lines, but I think you're on the right track.
  12. Dec 13, 2007 #11
    wait Wfriction=deltaSNcos180?? i used Cos18(9.81)
  13. Dec 13, 2007 #12
    The work done by friction is Δs⋅N⋅μ⋅cos180. Since friction is acting opposite the direction of motion, we must include cos180, effectively giving the work done a negative value. The normal force is cos(18)w = cos(18)mg, which is what you had for the work done by friction (once you eliminate the mass constants from both sides).
  14. Dec 14, 2007 #13
    I got [tex]\mu = 0.219[/tex]
  15. Dec 14, 2007 #14
  16. Dec 14, 2007 #15


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    Note that there's a bit of ambiguity in this post. What I mean is that the work done by the frictional force is equal to the total change in mechanical energy. Of course, the work done by component of the weight of the skier acting down the slope is equal to the potential energy gained by the skier on climbing the slope, and so does not need to included twice.

    Still, it appears you have the answer now, but I wanted to prevent any possible confusion that this may bring to other readers.
  17. Dec 14, 2007 #16
    Well, I'm not sure of my value... I was hoping that someone would confirm the answer that I arrived at.
  18. Dec 14, 2007 #17
    Yes, that's the correct answer. I don't really see the point in just posting the answer though...
  19. Dec 14, 2007 #18
    Haha, I don't really want to post my working if it's wrong. Furthermore, some textbooks provide answers for their questions without the working. I was thinking that, even if I give an answer, I'm not really doing his homework for him, just merely giving a guide... Or at least that is my opinion.

    So here's my working;

    Using the principle of conservation of energy, the initial kinetic energy of the skiier is converted into gain in gravitational potential energy and work done against friction between the skis and the snow.

    [tex] E_k = E_p + W_f [/tex]
    [tex] \frac{1}{2}mv^2 = mgh + (mg\ cos\ \theta) \mu s [/tex]

    where h is increase in vertical height of the skiier, [tex]\mu[/tex] is the coefficient of friction and s is the distance moved along the inclined plane of the hill.

    Cancelling m from the equation and solving for [tex]\mu[/tex],

    [tex] \frac{1}{2}(12.0)^2 = (9.81)(14.2\ sin\ 18^o) + (9.81\ cos\ 18^o) \mu (14.2) [/tex]
    [tex]\mu = 0.219[/tex]
  20. Dec 15, 2007 #19


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    This is all well and good, but some people come on here asking for help with questions that part of an online assessed test/coursework. These things only request the answer, and so by providing that you are giving them the answer.

    As a general rule, full solutions, or answers, should not be given in the homework forums; rather you are encouraged to try and guide the student towards the correct answer, in the same way a tutor would.
  21. Dec 15, 2007 #20
    Okay, I'll bear that in mind.
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