5. Work-Energy Theorem and Law of Conservation of Energy

Uploaded by YaleCourses on 22.09.2008

I was going to do a brand new topic but I didn't want to go
without making sure everyone understands the loop-the-loop
problem. Because that's one problem
where if you didn't learn Newton's laws and apply them
properly, you just couldn't figure it out,
okay? All the other things we do,
you know, blocks sliding down planes, friction,
pulleys, you have an intuitive feeling
for that, but this one really is very non-intuitive right?
Here is a trolley at the top of the track.
The forces on the trolley, both gravity and the force of
the track, are both pushing down.
And the question is, "Why isn't that falling down?"
Everything is pushing down. Why isn't it falling down?
That's the mystery, so what's the answer you'd give
to somebody who says, "Why isn't it falling down?"
Yes? Student: [inaudible]
Professor Ramamurti Shankar: Yes,
so they might say, "Well, if you say it's falling
down, how come it doesn't end up here to the center of the

Yes, somebody have an answer to that, yes?
Student: [inaudible] Professor Ramamurti
Shankar: Pardon me? Student: [inaudible]
Professor Ramamurti Shankar: No,
there are no forces in the horizontal direction;
the forces are all--yes? Student: [inaudible]
Professor Ramamurti Shankar: It's not
negligible. There is an acceleration that's
producing a change in velocity. But when we say an apple has a
downward acceleration, you let it go and it falls
down; that's what we think is
downward acceleration. Here is something which is
downward acceleration, but it's not falling down.
Yes? Student: Force changes
the direction of velocity vectors over time,
acceleration times time, it just changes the direction
of the velocity vector. Professor Ramamurti
Shankar: Ok. Yes, I'll take one more answer,
yes? Student: The downward
component and the horizontal component of the velocities are
independent of each other so at the same time that the cart can
be falling down it can be moving so much in one direction
[inaudible] Professor Ramamurti
Shankar: Alright. So, let me summarize all the
different answers. First of all,
one thing you should remember, there are no horizontal forces
here. The forces are what I've drawn;
that's why you should be very, very careful.
Just because something has a tendency to move to the left
doesn't mean there's a force to the left.
That's the whole point of Newton's law,
force is connected to acceleration not to velocity.
So yes, this will have an acceleration,
you cannot avoid it. F = ma will tell you
T + mg is ma, and the a that you want
in this problem, if it's moving in a circle,
is mv^(2)/R. So, it is accelerating,
and the difference between this one accelerating,
let us say--If I go to the loop-the-loop and I drop an
apple, forget the track it's got
mg acting, it's accelerating too.
What's the difference between these two?
Here, we start with zero velocity;
you accelerate downwards. That means you pick up a
downward velocity. As it accelerates you keep
falling down and picking up more and more speed.
The same acceleration's experienced by the cart also,
and a little more because T is also pushing down.
But there the change in velocity over a tiny amount of
time is not added to zero velocity, but to this large
velocity it has at that instant. And the tiny increment in
velocity if added this way will produce a velocity in that
direction, and what that really means if
you wait a small amount of time, this thing is going along the
new velocity direction. So, that's the difference
between--So, the acceleration is the same for a falling apple as
for the loop-the-loop cart. But in the apple's case,
it is added to zero initial velocity so the acceleration and
the velocity are the same, namely down.
If you have a huge velocity to which you add a tiny change in
velocity, the downward direction you rotate the velocity vector
that just means that as it goes around the circle,
the velocity vector is constantly tangent to the
circle. There's another way in which
people understand this. If you are on the Earth and you
are standing, say, on a building and you fire
a gun, it goes like that and it hits
the ground because of acceleration due to gravity.
If you fire it a little faster it'll go there.
There'll be a certain speed at which it keeps falling but
doesn't get any closer to the center.
That is in fact how you launch a bullet into orbit.
So, what's the first thing you should do when you fire this

