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

[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:

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

 

[Orodruin]

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

By the problem’s formulation, x is a dimensionless number.

2(x)^0.5 m/s

That people want to name a displacement the same thing is a problem for them. It is already taken. I did the problem for fun, I used $\delta$ for the displacement.

 

[erobz]

By the problem’s formulation, x is a dimensionless number.

That people want to name a displacement the same thing is a problem for them. It is already taken. I did the problem for fun, I used $\delta$ for the displacement.

Fair enough, I allowed myself to be seduced by standard presentation of spring force. But now that I see $\delta$ it’s a little more curvy!

 

[Orodruin]

Fair enough, I allowed myself to be seduced by standard presentation of spring force. But now that I see $\delta$ it’s a little more curvy!

Not just you, to be clear. Apart from that it is also a problem-issue. The typical use of $x$ would be for a distance, but alas that is what we have to work with.