# Practice English Speaking&Listening with: Module -6 Lecture -5 WORK AND ENERGY - IV

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In the previous lecture on force and energy, we saw how a force is obtained from

corresponding potential energy by taking its gradient. In this lecture we explode this

little further and then go on to discuss non conservative

forces as a real life example of calculating the force from a potential. Let

us now look at the electric field and the potential or equivalently the potential energy

and the force on a charge an a in an electric dipole.

.

So, I am going to consider a dipole sitting at the origin with charge with minus q on

negative y x axis and charge plus q on the positive side of the x axis with the distance

between them being 2 a. I want to find the electric field at a general

point r form the origin, I will do, so by taking

superposition of fields due to charge minus q charge plus q adding and then taking the

limit a going to 0. So, that limit is taken in such a manner that p equals 2 q a remains

unchanged that defines for me point dipole setting at the origin. So, I could find the

field at r by superposition of field due to minus

q at r plus field due to plus q at r and get my

.answer in an easier way. This will tell you how talking about potential helps at times

could be that I find the potential energy of a unit charge at point r and take its gradient.

Negative of that will give me force on that unit charge which by definition is the electric

field at r, let us do this as an exercise. .

So, recap I have dipole sitting here I want to find the field at distance r which is x

from the origin and y this way. Finding potential

in this case is easier because I am not doing a

vector sum, therefore potential energy for unit charge q equals 1 at r. The r is given

in 2 dimensional spaces this x and y point is going

to be potential due to the first charge plus the potential due to the second charge. It

is going to be k q over square root of x minus a

square plus y square minus due to the other charge this one, k q over square root of x

plus a square plus y square. Here, I just want to remained to this distance between

the charges is 2 a and k is one over four pi epsilon

of this. Since a is very small, so far away from the

dipole I can write 1 over square root of x minus a square plus y square as 1 over square

root of x square plus y square minus 2 ax plus a square. I am going to neglect this

terms because finally I am going to take the limit

going to 0 and keep only q times a term, so I neglect that.

..

Making the figure again, I am trying to find the potential due to a dipole at a far away

point which is x from the origin this way y this way this is q minus q. I made the

approximation that 1 over the square root of x minus a square plus y square is equal

to one over the square root of x square approximately

plus y square minus 2 a x which I can write as 1 over square root of r square minus

2 ax. This is approximately equal to one over r one plus ax over r square, similarly

I can write 1 over square root of x plus a square plus y square approximately as 1 over

r 1 minus ax over r square. .

.Combining the two, I get the potential for this dipole separated by 2 a minus q plus

q as U at r. Here, r is this point distance y distance

x from the origin as k q over r 1 plus ax over r square I am working under those approximation,

where a is very small minus k q over r 1 minus ax over r square. That gives

me U r to be 2 k q ax over r curbed, now when I take the limit a going to 0 such that

2 q a remains constant equal to the dipole moment that I want to have. So, then in that

limit becomes 2 q a gives me the dipole moment k p x over r curbed, you can check

that the dimensions are alright p is q times the distance this distance divided by this

distance q. So, this is the potential, now I want

to find the electric field due to this dipole at point r.

.

So, I am finding, so there is a dipole is here of the magnitude p pointing in x direction.

So, let me now make this dipole I am making slightly big is in x direction its magnitude

is p. I am trying to find the electric field at this point r distance away which is x this

way and y this way, I found that U of r or a single

charge. So, this is a potential is equal to k p

x over r cubed, so I can find the x and y components of the force of the electric field

because electric field is same as force on a unit charge is equal to F x is equal to

minus d U d x which is equal to minus d over d x.

I am taking a partial derivative, so when I do that y remains a constant of k p x over

x square plus y square raise to three halves.

When I do that, I find I differentiate this, first I

will get three over 2, this minus sign goes with that k p x over x square plus y square

.raise to 5 halves times 2 x minus k p I differentiate x, so I get divided by x square plus y

square raise to 3 halves, this two cancels. .

So, the expression for the electric field x component that I get is going to be x which

I took to be the partial derivative of U with

respect to x comes out to be 3 k p x square divided by r raise to 5 minus k p over r cubed.

.

Let me see if it is correct going back to the previous slide this 3 k p x square over

r raise to 5 minus k p over r cubed.

..

