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Newton's Second Law Particle Problem

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    4. A particle of mass m is experiences a Force that attracts the particle following
    F(x) = -k*x^-2
    (an inverse square relationship) where x is the distance from an origin at x = 0 and k is a
    positive constant. If the particle is released from rest at x = d, how long does it take to reach the origin?

    2. Relevant equations

    F = ma = m(dv/dt)

    3. The attempt at a solution

    Equating newton's second and the Force equations yields -k*x^-2 = m(dv/dx)(v).

    After seperating the variables and integrating once, I got V = sqrt(2k/m)*sqrt((1/x)-(1/d)).

    My boundary condition was that at x = d, v =0 for the integration. Setting V = dx/dt and trying to solve for x(t) is unfathomable, so I'm not sure how to continue here, or if I have been taking the wrong approach. I've double checked my math as well, so I don't believe the error is there, but I could always be wrong.

  2. jcsd
  3. Feb 10, 2010 #2


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    Let the velocity of the particle be v when it passes the origin.
    The kinematic equation becomes
    v = vo + at. But a = -k/m*1/x^2 and vo = 0
    dx/dt = -k/m*1/x^2 or
    x^2*dx = -k/m*t*dt
    Find integration between the limits x = d to x = 0 and find t.
  4. Feb 10, 2010 #3


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    Hi skyrolla! :smile:

    (have a square-root: √ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)

    If you need to solve ∫ √x/(√x - a) dx, just make the obvious substitution. :wink:
  5. Feb 10, 2010 #4
    I thought that kinematic equation was only applicable to a constant force, not a position-dependent varying one?
  6. Feb 10, 2010 #5


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    In the kinematic equation we have taken into account the position dependence of the acceleration. And we have taken the integration to find t. So it must work.
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