A block slides down an inclined plane with an acceleration $g/2$ as shown in fig. Then coefficient of kinetic friction is
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$\mathrm{mg} \sin \theta-\mathrm{f}=\mathrm{ma}$
$\operatorname{mgsin} \theta-\mu \operatorname{mg} \cos \theta=\mathrm{m} \mathrm{a}$
$\theta=60^{\circ} \quad a=\mathrm{g} / 2$
$\mathrm{g}\left(\sin 60^{\circ}-\mu \cos 60^{\circ}\right)=\mathrm{g} / 2$
$\frac{\sqrt{3}}{2}-\mu\left(\frac{1}{2}\right)=\frac{1}{2}$
$\sqrt{3}-1=\mu $
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