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Potential Problem

  1. Jan 15, 2006 #1
    My problem is a question stated as follows:

    A particle of mass m moves along the positive x-axis with a potential energy given by

    V(x) = C + x^2 + 4/(x^2)

    where C is a positive constant.

    Calculate the equilibrium position X0 of the particle.

    Now, I so far have considered that the equilibrium position is where the potential energy is zero, however I do not know how to solve for:

    C + x^2 + 4/(x^2) = 0

    without knowing the value for the constant C.

    Is my reasoning correct that the particle is in equilibrium when the potential energy is zero?

    Any help is much appreciated
     
  2. jcsd
  3. Jan 15, 2006 #2

    arildno

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    Why should a position with zero potential energy be an equilibrium??

    What dynamical feature characterizes objects at rest (or in uniform motion)?
     
  4. Jan 15, 2006 #3
    I guess that there is no resultant force acting on the object. The problem now is therefore the relationship between the potential energy at distance (x) and the resultant force on the particle. How do i find that relationship?
     
  5. Jan 15, 2006 #4

    arildno

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    How is potential energy related to the force that "provides" the potential energy?
     
  6. Jan 15, 2006 #5
    The basic relationship is Force x Distance = Potential Energy. Is it then the case in this question that the particle is in equilibrium when that force is zero? So the expression for Potential energy divided by distance (x) gives an expression for that force?
     
  7. Jan 15, 2006 #6

    arildno

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    Sort of, but:
    You HAVE learnt about derivatives, right?
     
  8. Jan 15, 2006 #7
    Yes, but I cant really see the application here, can you explain it a bit more please? To differentiate the expression for Potential Energy w.r.t distance (x). I think I know the mthod now;

    d (P.E)/dx gives 2x - 8/x as the expression for force

    Now the particle is at rest (equilibrium) when the force is zero

    so solving 2x - 8/x = 0 gives x =2

    I think this is right but can you explain why the differentiation w.r.t distance is neccesary in this case?
     
  9. Jan 15, 2006 #8

    arildno

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    Do you know what a "conservative force" is?

    How is energy conservation derived from Newton's laws of motion?
     
  10. Jan 15, 2006 #9
    If a force is conservative then the work done by that force over a distance is stored as potential energy. It is also independant of the path taken whereas work done by a non-conservative force is dependant on the path. As the particle moves along the path of the x-axis the work done by the force is stored as potential energy which is defined by the given expression. This means that the potential energy is a cumulative effect of work done by the force over distance. Therefore a differential over distance shows how the accumulation occurs over distance and where this accumulation is zero the resultant force must also be zero. Now if the force was a non-conservative force then the differential of energy over distance could be zero with a resultant force still acting, correct? As the work done by the force does not add to the potential energy of the particle?
     
  11. Jan 15, 2006 #10

    arildno

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    Well, seems you have some idea about it, so I'll end it by stating that given a potential V(x), the force F acting upon the particle due to this potential is:
    [tex]F=-\frac{dV}{dx}[/tex]

    I'm sure you've seen this; use it for what it's worth.
     
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