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(A) Let $a$ be the acceleration of the system $(M_0 + M + m)$ in the horizontal direction. The force $F$ is given by $F = (M_0 + M + m)a$.For block $M

Vedclass · Accepted Answer

$(M_0 + M + m) \frac{g}{\mu}$

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(A) Let $a$ be the acceleration of the system $(M_0 + M + m)$ in the horizontal direction. The force $F$ is given by $F = (M_0 + M + m)a$.For block $M$ to be stationary with respect to $M_0$,the tension $T$ in the string must balance the pseudo force acting on $M$. Thus,$T = Ma$.For block $m$ to be stationary with respect to $M_0$,the vertical forces must balance. The forces acting on $m$ are gravity $mg$ downwards,friction $f = \mu N$ upwards,and the normal force $N$ from the vertical surface of $M_0$ due to the pseudo force $ma$. Thus,$N = ma$ and $f = \mu ma$.For equilibrium in the vertical direction,$f = mg$,which implies $\mu ma = mg$,so $a = \frac{g}{\mu}$.Substituting $a = \frac{g}{\mu}$ into $F = (M_0 + M + m)a$,we get $F = (M_0 + M + m) \frac{g}{\mu}$.However,if the question implies a range or specific conditions where $M$ is also constrained by friction or other forces,we re-evaluate. Given the options,if $M$ is on a smooth surface,$T=Ma$ is the only horizontal force. If $m$ is held by friction,$a=g/\mu$. If $M$ were also subject to friction,the result would change. Based on standard physics problems of this type,option $(A)$ is the primary solution.

Imagine a situation in which the horizontal surface of block $M_0$ is smooth and its vertical surface is rough with a coefficient of friction $\mu$. In the above problem,choose the correct value$(s)$ of $F$ for which the blocks $M$ and $m$ remain stationary with respect to $M_0$.

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