Similarly, I can find E y which is minus partial of U with respect to y and this comes out

to be minus partial of U with respect of y k p x over x square plus y square raise to

3 halves and that gives you three k p x y over

r raise to 5. These are the two components with electric field you can check for yourself

that this is consistent with the general expression of the electric field which is

given as 1 over r cubed rather k over r cubed p

dot unit vector in r direction times 3 r minus p.

Getting this directly by superimposing the fields due to 2 charges would be slightly

more difficult. You see that taking gradient of

the potential which is slightly easier to find is an

effective way of finding the field. Having talked about all these quantities, you may

be now wondering what happens to the minimum

maximum of the potential in three dimensional cases.

..

In one d cases, I knew that the force was 0 when d U d x was 0 if d 2 U dx square was

greater than 0, then I had a minimum of the potential. Similarly, d 2 U dx square less

than 0 gave me maximum of the potential and we have talked in the

past this gave me an equilibrium point

or I should say a stable equilibrium point. If I moved a body from this

point it will come back and this gave me an unstable equilibrium point, the body had

force 0, but if I moved it away from there, we will just take off it will not come back

what about in 3 d cases? .

.In 3 d cases, I am talking about a three dimensional force which is the gradient of

U off course if gradient of U is 0 it implies that

the force is also going to be 0. As far as the

minimum and maximum and such things are concerned, these are slightly more

complicated in 3 d cases. For example, although a point would be where the gradient is

0, it does not mean whether that it will be a maximum or a minimum for sure in three

dimensions. For example, if I come along the x line suppose at the origin the gradient

is 0 if I come along the x line x axis U could

go through a maximum. On the other hand, you could have a minimum

along the y direction, so to find the minimum maximum in 3 d, case one really has

as to see for example, for a minimum whether in all directions that point the U

is minimum. For a maximum, whether coming from all directions that point is a maximum

there could be more complicated cases, I just discussed coming in one direction would be

a maximum and the other could be minimum the which is known as a settle point.

You will learn about these things more in your electrodynamics course I thought I would

just give you a gist of it. .

Now, let us talk about an issue that we have been avoiding. So, far conservative forces

and non conservative forces how to deal with non conservative forces this you already

dealt with. To work with non conservative forces, we said earlier I should not be able

to define a potential energy since the work depends

on the path one simplest example in one dimension is frictional force.

.More you move more work you do you come back to the same point the work done is

not 0 and therefore you cannot define a potential energy for a frictional force in one

dimension how about in three dimensional case. Let us take to 2 examples and learn

through examples, let us take 2 forces F equals x I plus y j and try to construct potential

energies for these and this equals minus y I plus x j.

.

So, I am taking 2 forces let me just draw a line here F equals to x I plus y j and F

equals minus y I plus x j in this case you see that

the force lines are going radially outward. If

we do not see directly write F as x is r cosine theta y is r sin theta, so I can write this

as cosine theta I plus sin theta j and by now

having learnt polar coordinate this is nothing but r in the radial direction. On the other

hand, in this case the force line is going to be in

theta direction at all points, so they are going to look like this.

Again, if you do not go see them clearly write this in polar coordinates and you will see

that this is nothing but r minus sin theta I plus cosine of theta j which is nothing

but r in theta direction. So, further out you go the

force magnitude increases with respect to r and

this direction is in theta direction, let us try to find the potential energy for the

2 cases.

..

Again, let me draw a line F equals to x I plus y j which is minus partial of U with

respect to x I minus partial of U with respect to

y j and this implies partial of U with respect to x

is equal to minus x or U could be equal to minus x square plus some function of y.

Therefore, d U d y is equal to d f d y is equal to minus y and this after integration

implies that U is equal to minus x square minus y

square plus some constant of integration C. Let

us try to do the something for F equals minus y I plus x j, so partial of U with respect

to x is equal to minus y.

I can therefore write U as minus y x plus some function of y and partial of U with respect

to y therefore, is going to be equal to minus x plus some d f d y, which is a strictly a

function of y, but, d U d y cannot be minus x because d U d y is given to be plus x.

Therefore, U cannot be defined, this is a conservative force this is a non conservative

force, let me give the same take the arguments further.

..

