Hi sir! @JohnRennie

I assumed they asked work done on the block

I am getting wrong answer if i do it by W = F.ds

I am getting the correct answer if i do it by change in total energy = work done by the force

@PinkAura What is the answer given ?

@KavinIshwaran -800

@PinkAura I guess the point of the question is that in the observer's frame the ramp is moving left at 20 m/s so the point of application of the normal force is moving left.

yes

@JohnRennie i found the normal reaction and the displacement relative to observer in the direction of normal reaction and multiplied the two, answer im getting is -800/(root 2)

Give me a few minutes to make a coffee and I'll have a think about it.

ok sir

@PinkAura Hmm. I am missing a factor of root2

What I'd do is work out the power, where power = force . velocity

Then the work done is ∫power dt

@PinkAura Does this sound reasonable?

@JohnRennie yes sir

i did the same , i multiplied the total time that is 2 root 2

since power is constant?

Yes, so we get W = 400√2

yes sir

but the answer is -800

you would get this if you find change in total energy

in the frame of observer

initially the box was having 1/2 *2 * (20)^2 =KE and final KE =0 , similarly change in G.PE=-400

Hmm, the final KE is zero only after the box has turned the corner at the bottom of the ramp. It is not zero just before the box has turned the corner.

yes sir

why the answers aren't matching?

I think it's the difference between just before the end of the ramp and just after the end of the ramp.

In the lab frame the speed just before and just after is the same, but in the observer's frame it is not.

are we getting the same answer if we do it by energy method? just before the end of the ramp

In the lab frame the velocity just before the end of the ramp is 20m/s at an angle of 45°

yes sir

So in the observer's frame the velocity is the vector sum of 20 m/s left + 20 m/s at 45°

yes

root ( 800 + 400 root 2)

@PinkAura v² = 800 - 400√2

minus not plus

ohh sorry

yes sir

So the change in KE is 400 - (800 - 400√2) and that's not 800J

-400 + 400√2 + change in PE

wait shouldnt it be final -initial?

OK, yes, so we get the same as our power calculation.

yes sir but what is it that is doing work in turning it

That makes sense since our power calculation was only up to the end of the ramp.

hmm

So it's the change of direction at the end of the ramp that makes up the difference between our answer and the key.

yes

can we calculate that?

The problem is that the behaviour at the end of the ramp is undefined as if it's an instant change we get an infinite force acting over zero distance.

We'd have to assume some gradual change e.g. an arc of a circle, but that would get long and messy to calculate.

i see sir, but just qualitatively define that force, isn't that force centripetal force which is doing work

In the lab frame it's a centripetal force so it does no work, and that's correct because in the lab frame the KE doesn't change.

But in the observer's frame the force wouldn't be centripetal and would do work.

But doing the calculation in the observer's frame seems hard to me.

why wouldn't it be centripetal? can't we still call it centripetal ? i believe the forces are frame independent

@JohnRennie i agree

yes i do give up on calculating the force, i was just concerned about the name of that force which is doing that work (is it the normal force as the question says or the centripetal force)

Well gravity doesn't do any work because the height doesn't change, and since the ramp is frictionless there is no component of force parallel to the ramp so there is only a normal force.

So it must be the normal force doing the work.

hmm

sir just one more thing

if we consider the ramp + block as a system

the work done by normal force is zero

so does the static friction which is holding the ramp in place doing that work

observer's frame

Good point! Yes, in the observer's frame the ramp is moving so the frictional force holding it in place does work.

Yes, that's a nice way to look at it.

okay, thank you sir! :)

You're welcome :-)

@JohnRennie sir is current frame independent? I think it is, bcoz the heat releases is i^2Rdt and that would be frame independent so current too has to move. And another way to look at it would be that current is movement of electron with respect to the lattice/protons and as relative velocity is frame independent so current too has to be frame independent. Are these reasonings current?

This was the question that made me think about the frame of reference of an electric current

***so current too has to be frame independent ~~has to move~~

@Swan Current depends on the speed of the electrons relative to the observer, but heat released depends on the speed of the electrons relative to the wire. So current and heat do not transform in the same way.

@JohnRennie I don't get how current depends on the speed of electrons relative to the observer. Is the u_d

**u_d i.e. the drift velocity in I=neAu_d wrt an observer?

If that is the case then why doesn't the current change in the given question. Does the question only works bc the person is moving backwards with drift velocity itself?

In the motional emf derivation I can see how u_d is not wrt the lattice and is infact wrt a frame of our choice. I will have to think more about this

@JohnRennie sir don't waste your time on my doubt now. I will think through them and when I'm able to articulate them better, I'll ask them to you. :)