Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as m_{1}  and m_{2} respectively

Let m_{1} =1 kg and m_{2} =2 kg

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             m_{1} g-T=m_{1} a        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction [mu]=0.13

Hence the frictional force F=mu N

                                                     F=mu m_{2} g

Hence the net force acting on the body having mass 2 kg-

                                  T-mu m_{2} g=m_{2} a  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           m_{1} g-T=m_{1}a             [1]

                           T-mu m_{2}g= m_{2} a   [2]

                           —————————————————

                           [m_{1} -mu m_{2} ]g=[m_{1} +m_{2} ]a

                           a=frac{m_{1}-mu m_{2}} {m_{1}+ m_{2}}*g

                           a=frac{1-[2*0.13]}{1+2} *9.8 m/s^2

                           a=frac{0.74}{3} *9.8 m/s^2

                           a=2.417 m/s         [ans]