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Finding Δx of a box sliding down an inclined plane using Work theory.

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data

    A box of mass m with an initial velocity of v0 slides down a plane, inclined at θ with respect to the horizontal. The coefficient of kinetic friction is μ. The box stops after sliding a distance x. How much work is done by friction?


    2. Relevant equations

    Work=Force*x*cosθ
    W=ΔKE
    KE0=(.5)(m)(v02)


    3. The attempt at a solution


    FF=(μ)(FN)

    FNety=FN-Fgy
    0=FN-Fgy
    FN=Fgy

    Fgy=mgcosθ

    FFK=(μ)(FN)
    FFK=(μ)(mgcosθ)

    Work=Force*x*cosθ
    Work=(μ)(mgcosθ)*x*cosθ

    Something tells me this is not correct... To be honest I'm completely lost on this one =( Thats as close as I could get, and its not one of the answer choices. I'd be really thankful for anyone who could help show me the process of deriving the work done by friction!
     
    Last edited: Oct 26, 2011
  2. jcsd
  3. Oct 26, 2011 #2
    From your line here:
    Work=(μ)(mgcosθ)*x*cosθ

    (μ)(mgcosθ) is the Force of friction.

    I take it that xcosθ is supposed to be the distance. The distance should be the length of the hyp of the inclined plane that the block has traveled. Since you have μmgcosθ as the force of friction, it looks to me like your axis is parallel with the top/side of the box. In that case, if down the slope was positive, wouldn't the distance traveled just be x?
     
  4. Oct 26, 2011 #3
    Oh your saying the second cosθ=1 because the angle between the force and displacement is 0°! So that would give me:

    Work = (μ)(mgcosθ)*x*1

    Now That seems more reasonable haha!
     
  5. Oct 26, 2011 #4
    Oh wait >.< My bad, I typed up the wrong title for my question. Yes, your right haha, the distance traveled would be represented by x. Sorry, its late. But you did help me indirectly =P
     
  6. Oct 26, 2011 #5
    Sorry for the late reply then, glad I could help.
     
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