So, in the case of F equals x I plus y j I found that U is equal to minus x square minus

y square plus C. I can take as a reference point

U at 0, 0 to be 0, then the C becomes 0 and the constant and the surfaces are like this

U becomes more and more negative as you go further and further out. Therefore, the force

is in this direction is v you can already it is

radially out, on the other hand when I take F to be minus y I plus x j I cannot really

find a function that will give me gradient of U to

be equal to minus F. Another way of looking at it is you see from d U dx equals minus

y I find U should be of the form of minus x y.

On the other hand, if I look at d U d y to be equal to x I find U to be of the form of

x y both cannot be true, so this is a non conservative

force I cannot really get any potential for this. Now, the question we asked do I

always have to try to find the potential energy and if I cannot find it, then I say it is

non conservative or is there any easier way to look

at it. If I all the time I have to integrate the force it becomes a difficult task at it,

so happens that they happens to be a differential

way of doing the same thing and now I like to talk about that it also gives me the definition

of a quantity called curve.

..

So, go back to the definition restricting myself to 2 dimensions and generalization

into three dimension is quite easy I go from point

one to point 2 and if I find that the work done is independent of which two paths I choose

is going to be a conservative force. If it is not independent or it depends on the path,

which I choose, then it is going to be non conservative force also put another way. If

I go from point one, go around this path and come back to the same point if the work done

is 0, then it is a conservative force if it is

non 0, then it is not a conservative force. So, let us try to do that let me go from point

x y to point 2 which is x plus delta x y plus delta y and without any loss of generality,

let me take this path to b 1 a 1 be let me take

this path to be 2 a 2 b. So, work along path 1 a since I am moving along the x axis, this

is path one a is going to be F x at x y times

delta x plus some correction because F x itself might change I am I am going to go all the

way to second order. So, some correction let us call it delta F as F x changes along the

x axis delta x square and some constant, this is

going to cancel. So, I am just putting some constant k k could to be one half or

something. As I move along path one b that is from here

to here only thing that is changes y, so this is going to be F of y at x plus delta x plus

y has not change plus some correction, let us

call it k one k 2 as I move along y F y may change delta y square. I am going to do this,

there should be a delta y here because I am let has a distance, so let me expand this.

..

So, I get delta w along one a is equal to F x x x y delta x plus some constant k 1 partial

of x delta x square is going to be F x at oh

sorry F y at x plus delta x y delta y plus some k 2

d F y d y delta y square. If I expand this further I get F y at x and y delta y plus

delta F partial of y component of F with respect to

x changes. Since I moved by delta x delta y

by Taylor series expansion plus a second order term k 2 y square, so I am keeping all the

terms to second order and there could be some higher term. Let me remind you again I

am calculating this work in going from point 1 to point 2 along path 1 a and 1 b. I could

do similar excise while going from path 2 a in 2 b liking go this you can take similar

steps and I leave this as an exercise.

..

You would find that the work done when I go from along this path comes out be along 2

a comes out be F y at x y delta y plus you can write k 2 delta F y del y del y square

and along 2 b this is 2 a this is 2 b. It comes

out be F x at x y delta x plus partial of F x with

respect of y delta x delta y plus k 1 partial of x with respect to x delta x square plus

higher order terms. So, let me flash the previous expressions again this is delta W 1 a one

b expanded it looks like this. This flash back and you see when I did the work along

path one I got these contributions and I worked

along path 2 I got these contributions comparing you will see that this term appears

in this case. Similarly, this term and this term, so I will

remove this I will focus only on this this term

and this term appears in this case also. This term appears and this term appears only

terms that I am left with are this and this, so if I were to take the difference moving

along path one a and one and path 2.

..

I will find that work done along path 1 minus work done along path 2, which is delta W

1 a plus delta W 1 b minus delta W 2 a minus delta W 2 b and let me flash those things

back again. This was delta W 1 a this delta W one b these terms canceled, similarly in

this case this is delta W 2 a delta W 2 b the only terms which were left were these

this also canceled. Therefore, I get this to be

equal to partial of F y with respect to x minus

partial of F x with respect to y delta x delta y. If this were to be 0 that is if force is

conservative, then this term must be 0 this implies partial of F y with respect to x minus

partial of F x with respect to y must be 0 otherwise not.

.

