A piece of ice slides down a rough inclined p | vedclass

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(C) Given,$	heta=45^{\circ}, s_{1}=s_{2}, u=0$.On the rough incline,the acceleration is $a_{1}=g(\sin 	heta-\mu \cos 	heta)$.On the frictionless in

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$\frac{3}{4 \cot 	heta}$

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(C) Given,$	heta=45^{\circ}, s_{1}=s_{2}, u=0$.On the rough incline,the acceleration is $a_{1}=g(\sin 	heta-\mu \cos 	heta)$.On the frictionless incline,the acceleration is $a_{2}=g \sin 	heta$.Let $t_{1}$ be the time taken on the rough plane and $t_{2}$ be the time taken on the frictionless plane. Given $t_{1}=2 t_{2}$.Using the equation of motion $s=ut+\frac{1}{2}at^{2}$,for $u=0$:$s_{1}=\frac{1}{2}g(\sin 	heta-\mu \cos 	heta)t_{1}^{2}$$s_{2}=\frac{1}{2}g \sin 	heta t_{2}^{2}$Since $s_{1}=s_{2}$,we have:$\frac{1}{2}g(\sin 	heta-\mu \cos 	heta)t_{1}^{2}=\frac{1}{2}g \sin 	heta t_{2}^{2}$$\frac{\sin 	heta-\mu \cos 	heta}{\sin 	heta}=\frac{t_{2}^{2}}{t_{1}^{2}}$Substituting $t_{1}=2t_{2}$:$1-\mu \cot 	heta=\frac{t_{2}^{2}}{(2t_{2})^{2}}=\frac{1}{4}$$\mu \cot 	heta=1-\frac{1}{4}=\frac{3}{4}$$\mu=\frac{3}{4 \cot 	heta}$. Since $	heta=45^{\circ}$,$\cot 45^{\circ}=1$,so $\mu=\frac{3}{4}=0.75$.

$A$ piece of ice slides down a rough inclined plane at $\theta=45^{\circ}$ inclination in twice the time that it takes to slide down an identical but frictionless inclined plane. What is the coefficient of friction between ice and incline?

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