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Math Help - Springs -lost

  1. #1
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    Springs -lost

    Suppose that 2 J of work is needed to stretch a spring from it's natural length of 30cm to a length of 42cm. How much work is needed to stretch the spring from 35cm to 40cm?

    The answer is \frac{25}{24}.

    The difference is 12cm, but Hooke's law is in meters, so I need to use .12. Accordingly I get 2=.12k just plugging into Hooke's law. k is \frac{50}{3}.

    For the setup, I get:

    \int(\frac{50}{3})x  dx The interval is supposed to be .05 to .10, but I can't get it to work for some reason. I get 1/16 as the answer, which is wrong.

    If I don't convert to meters, I get k=\frac{1}{6} with an interval of 5 to 10, and that yields an answer of \frac{25}{4}, also wrong.

    What am I doing wrong?
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Springs -lost

    First, your value for k is incorrect. The work required to stretch the spring from rest to 42 cm is:

     W =2J = \int F dx = \int kx dx = \int_0 ^ {0.12} x dx = \frac 1 2 k (0.12^2 - 0^2).

    Hence  k = \frac {2J}{0.12m^2} = 277.8\ N/m

    Then use this value of k to evaluate the work to stretch from 35 to 40 cm:

    W =  \int _{0.05} ^{0.10} kx dx.

    Remember rhat the deflections are always measured relative to the rest position of the spring.
    Thanks from baldysm
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