.So, I get delta W 1 minus delta W 2 is equal to partial of Fy with respect to x minus

partial of Fx with respect to y delta x delta y. This delta W 1 minus delta W 2 is nothing

but this is delta W 1 is the work done in moving from this point to this point and minus

delta W 1 is the work done and moving back along path 2. So, this is a work done in

moving the entire closed path and if this is non zero, then the force is non conservative.

I can give similar arguments for other dimensions

movement along the plane of y z or movement along the plane of x z.

I find in general that if delta F y over delta x minus delta F x over delta y or delta F

z over delta y partial y minus partial y over

partial z. Let me write it here partial F x with

respect to z minus partial of F z with respect to x if they are any one of these is non zero

you have a non conservative force. If these are all 0, you have a conservative force is

there a compact way of writing this the answer is yes.

.

These are all components of a vector form by this which is known as curl of F and you

can see this I j k this is by definition F x F y F z determinant, this gives me all

three components that I wrote in the previous slide.

..

This is a z component this is the x component this is the y component of that curl that

I wrote here.

.

If any of these components is non zero, force is non conservative, if they are all 0 the

force is conservative. So, I found given a force I do not really have to calculate the

potential energy I can straight away take its curl if it is non zero non conservative

force if it is 0 conservative force. This also tells

you that by the way that curl of a gradient must

be 0 because if I define a gradient; that means, I am taking about a conservative force.

.What does this curl mean, can I get more feeling for it, the answer is yes, as a name

suggest curl is something that describes a curly sort of thing.

.

Let us again go back to our two examples F equals x I plus y j and F equals minus y I

plus x j and take the curl, curl is going to be I j k partial with respect to x partial

with respect to y partial with respect to z of

x y and z and you can see this 0.This is 0 here, on

the other hand if I look at the curl of the other example j k partial of x partial of

y partial of z this is minus y x 0. You will see that

it gives k component or z component which is

equal to 2, so this is non zero, let me again make how these force will looking.

..

So, F equals x I plus y j the force was looking something like this and in this case F

equals minus y I plus x j the force field was looking something like

this in this case curl of F is equal to 0. In this curl of F is not

equal to 0, so this quantity curl seems to give

you a non zero value when things seem to be moving around in circles or they are curling

they turning around. On the other hand, if they do not they go in straight line curl

is 0 you can see in this case very easily if I

take a part and then go around in a circle work

done is going to be non zero. So, it is a non conservative force, this is a conservative

force to get more feeling for this let us look at some more examples.

.

.Let us take a force field F I just take make pictures which is something like this as you

go out it increases as you go on the negative side it again increases. Let us take another

one as you go up it increases as you go down it increases when the same direction. You

can see if I take a particle around here like this I do positive work when I go this way,

no work same positive work and come back here

I have gathered some work. Work done in going around in a closed path is non 0 that

means, curl of F is not equal to 0. Similarly, here if I go around the origin

in symmetric way you will see I pick up some positive work 0 work negative work equal to

this positive work and 0 again. So, around the origin the work done seems to be 0, but

if I take a closed path here, I do some work here 0 work less work because of force is

become less and come back here. So, I picked up some work in going around a circle, so

on this point curl of F is not equal to 0 around

the origin near the origin curl of F is equal to 0. Again, you see just looking at the

pictures the curl seems to give you a feeling for this moving feeling for a rotation on

this point this seems to be the lines are becoming

smaller and smaller. So, this sort of curling around they are curling

around here, they are not curling around they are changing the same way on both sides

and this is really the meaning of curl just to quantify this. Let us say this could be

something like F is equal to as you increase y F

magnitude increases in x direction. So, this could be a function that is described by this

sort of geometric shape, in this case as you no matter which way you go on the y

direction F could be y square i, so let us see if really our feeling of about curl is

correct or not.

..

If I make it again in this case I had force field like this and I wrote F equals to y

I, then I also had F equals y square I and the force

filed looks something like. If I calculate the

curl is I j k partial with respect to x partial with respect to y partial with respect to

z this is y this is 0, 0 and you will see that it

has a k component which is non 0, this gives you

minus 1 k, this is 1. So, this is non 0 all the time, on the other hand in this case if

I take y square, so I j k partial with respect to x

partial y partial z y square 0 0 it gives me a curl

which is equal to 2 y k. I am calculating curl of F I am calculating

curl of F and this is nonzero for nonzero y,

but, 0 at y equals 0. So, our feeling that at the origin curl is 0, so this gives you

curl of F is equal to 0 for y equal to 0, but non 0

at any other y for y not equal to 0 curl of F is not

equal to 0.

