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 theres 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...

 

[tellmesomething]

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...

Also sorry another doubt, if we draw the FBD of the string we would have Spring force acting on it upwards, the reaction pair of tension on the spring and reaction pair of tension on the block. I know this is impossible as tension throughout the string should be the same for the sake of it being massless...... I feel like im stuck in a circular loop

 

[haruspex]

Do you mean we will not include the tension by the string in the FBD of the spring

That's not what I wrote.

if we draw the FBD of the string we would have Spring force acting on it upwards,

What do you mean by "spring force" here?

 

[tellmesomething]

That's not what I wrote.

What do you mean by "spring force" here?

Spring force as in the force exerted by the spring after elongation on the string to bring it back to its original length

 

[haruspex]

Spring force as in the force exerted by the spring after elongation on the string to bring it back to its original length

Your diagram shows two forces acting on the spring, T exerted by the string and T' by the suspended mass.
If the spring is massless, what is the relationship between those two forces according to Newton II?

What is the relationship between T and what you call the spring force in post #33 according to Newton III?

 

[tellmesomething]

Your diagram shows two forces acting on the spring, T exerted by the string and T' by the suspended mass.
If the spring is massless, what is the relationship between those two forces according to Newton II?

What is the relationship between T and what you call the spring force in post #33 according to Newton III?

My diagram also shows the reaction pair of the spring force...why are we excluding that.....
To answer your questions
T=T' i.e if we are neglecting the reaction pair of the spring force
I dont see any relation between the tension with which the upper string pulls on the spring and the spring force being exerted on the lower string by the spring

 

[erobz]

Not to derail this approach with the forces, but if indeed at the instant $m_2$ begins to accelerate, $m_5$'s velocity is given by:

$$ v = 2 \sqrt{x} \left[ \frac{ \sqrt{m}}{s} \right] $$

As is stated in the problem, then we can skip all the all the tension issues (with the spring and the string -not that is shouldn't be understood by the OP) and just use conservation of energy (only involving $m_5$).

Are any of the other helpers concerned about this seeming "excess" of information, or am I missing something? The math doesn't jive. Even if I write the force equation and apply $v \frac{dv}{dx}$ I find an $x$ when multiplied by $k$ that does not yield $20 ~N$...

 

[tellmesomething]

Not to derail this approach with the forces, but if indeed at the instant $m_2$ begins to accelerate, $m_5$'s velocity is given by:

$$ v = 2 \sqrt{x} \left[ \frac{ \sqrt{m}}{s} \right] $$

As is stated in the problem, then we can skip all the all the tension issues (not that is shouldn't be understood by the OP) and just use conservation of energy(only involving $m_5$).

Are any of the other helpers concerned about this seeming "excess" of information, or am I missing something?

Why is my argument so circular though...I cant point out what is wrong... Thanks for this alternative approach too

 

[tellmesomething]

Why is my argument so circular though...I cant point out what is wrong... Thanks for this alternative approach too

I think its sinking in slowly .Please check
Firstly for the FBD of the spring we do not include the reaction force of the spring force since spring force comes only when tension is exerted its not instantaneous.



Though I still do not understand for the FBD of the lower string why we would not include the reaction pair of the tension force on the spring above it... Maybe because the spring force is something like a response to the tension force already.

 

[erobz]

I think its sinking in slowly .Please check
Firstly for the FBD of the spring we do not include the reaction force of the spring force since spring force comes only when tension is exerted its not instantaneous.



Though I still do not understand for the FBD of the lower string why we would not include the reaction pair of the tension force on the spring above it... Maybe because the spring force is something like a response to the tension force already.

Start with a FBD of the block. Then above it segment out the spring with its forces, the above that the string. The force pairs should become clear.

 

[haruspex]

My diagram also shows the reaction pair of the spring force...why are we excluding that.....

If your diagram is a FBD of the spring then it should only show forces acting on the spring, not forces the spring exerts.
You explained what you meant by spring force, but I am not sure what you mean by "reaction pair of the spring force". If you mean that force which is the action-reaction pair of your "spring force", that is T'. If you mean the pair of forces exerted by the spring, they do not act on it.

I dont see any relation between the tension with which the upper string pulls on the spring and the spring force being exerted on the lower string by the spring

Small misunderstanding… when you wrote that the spring force is the force exerted on the string I thought you meant the upper string. So I reword my question:
What is the relationship between T' and what you call the spring force in post #33 according to Newton III?

 

[tellmesomething]

If your diagram is a FBD of the spring then it should only show forces acting on the spring, not forces the spring exerts.
You explained what you meant by spring force, but I am not sure what you mean by "reaction pair of the spring force". If you mean that force which is the action-reaction pair of your "spring force", that is T'. If you mean the pair of forces exerted by the spring, they do not act on it.

Small misunderstanding… when you wrote that the spring force is the force exerted on the string I thought you meant the upper string. So I reword my question:
What is the relationship between T' and what you call the spring force in post #33 according to Newton III?

