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Max distance up incline plane with variable friction

  1. Dec 16, 2012 #1
    1. The problem statement, all variables and given/known data
    A block with initial velocity of 10 m/s is sent up an incline at 30 degrees from the horizontal. The coefficient of friction is 0.2x where x is the displacement. Find the maximum distance the block moves up the incline.


    2. Relevant equations
    F = ma
    (1/2)mv^2
    mgh

    3. The attempt at a solution
    I decided to try and use energy for this one.

    Initial energy is (1/2)m(10)^2 + friction and final is mgh because the block comes to rest eventually.

    So: (1/2)m(10)^2 + friction = mgh

    Friction = coefficient * Normal force, so Friction = 0.2x * mgcos(30). Therefore:

    (1/2)m(10)^2 + (0.2x*mgcos(30)) = mgh

    Masses cancel out:

    50 + (0.2x*gcos(30)) = gh

    We want to find x because x is along the incline, and sin(30) = h / x, so h = xsin(30) and substitute back in:

    50 + (0.2x*gcos(30)) = g(xsin30)

    Simplify:

    50 + 1.697x = 4.9x
    50 = 3.203x
    x = 15.6 meters

    The solution is 5.31 meters, though. What have I done incorrectly?
     
  2. jcsd
  3. Dec 16, 2012 #2

    TSny

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    By "friction" do you mean the force of friction or the work done by the friction force? Note that you are setting up an energy expression.
     
  4. Dec 16, 2012 #3
    Oh...the work done by friction would be 0.2x * mgcos(30) * x * cos(180) I think. So is that what I'm supposed to use for the "friction" part in my equation?
     
  5. Dec 16, 2012 #4

    TSny

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    Right, you need the work done by friction. But the force of friction is a variable force (it depends on x). So, how do you get the work done by a variable force?
     
  6. Dec 16, 2012 #5
    Ohh I just learned this a few minutes ago, lol. Integrate the Friction Force, not multiply Friction Force * displacement.
     
  7. Dec 16, 2012 #6

    TSny

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    Yes. Good.
     
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