The scheme presents two blocks:
The first, Block A, is on a table. Block A is attached to an end of a string. At the end of the table, there is a pulley; the string passes by this pulley, then becomes vertical. At the other end of this string (end of this vertical string) there is Block B. So Block A is attached to Block B by this string. One on the table, the other (B) attached to the string (it is a very classical scheme).
Pulley has radiusR and a moment of inertia I.
The rope doesn't slip over the pulley, and the pulley spins on a frictionless axle.
The coefficient of kinetic friction between the block A and the table is (Mu).
The system is released from rest, and block B descends.
Block A has mass M_A, B has mass M_B.
Note 1. Distance d is the distance MB descends and also the distance MA slides.
Note 2. At the end of the descent, both blocks are moving with final velocity V and ...
This solution provides notes and steps to explain the problem. It uses conservation of energy calculations to find the final velocity equation.