Move away because it's going to come back and get you from
behind, okay [students laughing].
Alright that's a very useful lesson.
So, you might think, "Hey I shot it in that
direction, it's not my problem." Like everything else it'll come
back and get you when you least expect it.
This is how you have--;So, is this thing falling or not?
It is constantly falling but the Earth is falling under it so
it's endless search to come to the Earth.
Whereas if you drop the bullet, it'll just fall and it'll hit
the ground; that's what we always think as
drop. Okay, so this is something I
did in a rush, and I didn't want that because
this is really the part of physics which is not intuitive,
and I think we all know what the main point is.
The main point is velocity is a vector and its value can change
due to change in direction, not change in magnitude.
Yes sir? Student: That negative
or that downward change in velocity, what is that equal to?
Professor Ramamurti Shankar: Okay,
if you wait a certain time Δt,
then the acceleration is v^(2)/r,
put the v, put the r,
get some number, multiply by the Δt,
1/10^(th) of a second. That'll be the change in that
time. So, in that time also the guy
would have moved over in the circle a little bit,
just precisely so that at that point your new velocity vector
is tangent to your new location. So this little Δv I
drew is not a fixed number; it depends on how long you want
to wait. Calculus says wait an
infinitesimal amount of time, in practice wait 1/10^(th) of a
second and it'll be fine. Okay, today there is a very,
very important concept I want to introduce.
It's a very robust and powerful concept;
it has to do with energy.

It is powerful because after the laws of quantum mechanics
were discovered in the sub atomic world,
we basically gave up on many cherished notions.
You know, you must have heard that particles cannot have a
definite position and definite velocity at a given time.
They don't have any trajectories.
You see a particle here now, then you see it there.
That's true because they're observed events.
What happens in between? You might think they have a
trajectory but they don't. The forces are not well defined
because if you don't know where the guy is, you don't know what
the velocity is, you don't know what the
acceleration is. So most of the ideas of
Newtonian mechanics are basically surrendered,
but the notion of energy turns out to be very robust and
survives all the quantum revolution.
In fact, there is a period when people are studying nuclear
reactions and the energy you began with didn't seem to be the
energy you ended up with. So something was missing.
So Neils Bohr, father of the atom said,
"Maybe the Law of Conservation of Energy is not valid."
Then Pauli said, "I think I put my money on the
Law of Conservation of Energy. I would postulate that some
other tiny particles that you guys cannot see,
that's carrying away all the missing energy."
That was a very radical thing to do in those days when people
did not lightly postulate new particles.
Nowadays, if you don't postulate a particle you don't
get your PhD in particle physics [students laugh].
In those days it was very radical, and the particle was
not seen for many, many years.
It was called a neutrino. Nowadays, neutrinos are one of
the most exciting things one could study.
There's a lot of mysteries in the universe connected with
neutrinos. Okay, so energy is a concept
you're going to learn. Here's again a place where I
ask you, don't memorize the formulas and start plugging them
in. Try to follow the logic by
which you are driven to this notion, because then it'll be
less luggage for you to carry in your head.
And I'll try to make the notion as natural as possible rather
than saying, "This is my definition of energy,
this is the definition of kinetic energy."
Let's ask where does all that come from.
How many people have seen kinetic and potential energies
in your past? Okay, if you haven't seen it
that's fine, this is not something you're supposed to
know. If you know it and you learned
it properly it'll be helpful, but you don't have know that.
You just have to know what we have done so far.
So let's go back and ask a simple question.
When a force acts on a body, what is its main effect?
What does it do? Yes?
Student: It changes its motion.
Professor Ramamurti Shankar: Be more precise by
it when you say "changes its motion."
What does it mean to change its motion?
Yes? Student: It accelerates
the object [inaudible] Professor Ramamurti
Shankar: Right, that's from Newton's law;
that means it changes the velocity.
So let's for a moment, in fact for the entire lecture
today, go back to one dimension. Then, we will do the whole
business in higher dimensions. In one dimension changing the
velocity simply means speeding up.
So bodies are going to speed up and forces act on them.
That we know. We're going to find the
relation between how much speed is accumulated when a force acts
on a body; we're going to derive that.
So, let us take an example where the force is in the
x direction and it is some constant.
"Constant" meaning it's not varying with time;
somebody just has to just apply 10 Newtons constantly to a body.
What does that do? That produces an acceleration
which is F/m, right, that's Newton's law.
That's a constant acceleration. So, let's go back to lecture
number one. Lecture number one we learned
that if a body has constant acceleration,
then the velocity at the end is the velocity squared at the
beginning plus 2 times a times the distance traveled.
That's what we learned long back, this is just kinematics.
In those days we didn't ask, "Why is it having a constant
acceleration?" You were just told,
"It's a constant acceleration, just go ahead."
Now that we learned dynamics, we know acceleration has a
cause, namely a certain force. So, I'm going to write here as
you can imagine F/m in the place of a,
but I'm going to make one more cosmetic change in notation.
In all these energy formulas, rather than calling the initial
velocity or initial anything with subscript zero and the
final is no subscript, we are going to use a notation
where the final quantities are denoted with a subscript 2,
initial quantities are denoted by a subscript 1.
In other words, initially, the body had a
velocity of v_1,
and then it had an acceleration a for a distance
d, then the relation between final
initial velocities is this. So let's put that F/m
here and let's put a d.

