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 str

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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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Ngọc Hoa 4 years 2021-08-17T19:16:35+00:00 1 Answers 9 views 0

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  1. Amity
    0
    2021-08-17T19:17:46+00:00
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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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