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Kinetic Friction of a released block

  1. Sep 5, 2004 #1
    Kinetic Friction !!!

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    Ok guYS, Really need help for this... i have been staring at this question blankly...

    The ONly solution i came up was to integrate the net acceleration with respect to x. But i what do u get when u integrate acceleration... the answers for part a is 4 m and part b is 3.72m/s.

    Qns as follows... :confused: :confused:
    A block is released from rest at the top of a plane inclined at an angle of 45o. The coefficient of kinetic friction varies along the plane according to the relation µk= 0.5x, where x is the distance along the plane measured in meters from the top of the incline. Determine
    a. how far the block slides before coming to rest and
    b. the maximum speed it attains.
    :zzz:
     
  2. jcsd
  3. Sep 5, 2004 #2
    You get

    gSin45 - 0.5x*Cos45=acceleration

    Express acceleration as a diffential involving x as x is already in the problem (and I dont know to solve differential eqns in 3 variables)
    Solve!

    Hint(what is v*dv/dx??)

    EDIT:I assumed that µ=.5x what is k?
     
    Last edited: Sep 5, 2004
  4. Sep 5, 2004 #3
    I think they meant k was a subscript to indicate coefficient of kinetic friction.
     
  5. Sep 5, 2004 #4
    Yeah THANKS

    ThAnks Dude...

    i see v*dv/dx is actually dx/dt * dv/dx =a Hmmm Nice one.. yeah managed to find a reason to integrate the equation wif no worries... and for the 2nd question i reckon that i could find out the value of x and integrate Fk to get work done then equate it to 1/2mv^2 to get max speed... =) :rolleyes:
     
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