Like with our atwood's machine example we solve this by looking at each mass individually. The forces acting on M are friction due to the table, opposing the tension T of the rope. The forces acting on m is the weight of m and the tension T of the rope. Now for atwood's situation we started with
m1 g - T = m1 a
T - m2 g = m2 a
and arrived at
a = g(m1 - m2 )/(m1 + m2 )
For this situation, we can see that m1 can be substituted with M as we define positive acceleration as M going left and m going up, and negative acceleration as M to the right and m going down matching the convention of gravity's downward acceleration. The only difference is that instead of m1 g being the force opposing T we have μMg which is the friction force due to the table. Making this substitution we arrive at
a = g(μM - m )/(M + m )
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