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Work and Energy

  1. Dec 7, 2003 #1
    A particle of mass m is attatched to 2 identical springs on a horizontal frictionless table. both springs have spring constant k and an unstretched length L the particle is pulled a distance x along a direction perpindicular to the initial configuration of the springs, show that the force exerted on the particle due to the springs is

    F= -2kx(1-(L/(x^2+L^2)^.5))i

    please help us, we've tried to use f=-kxcos(theta), where cos(theta) = x/(x^2+L^2) but we are unable to determine why x = (x^2+L^2)^.5 - L which is needed to solve the problem
  2. jcsd
  3. Dec 8, 2003 #2


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    The problem itself is not clear. I could interpret this as two springs in series with the "particle" attached at the end or with the two springs both attached to a wall and the particle between them.

    Since in the first case it would be simpler to treat the two springs as a single spring, I suspect the second is intended.

    It also is not stated clearly, but I will assume it, that the two springs are attached at points 2L apart so that the springs start in equilibrium position with the particle between them.

    Now, as the particle is moved perpendicular to the wall, a distance x, we see two congruent right triangles. One leg is of Length L (from point of attachment to center) and the other is of length x so the hypotenuse has length √(L2+ x2). The hypotenuse is, of course, the full stretched length of one spring so the amount of stretch is √(L2+ x2)-L.

    The total force due to one spring is k√(L2+ x2)-L but this is directed toward the point of attachment. If we break that into components, parallel and perpendicular to the wall, we see that the parallel forces of the two springs cancel each other while the ones perpendicular to the wall add.

    You are correct that the pependicular force for each spring is F (x/√(L2+ x2)so the total force is twice that: (2k√(L2+ x2)-L)(x/√(L2+ x2).

    The answer you give is that, with -i since it is directed back toward the wall.
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