Answer
The hanging block falls a maximum distance of $~~0.39~m$
Work Step by Step
The blocks will fall until the change in gravitational potential energy of the hanging block is equal in magnitude to the elastic potential energy stored in the spring. Note that the kinetic energy must be zero at this point.
We can find the distance $x$:
$\frac{1}{2}kx^2 = (2M)gx$
$x = \frac{4Mg}{k}$
$x = \frac{(4)(2.0~kg)(9.8~m/s^2)}{200~N/m}$
$x = 0.39~m$
The hanging block falls a maximum distance of $~~0.39~m$