..

Let me make it clean once more F equals y I curl of F as we can calculated

previously is not equal to 0 F equal to y square I curl

of F is equal to 0 for y equal to 0 not 0 for y not

equal to 0. So, you can see that the curl seems to give you a feeling here again the

some sort of rotation involved here, there is no

rotation involved it gives you a feeling for a

rotational kind of thing, these quantities curl and so on. What introduced initially

in the studies of fluids that is where the origin

lies and you can get a feeling again a physical sort of feeling if you think in those terms

think of any given force field. .

.For example, the one that we took this or this or this or the radial one and imagine

these lines to be representing velocity of a fluid

flowing and think take a small match stick put

it somewhere. So, I take a match stick put it somewhere here if you feel that the match

stick rotate then curl is not 0 if you feel match stick would not rotate curl is 0. So,

for example, in this case no matter where you

put the match stick it will tend to rotate if this

is the velocity of the fluid, curl is non 0 anywhere in this case. No matter where I

put the velocity is becoming larger and larger as

you go outside the match stick would tend to

rotate curl is non 0. In this case, if I put the match stick at

the centre it is being pushed by the same force on

the same velocity on both the sides it will not rotate curl is 0, but if I put it somewhere

here, it will tend to rotate like this curl is non 0 in this case. No matter where I put

it will not rotate curl is non 0, so this sort of

physical way of looking at the quantity curl having

learnt about curl and getting some feel for it. Let us talk about little more, so that

you get a better feeling for it, I will first discuss

a fantastic theorem about curl that really arises

from the way we derived it. .

So, this is known as Stokes theorem and what it states is that if I take a closed path

and given a force or a vector function F which

I multiply over dl where dl is a line element path and integrate over this closed path.

.For example, I could be calculating the force in sorry this should be the other way force

in going around this close path from this point and coming back to the same point. So,

I am going this way and coming back to this

point and I want to calculate the work done of course is the force is conservative that

means, a curl is 0 the work done is 0. If it is not

conservative curl is non 0 and this integral is given as the curl of the force field is

perpendicular component to this the this surface on which the curve is being made times

the area d a. Let me elaborate a little bit I am taking

a close path, so I am calculating F dot dl along

this path this is related to the curl. So, I can think of a small area da on the surface

find the curl there multiply with a component of

curl perpendicular to x surface with that delta area small area and added up. It is

a very useful theorem and you can see right away how curl becomes very important quantity

for us. The proof is also very nice and I would like to go where it, so that you sort

of get a nice feeling about curl. .

For proving, I will go back to the way we obtain the quantity curl where we obtained

was it was by taking a small rectangular path

going around it starting from x y going to x plus

delta x y going up to x plus delta x y plus delta y coming back to x y plus delta y. Then,

getting back to where we restarted from, in that case we found when I go counter

clockwise then the work done F dot dl along this path close path was equal to delta F

y delta x minus delta F a partial F x partial

y times delta x delta y.

.This is nothing but the perpendicular component of F on the x y surface times the small

area delta a sort of Stokess theorem and a miniature and infinitesimal version. Now,

would I can do is take a big path and divide it into this small infinitesimal areas and

for each area I can go around like this for this

area I will go like this. For area I go this I will

go like this and in doing, so finally, when I add

F dot dl around each area you will see that opposite going paths cancel each other.

Finally, the only place where they would not cancel is this outside line, so finally, if

I add all these F dot dls, I am going to get F dot

dl around this path, let me call this C one to distinguish it from this one.

For each path, I know it is delta F y over delta x minus l partial F x over partial y

or perpendicular to of the curl times delta.

So, this is going to be a sum over curl of F

perpendicular times delta x or delta x delta y or delta a, which is nothing but I can write

this in integral forms it is sum over small infinitesimal areas F delta x. You have to

be careful in these because we are talking about

directional quantities in that you follow the

convention when I go counter clockwise and then the thumb. If I curl my fingers around

the counter clockwise direction, the thumb gives me the direction along which I should

take the component of the curl of F. .

