Rigid question

Let the mass of the large block be $M$ , the mass of the small block be $m$ , the length of the large block be $l$ , and the position of the small block be $x$ .

First, recognize that the force on the large block is balanced no matter where the small block is placed, What changes is the partition of force between the hinge at C, and the small block at B.

Because the large block stays horiztonal, we have a balance of torque about the hinge at C. The clockwise torque from the large block is $\tau_A = \frac{l}{2}Mg$ , and the torque from the small block is $\tau_B = x F_N g$ . Obviously, $\tau_A = \tau_B$ , so $F_N = \frac{Ml}{2x}$ . This normal force contributes to the friction between the large and small block, and also to the friction between the small block and the floor.

Additionally, the weight of the small block weighs on the floor, adding to the friction between them (but not to the friction between the large and small block.

Thus, the total force of friction as a function of $x$ is given by $F_d = \mu\left(2F_n(x) + mg\right)$ .

The work dissipated by friction is given by

$W_d = \displaystyle\int\limits_{\frac12 l}^{\frac34 l} F_d(x) dx$

or

$W_d = gl\mu\left(\frac{m}{4}+M\log \frac32\right)$

The kinetic energy of the small block is then equal to the reversible work, defined as $W_r=W - W_d = Fl/4 - W_d$ , and the velocity of the small block is $v_B = \sqrt{2W_r/m}$ which, given the provided values, gives $v_B \approx 18.03$ m/s.