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Spring Question

  1. Dec 9, 2009 #1
    I'm given the force of a spring in the form F(x) = 3.133x - 4.333x^2 +5.333x^3 - 2.667x^4+.5333x^5 and asked to find the maximum velocity of an object so that the force does not exceed 12 lbs

    My approach was to find x from the force equation which was about 2.5 inches and then set the kinetic energy of the object equal to the potential of the spring and solve for velocity but I am not sure how to change the force equation to fit in 1/2 kx^2
    could someone tell me if this is the wrong approach or how to change the equation please
     
  2. jcsd
  3. Dec 9, 2009 #2

    Doc Al

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    Staff: Mentor

    The 1/2kx^2 applies to a spring that obeys Hooke's law (F = -kx), which is not the case for this spring. Derive a potential energy function for this spring in a similar manner, using integration to find the work required to stretch the spring.
     
  4. Dec 9, 2009 #3
    It seems to me you may have to take an integral to find the work done in compressing the spring to where the force equals 12lbs.

    Then solve for velocity using the work-energy theorem. (This spring does not obey Hooke's law, so (1/2)kx^2 is irrelevant.

    (Sorry Doc, I was typing before I saw your post).
     
  5. Dec 9, 2009 #4
    What I don't understand is that the force doesn't seem to be restoring. If the object is at say +1.0, then your function means the force will also be positive. I don't understand how any spring could increase its force directed away from x=0, the farther you are from x=0.
     
  6. Dec 9, 2009 #5

    Doc Al

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    Staff: Mentor

    I agree. Looks like there's a minus sign missing.
     
  7. Dec 9, 2009 #6
    it isn't an actual spring the equation is for package cushioning in a box the equation was right thanks for the help I got it
     
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