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Rectilinear motion of a particle

  1. Oct 4, 2011 #1
    A particle of mass m moves without friction subject to a force F(x) =
    −kx + [itex]\frac{kx3}{A2}[/itex], where k and A are positive constants. It is projected
    from x = 0 to the positive x direction with initial velocity v0 =
    A[itex]\sqrt{\frac{k}{2m}}[/itex]. Find:
    (a) the potential energy V (x),
    (b) the kinetic energy T(x),
    (c) the turning points of the motion.

    So I don't know if I am on the right track, I feel I am missing something. For V(x) (potential energy), i got:
    [itex]\frac{1}{2}[/itex] kx[itex]^{2}[/itex] - [itex]\frac{1}{4}[/itex] [itex]\frac{kx^{4}}{A^{2}}[/itex]

    Then I found T(x) (kinetic energy) to be:
    T[itex]_{o}[/itex] - [itex]\frac{1}{2}[/itex]kx[itex]^{2}[/itex] - [itex]\frac{1}{4}[/itex] [itex]\frac{kx^{4}}{A^{2}}[/itex]

    I know I am given an initial V[itex]_{o}[/itex], but where would I plug that in to find V(x) or T(x)?

    Any help is very much appreciated.

    Regards
     
  2. jcsd
  3. Oct 5, 2011 #2
    Anyone? Do I need to include other attempts?
     
  4. Oct 5, 2011 #3
    Hi,
    It seems to me you've done very well so far.
    You did correctly work out, that under a central force expressible as:
    [itex]
    \vec{F} = -\vec{\nabla}{V}
    [/itex]
    So that in your case,
    [itex]
    V(x) = -\displaystyle \int_0^x{Fdx}
    [/itex]
    Here, under the constraints on the given question, it is okay to take the potential energy equal to some base/initial value of your choice(at x = 0), say even zero.
    But recall that m*x''t = F(x), so it may be best to solve a differential equation for x(t), in order to find out the kinetic energy, which will be, in this case K = mx'(t)^2/2;
    Try that,
    Daniel
     
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