In the arrangement shown in the figure,the bl | vedclass

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(D) Let the acceleration of the wedge be $a$ towards the right and the acceleration of the block $m$ relative to the wedge be $a_{rel}$ down the incli

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None of these

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(D) Let the acceleration of the wedge be $a$ towards the right and the acceleration of the block $m$ relative to the wedge be $a_{rel}$ down the incline. The absolute acceleration of the block $m$ is the vector sum of the wedge's acceleration and the relative acceleration.Using the constraint relation for the string length $L$,we find the relationship between the accelerations. The length of the string is $L = x_m + x_p$,where $x_m$ is the length from the block to pulley $P_1$ and $x_p$ is the length from $P_1$ to $P_2$. Given the geometry,the acceleration of the block relative to the ground along the incline is $a_{rel} - a \cos(60^\circ)$.For the wedge of mass $M=8\,kg$: The forces acting horizontally are the tension $T$ from the wall and the horizontal component of the normal force $N \sin(60^\circ)$. $T(1 + \cos(60^\circ)) - N \sin(60^\circ) = Ma$.For the block of mass $m=2\,kg$: Along the incline,$mg \sin(60^\circ) - T = m(a_{rel} - a \cos(60^\circ))$.Normal force $N = mg \cos(60^\circ) + ma \sin(60^\circ)$.Solving these equations with the constraint $a_{rel} = 2a$,we find the acceleration of the wedge $a = \frac{mg \sin(60^\circ)(1 + \cos(60^\circ))}{M + m(1 + \cos(60^\circ))^2 + m \sin^2(60^\circ)}$.Substituting the values $m=2\,kg, M=8\,kg, g=10\,m/s^2$,we get $a = \frac{20 \cdot \frac{\sqrt{3}}{2} \cdot \frac{3}{2}}{8 + 2(\frac{9}{4}) + 2(\frac{3}{4})} = \frac{15\sqrt{3}}{8 + 4.5 + 1.5} = \frac{15\sqrt{3}}{14} \approx 1.85\,m/s^2$. Since this does not match the options,the correct choice is $D$.

In the arrangement shown in the figure,the block of mass $m=2\,kg$ lies on the wedge of mass $M=8\,kg$. Find the initial acceleration of the wedge if the surfaces are smooth and the pulley and strings are massless.

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