Answer
a). $6kg$
b). $1.225m/s^{2}$
Work Step by Step
a). Since the bodies are at rest, tension is equal to weight of the second body and maximum friction force acting on first body.
Therefore, $coefficient\, of \,static \,friction\times m_{1}g=m_{2}g$
$m_{2}=0.6\times10kg=6kg$
b). For first body,
$T=m_{1}a+coefficient \,of\, kinetic\, friction\times m_{1}g$
For second body,
$T=m_{2}(g-a)$
From above 2 equations,
$a=\frac{g(m_{2}-coefficient\, of \,kinetic \,friction \times m_{1})}{m_{1}+m_{2}}=9.8\times\frac{6-0.4\times10}{16}=1.225m/s^{2}$
