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Stopping distance using velocity and friction coefficient on an incline plane

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data
    A skier skiing downhill reaches the bottom of a hollow with a velocity of 20m/s, and then coasts up a hill with a 10° slope. If the coefficient of kinetic friction is 0.10, how far up the slope will she travel before she stops?


    2. Relevant equations
    aΔt=v2-v1
    d= 1/2(v1 + v2)∆t
    d=v1Δt + 1/2aΔt^2
    v2^2=v1^1 + 2ad
    F=ma
    μ=Ff/Fn


    3. The attempt at a solution


    d=20^2 / 2(9.8)(sin10)(0.1)


    I tried to substitute for a in a formuala I found online that worked previously for stopping distance on a horizontal plane.
     
  2. jcsd
  3. Nov 5, 2012 #2

    PBV

    User Avatar

    First you have to take into account the force of gracity slowing her down on the slope as well. Which is Fgrav= - mg sin(10).

    Secondly the friction, which is the normal force times the constant of friction. That gives Ffriction = - mg cos (10) * 0.1

    The resulting force slowing her down would then be F = -mg ( 0.1cos(10) + sin(10) = ma, implying a = -g(0.1cos(10)+sin(10)).

    Now using your top formula we get Δt = (v2-v1)/a = -20/-[g(0.1cos(10)+sin(10))].

    Then using the second one gives:

    d = 20^2 / 2[g(0.1cos(10)+sin(10))]
     
  4. Nov 5, 2012 #3
    Thank you so much! That gets the exact answer in the book (75m) Much appreciated :)
     
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