Answer
The maximum kinetic energy is $~~2.25~J$
Work Step by Step
In part (b) we found that the block reaches a point $0.10~m$ above the unstretched point, and the block reaches a point $0.20~m$ below the unstretched point.
The maximum kinetic energy occurs at the point that is in the center of the motion. Thus the maximum kinetic energy occurs when the block is $5.0~cm$ below the unstretched point.
We can find the maximum kinetic energy:
$K+U_e+U_g = 2.00~J$
$K+\frac{1}{2}ky^2+mgy = 2.00~J$
$K = 2.00~J-\frac{1}{2}ky^2-mgy$
$K = (2.00~J)-\frac{1}{2}(200.0~N/m)(-0.050~m)^2-(10.0~N)(-0.050~m)$
$K = 2.25~J$
The maximum kinetic energy is $~~2.25~J$
