A given object takes n times more time to sli | vedclass

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(B) Let $L$ be the length of the incline and $	heta = 45^\circ$. The time taken to slide down a smooth incline is $t_1 = \sqrt{\frac{2L}{g \sin 	het

Vedclass · Accepted Answer

$1 - \frac{1}{{{n^2}}}$

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(B) Let $L$ be the length of the incline and $	heta = 45^\circ$. The time taken to slide down a smooth incline is $t_1 = \sqrt{\frac{2L}{g \sin 	heta}}$.For a rough incline,the acceleration is $a = g(\sin 	heta - \mu \cos 	heta)$. The time taken is $t_2 = \sqrt{\frac{2L}{g(\sin 	heta - \mu \cos 	heta)}} = n t_1$.Squaring both sides: $\frac{2L}{g(\sin 	heta - \mu \cos 	heta)} = n^2 \frac{2L}{g \sin 	heta}$.This simplifies to $\sin 	heta = n^2(\sin 	heta - \mu \cos 	heta)$.Rearranging for $\mu$: $\mu \cos 	heta = \sin 	heta (1 - \frac{1}{n^2})$.$\mu = 	an 	heta (1 - \frac{1}{n^2})$.Since $	heta = 45^\circ$,$	an 45^\circ = 1$,so $\mu = 1 - \frac{1}{n^2}$.

$A$ given object takes $n$ times more time to slide down a $45^o$ rough inclined plane as it takes to slide down a perfectly smooth $45^o$ incline. The coefficient of kinetic friction between the object and the incline is

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