The coefficient of static friction,mu_s betwe | vedclass

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(D) Let the mass of block $B$ be $M$ and the mass of block $A$ be $m = 2\,kg$.For block $B$ to be in equilibrium,the tension $T$ in the string must ba

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$0.4$

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(D) Let the mass of block $B$ be $M$ and the mass of block $A$ be $m = 2\,kg$.For block $B$ to be in equilibrium,the tension $T$ in the string must balance its weight:$T = M g \ldots(i)$For block $A$ to remain stationary,the tension $T$ must be balanced by the limiting static frictional force $f_s$:$T = f_s$Since $f_s = \mu_s N$ and the normal force $N = m g$ (as the table is horizontal),we have:$T = \mu_s m g \ldots(ii)$Equating $(i)$ and $(ii)$:$M g = \mu_s m g$$M = \mu_s m$Given $\mu_s = 0.2$ and $m = 2\,kg$:$M = 0.2 	imes 2 = 0.4\,kg$Thus,the maximum mass of block $B$ is $0.4\,kg$.

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