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1 August, 16:28

If the work required to stretch a spring 3 ft beyond its natural length is 15 ft-lb, how much work is needed to stretch it 27 in. beyond its natural length?

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  1. 1 August, 16:54
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    8.4 ft-lb

    Explanation:

    Work = change in energy

    W = ½ kx²

    When x = 3 ft, W = 15 ft-lb:

    15 ft-lb = ½ k (3 ft) ²

    k = 30/9 lb/ft

    When x = 27 in = 2.25 ft:

    W = ½ kx²

    W = ½ (30/9 lb/ft) (2.25 ft) ²

    W = 8.4375 ft-lb

    Rounding to 2 sig-figs, it takes 8.4 ft-lb of work.
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