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(D) Let the trolley accelerate with $a_0$ to the right. In the frame of the trolley,a pseudo force $ma_0$ acts on block $m$ to the left.For block $m$,

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All of the above.

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(D) Let the trolley accelerate with $a_0$ to the right. In the frame of the trolley,a pseudo force $ma_0$ acts on block $m$ to the left.For block $m$,the tension $T$ in the string is $T = m(g^2 + a_0^2)^{1/2}$ if we consider the effective gravity,but here the string is vertical,so $T = m(g)$ is not correct. Actually,the block $m$ is in contact with the vertical wall. The pseudo force $ma_0$ acts on $m$ horizontally,so the normal force $N_m = ma_0$. The friction on $m$ is $f_m = \mu N_m = \mu ma_0$ (if sliding). However,the tension $T$ is determined by the equilibrium of $m$ in the vertical direction. Since $m$ is hanging,$T = mg$.For block $M$,the forces are tension $T$ to the right and friction $f$ from the surface. The pseudo force $Ma_0$ acts to the left. The net force on $M$ is $f_{net} = T - Ma_0 = mg - Ma_0$.When $a_0 = g(m/M) = \beta$,the friction force $f = 0$.Since the friction force $f = |mg - Ma_0|$,it is symmetric about $a_0 = \beta$. Thus,for $a_0 = \beta \pm \alpha$,the magnitudes of friction are equal.The maximum static friction is reached at the limits of the range of $a_0$ where the block does not slip. These limits $a_1$ and $a_2$ are also symmetric about $\beta$,so $a_1 + a_2 = 2\beta$.Therefore,all statements are correct.

Imagine the situation in which the given arrangement is placed inside a trolley that can move only in the horizontal direction,as shown in the figure. If the trolley is accelerated horizontally along the positive $x$-axis with $a_0$,then choose the correct statement$(s)$.

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