Answer
a). $0.72kg \leq m_{2}\leq 1.68kg$
b). $0.88kg \leq m_{2}\leq 1.52kg$
Work Step by Step
$F_{f}=coefficient\, of\, static\, friction\times N=coefficient\, of\, static\, friction\times m_{1}gcos\theta$
a). Now, $T=m_{2}g$
$T=m_{1}gsin\theta+/-F_{f}$
from above 2 equations,
$m_{2}=m_{1}(sin\theta+/-coefficient\, of\, static\, friction\times cos\theta)$
Calculating the extreme values,
$0.72kg \leq m_{2}\leq 1.68kg$
b). Here we need to use the coefficient of kinetic friction.
In the similar way, the result is
$0.88kg \leq m_{2}\leq 1.52kg$
