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(A) Let the acceleration of the system be $a$. Since the block of mass $m$ is stationary with respect to the block of mass $M$,both move with the same

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$(M + m)g 	an \beta$

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(A) Let the acceleration of the system be $a$. Since the block of mass $m$ is stationary with respect to the block of mass $M$,both move with the same acceleration $a$.For the whole system of mass $(M + m)$,the force $P$ is applied:$P = (M + m)a$  --- $(1)$Now,consider the free body diagram of the block of mass $m$. The forces acting on it are its weight $mg$ (downwards),the normal reaction $R$ (perpendicular to the incline),and the pseudo force $ma$ (acting horizontally to the left in the frame of $M$).Since $m$ is stationary relative to $M$,the net force along the incline must be zero:$ma \cos \beta = mg \sin \beta$Dividing by $m \cos \beta$,we get:$a = g 	an \beta$  --- $(2)$Substituting the value of $a$ from equation $(2)$ into equation $(1)$:$P = (M + m)g 	an \beta$

Two wooden blocks are moving on a smooth horizontal surface such that the block of mass $m$ remains stationary with respect to the block of mass $M$ as shown in the figure. The magnitude of force $P$ is:

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