Answer
(a) W = 0.0625 J
(b) v = 0.177 m/s
Work Step by Step
(a) The work done on the block will be equal to the elastic potential energy stored in the spring.
$W = \frac{1}{2}kx^2$
$W = \frac{1}{2}(200~N/m)(0.025~m)^2$
$W = 0.0625~J$
(b) The work done by the spring will be equal to the block's kinetic energy as it leaves the spring.
$\frac{1}{2}mv^2 = 0.0625~J$
$v^2 = \frac{(2)(0.0625~J)}{m}$
$v = \sqrt{\frac{(2)(0.0625~J)}{4.00~kg}}$
$v = 0.177~m/s$