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28 November, 14:24

A force of 10 lb is required to hold a spring stretched 2 ft. beyond its natural length. How much work is done in stretching the spring from 2 ft. beyond its natural length to 5 ft. beyond its natural length

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  1. 28 November, 15:30
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    workdone = 30.5J or 22.5 Ib. ft

    Step-by-step explanation:

    From Hookes law, F=ke

    k=F/e

    F=10 Ib, e = 2f

    K=10/2ft = 5 Ib/ft

    when e = 5ft

    F=ke

    F = 5 Ib/ft * 5 ft

    F=25 Ib

    force required for stretching from 2ft to ft beyond its natural length equals = 25-10 = 15 Ib

    Workdone in stretching a spring is = 1/2Fe

    = 1/2 * 15 Ib * 3ft

    =22.5 Ibft

    1 Ibft is equal to 1.35582joules.

    22.5 Ibft = 22.5*1.35582 = 30.5059J

    workdone = 30.5J
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