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Spring equation proof and work question

  1. Oct 19, 2004 #1
    A particle is attached between two identical springs on a horizontal frictionless table. Both springs have spring constant k and are initially unstressed.

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    a) If the particle is pulled a distance x along a direction perpendicular to the initial configuration of the spring, show that the force exerted by the springs on the particle F=-2kx(1-L/(sqrt(x^2+L^2)))i (where L is the vertical distance of each spring at rest)

    b) Determine the amount of work done by this force in moving the particle from x=A to x=0

    On part a)

    I got that the 2 is there since there are two springs. The L/(sqrt(x^2+L^2) is the sin of the angle between the a spring and the vertical. Where does the (1-) comes from? Also how do I tie the equation together, after figuring out all the parts?

    On part b)

    Do I have to integrate f(x) as x=A goes to x=0?

    Thank you
  2. jcsd
  3. Oct 19, 2004 #2


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    Calculate the stretch of one spring from its equilibrium length of L. (This is where you'll see the "1-" term.)
    Only the horizontal component of that force survives in the net force.
  4. Oct 19, 2004 #3
    Thank You ! I did not realize that the vertical components cancel out!

  5. Dec 15, 2004 #4
    Need Explanation

    I still don't understand how the problem will work.... could you explain with a few steps how to solve both parts? :redface:
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