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(A) When the box is dropped from height $h$,its speed when it reaches the ground is given by the conservation of energy: $v = \sqrt{2gh}$.When the blo

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$\frac{8}{9}$

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(A) When the box is dropped from height $h$,its speed when it reaches the ground is given by the conservation of energy: $v = \sqrt{2gh}$.When the block slides down the rough inclined plane with angle $	heta = 45^{\circ}$,the net acceleration $a$ is given by:$a = g(\sin 	heta - \mu \cos 	heta)$.Since $	heta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.Thus,$a = g \left( \frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}} ight) = \frac{g}{\sqrt{2}}(1 - \mu)$.The length of the inclined plane $s$ is related to height $h$ by $s = \frac{h}{\sin 	heta} = h \sqrt{2}$.The final velocity $v'$ at the bottom of the incline is $v' = \sqrt{2as}$.Substituting $a$ and $s$:$v' = \sqrt{2 \cdot \frac{g}{\sqrt{2}}(1 - \mu) \cdot h \sqrt{2}} = \sqrt{2gh(1 - \mu)}$.Given that $v' = \frac{v}{3}$,we have:$\sqrt{2gh(1 - \mu)} = \frac{1}{3} \sqrt{2gh}$.Squaring both sides:$2gh(1 - \mu) = \frac{1}{9} (2gh)$.$1 - \mu = \frac{1}{9}$.$\mu = 1 - \frac{1}{9} = \frac{8}{9}$.

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