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Horizontal motion with quadratic resistance

  1. Oct 12, 2004 #1
    so i got a block with mass=m traveling on an oiled surface. the block suffers a viscous resistance given:
    [tex] F(v)= -cv^{3/2} [/tex]
    the initial speed of the block is [tex] v_{o} [/tex] at x=0, i have to show that the block cannot travel farther than [tex] 2mv_{o}^{1/2} /c [/tex]
    so far i have;
    [tex] ma=-cv^{3/2} [/tex]
    [tex] m \frac{dv}{dx} \frac{dx}{dt} = -cv^{3/2} [/tex]
    [tex] mvdv=-cv^{3/2} dx [/tex]
    [tex] dx= \frac {mvdv}{cv^{3/2}} [/tex]

    where should i go from here?
  2. jcsd
  3. Oct 12, 2004 #2


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    Homework Helper
    Gold Member

    Put back the "-" sign you have lost,


    [tex] dx= \frac {-mdv}{cv^{1/2}} [/tex]

    Integrate from x=0 to x(final) and from v=v0 to v=0 and you get the desired result.

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