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Failure of wrok energy theorem

Failure of wrok energy theorem

by Vinay Arya -
Number of replies: 8

Suppose a disc of radius a placed on the ground with uniform surface charge density s with its axis vertical.A charged particle is dropped from a height H from the ground along the axis of the disc.Initially the particl is at rest so initial kinetic energy is equal to zero.Then see the point when the electric force on the particle is equal to the gravitational force.At that point velocity of the particle is zero.Hence the change in kinetic energy is equal to zero.So the work done by the net external force should be zero.But the resultant force is causing a net displacement in the downward direction.So there has to be some work done by the net external force.Please explain this.

In reply to Vinay Arya

Re: Failure of wrok energy theorem

by Braj Arora -

The pt. where electrical force on the particle is equal to the gravitational force, is not the pt. of zero velocity due to the momentum of the particle.

In reply to Braj Arora

Re: Failure of wrok energy theorem

by Vinay Arya -

Can you tell me which momentum is there at that point?

In reply to Vinay Arya

Re: Failure of wrok energy theorem

by Braj Arora -

Due to the motion momentum is not 0. Think about a pendulum performing SHM. At the mean position the force is 0 but due to the momentum the bob does not stop.

In reply to Braj Arora

Re: Failure of wrok energy theorem

by Vinay Arya -

In an SHM the particle goes to the extreme positions,but here,could  you tell me which extreme positions are there?If you think that the force is zero but the velocity is maximum then it means that the particle has is at its mean position,then it means that the particle will go to the other extreme position after crossing that point.But as it will cross the point it will experience a force in the upward direction so it will move in the upward direction.Can you now explain it me this?

In reply to Vinay Arya

Re: Failure of work energy theorem

by Vinay Arya -

Let me give you the proof that velocity at the final point is zero.

What is the net force on the particle?

F=mg-qE(x)

Where E(x) is the electric field due to the disc.

From Newton's Second Lwa of motion

ma=mg-qE(x)

=>a=g-qE(x)/m

From first equation of motion

v=u+at

u=0

v=(g-qE(x)/m)...........(1)

At the equilibrium point

mg-qE(x)=0

=>mg=qE(x)

=>g=qE(x)/m............(2)

From equation (1) and (2)

v=0

Is this proof enough?

In reply to Vinay Arya

Re: Failure of work energy theorem

by Namitha Kumar -

a isn't constant here as E is depending on x. That means you can't use the formula v=u+at.

In reply to Namitha Kumar

Re: Failure of work energy theorem

by Vinay Arya -

So sorry.I did not see that.But would you like to prove that v=some quantity.

In reply to Vinay Arya

Re: Failure of work energy theorem

by Vinay Arya -

From Newton second law

ma=mg-qE(x)

a=g-qE(x)/m

vdv/dx=g-qE(x)/m

v2/2=gx-q(int.)E(x)dx/m

At equilibrium

g=qE(0)/m

v2/2=qx(E0)/m-q(int.)E(x)dx/m

yes.Velocity is not equal to zero.Hence work-energy theorem is still valid.