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18 October, 15:31

A spring on a horizontal surface can be stretched and held 0.6 m from its equilibrium position with a force of 78 N. a. How much work is done in stretching the spring 5.5 m from its equilibrium position? b. How much work is done in compressing the spring 1.5 m from its equilibrium position?

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  1. 18 October, 15:50
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    Step-by-step explanation:

    x = 0.6 m

    F = 78 N

    Let k be the spring constant.

    F = k x

    78 = 0.6 k

    k = 130 N/m

    (a)

    x = 5.5 m

    W = 0.5 kx²

    W = 0.5 x 130 x 5.5 x 5.5

    W = 7865 J

    (B) x = 1.5 m

    W = 0.5 kx²

    W = 0.5 x 130 x 1.5 x 1.5

    W = 146.25 J
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