Two smooth inclined planes A and B each of he | vedclass

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(C) The length of an inclined plane of height $h$ and angle $	heta$ is $L = \frac{h}{\sin 	heta}$.The acceleration of a block sliding down a smooth

Vedclass · Accepted Answer

$4\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) \ s$

Vedclass · Answer

(C) The length of an inclined plane of height $h$ and angle $	heta$ is $L = \frac{h}{\sin 	heta}$.The acceleration of a block sliding down a smooth inclined plane is $a = g \sin 	heta$.Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $s = L$:$L = \frac{1}{2} (g \sin 	heta) t^2 \implies \frac{h}{\sin 	heta} = \frac{1}{2} g \sin 	heta \ t^2$.Thus,$t = \sqrt{\frac{2h}{g \sin^2 	heta}} = \frac{1}{\sin 	heta} \sqrt{\frac{2h}{g}}$.Given $h = 20 \ m$ and $g = 10 \ m \ s^{-2}$,we have $\sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 20}{10}} = 2 \ s$.For plane $A$ $(	heta_1 = 30^{\circ})$: $t_1 = \frac{1}{\sin 30^{\circ}} 	imes 2 = \frac{1}{0.5} 	imes 2 = 4 \ s$.For plane $B$ $(	heta_2 = 60^{\circ})$: $t_2 = \frac{1}{\sin 60^{\circ}} 	imes 2 = \frac{1}{\sqrt{3}/2} 	imes 2 = \frac{4}{\sqrt{3}} \ s$.Therefore,$t_1 - t_2 = 4 - \frac{4}{\sqrt{3}} = 4 \left( 1 - \frac{1}{\sqrt{3}} ight) = 4 \left( \frac{\sqrt{3}-1}{\sqrt{3}} ight) \ s$.

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