Answer
$A_{new}=A+\frac{mg}{K}$
Work Step by Step
We know that
$A_{new}=x_{break}-x_{new}$
We plug in the known values to obtain:
$A_{new}=\frac{2mg}{K}+A-\frac{mg}{K}$
$\implies A_{new}=A+\frac{mg}{K}$
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