@JohnRennie Ahh yes! This type of approach was new to me. I will use this in future ! [ps: I replied late 'cause I went to bed too early :-( ]

I made a lot of progress on a similar question I had, but I never got the answer :-(

[ will share it here once my school gets over ]

@PandaScientist OK :-)

@JohnRennie Will you around at 9 PM IST ?

@KavinIshwaran Nearer 9:45, but I will be around.

Ok :-)

@JohnRennie Hi !

Are you free ?

@KavinIshwaran Hi, yes I'm free :-)

Can someone answer a few basic questions I have about work relating to ideal springs? I'm trying to check my comprehension.

Yes, what are the questions?

@JohnRennie I solved this question by taking Kirchoff's law by taking the current to be flowing from positive terminal of the battery to negative terminal

and I also got the correct answer

OK, that sounds fine.

but I am having doubt as I get the current to be negative which says the current to be flowing from the negative terminal of the battery to the positive terminal.

Is it fine ?

The current in the submerged resistor came out negative?

Total current

Yes

Do you want to post your working?

Just few minutes I will copy my working neatly :-)

OK :-)

@AngelPray You do work to extend the spring by 5cm, then when you let the spring shrink back by 2cm, you get work back out again.

The net work i the same as if you had just stretched the spring by 3cm.

For springs net work is conservative so only the start and end point matter.

But does the sign of the position matter?

You mean the sign of the displacement?

If we say the equilibrium state is 0 is there a difference is the final position is positive and the initial value is negative? Or if we have the same distances but with the same sign?

Because if I understand correctly, work done by a spring is always positive.

That's because if you allow a spring to relax the force and displacement are always in the same direction.

e.g. if a spring is compressed then if you let it go it will expand and exert a force in the direction it is expanding. So ∫F ds is always positive.

So when solving for work do we use distance instead of displacement?

Solving for work when it comes to ideal springs I mean.

Well work is always ∫F ds.

So you need to figure out the direction the spring moves and the direction of the force it exerts.

Always the same, I assume?

If it's the spring doing the work then the force and the displacement have the same direction. If you are doing work on the spring you'll find they have different directions.

I feel like I have a mental block so I really want to go back to the original question I formulated. Let's ignore how the spring got to its initial state. Let's say the spring is doing the work. If we are only concerned with the work involved from going from its initial state to its final state is there a difference in the work involved if it starts out extended 5cm and end up extended 3cm and if it starts out extended 5cm and ends up compressed 3 cm?

Based on how I understand the math, my answer currently would be "yes", but this seem unintuitive so I feel I might be wrong.

And by "yes", I mean the work is the same.

Silly me.

Suppose we fix one end of the spring to the floor and pull or push the other end. Then we take upwards to be the positive direction. I'm doing it this way so we have a well defined sign convention to decide what the signs of forces and displacements are. The wy I have described it upwards is positive and downwards is negative.

OK so far?

Yes.

Suppose the spring starts out extended by 5cm, then it is pulling downwards so the force the spring exerts is negative. Yes?

Yes.

If we now let the spring shrink back to 3cm extension then the displacement is also -ve because the length of the spring is decreasing. Yes?

Yes.

So the force and the displacement are both negative so F.ds is positive. That means ∫F.ds is positive so the spring does work.

We'd have to do an integral to find the actual work done, or we could could take a short cut and just use the change in the energy stored in the spring.

Anyhow that's the situation when the spring starts out extended by 5cm and ends up extended by 3cm.

So now we need to consider starting out extended by 5cm and ending up compressed by 3cm. Yes?

Yes.

Suppose we do this in two stages. We start extended by 5cm and let the spring shrink to extended by 3cm. So the work for this stage is the same as above. yes?

Yes.

@KavinIshwaran a couple of mins ...

Yes :-)

@AngelPray Now we let the spring shrink back to an extension of zero, and the spring does work as it shrinks back to zero.

Yes.

But then to compress the spring by 3cm we need to do work on the spring. Yes?

This time the spring is not doing work, it's having work done on it.

Can't springs idk, bounce back a bit?

We are assuming we are doing this slowly so the spring stays in equilibrium the whole time.

Ok.

If we now compress the spring by 3cm then the work we have to put into the spring is the same as the work we got out of the spring as is shrunk from 3cm extension to zero.

So the work we did on the spring is the same as the work the spring did on us.

i.e. the net work in going from 3cm extended to 3cm compressed is zero.

Does this make sense so far?

Yes.

So in fact the total work we got out of the spring starting from 5cm extended is the same whether we end up at 3cm extended or 3cm compressed.

That's because the total work going from 3cm extended to 3cm compressed is zero!

Yes?

That makes sense, but still screws with my head.

@AngelPray It takes practice to get used to this. The point is that you need to be clear about the difference between the spring doing work, and the spring having work done on it.

I need to get back to @KavinIshwaran s question now.

@KavinIshwaran Hi

No problem, thank you very much.

You're welcome :-)

@JohnRennie Hi !

So I3 = -2

This is enough to solve this question. But simply I substituted it to see what I get for I1 and I2 but they turn to be negative too.

Total potential drop through resistors = emf of cell right ?

ABD = 70 and ACD = 70

ABCD = 70 ?

So 5I₁ + 10(I₁ - I₃) = 70

Ah, OK, yes from A to B then C then D has to be 70V

Yes That is how I got the equation for the first loop

OK, in that case yes I agree.

Then you did A to C to B to D?

Yes

and then ACD

ABD?

Oops, Yes :-)

I wouldn't have done it that way, but it all seems correct.

@JohnRennie How you would have done it ?

Oh

That just seems simpler.

But it should end up the same.

Yes :-)

Are you getting I₁ and I₂ coming out negative?

Yes

I think that's because the voltage is dropping over the resistors i.e. ΔV = -70V

I don't think it matters. Your working is fine.

Ok :-)

@JohnRennie Thank you for the clarification :-)

Will see you tomorrow !

Bye :-)