Failure of work energy theorem

But why did you not tell me it before.

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

account blocked
W=∫F.dx
I have written there that it is the net external force.And please tell me if you can that work energy theorem is still working.
As we know that W=0 then either F=0 or d=0.But d is not equal to zero and F is also not equal to zero.Can you explain me now?
swordfish
Net work done is not equal to zero.As you can see there is vertical displacement not the circular one.If there is no net work done we have final configuration equal to the initial configuration.So you are absolutely wrong.The net force in the final configuration is zero that is also I have said two times and this the third time.
Now your last paragraph
During a part of the displacement,the work done at that instant by the external force is not zero................because the particle has some KE at that instant.At that instant,there is some net external force........which gave the particle KE.
Now my view on this statement is as follows:
Yes you are right.But as you can see you could not explain the thing which I asked earlier.Is there any statment which tells us that work energy theorem is working here still?As you say that work is done then where does it go according to work energy theorem.Consider a small displacement dx in that the work done isdW.Now keep on adding such works.You will get that it is not equal to zero.Can you show me if according to work energy theorem the net work done is not equal to zero?No body has explained it yet.As you can see here.I am not saying false.

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

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.

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,
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 ?

Dear Anirudh
The charges and masses are such that the particle just reaches the disc.When it is at the point just near the disc then it has to be in equilibrium.Because as the particle was falling initially then its accelration increased due to gravity but as there is electrostatic force of repulsion the charged particle will experience a force opposite to the motion of the charged particle.And as the given condition is that it stops at the point near the disc so the electrostatic force of repulsion is more than the gravitational force.So the charged particle will experience an acceleration in the upward direction.But as the velocity is in the downward direction.We have v=u-at taking downward to be positive and upward to be negative.Let me right it again at the bottom
v=u-at
The initial speed of the particle is zero as it is dropped.So u=0 then we have
v=-at
The electrostatic force of repulsion is given by the equation
F=KQ₁Q₂/r²
As the particle is falling in the downward direction r has to reduce.So F has to increase.As we know from Newton's second law
F=ma
If F increases then a increases.So v goes more negative.So the total v will go on reducing.And at a point in the space it will reduce to zero.Now consider the final and the initial point.We have both the velocities equal.So the change in the kinetic energy is equal to zero.
From wrok energy theorem we have
W=K₂-K₁
So the net work done is equal to zero.
Now consider this equation W=f.d
Consider f to be the net force.
0=f.d
d is not equal to zero.
so f=0
But as the particle is moving in the downward direction there has to be some net force.So f should not be equal to zero.Hence here the work energy theorem gets violted.And I am also saying that see the point when the particle is equilibrium.Now I hope that you can understand it better.And I advise you to read the question again.

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.

how does net force =0 imply velocity = 0 ?????????????????

btw if u r talking abt the situation when vel = 0 ,

then W_gravity + W_{electric force} = ΔK = 0

and W_gravity = - ΔU_{gravitational}

W_{electric force}= - ΔU_electrical

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 ???

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