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23 September, 09:57

A force of 1 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. 23 September, 10:08
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    Given Information:

    Force = f = 1 pound

    Stretched length = x = 0.1 ft

    Required Information:

    Work done = W = ?

    Answer:

    Work done = 6.05 ft. lb

    Explanation:

    From the Hook's law we know that

    f (x) = kx

    Where f is the applied force, k is spring constant and x is length of spring being stretched.

    k = F/x

    k = 1/0.1

    k = 10 lb/ft

    f (x) = 10x

    The work done is given by

    W = ∫ f (x) dx

    Where f (x) = 10x and limits of integration are (1.1, 0)

    W = ∫ 10x dx

    W = 10*x²/2

    W = 5x²

    Evaluating the limits,

    W = 5 (1.1) ² - 5 (0) ²

    W = 6.05 - 0

    W = 6.05 ft. lb

    Therefore, 6.05 ft. lb work has been done in stretching the spring from its natural length to 1.1 feet beyond it's natural length.
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