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12 October, 01:14

A force of 6 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it from its natural length to 8 in. beyond its natural length?

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  1. 12 October, 01:40
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    Explanation and answer:

    Given:

    6 lb is needed to stretch 4 inches beyond natural length.

    Need work done to stretch same string from natural length to 8 inches.

    Solution:

    string stiffness, K

    = Force / stretched distance

    = 6 lb / 4 inches

    = 1.5 lb/inch

    Work done on a string of stiffness K

    = (Kx^2) / 2 lb-in

    = 1.5 lb/in * (8 in) ^2) / 2

    = 48 lb-in.
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