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Potential Energy/Work done paradox?

  1. Apr 1, 2013 #1
    We all know that whenever an object is subjected to move against a conservative force field by an external agent,work should be done to do that.Let that work be W(applied). This is stored as Potential energy of the system.
    If F is conservative force acting on a body and if we move that body against that force,then potential energy change ΔU equals W(applied) or negative of work done by the force(W(force).
    ΔU = W(applied)= -W(force).


    If there are two bodies in space,A and B. A exerts an attractive conservative force F and similarly B exerts the same force F on A. Just like "unlike" charges in electrodynamics. If A is fixed and B is allowed to move away from A ,so that attractive forces of each other pull each other. To move B away from A by a small distance ds, work should be done.And this is stored as potential energy of the system and is equal to the negative of the work done by the attractive force F of A.
    ΔU=-Fds

    This is common.Very common to everyone.Now my question.
    If A exerts a force F on B and B exerts the same force F on A ,then we really have to do work against the attractive force of A on B and at the same time,attractive force of B on A,right?
    I think the above works if above bodies are unlike charged particles.Because,there should be a central force of magnitude 2F pulling each other.
    Hence , ΔU=-2Fds
     
  2. jcsd
  3. Apr 1, 2013 #2

    A.T.

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    Gold Member

    In a frame where they both move you do work on both, but the distance is less.
     
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