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Coefficient of kinetic friction and constant acceleration

  1. Apr 25, 2013 #1
    1. The problem statement, all variables and given/known data
    A child's slide is flat, 5 meters long, and sloped down so that the top is 3 meters higher than the bottom. A child slides down, starting from rest, and has v = 6 m/s at the bottom. Find the coefficient of kinetic friction, μk is:



    2. Relevant equations
    F=ma
    f=μkFN
    v2=v02+2aΔx

    3. The attempt at a solution
    use above equation to find constant a=3.6m/s2

    3-4-5 triangle, so θ= 30°

    free body diagram gives:
    FN=mgcos30

    and using f=ma,
    Fnet=3.6m = mgsin30-μkFN

    the masses cancel out to give 3.6 = gsin30-μkgsin30

    solving for μk gives 0.153

    my key says the answer is 0.3
     
  2. jcsd
  3. Apr 26, 2013 #2

    haruspex

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    Think that bit through again.
     
  4. Apr 26, 2013 #3
    The angle here is not correct.
    You don't even need to calculate the value for the angle since you already have the length of the sides.

    The second sin here is not correct - it should be cos as you worked out previously.
     
  5. Apr 26, 2013 #4
    θ=36.9°

    ok, since we already know the length of the sides, we get

    3.6 = 5 - 5cos(36.9)μk

    5 is the force from gravity parallel to the slide

    5cosθ is the normal force perpendicular to the slide

    this gives me μk=.35
     
  6. Apr 26, 2013 #5
    3.6 = gsin(θ)-μkgcos(θ)
    gives
    3.6 = (10)*(3/5) - μk*(10)*(4/5) (don't need to work out θ)
    gives 0.3
     
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