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

[tellmesomething]

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....

 

[PeroK]

You must calculate instead of just assume.

 

[PeroK]

Draw a new diagram when the 5kg mass has dropped a small distance.

 

[tellmesomething]

Draw a new diagram when the 5kg mass has dropped a small distance.

OK

 

[tellmesomething]

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

 

[erobz]

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.

 

[Orodruin]

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.

 

[tellmesomething]

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]

I am unsure about the forces as I peeked at a solution and saw that the tension is apparently the spring forceView attachment 342443

In effect, you are treating the spring like just part of the inextensible string?

What if the spring were a bungee cord?

 

[tellmesomething]

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 ....

 

[tellmesomething]

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

 

[PeroK]

The 2kg mass is not massless. M1 in that diagram is massless. There are no massless supports in this new problem.

 

[tellmesomething]

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

 

[SammyS]

So does that mean the spring and the string in this problem will not have the same acceleration

Which end of spring has the same acceleration as the string ?

 

[Orodruin]

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.

 

[tellmesomething]

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...

 

[tellmesomething]

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..

 

[Orodruin]

No by support I meant the string attached to the upper end of the spring not the mass...

Why are you quoting my text attributing it to someone else?

 

[Orodruin]

No by support I meant the string attached to the upper end of the spring not the mass...

The same thing applies.

 

[tellmesomething]

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..

 

[Orodruin]

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!

 

[tellmesomething]

Why are you quoting my text attributing it to someone else?

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?

 

[erobz]

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.

 

[tellmesomething]

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

 

[tellmesomething]

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]

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

 

[tellmesomething]

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....

 

[erobz]

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?

 

[haruspex]

Considering what you both said...

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

Yes

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?

 

[tellmesomething]

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...