Answer
Please see the work below.
Work Step by Step
We know that
(a) $K=\frac{F}{x}$
$K=\frac{mg}{x}$
$K=\frac{(125)(9.8)}{0.00266}=4.6\times 10^5\frac{N}{m}$
(b) We know that
$W=\frac{1}{2}kx^2$
This can be rearranged as:
$x=\sqrt{\frac{2W}{x}}$
We plug in the known values to obtain:
$x=\sqrt{\frac{(2)(50.0)}{4.6\times 10^5}}$
$x=0.015m$
