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Potential Energy in two springs system problem

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m= 1.18kg is attached between two identical springs on a frictionless, horizontal tabletop. Both springs have spring constant k and are initially unstressed, and the particle is at x=0.

    a) The particle is pulled a distance x along a direction perpendicular to the initial configuration of the springs as shown. Show that the force exerted by the springs on the particle is F = -2kx(1- L/ (sq rt (x2+L2)) in the i (or x) direction.
    http://imageshack.us/a/img690/1021/20130308194200.jpg [Broken]
    2. Relevant equations
    ΔU= - W = -1/2kx^2
    Partial derivatives of U
    Force = dU/dX

    3. The attempt at a solution

    Considering that the length of an unstressed spring is L, and that the particle was pulled a distance x away from x=0, I calculated the length of a stretched spring to be √(x^2 + L^2).
    Additionally, in the initial position of the system, the particle was not moving, and the springs were unstressed, so the initial kinetic energy and potential energy both are 0. As the particle was moved away, the potential energy increased. The change in potential energy would be -1/2kx^2. I wasn't sure if the particle being pulled away would be considered to be kinetic energy, so I assumed that the final kinetic energy was also zero.

    Seeing as force = dU/dX (or dU/dx + dU/dy for two dimensional problems like this problem), I performed a partial derivation on -1/2 kx^2, which resulted in -kx i + 0 j .(would this be why there is no j-component force acting on the particle?)

    So now I have -kx i, which I multiplied by 2 since there are two springs in the system, which leads to -2kx i. This is where I am stuck. I don't understand where (1 - L/√(x^2 + L^2)) came from, although I do recognize L/ √(x^2 + L^2) to be the sine of the triangle. (I hope the picture will help clarify what I am talking about)

    Any hints and corrections to my reasoning would be appreciated. :)
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 8, 2013 #2

    ehild

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    The potential energy of a spring is connected to the change of its length. The initial length of the springs is L, the final is √(x^2 + L^2). How much does the length change if the particle displaces by x?

    ehild
     
  4. Mar 8, 2013 #3

    SammyS

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    You can solve this problem in either of the two following ways.
    1. Calculate the force directly, using Hooke's law to find the force (as a vector) exerted by each spring, the adding the two forces vector-wise.

    2. (As you have attempted) find the potential energy, U, of the system, then use:

    [itex]\displaystyle F_x=-\frac{\partial U}{\partial x}[/itex]

    [itex]\displaystyle F_y=-\frac{\partial U}{\partial y}[/itex]

    [itex]\displaystyle F_z=-\frac{\partial U}{\partial z}[/itex]​
     
    Last edited by a moderator: May 6, 2017
  5. Mar 9, 2013 #4
    Hi, I understand where (1-L/sqrt (x^2 + L^2) ) comes from now.. I subtracted the unstressed length from the stretched length, which is (sq rt (x^2 + L^2) - L ) which simplifies to the former equation. But, now I am confused. Since the variable x in -kx is the displacement of the spring, shouldn't it be -2k(1- L/ (sqrt (x^2+L^2))? The problem is asking for force, which has a SI unit of newtons. But with -2k x (1 - L/ sqrt( x^2 + L^2)), wouldn't the result be in Joules instead?
     
  6. Mar 9, 2013 #5

    ehild

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    The contribution of a single spring to the potential energy is k/2(ΔL)2,the total potential energy is twice that. Find the negative gradient of U=k(sqrt (x^2 + L^2) - L )2.

    ehild
     
    Last edited: Mar 9, 2013
  7. Mar 9, 2013 #6

    SammyS

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    What do you get for the potential?

    -2k(1- L/ (sqrt (x^2+L^2)) has the same units as k -- N/m .

    Potential has units of N∙m .

    That's a big problem !
     
  8. Mar 9, 2013 #7
    I feel really silly now

    I feel silly now. I finally realized that (1/2)k x2 should be (1/2)k (√(x2+L2), expanded it, and multiplied the result by 2. The result is kx2 + 2kL(L-√(x2 + L2)) which is the potential energy of the system! Knowing the potential energy to be a negative integral of the force, I derived the potential energy, and got -2kx(1-L/√(x2 + L2)).

    I was confused because in this problem, part a) asked me to find the force exerted by the springs on the particle, but part b) asked me to find the potential energy of the system. Because of how the question ordered those parts, I mistakenly thought I HAD to find the force first, THEN integrate it to find the potential energy. Doh! That's what I get for thinking in an inflexible manner!

    Thank you, ehild and SammyS! :)
     
    Last edited: Mar 9, 2013
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