A body takes 1frac{1}{3} times as much time t | vedclass

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(A) Let $t$ be the time taken to slide down a smooth inclined plane and $t'$ be the time taken to slide down a rough inclined plane.Given that $t' = n

Vedclass · Accepted Answer

$\frac{7}{16}$

Vedclass · Answer

(A) Let $t$ be the time taken to slide down a smooth inclined plane and $t'$ be the time taken to slide down a rough inclined plane.Given that $t' = n t$,where $n = 1\frac{1}{3} = \frac{4}{3}$.For a smooth inclined plane,the acceleration is $a = g \sin 	heta$. The time taken is $t = \sqrt{\frac{2L}{g \sin 	heta}}$.For a rough inclined plane,the acceleration is $a' = g(\sin 	heta - \mu \cos 	heta)$. The time taken is $t' = \sqrt{\frac{2L}{g(\sin 	heta - \mu \cos 	heta)}}$.Since $t' = n t$,we have $\frac{t'}{t} = n = \sqrt{\frac{\sin 	heta}{\sin 	heta - \mu \cos 	heta}}$.Squaring both sides,$n^2 = \frac{\sin 	heta}{\sin 	heta - \mu \cos 	heta}$.Rearranging for $\mu$,we get $\mu = 	an 	heta \left(1 - \frac{1}{n^2}ight)$.Substituting $	heta = 45^\circ$ and $n = \frac{4}{3}$:$\mu = 	an 45^\circ \left(1 - \frac{1}{(4/3)^2}ight) = 1 	imes \left(1 - \frac{9}{16}ight) = \frac{7}{16}$.

$A$ 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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