Since we are interested in the effect of the force,
we want to see what the force did to the velocity.
What you find here is that if you want to find the change in
the velocity, of course you should take this
guy to the left-hand side, right?
Take it to the other side and you want to isolate the force,
so bring m/2 to the left-hand side.
So, do that in your head and what do you get?
m/2V2^(2) -
m/2v1^(2) is F
times d.

This is a very important concept.
It says, when the force acts on a body it changes the velocity,
and it depends on how far the force has been acting,
how many meters you've been pushing the object.
It's clear that if you didn't push it at all,
then even the forces acting, velocity hasn't had time to
change. So, it seems to depend on how
far the force acted, and the change is not simply in
velocity but in velocity squared.
That's what comes out; then, you realize that it's the
most natural thing in the world to give this combination a name
because that's what comes out of this.
That combination is called kinetic energy,
and I'm going to call it K_2,
meaning K at the end minus K at the beginning
is F times d. So F times d is
denoted by the symbol W, and that's called the work done
by F. That's how we introduce a
notion of a force times distance as the definition of work.
This is how people--you can say, "Why did anyone think of
taking the combination F times d?"
Well, this is how it happens. So, you need to have a unit for
this, which is actually Newton meters, but we are tired of
calling everything a Newton meter,
so we're going to call it a joule.
So joule or J, is just a shorthand for Newton

Alright, so let us now write down this following expression
here. ΔK, Δ in anything we
do is the change in any quantity, it's called delta of
that quantity, ΔK is a change in
K is equal to the F times the distance
traveled. Let the distance traveled be
some tiny amount Δx. This formula is valid even if
d is 1km, as long as force is constant.
But I'm saying, let's focus on a problem,
whether the force is constant or not, over a tiny interval in
which the body moves the distance Δx.
Then, we find that the change in kinetic energy is force times
Δx. Now, what if there are 36
forces acting on the body? Which one should I use?
I'm pulling and you're pushing and….
It's got to be the net force because Newton's law connects
the net force to the acceleration.
If you and I have a tug of war and we cancel each other out to
zero then there's no acceleration.
So, you've got to go back to the definition;
this is very, very important. This F better be the
total force. I should really use a subscript
T for total, but I don't want to do that and
we write F = ma; it is understood F is
everything. We'll come back later to the
notion of what happens if F is made up of two
parts. So, maybe I'll just give you a
simple example. Here's a body,
right, and I'm pushing it this way and you're pushing it that
way, let's say with equal force. And the body could be at rest
or if the body had initial velocity it'll maintain the
initial velocity because velocity is for free.
Then, you can ask, "What does this theorem say?"
This is called a Work Energy Theorem.
The Work Energy Theorem says, "The change in energy is equal
to the work done by all the forces."
In this case, there is no change in energy,
but there are two forces at work that cancel.
Then, it turns out that it is sensible to define the work done
by me, which is equal to F times the distance
traveled, and the work done by you will
be in fact given a minus sign. So, when do we attach a plus
sign and when do we attach a minus sign?
If you go back to the whole derivation, it was understood
that a was a positive quantity.
Then everything works. But if the body is moving to
the right, and I'm pushing to the right, then the work done by
me is positive. And if you were pushing to the
left and the body still moved to the right, the work done by you
is negative. In other words,
if you get your way, namely, things move the way
you're pushing, the work done by you is
positive. If people are pushing you
counter to your will, in the opposite direction to
your force, work done by you is negative.
See, here is a simple example. I have a piece of chalk and I'm
lifting it at constant velocity from the ground.
Its kinetic energy is not changing.
That means the total work done on it is zero,
but it's not because there are no forces on it.
There is gravity acting down and there is me countering
gravity with exactly the right amount I'm applying the force
mg. So, work done by me would be
positive because I want it to go up, and it goes up.
So, if it goes up by an amount h work done by me is
mg times h. Work done by gravity in the
meantime is minus mg times h, so the total
work done is zero. So, this formula says this is
the total force, but I'm telling you if you take
the work done and call it ΔW it maybe a result of
many works, work done by me,
work done you, work done by somebody else,
some are positive, some are negative,
you add them all up. So, that is the notion of work
done by one of many forces acting on a body.
And the total work done is the algebraic sum of the work done
by all the forces. Alright, so we've got this
result; let's do the following.
Let us imagine all of this occurs in a time Δt.
Then ΔK/Δt, which will eventually become
dK/dt if you take all the proper limits becomes F
times dx/dt, which is force times velocity
and that's called power. So, power is the rate at which
work is done. For example,
if I climb a 12-story building I've done some work;
my m times g times the height of the
building. I can climb the building in one
minute, I can climb the building in one hour,
the work done is the same, but the power is a measure of
how rapidly work is done. That's why it's the product of
force and velocity. So, the units for power are
joules per second; now, that also has a new name,
which is watts or just W.
You may use a kilowatt, which is kW,
little k then W is thousand watts.
So, if you've got a 60-watt bulb, it's consuming energy at
the rate of 60 joules per second.
That's the meaning of a 60-watt bulb.
Okay, now I'm going to do the next generalization,
which is when the force is not a constant but varies with

