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Spring Compression Question

  1. Mar 27, 2004 #1
    I'm stuck on the following application of integrals:

    Question: It requires .05 joule (newton-meter) of work to stretch a spring from a length of 8 centimeters to 9 centimeters and another .10 joule to stretch it from 9 centimeters to 10 centimeters. Evaluate the spring constant and find the natural length of the spring.

    My work so far:
    I know that the spring constant is the k in the equation F(x) = kx.
    I know that work is: W=Integral from a to b of F(x).
    I got W=FD
    .05 = F(1)
    F = .05

    F = kx
    .05 = k(1)
    k = .05 ----> spring constant?

    My teacher said that this isn't right....I think I'm way off and I just want to know the technique to solving these types of problems.

    I don't understand how to get the natural length of the spring. In other word problems similar to this one they don't ask for the natural length, just the work or the force required to stretch or compress the spring an addition blah units. The natural length is usually given in 90% of the problems in my book.


    HELP!
     
  2. jcsd
  3. Mar 27, 2004 #2

    jamesrc

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    Gold Member

    You can't go from [tex] W = \int_0^D{ \vec F\cdot d\vec s [/tex] to W = FD unless the force is constant (independent of displacement). That is not the case here, since F = kΔx

    [tex] W_{12} = \int_{8 \rm cm }^{9 \rm cm}{k(x-x_o)dx} [/tex]
    [tex] W_{12} = \frac 1 2 k(x - x_o)^2|^{9\rm cm}_{8\rm cm} = 0.05 {\rm J} [/tex]

    Similarly:

    [tex] W_{23} = \frac 1 2 k(x - x_o)^2|^{10\rm cm}_{9\rm cm} = 0.1 {\rm J} [/tex]

    (which can all be written directly from the work-energy theorem if you remember the expression for the potential energy of a linear spring, but anyway...)

    You see that you've got 2 equations and 2 unknowns (k and xo), so you know you should be able to solve it. It's just a few lines of algebra from there. Let me know if you get stuck.
     
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