Can you give me an exact proof of your statement of the spring force being the reaction force to Tension?
I thought of tension acting here as the the external constant force which is applied to a spring to stretch it. (The way there is a constant external force being applied to a normal spring and force problem).

It appears to me that tension here instead of acting as that constant external force, it is actually the constant external force - the net force which is definitely the reaction pair of the spring force.....

Is this right?

 

[tellmesomething]

Can you give me an exact proof of your statement of the spring force being the reaction force to Tension?
I thought of tension acting here as the the external constant force which is applied to a spring to stretch it. (The way there is a constant external force being applied to a normal spring and force problem).

It appears to me that tension here instead of acting as that constant external force, it is actually the constant external force - the net force which is definitely the reaction pair of the spring force.....

Is this right?

This is also consistent with our very first equation, that T'=50-5a

 

[tellmesomething]

Can you give me an exact proof of your statement of the spring force being the reaction force to Tension?
I thought of tension acting here as the the external constant force which is applied to a spring to stretch it. (The way there is a constant external force being applied to a normal spring and force problem).

It appears to me that tension here instead of acting as that constant external force, it is actually the constant external force - the net force which is definitely the reaction pair of the spring force.....

Is this right?

I have edited this response there was an "is" instead of an "of", which might have caused some confusion. Sorry for the inconvenience

 

[haruspex]

Can you give me an exact proof of your statement of the spring force being the reaction force to Tension?

You defined the "spring force" as

Spring force as in the force exerted by the spring after elongation on the [lower] string

and T' as the force the lower string exerts on the spring.
That makes them an action-reaction pair.

I thought of tension acting here as the the external constant force which is applied to a spring to stretch it.

No, tension is something that happens inside a stretched object. It appears at its ends as a pull on some adjacent object.

 

[tellmesomething]

Can you give me an exact proof of your statement of the spring force being the reaction force to Tension?
I thought of tension acting here as the the external constant force which is applied to a spring to stretch it. (The way there is a constant external force being applied to a normal spring and force problem).

It appears to me that tension here instead of acting as that constant external force, it is actually the constant external force - the net force which is definitely the reaction pair of the spring force.....

Is this right?

Also if I did not make it clear already, I understand now why we should not include both the tension and the reaction pair of the spring force in the FBD of the spring since they are the one and the same force. Similarly why we shouldn't include both the action pair of the spring force and the reaction pair of the tension in the FBD of the string as they are the one and the same force. :)

 

[tellmesomething]

You defined the "spring force" as

and T' as the force the lower string exerts on the spring.
That makes them an action-reaction pair.

No, tension is something that happens inside a stretched object. It appears at its ends as a pull on some adjacent object.

Yes got it. Thankyou very much:) You and all the helpers here been the best possible help I could have recieved

 

[erobz]

Continuing on with the path the OP was taking. I think there is an issue in the end of problem?

At the instant the 2kg mass lift off $\downarrow^+$:

$$ -m_2g + m_5g = m_5 \ddot x $$

$$ \implies \ddot x = \frac{(m_5 - m_2) g}{m_5} = 6 \left[ \frac{ \text{m}}{\text{s}^2} \right]$$

So:

$$-kx + m_5 g = m_5 \ddot x $$

$$ \implies x = 0.5 [\rm{m}] $$

( I know with understanding about the tensions we can go directly to that result in one step )

Now, I think the problem should have just ended there, but they are trying to be fancy about it. Are they saying that at that instant:

$$ v = 2 \sqrt{x} \left[ \frac{ \sqrt{ \text{m}}}{\text{s}}\right] $$

and asking the solve to solve for $x$???

Using Conservation of Energy:

$$ 0 = -m_5 g x + \frac{1}{2}kx^2 +\frac{1}{2}m_5v^2$$

Subbing for $v^2$

$$ 0 = -m_5 g x + \frac{1}{2}kx^2 +\frac{1}{2}m_5 4 x \left[ \frac{ m}{s^2}\right] $$

From which I find that:

$$x = \frac{-m_5( 4 [\frac{ \text{m}}{\text{s}^2}] - 2 g )}{k} = 2 [\text{m} ] $$

???

 

[PeroK]

Note that $x$ is a speed squared. The question looks a bit odd. Why not just ask for the speed of the 5kg mass when the 2kg mass leaves the ground.

 

[erobz]

Note that $x$ is a speed squared. The question looks a bit odd. Why not just ask for the speed of the 5kg mass when the 2kg mass leaves the ground.

Oh, so $x$ is not the extension of the spring. Strange.

 

[erobz]

Note that $x$ is a speed squared. The question looks a bit odd. Why not just ask for the speed of the 5kg mass when the 2kg mass leaves the ground.

If reading like $x$ is some speed squared, they are saying the speed is two times some speed...! SMH.

For the OP's reference in terms of $x$ as the displacement of the 5kg mass:

$$ v = \sqrt{x \left( 2g - \frac{k}{m_5}x\right)} \approx 2.8 [ \text{m/s}] $$