Let T be the tension acting throughout the string. And also a and b be the accelerations of block A and B respectively.
So considering the forces on A we can write
F-2T= 2ma.............(1)
Again considering the forces on B we get
3T = 4mb.............(2)
Due to application of force for small interval of time t ,
the displacement of A will be =1/2at^2
and the displacement of B will be =1/2bt^2.
During this time work done by 2T and 3T will be same
Hence 2T×1/2at^2=3T×1/2bt^2
=> 2a= 3b..........(3)
From (1) and (2) we get
F - (8/3)mb= 2ma......(4)
Combining (3) and (4) we get
F -(8/3)mb = 3mb
=>(17/3)mb =F
÷> b = (3F)/(17m)
Inserting F =17N and m =1kg we get
Acceleration of B , b= 3m/s^2
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