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Physics Work and Energy problem

  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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    Science Advisor

    This is getting out of hand!

    This is the THIRD post of exactly the same question that I have found. The other two were under "General Physics" and "K-12 Homework".

    PLEASE, PLEASE, PLEASE do not post the same question repeatedly!!!!
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