So, do we know any example of a force that varies with location?
Student: A spring. Professor Ramamurti
Shankar: A spring is one example.
Even gravity I think we know is not a secret,
the force of gravity is mg near the Earth,
but if you go sufficiently far you will notice gravity itself
is getting weaker. It'll still look like mg
but g won't be 9.8, g will become 9.6 and
9.2 and 9.1, it'll decrease. But the variation of gravity in
terrestrial experiments is pretty small so we won't worry
about it. But the force of a spring
definitely varies with x in a noticeable way.
The force of a spring is one concrete example;
it's what you all seem to recognize;
it's this one [(-kx)]. So then, we have to ask what is
the Work Energy Theorem when the forces itself vary?
So, let's draw ourselves a force which is varying,
here is x, here is my force;
it does something. F = kx or –kx
is a particular graph, I'm just taking any function of
x that I want. Now, the acceleration is not
constant because the force is not a constant.
So, if you wait for one hour the change in kinetic energy is
not simply force times distance because what force are you going
to use? It was something now,
it is something later, it is positive,
now it's negative later. So, we cannot apply this
formula because F is not a constant.
So, the usual trick in calculus is find me an interval in
x, which is so narrow, and I'll make it as narrow as
you insist, so that during that period I'll think that F
is a constant and the constant value of F is just the
function F at that time x,
at that time where the location is x.
So, for that tiny interval I can still say that F is
constant; F over m is some
constant over the tiny interval, then the change in kinetic
energy will be then F(x) times Δx.
Then, you wait a little longer, you go to a new place,
then you've got a new value of the force, maybe somewhere here.
Then, that times Δx is the work done in that region.
But geometrically, F(x) times Δx is
the area of the little rectangle whose base is Δx and
whose height is the function F at that time.
So, if you eventually went from here to here [distance on
graph], let me call that point as x_1 and let
me call the ending point as x_2,
then the work done by you is really given by the area under
that graph. And in every segment you pick
up the change in kinetic energy, you add it all up.
When we say add it all up on the left-hand side,
all the ΔKs will add up to give me the final minus
initial kinetic energy. This one is what is written in
the notation of calculus as F(x)dx from
x_1 to x_2.
This is called the integral of the function F(x).
Now again, in a course at this level I'm assuming that you guys
know what an integral is. Even if you've never heard of
an integral, if I give you a function and you never heard of
calculus, you can still deal with this problem.
Because you'll come to me and say, "Give me your function."
I'm going to plot it in some kind of graph paper with a grid
on it, and I'm just going to count the number of tiny squares
enclosed here. That's the area and that's the
change in kinetic energy. So, integration is finding the
area under graphs bounded by the function of the top,
x axis below, and two vertical lines at the
starting and ending points.

Now, a little digression. Even though you're supposed to
know this, I want to make sure--part of my "No Child Left
Behind" program, that you all follow this.
If you never heard calculus or integral calculus this is going
to be a two-minute, three-minute introduction.
There is a great secret, which is that if you give me a
function and you tell me to find the area under this from
x_1 to x_2,
I can go to the graph paper and draw that, or there is quite
often a trick. You don't have to draw
anything, you know. And I will tell you a little
more about what it is. If you give me a function
F(x), the area under this graph happens to be the
following. There is another function
G(x) that will be specified in a moment.
And you just have to find G at the upper limit
minus G at the lower limit, and that will be the
area. And who is this mystery
function G? It is any function whose
derivative is the given function F.

