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Friction and stopping distance

  1. Oct 6, 2004 #1
    I'm trying to work out a problem involving the stopping distance of a skiier, and I thought it was a straightforward problem but I've gone wrong somewhere.

    Determine the stopping distance for a skier moving down a slope with friction with an initial speed of 22.2 m/s.
    Assume μk=0.173 and θ=4.9

    To solve, I tried the formula:

    Xinitial= Vsquared/2g(sinθ + μkcosθ)
    so i used the data given and got

    X=(22.2)squared/2(9.8)(sin4.9+0.173cos4.9)
    = 97.5 m

    but this did not work out.

    Anyone have any thoughts?
     
  2. jcsd
  3. Oct 6, 2004 #2

    Gokul43201

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    Here's your problem : why are you adding [itex]gsin \theta~and~\mu _k gcos \theta[/itex] ?
     
  4. Oct 6, 2004 #3
    that was the formula given in the textbook in the section on stopping distances.

    based on X1= - Vsquared/2a
    (all in the x direction)

    and 2a= 2g x (sinθ + μkcosθ)

    i'm ignorant-- why doesn't that work?
     
  5. Oct 6, 2004 #4

    Pyrrhus

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    Well doing the force analysis

    Y-axis
    [tex] N = mgcos \theta [/tex]

    Using the formula

    [tex] F_{f} = \mu N [/tex]

    [tex] F_{f} = \mu_{k} mgcos \theta [/tex]

    X- axis

    [tex] F_{f} - mgsin \theta = -ma [/tex]

    [tex] \mu_{k} mgcos \theta - mgsin \theta = -ma [/tex]

    [tex] mgsin \theta - \mu_{k} mgcos \theta = ma [/tex]

    [tex] gsin \theta - \mu_{k} gcos \theta = a [/tex]

    [tex] g(sin \theta - \mu_{k} cos \theta) = a [/tex]
     
  6. Oct 6, 2004 #5
    ok that definitely makes sense.
    the example in the textbook had a different co-ordinate system so I guess that was the difference.
    thank you both for the help!
     
  7. Oct 6, 2004 #6

    Gokul43201

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    Lannie, instead of trying to use formulas blindly, learn how to figure them out using free body diagrams.

    And the example in your textbook had the block slidingup the slope, not down it. That was the difference, not the co-ordinate system.
     
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