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

m

_{1}g - T = m_{1}a
T - m

_{2}g = m_{2}a
and arrived at

a = g(m

_{1}- m_{2})/(m_{1}+ m_{2})
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 m

_{1}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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