Answer
(a) $1.5mm$
(b) $0.46J$
Work Step by Step
As we know that
$\Sigma F_y=2F_{sole}-mg=0$
$\implies 2K_{sole}x=mg$
This simplifies to:
$x=\frac{mg}{2K_{sole}}$
We plug in the known values to obtain:
$x=\frac{62(9.81)}{2(2.0\times 10^5)}=0.0015m=1.5mm$
(b) We can find the required energy as follows:
$U=\frac{1}{2}Kx^2$
We plug in the known values to obtain:
$U=\frac{1}{2}(2.0\times 10^5)(0.0015)^2=0.23\times 2shoes=0.46J$
