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Math Help - Force to hold a spring

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    Force to hold a spring

    A force of 4 pounds is required to hold a spring stretched 0.3 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.9 feet beyond its natural length?
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    Quote Originally Posted by derekjonathon View Post
    A force of 4 pounds is required to hold a spring stretched 0.3 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.9 feet beyond its natural length?
     <br />
F = kx<br />

    use the 4 lb force and 0.3 ft to determine the spring constant k in lbs/ft.

    then integrate the variable force over the interval of the spring's displacement to find the work

     <br />
W = \int_{x_0}^{x_1} kx \, dx<br />
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    Quote Originally Posted by derekjonathon View Post
    A force of 4 pounds is required to hold a spring stretched 0.3 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.9 feet beyond its natural length?
    First you must find the spring constant, k, using F=-kx.
    You are given x_1=0.3 feet, and F= 4 lbs.

    k=\frac{-F}{x_1}.
    The negative sign indicates that the spring is in tension.

    You are also give x_2=0.9 feet, which is how much the spring is stretched beyond its natural length.

    Work = \int dW=\int_{0}^{x_2} F dx =-\int_{0}^{x_2} kx dx

    The negative sign indicates work exerted against the sprink energy.
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