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Retarding Force on a Particle

  1. Oct 12, 2007 #1
    1. The problem statement, all variables and given/known data
    A retarding force acts on a particle with mass m and initial velocity v0 moving on a horizontal surface; this force is proportional to the square root of the instantaneous velocity of the particle:

    Fr = -k*sqrt(v)

    Find an expression for the velocity of the rock as a function of time.


    2. The attempt at a solution
    I believe that this has to be solved through the integration of the force expression, possibly setting the right side equal to ma or m(dv/dt):

    m*dv/dt = -k*sqrt(v)
    dv/dt = -(k/m)*sqrt(v)

    I don't know how to go about integrating this though, especially considering the v is present on both sides. I tried and this is the answer I got, though I doubt that it is correct:

    v = (-3kt/m)^2 + vo

    Thanks for any help!
     
    Last edited: Oct 12, 2007
  2. jcsd
  3. Oct 12, 2007 #2

    cristo

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    Where does the 3 come from? Could you show your integration please?
     
  4. Oct 12, 2007 #3
    Now that I think about it I think it should be done this way (though I'm not sure of this either because it ends with the square root of -kt/m... so unless k is negative it's not real):

    dv/dt = -(k/m)*sqrt(v)
    v = -(k/m)*int[sqrt(v)]dt
    v = -(k/m)*[t*sqrt(v)*(1/v)]
    v = -(k/m)*(t/v)
    v^2 = -kt/m
    v = sqrt(-kt/m) + v0
     
  5. Oct 12, 2007 #4

    cristo

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    No, that's not correct. You have a v on the right hand side, so cannot simply integrate the left hand side to give v. This is a separable equation: [tex]\frac{dv}{\sqrt{v}}=-\frac{k}{m}dt[/tex]. Can you solve this?
     
  6. Oct 12, 2007 #5
    So you would integrate both sides with respect to t? If that is the case, I know that the right side will be -kt/m, but I do not understand how to perform the left side's integral.
     
  7. Oct 12, 2007 #6
    Oh. You integrate the left side with respect to v and the right with respect to t? So
    int[v^-1/2]dv=-kt/m + c
    1/2v^1/2= -kt/m + c
    v=sqrt[-2kt/m+c]

    Would c somehow have to do with v0 then? Would it be

    v=sqrt[-2kt/m] + v0 ?
     
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