So, the whole business of finding area which was solved
around Newton's time, is if you have a function they
want the area under it, you go and think of a function
whose derivative is the given function.
So, it is the opposite of taking derivatives.
Differentiation is an algorithmic process;
you take F(x), you add a Δx to the
x, you find the change in F divide by Δx.
Given any function you can do that.
This is the opposite; this is guessing the function
G whose derivative is F.
So, if I say F(x) equals x^(3),
then G(x) is that function whose derivative is
x^(3). So, I know I've got to start
with x to the 4^(th) because when I take derivatives
it'll become an x^(3), but with the 4 in it.
So, I fight that by putting a 4 downstairs and that's the
function G. Yes?
Student: Shouldn't there be a c?
Professor Ramamurti Shankar: Yes,
that is the plus c. The c is,
of course, velocity of light [students laugh].
Now, why is this plus c there?
She's quite right, because if you've got a
function whose derivative is something, adding a constant
doesn't change the derivative. Then you can say,
"Well, we are in trouble now because you told me to find the
function G, and you're telling me the world
cannot agree on what G is because I've got one G,
you've got another one with a different constant."
But the beauty is that when you take this difference,
I may not put the constant, and you may put the constant,
but it's going to cancel out. So, most of the time,
people don't bother with the constant.
Sometimes it's very important to keep the constant in place --
it plays a special role -- but we don't want that here.
We'll just call this the integral.

So, maybe it's again worth asking for a second,
why is this true, you know.
Why is G the function whose derivative is F?
Why is that the answer to the problem?
Well, one way to think about it, this is really more for your
completeness than anything else. Suppose there is a function
F, and I'm finding the area, let me find it up to the
point that I'm going to call x_2.
For the starting point let me just pick zero.

This is the area I'm looking for, and I write it as
F(x)dx from zero to the point x_2.
You agree that as I slide x_2 back and
forth, the area is varying with x_2,
so it's legitimate to call that a function of

A function is anything that depends on something else,
and we all agree that if the point actually moves,
the area moves. This is my shorthand for the
area. The starting point is nailed at
zero; this is G(x_2).
The left asks what is G(x_2+Δ).
That is the area to this point, that we denoted as
F(x)dx from zero to x_2 + Δ,
that's G(x_2+Δ).
So, what is the difference between the two?
The difference between the two is G(x_2 + Δ) -
G(x_2). If you look at that graph,
the extra area that I have is this thing, and that is clearly
equal to F at the point x_2 times
Δx. In other words,
by how much is the area changing when I change the upper
limit? Well, it's changing by the
height of the function at the very last moment times the
change in x. Well, we know what to do now,
divide by Δx and take the limit, and you can see that
F(x_2) is dG/dx at the point
x_2. But the point
x_2 is whatever point you want.
So, we drop the label, and we just say that F
as a function of x is dG/dx.
This is the origin of the result that the area can be
found if you can find the inverse derivative of the
function. So, let's keep a couple of
popular examples in mind. One, if F(x) equals the
constant, then the G(x) is equal to constant times
x. If F(x) = x then,
G(x) is equal to x^(2) over 2.
As you know, the integral of x^(n) is x^(n)
+1 over n + 1. There are integrals for all
kinds of functions, and you check them by taking
derivatives and make sure you get back your F.
Alright, suppose this is what we learned from the math guys,
then go back to the Work Energy Theorem and see what it says,
K_2-K _1.
Oh, by the way, I didn't complete one story;
let me finish that. This was, if
G(x_2) was the area from zero to the point
x_2, G(x_1) is the
area from zero to the point x_1.
That is perhaps clear to you, if I only wanted the area
between the point x_1 and the
point x_2, I can take the area all the way
up to x_2, subtract from it the area all
the way up to x_1,
and that's of course the claim that this integral F(x)dx
from x_1 to x_2 is
G(x_2) - G(x_1).

That's the reason the integral is given by a difference of this
G. So, let me use a shorthand
notation and call G(x_2) as
simply G_2, G(x_1) is
G_1. You should understand,
G_2 means G evaluated at the final
point, G_1 to G evaluated initial
point. So, you have got
K_2 - K_1 is
G_2 - G_1,
but G is some definite function of x.
You've got to understand that. For example,
it could be x_2^(2) over 2
- x_1^(2) over 2 in the problem where F
is equal to x. Then, let's rearrange this
thing to read K_2 - G_2 is equal to
K_1 - G_1.

