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

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Question

(B) Let the acceleration of the system be $a$. The total mass of the system is $(M + m_0 + m)$.The driving force is the weight of the hanging block $m

Vedclass · Accepted Answer

$\frac{m(m_0 + M)}{m + m_0 + M}g$

Vedclass · Answer

(B) Let the acceleration of the system be $a$. The total mass of the system is $(M + m_0 + m)$.The driving force is the weight of the hanging block $m$,which is $mg$.Using Newton's second law for the whole system: $mg = (M + m_0 + m)a$.Therefore,the acceleration $a = \frac{mg}{M + m_0 + m}$.Now,consider the block of mass $m$. The forces acting on it are its weight $mg$ downwards and the tension $T$ upwards.Applying Newton's second law to block $m$: $mg - T = ma$.Substituting the value of $a$: $T = mg - m\left( \frac{mg}{M + m_0 + m} \right)$.$T = mg \left( 1 - \frac{m}{M + m_0 + m} \right) = mg \left( \frac{M + m_0 + m - m}{M + m_0 + m} \right)$.$T = \frac{mg(M + m_0)}{M + m_0 + m}$.

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