The problem is as follows:

The figure from below shows two blocks joined by a rope on each end goind through a pulley. The masses of each block are as follows $m_1=5\,kg$, $m_2=4\,kg$. The coefficient of friction is $\mu=0.2$. The pulley can be considered as a thin disk and its mass is $2\,kg$. The acceleration due gravity is $10\,\frac{m}{s^2}$. Given this information find the acceleration in meter per second square of the system.

The alternatives are as follows:

$\begin{array}{ll} 1.&3\,\frac{m}{s^2}\\ 2.&4\,\frac{m}{s^2}\\ 3.&5\,\frac{m}{s^2}\\ 4.&6\,\frac{m}{s^2}\\ 5.&2\,\frac{m}{s^2}\\ \end{array}$

What I attempted to do to solve this problem was to relate the torque rotational inertia principle as follows:

$\tau=I\alpha$

This is illustrated in the diagram from below.

Since I’m given the masses of the objects and the acceleration due gravity and the coefficient of friction, then it is possible to obtain the acceleration as follows:

In this situation the tension acting in the pulley will generate a torque on it, hence.

For the block in the top:

$T-\mu N=ma$

$T=5a+0.2\times 50$

$T=5a+10$

For the block hanging from the other end in the pulley:

$T-mg=m(-a)$

$T=40-4a$

The rotational inertia for a disk with an axis going through the center is:

$I=\frac{1}{2}mr^2$

Therefore: (considering that a counterclockwise turn is positive and counterclockwise is negative)

$(R)(5a+10)-(R)(40-4a)=\frac{1}{2}mR^2 \times \alpha$

Simplifying:

$5a+10-40+4a=\frac{1}{2}(2)R \times \alpha$

Since the angular acceleration times the radius is the tangential acceleration this can be expressedas follows:

$9a-30=a$

$8a=30$

Therefore the acceleration should be:

$a=\frac{30}{8}$

Howevet this doesn’t check with any of the alternatives, supposedly the answer to this question is $3\frac{m}{s^2}$ What could I be doing wrong? Can somebody help me with this matter?.