A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch it an addit

Question

A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch it an additional 0.12 m

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Latifah 4 years 2021-08-31T08:16:27+00:00 1 Answers 21 views 0

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  1. Nem
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    2021-08-31T08:18:03+00:00
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    Answer:

    1.503 J

    Explanation:

    Work done in stretching a spring = 1/2ke²

    W = 1/2ke²……………………… Equation 1

    Where W = work done, k = spring constant, e = extension.

    Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

    Substitute into equation 1

    W = 1/2(26)(0.34²)

    W = 13(0.1156)

    W = 1.503 J.

    Hence the work done to stretch it an additional 0.12 m = 1.503 J

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