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Velocity as a function of distance [v(x)]

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data
    a body with mass M moves across a plane with friction

    friction constant:
    [itex]\mu = \lambda x^2[/itex]

    the body starts at x=0
    with velocity v0

    find at what x
    the body stops
    and what was the velocity half way there.

    2. Relevant equations

    [itex]v^2=v_0^2+2a\Delta x[/itex]

    3. The attempt at a solution

    obviously,
    [itex]F(x)=mg\mu = mg\lambda x^2[/itex]
    so
    [itex]a(x)=g \lambda x^2[/itex]

    so in the equation [itex]v^2=v_0^2+2a\Delta x[/itex]
    I get
    [itex]v^2=v_0^2+2g\lambda x^3[/itex]


    the Question is, can I use this equation? the acceleration is not constant and this equation
    depend on the fact that [itex]x=v_0t+ \frac{a}{2}t^2[/itex]
    and [itex]v=v_0+at[/itex]
    (and it's not true for non-constant acceleration)

    if I cant, how can I integrate the acceleration?
    or how do I get v(x)?

    Thanks.




    EDIT:
    I used [itex] a= v\frac{dv}{dx}[/itex]
    therefore
    [itex]vdv=adx[/itex]

    [itex]\int_{v_0}^{v(x)}{vdv} = g\lambda \int_{0}^{x}{x^2}[/itex]

    [itex]\frac{1}{2} ( v(x)^2- v_0^2) =\frac{1}{3} g\lambda x^3[/itex]

    [itex]v(x)^2=v_0^2+\frac{2}{3}g\lambda x^3[/itex]

    does that seem right?
     
    Last edited: May 3, 2012
  2. jcsd
  3. May 3, 2012 #2

    gulfcoastfella

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    Gold Member

    I don't see anything wrong with your edited solution.
     
  4. May 4, 2012 #3

    ehild

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    Homework Helper
    Gold Member


    Does the friction increase speed?
    Remember that velocity, acceleration and force are all vectors. You need to use proper signs with them.

    ehild
     
  5. May 4, 2012 #4
    ah, of course... it's with a minus :)
    on paper I actually did it with a minus. Thanks for pointing it out though!
     
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