So, let me rewrite this what I did was I took a close path divided this into small grids

or infinitesimal areas I took this F dot dl around

each path and added it all up when I added it all up all these counter going paths canceled

the contribution of F dot dl. Finally, I was

.left with the only this, but the area integral that is, so for each small one I had F dot

dl is small one let me say it C i when I summed

it over it gave me F dot dl around this path C

1. This is C i and when I summed over perpendicular component of F with the area it

gave me integral curl of F perpendicular delta a.

These two are equal and that is a Stokes, here again I remind you be careful about the

direction we are talking about directional quantities. I am going to follow right hand

rule when a fingers curl right hand figures curl

thumb gives me the direction along which I should take the component of the curl.

.

As I told you earlier, this Stokess theorem F dot dl is curl of F times delta a integral

over that surface really comes from the way we

defined initially the curl quantity. So, it is a

very useful quantity when I want very useful theorem when I want to see curl in different

situations or curl in different coordinate systems. As an example, let me take now what

will be the expression for curl in my polar coordinates, remember in the beginning of

the previous lecture. When I was talking about

gradients I had warned you that when I write gradient as I d by d x, let me just confined

myself to two dimensions. It does not mean, it does not imply that in

polar coordinates it is going to be r d partial with respect to r plus theta partial with

respect to theta. In fact, this is not even dimensionally correct, same thing is true

for all other quantities and as an example I will

.just derive curl for you using this definition or this theorem stokes theorem which really

initially introduced in infinitesimal way the quantity curl.

.

So, if I am talking about polar coordinates I am going to choose my path in polar

coordinates as this following right hand rule I go counter clockwise. So, for the curl the

component will be coming out of this screen, this point is r theta this point is r plus

delta r theta this point is r plus delta r theta

plus delta theta and this point is r theta plus delta

theta. I will go quickly over this if I want to calculate F dot dl over this path is going

to be Fr at r delta r that will be the contribution

from this plus F theta at r. However, as I go out theta component changes

it changes by partial of F theta by partial of r delta r and the distance I cover is this

one and this distance is r plus delta r times delta theta, now I am coming back. So, this

becomes minus F r, however, this is Fr at theta plus delta theta, so this is going to

be minus Fr at r theta. I should also have theta

here minus how the radial component of F changes with respect to theta delta theta and

the distance I cover. Let me make it plus put the brackets here delta r and finally,

when I come along this path is going to be minus

F theta at r theta delta theta times r. You can check for yourself these are the only

linear terms the other terms, if I really look

at the other variations of these quantities they will give the second order terms. So,

those higher order terms third order terms in delta

theta and delta r, so I am not worrying about those. Now, you can see this will cancel with

this and similarly, F theta r delta theta will

.cancel with this one part of this the other part I will get F theta at r theta delta r

delta theta. So, let me just write it here then

I will take it over to the next screen. I am going to get this equal to F theta delta

r delta theta the other component F theta r

delta theta cancels with this. Then, I am going to get plus partial of F theta with

respect to r delta r, r delta theta minus this term

will be left which is partial of F r with respect to

theta delta theta delta r. .

So, let me rewrite it what I have determined is F dl and this comes out to be I went along

this path, let me copy this comes out to be F theta delta r delta theta plus del partial

theta partial r r delta r delta theta minus partial

Fr with respect to theta delta r delta theta. This

is F dot dl by definition or by Stokes theorem this must be equal to curl of F coming out

of the screen this is x this is y. So, it will be z component times the area which is

going to are delta theta times delta r and that gives

me the z component of curl of F and z direction in terms of polar coordinates.

This comes out to be one over r F theta plus partial F theta over partial r minus 1 over

r partial F theta with respect to a partial

Fr with respect to theta, which in short I can write

as one over r partial r r F theta minus partial Fr over partial theta. So, you see I have

been able to derive using the basic definition

the curl in different coordinate system and use

the basic theorem of this.

.This is just to give you a feeling for these quantities you will obviously, be using the

advanced version of these or using it more and more in your coming courses in

electrodynamics advance mechanics and so on. So, you should practice a lot of problems

on this play around try to get may be the other components of curl in cylindrical

coordinates or in spherical coordinates. To conclude, I will give you potential energy

in three dimensional how potential energy and

force are related. .

How we can think of a conservative and non conservative force and how we can test

them using differential form of differential way of taking curl and this sort of some of

our introduction and some details of work energy potential energy and so on.

.

The Description of Module -6 Lecture -5 WORK AND ENERGY - IV