Answer:
The acceleration of
is 
Explanation:
From the question we are told that
The mass of first block is 
The angle of inclination of first block is 
The coefficient of kinetic friction of the first block is 
The mass of the second block is 
The angle of inclination of the second block is 
The coefficient of kinetic friction of the second block is 
The acceleration of
are same
The force acting on the mass
is mathematically represented as

=> 
Where T is the tension on the rope
The force acting on the mass
is mathematically represented as


At equilibrium

So

making a the subject of the formula

substituting values 
=> 