A force of 12 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 n

Question

A force of 12 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 11 in. beyond its natural length?

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Eden 1 week 2021-09-11T01:07:33+00:00 1 Answer 0

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    2021-09-11T01:08:53+00:00

    Answer:

    Work done in stretching the spring = 7.56 lb-ft.

    Step-by-step explanation:

    Normal length of the spring = 8 in or \frac{8}{12} ft

    = \frac{2}{3} ft

    If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in

    = \frac{11}{12} ft

    Force applied to stretch the spring = 12 lb

    By Hook’s law,

    F = kx [where k is the spring constant and x = length by which the spring is stretched]

    12 = k(\frac{2}{3})

    k = 18

    Work done (W) to stretch the spring by \frac{11}{12} ft will be

    W = \int\limits^\frac{11}{12} _0 {kx} \, dx

        = \int\limits^\frac{11}{12} _0 {(18x)} \, dx

        = 18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0

        = 9(\frac{11}{12}

        = 7.56 lb-ft

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