The time taken by an object to slide down a 4 | vedclass

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(D) Let $L$ be the length of the inclined plane and $	heta = 45^{\circ}$.For a smooth plane,the acceleration is $a_1 = g \sin 	heta = g \sin 45^{\ci

Vedclass · Accepted Answer

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

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(D) Let $L$ be the length of the inclined plane and $	heta = 45^{\circ}$.For a smooth plane,the acceleration is $a_1 = g \sin 	heta = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}$.The time taken is $t_1 = \sqrt{\frac{2L}{a_1}}$.For a rough plane with coefficient of kinetic friction $K$,the acceleration is $a_2 = g \sin 	heta - Kg \cos 	heta = \frac{g}{\sqrt{2}} - \frac{Kg}{\sqrt{2}} = \frac{g}{\sqrt{2}}(1-K)$.The time taken is $t_2 = \sqrt{\frac{2L}{a_2}}$.Given $t_2 = n t_1$,so $t_2^2 = n^2 t_1^2$.Substituting the expressions for $t_1^2$ and $t_2^2$: $\frac{2L}{a_2} = n^2 \frac{2L}{a_1} \implies a_1 = n^2 a_2$.Substituting the values of $a_1$ and $a_2$: $\frac{g}{\sqrt{2}} = n^2 \left[ \frac{g}{\sqrt{2}}(1-K) ight]$.$1 = n^2(1-K) \implies 1-K = \frac{1}{n^2}$.Therefore,$K = 1 - \frac{1}{n^2}$.

The time taken by an object to slide down a $45^{\circ}$ rough inclined plane is $n$ times the time it takes to slide down a perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the inclined plane is:

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