An arrangement of a spring,strings,a pulley,a | vedclass

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(B) Initially,the system is in equilibrium. For mass $M$,the spring force $F_s = Mg$. Since the pulley is massless and the string is continuous,the te

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the magnitude of the acceleration of $m$ is $(M-m)g/m$ and that of $M$ is zero

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(B) Initially,the system is in equilibrium. For mass $M$,the spring force $F_s = Mg$. Since the pulley is massless and the string is continuous,the tension $T$ in the string is equal to the spring force,so $T = Mg$.For mass $m$,the forces are tension $T$ upwards and gravity $mg$ downwards. Since it is in equilibrium,$T = mg$,which implies $Mg = mg$. However,the problem states $M > m$,which implies the initial setup must have been held in equilibrium by the string connected to the ground.Immediately after the string connected to $m$ is detached,the tension $T$ in the string becomes zero. The spring force $F_s$ cannot change instantaneously,so $F_s = Mg$ remains acting on $M$ upwards.For mass $M$,the forces are $Mg$ downwards and $F_s = Mg$ upwards. Thus,the net force on $M$ is zero,and its acceleration is $0$.For mass $m$,the only force acting is gravity $mg$ downwards. Thus,its acceleration is $g$ downwards. Wait,let's re-evaluate: The tension $T$ was supporting $m$. If the string to the ground is cut,the tension $T$ in the string connected to the pulley becomes $Mg$. The forces on $m$ are $T = Mg$ upwards and $mg$ downwards. The net force is $Mg - mg$ upwards. Therefore,the acceleration of $m$ is $a_m = (Mg - mg)/m = (M-m)g/m$ upwards. The acceleration of $M$ is zero because the spring force $Mg$ balances its weight $Mg$.

An arrangement of a spring,strings,a pulley,and masses is shown in the figure below. The pulley and the strings are massless and $M > m$. The spring is light with spring constant $k$. If the string connecting $m$ to the ground is detached,then immediately after detachment,

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