So, the Work Energy Theorem says that, "If a force is acting
on a body, a variable force associated with the force is the
function G, a variable function of
x, whose derivative is the given force."
Then, at the beginning, if you take the difference of
kinetic energy minus the value of G_2,
it'll be the same. I'm sorry, K_1 -
G_1, it'll be the same as
K_2 - G_2.
Now, we don't like the form in which this is written.
We have to make a little cosmetic change,
and the cosmetic change is to introduce the function
U(x), which is minus this function G(x).
So, once you do that, you've got to remember now the
force is equal to minus the derivative of this function,
because of the minus sign relating G to U.
Then, we can write it as K_2 + U_2
= K_1 + U_1.
And that combination, if you call that as
E_2 and you call this combination as
E_1, we are saying,
E_1 = E_2,
and that's called the Conservation of Energy.

So, what does conservation mean? Conservation of Energy in
physics has a totally different meaning from conservation of
energy in daily life. Here it means,
when a body's moving under the effect of this force
F(x), you agree it's gaining speed or
losing speed or doing all kinds of things.
So, as it moves around it's speeding up and slowing down,
so its speed is definitely a variable,
but a certain magic combination connected to the velocity of the
object and to its location, U depends on x,
that does not change with time. And it's very useful to know
that does not change with time, because if you knew the value
of E, even if E were to equal
some fixed number E at one time you know E at
all of the times. So, let's take a simple example.
So, we take a rock and we drop it.
We know it's picking up speed; we know it's losing height.
So, you may think maybe there is some combination of height
and speed, which does not change in this exchange,
and we find the combination by this rule now.
In the case of gravity, F = -mg;
therefore, U = mg times y, because
–du/dy will now be my F.
Then, we may conclude that ½mv^(2) + mgy at
the initial time will be equal to ½mv_2^(2) +
mgy_2 at the final time.

And that is the--And the whole thing can be evaluated any time.
You take any random time you want.
The sum of these two numbers doesn't change.
And that's a very, very powerful result.
So, I'm going to give you one more example later when you can
see the power of this. But in the case of gravity,
this is what it means. So, this agrees with the
intuitive notion that when you lose height you gain speed,
when you lower your y you gain speed,
but it's much more precise than that.
Instead of saying vaguely, "you lose and gain,"
½mv^(2) is the part connected with the speed and
mgy is the part connected with the height and that sum
does not change with time. So, let's go back to the mass
and spring system. In the case of the mass and
spring system, if you have a mass connected to
a spring, then we say here
½mv^(2) + ½kx^(2) equal to
E is a constant. That means, find it at any
initial time one, find it at any final time two,
add these two numbers, the total will not change.

So, let me put that to work. Let me use that in a
calculation. So, I say here is a spring;
here is where it likes to be. I'm going to pull it to a new
location a. x = 0 is where the
spring likes to be. I'm going to pull it by an
amount A and I'm going to let it go.
And I want to know how fast the guy is moving when it comes back
to, say, x = 0.

