A block of mass m slides on a wooden wedge,wh | vedclass

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(D) Let the acceleration of the wedge be $a_1$ and the acceleration of the block with respect to the wedge be $a_2$.For the wedge of mass $M$,the hori

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$\frac{2}{3} g$

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(D) Let the acceleration of the wedge be $a_1$ and the acceleration of the block with respect to the wedge be $a_2$.For the wedge of mass $M$,the horizontal force is the horizontal component of the normal force $N$ exerted by the block on the wedge:$N \sin 30^{\circ} = M a_1 = 16 a_1$$N (0.5) = 16 a_1 \Rightarrow N = 32 a_1$For the block of mass $m$ with respect to the wedge,we apply Newton's second law in the direction perpendicular to the incline:$N = m g \cos 30^{\circ} - m a_1 \sin 30^{\circ}$$32 a_1 = 8 g (\frac{\sqrt{3}}{2}) - 8 a_1 (\frac{1}{2})$$32 a_1 = 4 \sqrt{3} g - 4 a_1$$36 a_1 = 4 \sqrt{3} g \Rightarrow a_1 = \frac{\sqrt{3}}{9} g$Now,applying Newton's second law along the incline for the block:$m g \sin 30^{\circ} + m a_1 \cos 30^{\circ} = m a_2$$g \sin 30^{\circ} + a_1 \cos 30^{\circ} = a_2$$a_2 = g (\frac{1}{2}) + (\frac{\sqrt{3}}{9} g) (\frac{\sqrt{3}}{2})$$a_2 = \frac{g}{2} + \frac{3g}{18} = \frac{g}{2} + \frac{g}{6} = \frac{3g + g}{6} = \frac{4g}{6} = \frac{2}{3} g$

$A$ block of mass $m$ slides on a wooden wedge,which in turn slides backward on a horizontal surface. The acceleration of the block with respect to the wedge is: Given $m = 8 \, kg, M = 16 \, kg$. Assume all surfaces shown in the figure to be frictionless.

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