Answer
The kinetic energy of the hanging block is $~~1.8~J$
Work Step by Step
The decrease in gravitational potential energy is equal in magnitude to the sum of the elastic potential energy in the spring and the kinetic energy of the two blocks.
We can find the combined kinetic energy of the two blocks:
$K+\frac{1}{2}kx^2 = 2Mgh$
$K = 2Mgh-\frac{1}{2}kx^2$
$K = (2)(2.0~kg)(9.8~m/s^2)(0.090~m)-\frac{1}{2}(200~N/m)(0.090~m)^2$
$K = 2.718~J$
Let $K_1$ be the kinetic energy of the block of mass $M$
Let $K_2$ be the kinetic energy of the block of mass $2M$
Note that since both blocks have the same speed, then $K_2 = 2~K_1$
We can find the kinetic energy $K_2$ of the hanging block:
$K_1+K_2 = K$
$\frac{K_2}{2}+K_2 = K$
$\frac{3K_2}{2} = K$
$K_2 = \frac{2K}{3}$
$K_2 = \frac{(2)(2.718~J)}{3}$
$K_2 = 1.8~J$
The kinetic energy of the hanging block is $~~1.8~J$
