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Mechanics prove

  1. Jun 3, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle with mass m slides down from rest on a smooth plane inclined at an angle of theta with the horizontal . The particle, besides subjected to gravity experiences a resistive force of magnitude mkv ,with v as its velocity at time t and k as a positive constant. Show that

    [tex]v=\frac{g\sin \theta}{k}(1-e^{-kt})[/tex]

    2. Relevant equations



    3. The attempt at a solution

    i started with mg sin theta-mkv =ma

    g sin theta -kv=a

    do i integrate to get the velocity function? I tried but it didn't work.
     
  2. jcsd
  3. Jun 3, 2010 #2
    Yes, you do perform integration to get the velocity function. My attempt resulted in success in replicating the given equation, so you must have fumbled somewhere in your integration. Perhaps you would like to show us your steps so we can pinpoint the error?
     
    Last edited: Jun 3, 2010
  4. Jun 3, 2010 #3
    thanks Fightfish ,

    from a=g sin theta-kv

    [tex]v=\int g\sin \theta dt-\int kv dt[/tex]

    [tex]=tg\sin \theta-kvt+C[/tex]

    when t=0, v=0

    [tex]v=tg\sin \theta-kvt[/tex]

    and the furthest i can get

    [tex]v=\frac{tg\sin \theta}{1+kt}[/tex]
     
  5. Jun 3, 2010 #4
    The problem lies here. v is also a function of t and not a constant, so the integration of kv wrt t is not merely multiplying it by t. We need to consider the problem as a first-order differential equation:
    [tex]\frac{dv}{dt} + kv = g sin \theta[/tex]
    Then, solve it using the integrating factor method.
     
  6. Jun 4, 2010 #5
    thanks !
     
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