Failure of work energy theorem

again if you come up with some arguement , then this topic should be renamed to " failure of tiitians to make vinay understand "

haha abhi maza ane wala hai kyuki i m going to prove you wrong ....hahaha bohot maja ara hai type karne me

tujhe samjhate samjhate hum sabh apne concepts bhul jayenge :

your first mistake : acceleration isnt constant so v is not equal to acceleration into t

a = - vdvdx = g - E(x)/m

so vdv = gdx - E(x)dx / m

so v²/2= gx - ∫ E(x)dx / m

pehle ye E(x) nikal ke integrate karna padega

tune pata hai kya kiya hai ....

a = dvdt = g - E(x)/m

v = ∫dv = gt - ∫ E(x) dt m

tune E(x) ko integral se bahar nikal liye , that means tere hisab se x is independent of t ,

lekin tune v as a function of t nikala to x = ∫vdt , bhi as a function of t milega, so
isnt there a contradiction ????

ya fir main jo abh keh raha hu ye bhi galat hai ? aisa hai to main jee ka padai chod dunga

vinay have you even written even a single equation before asking such questions ?? you are saying that force is always downwards, then how the hell is the particle getting decelerated ??????????????????????????????????????????????????????
your arguement itselfis wrong

suppose electric field due to disc is E(x) ,it is obviously going to be function of x i.e distance of particle from discs centre

so net force = mg - qE(x) in downward direction

now vinay you are telling mg is always greater than qE(X) , first prove this then we will talk further

When you add all those works you will get W(gravitational force) + W(electrostatic force) which you already proved 0 from work energy theorem.

Ok. Let the particle covers a displacement dx for that W₁=fdx.Then for next dx we have W₂=f'dx.And keep on doing like this then Net W=W₁+W₂ +...Now how can you prove W to be equal to zero.

.But see from W=∫F.dx where F is net force.I want to add that we can keep net force to be constant for a small displacement dx.

Yes....it is constant for a small displacement dx....what next?

EmlnEm
I never said that force is constant.You can read my third post in which I have mentioned that force is not constant.And I mean constant for a small distance dx.ok.And you are saying that W is somewhere negative and somewhere positve.The resultant force is in the downward direction and the displacement is also in the downward direction.So it is always positive and is never negative.So all the work dones when added cannot be equal to zero.So there has to be some net work done.I think you might have understood by now.
And now what about your biggest nonsense.It is as follows:
You:dont think that if there is displacement work is done,because work will be done only if you have gained or lost energy and not if you have gained or lost displacement.

now dont tell that W=fd and W=0 then F=0 and d=0,this point of you is biggest nonsense,you should review basic calculus if you are going to post this point again.
Me:Work done is defined as
W=F.dx where F is the force and dx is the small displacement.(if you do not believe it then see HC verma textbook:work,energy and power)
If there is displacement then there is some work done.
The charged particle was at rest initially then it moved.
From Newton's first law motion:If a body is at rest it cannot move until or unless some external force is applied to it.So force on the particle cannot be equal to zero.And hence the work done cannot be equal to zero.So your statement is wrong.And your last statement is abject.Because you need to go to 10th or 11th class.If you think that you are the truth then tell me the proof of it being absolutely false.Also tell me if I need to revise my calculus again.If you all think that my all the statements are wrong can you prove tht the work energy theorem still works with complete proof.?That is my real and short and easy question.

the force is not of same magnitude for all dxthat you are adding up

F.dx=Fdxcostheta , also for few dx costheta is negative and for some it is positive

dont think that if there is displacement means work is done , because work will b edone only if you have gained or lost energy and not if you have gained or lost displacement

now dont tell that W=fd , and W=0 then F=0 and d= 0 , this point of your is nonsense, you should revise basic calculus if you are going to post this point again

I already agree with this.But see from W=∫F.dx where F is net force.I want to add that we can keep net force to be constant for a small displacement dx.

By Net work done I meant work done by external gravitational force + work done by external electrostatic force. ( which follows from W.E.T.)
The net work done by these forces is 0 after the whole process.

Do you agree with this?

@vinay:how much time did you take typing the posts?
real manuscripts these are,,,,you are more patient than rahul dravid

Vinay,
You need to understand the difference between net work done and work done at any instant.
Here the Net Work Done is 0. This net work done gives you the FINAL configuration of the system. The net Force in the final configuration is infact 0.

During a part of the displacement , the work done at that instant by the external forces is not 0....because the particle has some KE at that instant. At that instant, there is some net external force....which gave the particle that K.E.

f.d = work done ye kaise a gaya ??

W = ∫F.dx

now if W = 0 will u say F= 0 or dx = 0 ?

hey dude when the gravitational force is equal to the electrostatic force the velocity may not be zero it depends on charge,mass and both the fields.

work energy theorem will be valid.

And also that I am not taking velocity to be zero at any point.I am saying as it is.

Oh my dear friends please read the question again and please understand it with open mind.
Dear Daredev I am not talking about any potential energy I am talking about the change in kinetic energy.Ok.I think you might have understood by now.I cannot say anymore as I have said everything.Please read the problem correctly again.
Dear account blocked please read the problem again.You have jus written what I said into mathematical form.Let me explain this what you have said.The work done by the electric force is equal to the negative of the work done by the gravitational force.So the net work done is equal to zero.But as the there is net displacement there should be net work done.I have said this thing already in the question.Please read the question again.

hey de wrk dn by gravity is stored as electrostatic potential energy

Swaordfish the two results are differing.I am taking f to be net force.This is fifth time.
EmlnEm when the particle falls from top to bottom then shouldnt mg be more than qE(x)
ajoy if you want i can type in short but the thing is big.Please do not worry about my time.

to fir tere logic se when i throw a ball upwards mg is upwards

are bhai thoda thanda dimaaag se equations likh ke soch ..... did you try to find out E(x) ??

the particle when falling from top to bottom is also losing its KE which it had gained ,

so you can say that jab tak KE bad raha tha F.dx was +ve and the moment he was losing its KE F.dx was negative

are bhai soch na agar net force hamesha niche rahega to particle rukega kyu ???

EmlnEm
I have never seen such a weak student.As the particle is moving down should the net force be down or up?F=mg-qE(x)
F should be positive if it is moving down.
F>0
mg-E(x)>0
mg>qE(x)
hahahaha
What can be more big proof?
One more thing I should say for such a weak student thatE(x) increases as the particle moves down.Ok my Lord.

ohh I am weak you are strong !!! ok then please help me in this question !!

i am throwing a ball vertically up in the air , the ball is moving up so the net force on it is upwards na ? LOL

hahahahahhahahahha

kya logic hai yaar tera

take my case again , but see it upside down abhi bol

and one more thing you yourself said that "E(x) increases as the particle moves down."

so isnt that quantity " mg - qE(x)" decreasing as the particle comes near the disc ???? do
you know that direction of qE(x) is upwards ?? and during its motion qE(x) becomes greater
than mg and net force becomes upwards

dekh yaar tujhe explain karke fayda nahi hai ek baar hcv vagera ache se solve kar le ,
its pure timewaste to explain you
i again request you to please putforward your arguement only if you have some substantial mathematics backing it up , because mujhe to lag raha hai ki tu do din se hawe me hi baat kiye ja raha hai