Answer
$K = 4.5~J$
Work Step by Step
We can find the initial energy in the system:
$U_{e0} = \frac{1}{2}kd^2$
$U_{e0} = \frac{1}{2}(620~N/m)(0.25~m)^2$
$U_{e0} = 19.375~J$
We can use conservation of energy to find $K$ when $y = 0.050~m$:
$K+U_g+U_e = 19.375~J$
$K+mgy+\frac{1}{2}kd^2 = 19.375~J$
$K = (19.375~J) - mgy - \frac{1}{2}kd^2$
$K = (19.375~J) - (50~N)(0.050~m) - \frac{1}{2}(620~N/m)(0.20~m)^2$
$K = 4.5~J$
