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23 February, 18:15

A force of 18 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? W = ft-lb

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  1. 23 February, 18:28
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    11.54 ft - lb

    Explanation:

    F = 18 lb, x = 8 in = 8 / 12 = 0.66 ft

    F = k x

    k = F / x = 18 / 0.66 = 27.27 lb/ft

    y = 11 in = 11 / 12 ft = 0.92 ft

    Work done = 1 / 2 x k x y^2

    W = 0.5 x 27.27 x 0.92 x 0.92 = 11.54 ft - lb
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