An object takes n times as much time to slide | vedclass

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Question

(A) For a smooth inclined plane,the acceleration is $a_s = g \sin 	heta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin 	het

Vedclass · Accepted Answer

$\left(1-\frac{1}{n^2}\right)$

Vedclass · Answer

(A) For a smooth inclined plane,the acceleration is $a_s = g \sin 	heta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin 	heta}}$.For a rough inclined plane,the acceleration is $a_r = g(\sin 	heta - \mu \cos 	heta)$. The time taken is $t_r = \sqrt{\frac{2s}{g(\sin 	heta - \mu \cos 	heta)}}$.Given $t_r = n t_s$,we have $\frac{t_r}{t_s} = n$.Squaring both sides: $\frac{t_r^2}{t_s^2} = n^2 \implies \frac{\sin 	heta}{\sin 	heta - \mu \cos 	heta} = n^2$.Rearranging: $\sin 	heta = n^2 \sin 	heta - n^2 \mu \cos 	heta$.$n^2 \mu \cos 	heta = (n^2 - 1) \sin 	heta$.$\mu = \frac{n^2 - 1}{n^2} 	an 	heta$.Since $	heta = 45^{\circ}$,$	an 45^{\circ} = 1$.Therefore,$\mu = 1 - \frac{1}{n^2}$.

An object takes $n$ times as much time to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down a perfectly smooth inclined plane of the same inclination. The coefficient of kinetic friction between the object and the rough incline is given by:

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