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(A) For a smooth inclined plane,the acceleration is $a_2 = g \sin 	heta$. The time taken is $t_2 = \sqrt{\frac{2s}{g \sin 	heta}}$.For a rough incli

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$\frac{7}{16}$

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(A) For a smooth inclined plane,the acceleration is $a_2 = g \sin 	heta$. The time taken is $t_2 = \sqrt{\frac{2s}{g \sin 	heta}}$.For a rough inclined plane,the acceleration is $a_1 = g(\sin 	heta - \mu \cos 	heta)$. The time taken is $t_1 = \sqrt{\frac{2s}{g(\sin 	heta - \mu \cos 	heta)}}$.Given $t_1 = \frac{4}{3} t_2$,so $\frac{t_1}{t_2} = \frac{4}{3}$.Squaring both sides,$\frac{t_1^2}{t_2^2} = \frac{16}{9} = \frac{a_2}{a_1} = \frac{g \sin 	heta}{g(\sin 	heta - \mu \cos 	heta)} = \frac{\sin 	heta}{\sin 	heta - \mu \cos 	heta}$.Since $	heta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.$\frac{16}{9} = \frac{1/\sqrt{2}}{1/\sqrt{2} - \mu(1/\sqrt{2})} = \frac{1}{1 - \mu}$.$16(1 - \mu) = 9 \implies 16 - 16\mu = 9 \implies 16\mu = 7 \implies \mu = \frac{7}{16}$.

$A$ body takes $1\frac{1}{3}$ times as much time to slide down a rough inclined plane as it takes to slide down an identical but smooth inclined plane. If the angle of the inclined plane is $45^{\circ}$,the coefficient of friction is:

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