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24 July, 01:54

A force of 10 pounds is required to stretch a spring 4 inches beyond its natural length. Assuming Hooke's law applies, how much work is done in stretching the spring from its natural length to 6 inches beyond its natural length?

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  1. 24 July, 02:39
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    The work done is 5.084 J

    Step-by-step explanation:

    From Hooke's law of elasticity,

    F = ke

    F/e = k

    F1/e1 = F2/e2

    F2 = F1e2/e1

    F1 = 10 lbf, e2 = 6 in, e1 = 4 in

    F2 = 10*6/4 = 15 lbf

    Work done (W) = 1/2F2e2

    F2 = 15 lbf = 15*4.4482 = 66.723 N

    e2 = 6 in = 6*0.0254 = 0.1524 m

    W = 1/2*66.723*0.1524 = 5.084 J
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