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

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(C) Let the length of the inclined plane be $\ell$. In the presence of friction,the acceleration $a_1$ is given by:$a_1 = g(\sin 	heta - \mu \cos 	h

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(C) Let the length of the inclined plane be $\ell$. In the presence of friction,the acceleration $a_1$ is given by:$a_1 = g(\sin 	heta - \mu \cos 	heta)$Using the equation of motion $s = ut + \frac{1}{2}at^2$ with $u=0$:$\ell = \frac{1}{2} g(\sin 	heta - \mu \cos 	heta) t_1^2$  --- $(1)$In the absence of friction $(\mu = 0)$,the acceleration $a_2$ is:$a_2 = g \sin 	heta$$\ell = \frac{1}{2} g \sin 	heta \cdot t_2^2$  --- $(2)$Given that $t_1 = 2t_2$,we equate the two expressions for $\ell$:$\frac{1}{2} g(\sin 	heta - \mu \cos 	heta) (2t_2)^2 = \frac{1}{2} g \sin 	heta \cdot t_2^2$$4(\sin 	heta - \mu \cos 	heta) = \sin 	heta$$4 \sin 	heta - 4 \mu \cos 	heta = \sin 	heta$$3 \sin 	heta = 4 \mu \cos 	heta$$\mu = \frac{3}{4} 	an 	heta$Given $	heta = 45^{\circ}$,$	an 45^{\circ} = 1$:$\mu = \frac{3}{4} 	imes 1 = 0.75$

Starting from rest,a body slides down a $45^{\circ}$ 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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