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3 February, 23:36

A force of 6 pounds is required to hold a spring stretched 0.1 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 1.1 feet beyond its natural length?

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  1. 3 February, 23:39
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    36.3 lb-ft

    Explanation:

    F = force required to hold the spring stretched = 6 lb

    x = stretch in the spring = 0.1 ft

    k = spring constant

    Using the equation

    F = k x

    6 = k (0.1)

    k = 60 lb/ft

    U = work done in stretching the spring

    x' = 1.1 ft

    Work done in stretching the spring is given as

    U = (0.5) k x'²

    U = (0.5) (60) (1.1) ²

    U = 36.3 lb-ft
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