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(A) For block $A$ on a smooth incline:$a_A = g \sin 	heta$Given $	heta = 45^{\circ}$,so $a_A = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}$.For block $B$

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$0.5$

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(A) For block $A$ on a smooth incline:$a_A = g \sin 	heta$Given $	heta = 45^{\circ}$,so $a_A = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}$.For block $B$ on a rough incline:The forces acting along the incline are $mg \sin 	heta$ (downward) and $f_k = \mu_k N = \mu_k mg \cos 	heta$ (upward).$m_B a_B = m_B g \sin 	heta - \mu_k m_B g \cos 	heta$$a_B = g(\sin 	heta - \mu_k \cos 	heta)$$a_B = g(\sin 45^{\circ} - \mu_k \cos 45^{\circ}) = \frac{g}{\sqrt{2}}(1 - \mu_k)$.Given the ratio $\frac{a_A}{a_B} = \frac{2}{1}$:$\frac{g/\sqrt{2}}{g/\sqrt{2}(1 - \mu_k)} = 2$$\frac{1}{1 - \mu_k} = 2$$1 = 2 - 2\mu_k$$2\mu_k = 1$$\mu_k = 0.5$.

The ratio of acceleration of block $A$ placed on a smooth incline to block $B$ placed on a rough incline is $2:1$. The coefficient of kinetic friction between block $B$ and the incline is .........

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