There are two inclined surfaces of equal leng | vedclass

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(D) For a body sliding down an inclined plane of length $L$ starting from rest,the time taken is given by $L = \frac{1}{2} a t^2$,where $a$ is the acc

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(D) For a body sliding down an inclined plane of length $L$ starting from rest,the time taken is given by $L = \frac{1}{2} a t^2$,where $a$ is the acceleration.For the smooth surface,acceleration $a_S = g \sin 	heta$. Thus,$L = \frac{1}{2} (g \sin 	heta) t_S^2$.For the rough surface,acceleration $a_R = g \sin 	heta - \mu_k g \cos 	heta$. Thus,$L = \frac{1}{2} (g \sin 	heta - \mu_k g \cos 	heta) t_R^2$.Since $L$ is the same for both,$\frac{1}{2} a_S t_S^2 = \frac{1}{2} a_R t_R^2$,which implies $\frac{a_R}{a_S} = \left(\frac{t_S}{t_R}ight)^2$.Given $t_R = 2 t_S$,we have $\frac{a_R}{a_S} = \left(\frac{t_S}{2 t_S}ight)^2 = \left(\frac{1}{2}ight)^2 = \frac{1}{4}$.Substituting the expressions for acceleration: $\frac{g \sin 	heta - \mu_k g \cos 	heta}{g \sin 	heta} = \frac{1}{4}$.Since $	heta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.Therefore,$\frac{\frac{1}{\sqrt{2}} - \mu_k \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \frac{1}{4}$.$1 - \mu_k = \frac{1}{4} \Rightarrow \mu_k = 1 - 0.25 = 0.75$.

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