Does spring stop applying force just after we cut it? Like in this image will $F'$ be zero?
The question arises because i have solved a few problems in which it says we cut the spring but we completely remove it and I was confused if the small part of spring still attached to the block apply force since it should be stretched. (Considering ideal springs.)
Yes, if the spring is ideal, you can assume that.
However, on a more realistic situation where the spring has some mass somewhere, indeed it would be stretched and it would tend to contract, exerting force on the block. In fact, you can imagine putting a very small mass attached on top of the spring (right at the point where you cut it). This is the typical problem of two masses attached by a spring, which has two modes: one mode is pure translation (i.e. the whole system moves as a whole, in this case falling) and the second mode is oscillation around its center of mass (with some frequency and relative amplitude). However, the oscillatory mode should be negligible when the top mass is arbitrarily small compared to the block's mass. So you only have the dynamics of the system as a whole, which simply falls due to gravity. Hence this justifies assuming $F'=0$ instantly, as long as the mass of the spring is negligible.
If the spring had non-zero mass rather than exactly $0$, then the center of mass would fall freely, but the block would have some oscillation around the free-fall trajectory.
Does spring stop applying force just after we cut it?
No. Immediately after cutting, the spring still is stretched by the same percentage that it was before cutting, so it still applies the same force to the block as before. (But the upper part of the spring no longer applies a force to the lower part, and that is crucial for a different point of view, explained further below.)
This time span of force applied to the block will last only for a very short period of time, because the spring will quickly return to its "rest" shape, typically with some oscillation. And as soon as the spring reaches that shape, it no longer applies any force. Trying to calculate the behaviour of the spring during that oscillation period can turn out to be quite complex, so we happily accept the approximation that the timespan of this complex process is short enough to be ignored.
So, immediately after cutting, the spring still applies a force to the block, but shortly after cutting, this force is zero.
And if we change our point of view and talk about the system composed of the block plus the small part of the string still attached to it, then the force that the spring applies to the block is completely internal within the system and does not affect the motion of the system as a whole (exactly: its center of gravity).
What matters now, is that the upper part of the spring no longer applies a pulling force to the lower part (and thus the system), so the only remaining force is gravity, meaning that the system (exactly: its center of gravity) will begin a free fall. From that point of view, it does not matter if the spring-block system then develops some internal oscillation, rotation or whatever. And if we can neglect the mass of the still-attached spring part, the whole-system movement nearly perfectly describes the block movement.
The spring does not apply any force, i.e. F' is 0. (Considering no air resistance)
As soon as you cut the spring, the tension in it gets released and the small part doesn't exert a force and if the spring is massless, it doesn't matter if you keep the small part or remove it completely.
As you are doing an ideal case, I don't expect any oscillation, which would arise in a real case.
Hope that helps.
If we have a massless ideal spring, any non-zero spring force can accelerate the spring itself infinitely quickly - it takes exactly zero time for a massless spring to snap back to its rest length when cut, and carries no momentum when doing so. At $t=0$, we cut the spring and see that it contracts instantaneously, and immediately exerts zero force on the block below.
The intuitive trouble arises with the massless spring, since we normally see that the spring force is dependent on length, so how could there be zero force at the moment the spring is cut, when it is still extended? The answer is that this moment has literally zero duration - a massless spring is cut at $t=0$ and instantaneously reverts to its rest length. Exerting a finite force for zero duration is the same as exerting no force at all.
In a real-world case with real springs that have mass, the spring will not contract instantaneously, and will still exert a force for the brief period of time it is extended.
Does a spring stop applying force as soon as we cut it?
The real world answer is no. In the real world it is impossible to have a massless spring. You appear to want to analyse this assuming an unrealistic ideal spring and then the answer is yes, but that answer does not reflect what happens in the real world.
Here is the real world analysis: When the spring is cut at the half way mark, the half of the spring that remains attached to the block is stretched and after being cut it wants to shorten back to its rest length. Since in the real world it has mass, the centre of mass of the remaining attached portion accelerates towards the block. Since this acceleration is provided by the momentum of the block, this force acts in the opposite direction to the force of gravity acting on the block and the resultant total downward force on the block is reduced for a short period after the spring is cut until the spring returns to its rest length.
Check out this video of releasing a slinky with a weight attached.
It can be seen that weight does not move for a significant period after the slinky is released, due to the upward force of the attached slinky contracting back to its rest length. If instead, we had cut the attachment of the weight to the slinky, the weight would have started accelerating immediately after the cut because none of the slinky remains attached.
