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Math Help - How much work is done...

  1. #1
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    How much work is done...

    A force of 7 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it from its natural length to 12 in. beyond its natural length?

    Now I thought work was just force times distance. So 7 times 4...but my answer is wrong. What's the deal?
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  2. #2
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    You must find the spring constant and integrate, not just multiply 7 times 4.

    F=kx

    7=k(8)

    k=\frac{7}{8}

    \frac{7}{8}\int_{0}^{12}xdx
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  3. #3
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    Quote Originally Posted by galactus View Post
    You must find the spring constant and integrate, not just multiply 7 times 4.

    F=kx

    7=k(8)

    k=\frac{7}{8}

    \frac{7}{8}\int_{0}^{12}xdx
    But that's also incorrect. It'd be x^2/2 times 7/8, 12 squared is 144 divided by 2 is 72 times 7/8 is 63, and it's telling me 63 is incorrect...
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