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5 October, 16:13

You compress a spring by a certain distance. Then you decide to compress it further so that its elastic potential energy increases by another 50%

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  1. 5 October, 16:16
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    You can determine the percent increase in the spring's compression distance.

    The elastic potential energy of a spring is given by:

    PE = [1/2] k * X^2, where X is the deformation of the spring (compression is this case).

    Then if the initial deformation is A:

    Its PE is [1/2] k * A^2

    And the increased potential energy is PE' = 1.5 * PE = 1.5*[1/2] k*A^2.

    If you call B the new deformation (compression distance), the new potential energy is PE' = [1/2] k * B^2

    From which you can equal: 1.5*[1/2] k * A^2 = [1/2] k * B^2

    You can cancell the terms that appear in both sides = >

    1.5 A^2 = B^2 = > (B/A) ^2 = 1.5 = > B/A = √1.5 = 1.225

    Which means that the compression distance increased 22.5%
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