Initially, the system is in equilibrium. The tension $T$ in the main string supports the movable pulley and the connected loads.

For the movable pulley system connected to mass $M$, the upward force is $2T$ and the downward force is the weight $Mg$.

$$ 2T = Mg \implies T = \frac{Mg}{2} \quad \dots(1) $$

This relation holds for the tension $T$ supplied by the mass $M$.

The cord is cut at point P. We need to find the condition for the acceleration $a$ of the load of mass $2m$ to be greater than $g$ ($a > g$).

Using the Free Body Diagram and equations of motion for the system components:

• For the top mass $m$ on the right branch, the forces are Tension $T$ (up), Gravity $mg$ (down), and Spring Force $kx$ (down). $$ T – mg – kx = ma $$
• Rearranging for the spring force $kx$ based on the system’s coupled acceleration: $$ kx = m(3a – g) \quad \dots(2) $$

We are given the condition that the acceleration $a > g$. Substituting this condition into our derived equation (2):

$$ \text{If } a > g, \text{ then } 3a – g > 2g $$

Therefore:

$$ kx > m(2g) \implies kx > 2mg $$

Now, substitute this condition back into the force equation for the mass. The total upward tension $T$ must support the weight and the increased spring force required for this high acceleration:

$$ T – mg > 2mg $$ $$ T > 3mg \quad \dots(3) $$

Now we combine our result from Step 1 (Equilibrium Tension provided by M) with the condition from Step 3.

Substitute $T = \frac{Mg}{2}$ into the inequality $T > 3mg$:

$$ \frac{Mg}{2} > 3mg $$

Divide both sides by $g$ and multiply by 2:

$$ M > 2 \times 3m $$ $$ M > 6m $$