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

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Question

(A) Let $s$ be the distance, $	heta = 45^{\circ}$ be the angle of inclination, and $t$ be the time taken in the absence of friction $(\mu = 0)$.In th

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(A) Let $s$ be the distance, $	heta = 45^{\circ}$ be the angle of inclination, and $t$ be the time taken in the absence of friction $(\mu = 0)$.In the absence of friction, the acceleration is $a_1 = g \sin 	heta$.Using the equation of motion $s = \frac{1}{2} a_1 t^2$, we have $s = \frac{1}{2} (g \sin 	heta) t^2$.In the presence of friction, the acceleration is $a_2 = g \sin 	heta - \mu g \cos 	heta$.The time taken is $2t$. Thus, $s = \frac{1}{2} a_2 (2t)^2 = \frac{1}{2} (g \sin 	heta - \mu g \cos 	heta) (4t^2) = 2 (g \sin 	heta - \mu g \cos 	heta) t^2$.Equating the two expressions for $s$:$\frac{1}{2} (g \sin 	heta) t^2 = 2 (g \sin 	heta - \mu g \cos 	heta) t^2$$\frac{1}{2} \sin 	heta = 2 (\sin 	heta - \mu \cos 	heta)$$\frac{1}{4} \sin 	heta = \sin 	heta - \mu \cos 	heta$$\mu \cos 	heta = \sin 	heta - \frac{1}{4} \sin 	heta = \frac{3}{4} \sin 	heta$$\mu = \frac{3}{4} 	an 	heta$Since $	heta = 45^{\circ}$, $	an 45^{\circ} = 1$.Therefore, $\mu = \frac{3}{4} 	imes 1 = 0.75$.

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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