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Help for a dynamic problem

  1. May 26, 2007 #1
    an object with mass m=0.1kg is thrown with an initial velocity v0=20m/s in a viscous matter that exercises a resistant force of F=-Bv (B=2kg/s and v=velocity). ignoring the gravity force, find the distance covered by the object in the viscous medium.

    I tried this
    F=-Bv=ma => a=(-Bv)/m;
    a=(dv)/dt => dv/dt=(-Bv)/m --> v dv=(-Bv)/m *dt
    integrating I get

    [tex]-\frac{B}{m}t=\log v - \log 20[/tex]

    since ds/dt=v

    [tex]v=e^{-\frac{B}{m}t + \log 20}[/tex]

    then integrate again [tex]\int{ds}=\int {e^{-\frac{B}{m}t + \log 20}dt[/tex]
  2. jcsd
  3. May 26, 2007 #2
    Hint: e(a+b) = eaeb
  4. May 26, 2007 #3
    It would be much easier to use a = vdv/dx , than a=d2s/dt2
  5. May 26, 2007 #4
    sorry didn't get you. dv/dx? How is it ? Am I on the right track?

    [tex]s=\frac{C}{\alpha}e^{\alpha t} [/tex]

    where [tex]\alpha=-\frac{B}{m} \mbox{ and } C=e^{\log 20}[/tex]

    I know everything except t;
  6. May 26, 2007 #5
    No, i meant
    [tex] a =v\times\frac{dv}{dx}[/tex]

    If you don't know how this equation arises, just try dividing its RHS numerator and denominator by dt.
  7. May 26, 2007 #6
    Any hint that could help me solve this problem is appreciated.
  8. May 26, 2007 #7
    Okay let me make it a lot more simpler for you.
    = (dx/dt)*dv/((dx/dt)*dt) {Multiplying numerator and denominator by dx/dt}
    = v*dv/dx
    Do u get me now?
    Last edited: May 26, 2007
  9. May 26, 2007 #8
    Could be interresting to use the energy theorem.
    The energy dissipated by the friction force is easy to calculate.
  10. May 26, 2007 #9

    Doc Al

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    Staff: Mentor

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