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let mass of the block = m

then mass of rope     = m/2

 

total mass of system = (m + m/2)  = 3m/2

Total (net) force on the system = 6 N

therefore acceleration of the system =

since rope is uniform , it impliies mass of half of the rope =

now consider the system A :

bock + half part of the attached

 

In this system the only external force acting is due to the tension of the rope.

and the acceleration we found to be equal to 4/m

 

the net mass of this new system  M_A = m + m/4

                                                      = 5m/4

 

let tension force be T ( at middle of rope)

then 

    = 5m/4  *  4/m

    = 5 N