A block of mass M is placed on a horizontal s | vedclass

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Question

(B) Let $a$ be the acceleration of the system. Since the string is inextensible,the block $m$ moves downwards with acceleration $a$,and the blocks $M$

Vedclass · Accepted Answer

$\left[ {\frac{{m - \mu ({m_0} + M)}}{{m + {m_0} + M}}} \right]g$

Vedclass · Answer

(B) Let $a$ be the acceleration of the system. Since the string is inextensible,the block $m$ moves downwards with acceleration $a$,and the blocks $M$ and $m_0$ move horizontally to the right with the same acceleration $a$.For the block $m$: The forces acting are gravity $mg$ downwards and tension $T$ upwards. The equation of motion is $mg - T = ma$ ... $(1)$For the combined system of blocks $M$ and $m_0$: The normal force exerted by the horizontal surface on $M$ is $N = (M + m_0)g$. The kinetic friction force acting on $M$ is $f_k = \mu N = \mu(M + m_0)g$.The horizontal force acting on the system $(M + m_0)$ is the tension $T$. The equation of motion is $T - f_k = (M + m_0)a$,which gives $T - \mu(M + m_0)g = (M + m_0)a$ ... $(2)$Adding equations $(1)$ and $(2)$:$(mg - T) + (T - \mu(M + m_0)g) = ma + (M + m_0)a$$mg - \mu(M + m_0)g = (m + M + m_0)a$$a = \frac{m - \mu(M + m_0)}{m + M + m_0}g$Thus,the correct option is $B$.

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