Starting from rest,a body slides down a 45^o | vedclass

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(A) Let the distance be $s$ and the angle of inclination be $	heta = 45^o$.In the absence of friction,the acceleration is $a_1 = g \sin 	heta$. The

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(A) Let the distance be $s$ and the angle of inclination be $	heta = 45^o$.In the absence of friction,the acceleration is $a_1 = g \sin 	heta$. The time taken is $t_1 = t$.Using $s = \frac{1}{2} a_1 t_1^2$,we have $s = \frac{1}{2} (g \sin 	heta) t^2$.In the presence of friction,the acceleration is $a_2 = g(\sin 	heta - \mu \cos 	heta)$. The time taken is $t_2 = 2t$.Using $s = \frac{1}{2} a_2 t_2^2$,we have $s = \frac{1}{2} g(\sin 	heta - \mu \cos 	heta) (2t)^2 = 2 g(\sin 	heta - \mu \cos 	heta) t^2$.Equating the two expressions for $s$:$\frac{1}{2} g \sin 	heta t^2 = 2 g(\sin 	heta - \mu \cos 	heta) t^2$$\frac{1}{2} \sin 	heta = 2(\sin 	heta - \mu \cos 	heta)$$\sin 	heta = 4 \sin 	heta - 4 \mu \cos 	heta$$4 \mu \cos 	heta = 3 \sin 	heta$$\mu = \frac{3}{4} 	an 	heta$Since $	heta = 45^o$,$	an 45^o = 1$.Therefore,$\mu = \frac{3}{4} 	imes 1 = 0.75$.

Starting from rest,a body slides down a $45^o$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

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