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

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(D) Let $a$ be the acceleration of the system consisting of blocks $M$ and $m_0$ moving horizontally,and block $m$ moving vertically downwards. Since

Vedclass · Accepted Answer

none of these

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(D) Let $a$ be the acceleration of the system consisting of blocks $M$ and $m_0$ moving horizontally,and block $m$ moving vertically downwards. Since the string is inextensible,all blocks have the same magnitude of acceleration $a$.For the system $(M + m_0)$,the equation of motion is: $T = (M + m_0)a$.For the block $m$,the equation of motion is: $mg - T = ma$.Adding these two equations: $mg = (M + m_0 + m)a$,so $a = \frac{mg}{M + m_0 + m}$.For the block $m_0$ to remain stationary with respect to $M$,the static friction $f$ must provide the necessary pseudo force: $f = m_0 a$.The maximum static friction is $f_{max} = \mu N = \mu m_0 g$.For $m_0$ to not slide,we require $f \le f_{max}$,so $m_0 a \le \mu m_0 g$.Substituting $a$: $m_0 \left( \frac{mg}{M + m_0 + m} \right) \le \mu m_0 g$.Thus,$\mu \ge \frac{m}{M + m_0 + m}$.Since this expression is not among the given options,the correct choice is $(D)$.

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