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Qns on potential energy and forces

  1. Aug 8, 2006 #1
    Hello, i hope i posted this in the correct section..

    I have a question:
    the potential energy of a body when it is at point P a distance x from a reference point O is given by V = kx^2, where k is a constant. what is the force acting on the body when it is at P?

    the correct ans is 2kx in the direction of PO

    the ans i got is kx in the direction OP. as work = force x distance, so force required to bring the body to point p is kx^2 / x .

    could some please explain why is the correct ans so? thnk u so much!!
    Last edited: Aug 8, 2006
  2. jcsd
  3. Aug 8, 2006 #2


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    Work is "force times distance" as long as the force is constant- if the force is a variable then the work is given by [itex]\int f(x)dx[/itex]. Notice that this involves a "constant of integration". It is standard to choose that constant so that potential energy is 0 at some specific point; potential energy is always relative to some given point.

    Going the other way, if work is a constant times distance, W= Cx then force is that constant: F= Cx/x= C. But if a more general function then [itex]F= \frac{dW}{dx}[/itex].
  4. Aug 8, 2006 #3
    The potential energy is positive. So there is an attraction toward the origin.

    The force = - (dW/dx) = -2kx.
    So the force is directed from the point P to the point O, the origin.
  5. Aug 8, 2006 #4
    Thnx alot!!!!!
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