If you go back to Newton's laws, it's a pretty complicated
problem. You have got to think about why
it's a complicated problem. Because you start with a mass
at rest, if you pull it by an amount A,
a force acting on it, it would be –k times
A; that will produce a tiny
acceleration, will produce an acceleration,
right, kA/m, which will give it a new
velocity by the time it comes here.
But once it comes to the new location, a little later,
a different force is acting on it.
Because the x is now different, so the acceleration
during this interval is different,
and the gain and the velocity of that interval will be
different from that in the first interval.
You've got to add all these changes to find the velocity
here. That's a difficult proposition,
but with the Law of Conservation of Energy you'll do
that in one second. In fact, let's do it not just
to be at the mid point here, but at any point x
measured from equilibrium. Because we will say initially
½mv_1^(2) + ½kx_1² =
½m v_2^(2) + ½kx_2².
This is the truth; this is a constant that does
not change with time. Now, to get some mileage out of
this, you've got to know what the constant value is.
Well, we know what it is in the beginning because when you pull
this guy, at the instant we knew it had no velocity,
and we knew it was sitting at the point A,
so we know ½k A_1² is the total
energy of this mass and spring system.
That's not going to change with time.
We like to say that at the initial time all the energy is
potential energy; there is no kinetic energy
because there is no motion. At any subsequent time,
this number has to be equal to what you get.
Now, let me drop the subscript to there [pointing to equation].
Let it stand for some generic instant in the future.
This much we know. Now, do you see that we can
solve for the velocity on any location x?
You pick the x; if you want x to be zero
that's very easy, you kill that term,
you balance these two, you cancel the half and you can
see that's a formula for v^(2).
v^(2) is going to be ka^(2) over m;
that is going to be the velocity when it swings by this
midpoint. But you can pick any x
you like, if the x was not zero but .1;
well, put the .1, take it to the other side and
solve for v. This is one example of the Law
of Conservation of Energy. So, the power of the Law of
Conservation of Energy is if you knew the initial energy,
you don't have to go through Newton's laws,
and you'll find it in all the problem sets and many exams.
When a problem is given to you and you are told to find the
velocity of something, you should try whenever
possible to use the Law of Conservation of Energy rather
than going to Newton's laws. So, if you want to know what
have we done, how come finding the velocity
seemed so difficult when I described it to you earlier,
and would we have done it, the answer is,
if you tell me the force of the spring is a function of
x, I have done that integral in
the Work Energy Theorem, once and for all.
The integral of the spring force has been integrated once
and for all and that's what the potential energy is doing.
The potential energy comes from integrating that force.
Alright, so now lets take another problem where there are
two forces acting. Yes?
Student: Could you explain really quick how you got
the ½ in front of the kx^(2) for the energy?
Professor Ramamurti Shankar: Let me see,
here? Student: No,
down at the bottom of the center, yeah.
Professor Ramamurti Shankar: Here?
Student: Is it just because of the integration?
Professor Ramamurti Shankar: Yeah,
for a spring I think we said here,
if the force is x, the potential energy is G =
x^(2)/2, U will be
-x^(2)/2. So, the force is –kx,
then you have to find the U so that –du/dx
= –kx and the answer to that is U is
kx^(2)/2. So, if you knew the force,
you just find the function U by asking,
"Whose derivative gives me F with a minus sign?"
Then you are done. That's the potential energy for
that problem. Now, lets take another problem,
a mass is hanging from the ceiling.
Then, it's got two kinds of potential energy,
one is just a gravitational potential energy,
and the other is a spring potential energy,
and that's going to come because K_2 -
K_1 will be the integral of F(x)dx.
I'm sorry, let's call it dy and F(y)dy from
y_1 to y_2.
And this F(y) will have two parts, the force of gravity
-- I'm tired of writing the limits -- then,
the force of the spring, from 1 to 2 [adding the limits
to the integrals]. Right, there are two forces
acting on it. This one will be
mgy_2 - mgy_1,
this one will be ½ky_2^(2) -
½ky_1^(2). So, every force acting on the
body will turn into a potential energy, because the
force--Newton's law says Work Energy Theorem applies to the
total force. So, the total force is two
parts, and you can do two integrals;
each integral is an upper limit minus lower limit of some
function and this is what you get [pointing to formula].
Oh, I think I made one mistake here.

I think this should give me, yes, this overall sign is wrong
in these equations. Really, there's a minus in that
and minus in that. In other words,
the work done by the force, if there's only one force
acting, is really U(1) - (U2).
You see why I got the sign wrong?
When you do the integral in calculus, integral of F
is G_2 - G_1,
but G and U are minuses of each other.
So, integral of force will give U initially -U
finally. So, it should really be +
mgy_1 - mgy_2.
So, the Law of Conservation of Energy for this object will be
that, ½mv^(2) + mgy + ½ky^(2) is the total
energy and will not change. So, this mass can bump up and
down and go back and forth. If you knew the total of these
three numbers at one instant, you know the value at any other
instant. For example,
if you pulled it from its equilibrium by .2 centimeters
and you released it, and you say what's the speed
when it's .1 centimeters away, it's very easy because you know
the initial energy was all spring energy and gravitational
energy. At any later time if you knew
the y, if you knew the position,
you know the potential energy; that's the whole point.
Potential energy depends on where you are;
kinetic energy depends on how fast you're moving.
And some combination of where you are and how fast you're
moving is invariant, does not change with time.
So, if you tell me where you are, I will tell you how fast
you're moving. By the way, remember the
kinetic energy enters as v^(2).
You can imagine, when you solve for v
you're always going to get two answers.
Why is that? