If it requires 4.5 J of work to stretch a particular spring by 2.3 cm from its equilibrium length, how much more work will be required to st

Question

If it requires 4.5 J of work to stretch a particular spring by 2.3 cm from its equilibrium length, how much more work will be required to stretch it an additional 3.5 cm

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MichaelMet 3 years 2021-08-17T09:16:54+00:00 1 Answers 11 views 0

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  1. Giakhanh
    0
    2021-08-17T09:18:24+00:00
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    Answer:

    \Delta W=24.1162\ J

    Explanation:

    Given:

    • work done to stretch the spring, W=4.5\ J
    • length through which the spring is stretched beyond equilibrium, \Delta x=2.3\ cm=0.023\ m
    • additional stretch in the spring length, \delta x=3.5\ cm=0.035\ m

    We know the work done in stretching the spring is given as:

    W=\frac{1}{2} \times k.\Delta x^2

    where:

    k = stiffness constant

    4.5=0.5\times k\times 0.023^2

    k=17013.2325\ N.m^{-1}

    Now the work done in stretching the spring from equilibrium to (\Delta x+\delta x):

    W'=0.5\times k.(\Delta x+\delta x)^2

    W'=0.5\times 17013.2325\times 0.058^2

    W'=28.6162\ J

    So, the amount of extra work done:

    \Delta W=W'-W

    \Delta W=28.6162-4.5

    \Delta W=24.1162\ J

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