A cube of mass $m$ slides down an inclined right-angle trough. If the coefficient of kinetic friction between the cube and the trough is $\mu _k$ , then the acceleration of the block is
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$2 \mathrm{N} \cos 45^{\circ}=\mathrm{mg} \cos \theta$
$\mathrm{N}=\frac{\mathrm{mg} \cos \theta}{2 \cos 45^{\circ}}$
$\mathrm{mg} \sin \theta-2 \mu_{\mathrm{k}} \mathrm{N}=\mathrm{ma}$
$a=g \sin \theta-\frac{2 \mu_{k} m g \cos \theta}{2\left(\cos 45^{\circ}\right) m}$
$a=g\left(\sin \theta-\sqrt{2} \mu_{k} \cos \theta\right)$
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