Help again in interpreting a pulley with a spring and a block

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Homework Statement
The system as shown in the figure is released from rest. The pulley, spring and string are ideal and friction is absent everywhere. If the speed of 5 kg block when 2kg block leaves the contact with ground is 2(x)^0.5 m/s then value of x is: (spring constant k= 40N/m & g=10m/s)
Relevant Equations
None
I guess im still a bit rusty with spring problems. Heres the given figure:
View attachment 342442
Can someone walk me through what exactly would happen when the system is released from rest. Mainly im concerned about the 5kg block when it moves down by say a distance x, this means the gravitational force is more than the tension in the string above, but due to this tension the spring will elongate, but the upper part of the spring is not connected to a fixed support so it will accelerate downwards as well, making the elongation 0 hence the spring force 0? So what is the tension in the string connecting the 2kg block and the spring...

I have a lot of assumptions which seem very hand wavy and lead to nothing but more confusion....
 

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PeroK said:
Draw a new diagram when the 5kg mass has dropped a small distance.
OK
 
PeroK said:
Draw a new diagram when the 5kg mass has dropped a small distance.
I am unsure about the forces as I peeked at a solution and saw that the tension is apparently the spring force
17115743176276753582204215259021.jpg
 
tellmesomething said:
I peeked at a solution and saw that the tension is apparently the spring forceView attachment 342443
The (idealized) spring has zero mass. Apply Newtons 2nd to it as a free body to see why that is the case.
 
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Orodruin said:
The 5 kg moving a distance x is not equivalent to the 2 kg mass moving a distance x off the ground. Until the spring has elongated sufficiently to give enough tension to lift the 2 kg mass off the ground, the 2 kg mass won't move.
Thats what I dont get. The spring force is acting on the block in the upward direction....Yes?
The spring is pulling the support with the magnitude of force it itself is being pulled by I.e 50N no?
 
PeroK said:
In effect, you are treating the spring like just part of the inextensible string?

What if the spring were a bungee cord?
The same? The weight stretches the bungee cord and the cord pulls the support which is not fixed so ....
 
PeroK said:
In effect, you are treating the spring like just part of the inextensible string?

What if the spring were a bungee cord?
Also since sir you helped me in my previous post im thinking of this situation just like that, how we had a movable support so the spring didnt stretch.. Im talking about this situation
32337.png
 
PeroK said:
The 2kg mass is not massless. M1 in that diagram is massless. There are no massless supports in this new problem.
So does that mean the spring and the string in this problem will not have the same acceleration
 
tellmesomething said:
Thats what I dont get. The spring force is acting on the block in the upward direction....Yes?
The spring is pulling the support with the magnitude of force it itself is being pulled by I.e 50N no?
If the spring was pulling at the 5 kg mass with a force of 50 N, then the net force on the 5 kg mass would be zero and it would not accelerate. The force from the spring on the mass is determined by one thing and one thing only: the elongation of the spring.
 
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SammyS said:
If the spring was pulling at the 5 kg mass with a force of 50 N, then the net force on the 5 kg mass would be zero and it would not accelerate. The force from the spring on the mass is determined by one thing and one thing only: the elongation of the spring.
No by support I meant the string attached to the upper end of the spring not the mass...
 
SammyS said:
Which end of spring has the same acceleration as the string ?
I'm just spewing nonsense aren't I? I will review everything advised till now. Sorry for the baseless assumptions..
 
erobz said:
The (idealized) spring has zero mass. Apply Newtons 2nd to it as a free body to see why that is the case.
Missed this, I made a FBD of the spring not sure how this is supposed to tell me that its massless...to be fair SPRING FORCE reaction pair is more fitting..
17115772631021615989915149200983.jpg
 
tellmesomething said:
Missed this, I made a FBD of the spring not sure how this is supposed to tell me that its massless...to be fair SPRING FORCE reaction pair is more fitting..View attachment 342447
This is incorrect. Again, the 5 kg weight is not pulling the spring with 50 N. If it were, it could not accelerate!

edit: The forces on the spring are also not a reaction pair!
 
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Orodruin said:
Why are you quoting my text attributing it to someone else?
Orodruin said:
This is incorrect. Again, the 5 kg weight is not pulling the spring with 50 N. If it were, it could not accelerate!

edit: The forces on the spring are also not a reaction pair!
I dont know how that happened I apologize

I understand your part now the mass wouldn't have been able to accelerate if the net force becomes zero makes sense.. T' = 50-5a..
a being the acceleration of the block?
 
tellmesomething said:
Missed this, I made a FBD of the spring not sure how this is supposed to tell me that its massless...to be fair SPRING FORCE reaction pair is more fitting..View attachment 342447
If the spring had its own mass, whose mass center would be accelerating, there would have to be a net force on the spring. As it stands this spring is ideal, massless( otherwise we would have some more information). Because it has no mass the force balance ( even as it accelerates with the falling mass) is zero. This is not to say as @Orodruin points out that ##T## is 50 N though.
 
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Orodruin said:
This is incorrect. Again, the 5 kg weight is not pulling the spring with 50 N. If it were, it could not accelerate!

edit: The forces on the spring are also not a reaction pair!
When the spring elongates there's a restoring force acting on the mass of 5kg trying to pull it back to its original position....im talking about the reaction pair of this force which would be exerted by the block on the spring
 
erobz said:
If the spring had its own mass, whose mass center would be accelerating, there would have to be a net force on the spring. As it stands this spring is ideal, massless( otherwise we would have some more information). Because it has no mass the force balance ( even as it accelerates with the falling mass) is zero. This is not to say as @Orodruin points out that ##T## is 50 N though.
Considering what you both said...

T'= 50-5a (a being the acceleration of the block)
T= spring force + 50-5a
 
tellmesomething said:
Considering what you both said...

T'= 50-5a (a being the acceleration of the block)
T= spring force + 50-5a

If this is correct in the slightest...how do I find the acceleration of the block
 
erobz said:
If the spring had its own mass, whose mass center would be accelerating, there would have to be a net force on the spring. As it stands this spring is ideal, massless( otherwise we would have some more information). Because it has no mass the force balance ( even as it accelerates with the falling mass) is zero. This is not to say as @Orodruin points out that ##T## is 50 N though.
Similarly then we should be able to make the fbd of the string between the spring and the block....
 
tellmesomething said:
Similarly then we should be able to make the fbd of the string between the spring and the block....
Yeah, ideal massless string, frictionless pulley, implies what about the tension on ether side?
 
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tellmesomething said:
Considering what you both said...

T'= 50-5a (a being the acceleration of the block)
Yes
tellmesomething said:
T= spring force + 50-5a
No.
If the spring is massless then the net force on it is zero. (Do you see why?)
According to your diagram in post #20, what is the net force on it?
 
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haruspex said:
Yes

No.
If the spring is massless then the net force on it is zero. (Do you see why?)
According to your diagram in post #20, what is the net force on it?
Do you mean we will not include the tension by the string in the FBD of the spring..... IF so why? I get that the net force on the string is 0 but the tension should still be pulling on the spring...