Answer
The maximum speed of the block will be $1.57~m/s$
Work Step by Step
We can find the spring constant $k$:
$kx = mg$
$k = \frac{mg}{x}$
$k = \frac{(0.20~kg)(9.80~m/s^2)}{0.010~m}$
$k = 196~N/m$
The maximum kinetic energy of the block will be equal to the elastic potential energy stored in the spring when it is stretched a distance of 5.0 cm:
$\frac{1}{2}mv_{max}^2 = \frac{1}{2}kx^2$
$v_{max}^2 = \frac{kx^2}{m}$
$v_{max} = \sqrt{\frac{k}{m}}~x$
$v_{max} = (\sqrt{\frac{196~N/m}{0.20~kg}})~(0.050~m)$
$v_{max} = 1.57~m/s$
The maximum speed of the block will be $1.57~m/s$.
