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Skier Sliding Down Ski Slope

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data
    The record speed for grass skiing was set in 1985 by Klaus Spinka, of
    Austria. Suppose it took Spinka 6.60 s to reach his top speed after he
    started from rest down a slope with a 34.0° incline. If the coefficient of
    kinetic friction between the skis and the grass was 0.198, what was the
    magnitude of Spinka’s net acceleration? What was his speed after 6.60 s?
    The answer is:
    a=3.87m/s^2
    For the speed after 6.60s
    velocity final=25.5m/s^2

    What we Have
    Time: 6.60s
    Degree: 34
    mg:?
    n:?
    M:0.198
    From rest: V initial = 0 m/s
    a=?
    Speed after: 6.60s
    2. Relevant equations
    Friction=Mn
    Efx:
    Efy:

    3. The attempt at a solution
    This is what I did so far:
    EFx: Friction-mgsin(34)=max
    EFy:Normal Force-mgcos(34)=may
    Added the mg's to the other side
    Friction=mgsin(34)+max
    Normal Force=mgcos(34)+may

    Then plugged into the equation to find friction friction=(0.198)(normal force)
    mgsin34+max=(mgcos34+may)(0.198)
    The problem is that everything will cancel out so that is preventing me from finding acceleration
     
  2. jcsd
  3. Oct 25, 2014 #2

    haruspex

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    What is the value of ay?
     
  4. Oct 26, 2014 #3
    What do you mean? It didn`t specify a value is it zero?
     
  5. Oct 26, 2014 #4

    haruspex

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    It represents the acceleration perpendicular to the slope, right? Does the skier stay in contact with the slope? So what's the velocity perpendicular to the slope? Is it constant?
     
  6. Oct 26, 2014 #5
    He stays in contact with the slope, the velocity wouldn't be constant wouldn't it because acceleration is there to make it increase/decrease right?
     
  7. Oct 26, 2014 #6

    billy_joule

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    If the skier stays in contact with the slope what is his displacement in the y direction? What does that displacement say about the velocity and acceleration in that direction?
     
  8. Oct 26, 2014 #7

    haruspex

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    I asked about the velocity perpendicular to the slope.
     
  9. Oct 27, 2014 #8
    Your initial acceleration looks correct, you have assumed contant acceleration up to the time limit, whereas with air drag involved the acceleration rate would diminish during the journey, the terminal velocity never being reached.
    This involves maths of a greater sophistication than simply newtons rules.
     
  10. Oct 27, 2014 #9

    haruspex

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    I don't believe drag is supposed to be taken into account here. It would leave insufficient information.
     
  11. Oct 27, 2014 #10
    What these guys are asking is "If the skis are always in contact with the slope, what is their velocity component perpendicular to the slope?"

    Chet
     
  12. Oct 28, 2014 #11
    Can i go back to finding the acceleration ?
    g ive asumed at 9.81 m/s/s
    The root equation is :
    acceleration = f / m
    But :
    f = ( m * g * sine 34 ° ) - ( m * g * cosine 34 ° )
    f = ( 5.4856 m ) - ( 1.6103 m )
    Subtracting, you get :
    f = 3.8753 m
    Back to the root equation :
    acceleration = ( 3.8753 m ) / m
    The m's cancel, you get :
    acceleration = 3.87538 m/s/s

    If you then apply this acceleration constantly for 6.6 seconds youre velocity will be 25.58 m/s
     
  13. Oct 28, 2014 #12

    haruspex

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    You should know by now that you should not post complete solutions until the original poster has posted a solution. Until then, just helpful hints please.
     
  14. Oct 28, 2014 #13
    I sent an official Warning to Dean Barry about this earlier today.

    Chet
     
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