Take it step by step.
The clincer is that the system is at equilibrium in the beginning.Hence all the forces on all the balls are balanced. Looking at the lowermost ball, the force exerted by the spring must be equal to weight i.e 10g.
Taking the second ball, forces acting downwards are weight and force due to the lower spring, which is 10g as obtained from previous paragraph( tension developed depends on the lower ball only). So net force acting downwards on string is 20g+10g=30g. This is balanced by force due to upper spring, which must be 30 g as well.
Now the net force acting downwards on the uppermost ball is 10g + forced due to spring(30g)=40g. This is balanced by the string and hence the body is in equilibrium.
As a result, on cutting the spring, the net force is 40g downwards and hence acceleration at that instant is 4g.
In my opinion, the requirement that the string be nonextensible creates conceptual issues.
On the one hand, it is stated that the string is nonextensible. On the other hand, it is stated on the diagram the "External force $F$ is applied such that the block remains at rest". The problem is if the string is nonextensible, and initially has no slack, then the block cannot move to the left, i.e., it will always be "at rest", regardless of the force $F$ to the left.
But more importantly, it makes it difficult to explain (1) why the static friction force counters the applied force before the tension in the string and (2) why when equilibrium is reached the friction force no longer exists.
To facilitate the answers to these questions I will replace the string with an ideal spring (see the figures below). An ideal spring, like the string, is massless. But unlike the string, it is extensible to the degree allowed by the spring constant. The magnitude of the tension in the spring is equals the magnitude of its restoring force, or $T=k\Delta x$ where $k$ is the spring constant and $\Delta x$ is its extension beyond the "relaxed" stated.
Now consider the following where the block is considered a rigid (nonextensible) body:
The spring, like the string, is initially relaxed so there is no tension. FIG 1 shows the block with no external force and the relaxed spring attached to the wall
In Fig 2 we gradually apply an increasing external force $F$ that is less than the maximum static friction force. Since the block cannot move, the spring cannot extend and thus the tension in the spring is still zero. This explains, for a physically real scenario, why the applied force $F$ is countered first by the static friction force.
In FIG 3 the applied force reaches the maximum static friction force and the friction force becomes kinetic friction, which is generally considered constant. Since kinetic friction is usually less than static friction, if the applied force $F$ is maintained at the value of the maximum static friction there will initially be a net force to the left causing the block to move to the left. (Note that actual value of the friction force during the transition from static to kinetic is undefined for the standard model of friction). At the same time, however, the spring extends creating an opposing tension force. So during this phase before the extension of the spring is a maximum, we have
$$F-f_{k}> T$$
$$\mu_{s}mg-\mu_{k}mg>k\Delta x$$
and the block is moving to the left.
- When the extension of the spring is such that the tension in the spring equals the applied force $F$, it's extension is maximized and we have
$$F=T=k\Delta x_{max}$$
Substituting into the first equation,
$$f_{k}=0$$
Meaning there is no net force for friction to oppose.
Note that in this example, the stiffer the spring (the greater $k$ is) the less the block needs to move before the tension equals the applied force, i.e., the quicker the tension rises. The nonextensible string is simply a spring with an infinite $k$.
Hope this helps.
Best Answer
While I haven't seen the video, the description matches an old science trick using inertia: if you want the top string to snap, pull slowly. To snap the bottom string, pull suddenly - the inertia of the weight will “protect" the upper string for a brief moment.