A rough inclined plane BCE of height left(fra | vedclass

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(D) The block slides down the inclined plane $EC$. The acceleration $a$ of the block is given by $a = g(\sin 	heta - \mu_k \cos 	heta)$.Given $\sin

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$\frac{20}{3} 	ext{ m}$

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(D) The block slides down the inclined plane $EC$. The acceleration $a$ of the block is given by $a = g(\sin 	heta - \mu_k \cos 	heta)$.Given $\sin 	heta = 0.6$,so $\cos 	heta = \sqrt{1 - (0.6)^2} = 0.8$.Substituting the values: $a = 10(0.6 - \frac{1}{8} 	imes 0.8) = 10(0.6 - 0.1) = 5 	ext{ ms}^{-2}$.The length of the incline $EC = \frac{EB}{\sin 	heta} = \frac{25/6}{0.6} = \frac{25}{3.6} = \frac{125}{18} 	ext{ m}$.The velocity $v$ at point $C$ is $v = \sqrt{2as} = \sqrt{2 	imes 5 	imes \frac{125}{18}} = \sqrt{\frac{1250}{18}} = \sqrt{\frac{625}{9}} = \frac{25}{3} 	ext{ ms}^{-1}$.At point $C$,the block becomes a projectile with horizontal velocity $v_x = v \cos 	heta = \frac{25}{3} 	imes 0.8 = \frac{20}{3} 	ext{ ms}^{-1}$ and vertical velocity $v_y = -v \sin 	heta = -\frac{25}{3} 	imes 0.6 = -5 	ext{ ms}^{-1}$.Using the equation of motion for vertical displacement $h = v_y t + \frac{1}{2}gt^2$ (taking downward as positive): $10 = 5t + 5t^2 \Rightarrow t^2 + t - 2 = 0 \Rightarrow (t+2)(t-1) = 0$. Thus,$t = 1 	ext{ s}$.The horizontal distance $DF = v_x t = \frac{20}{3} 	imes 1 = \frac{20}{3} 	ext{ m}$.

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