Answer
$2.37Kg$
Work Step by Step
The required mass of each of the two cylinders can be determined as follows:
$T=K\Delta x$
We plug in the known values to obtain:
$T=100(2\sqrt{2}-2.5)=32.8N$
Now $\Sigma F_y=0$
$\implies Tsin\theta-mg=0$
$\implies 32.8N sin(45)-m(9.81)=0$
This simplifies to:
$m=2.37Kg$
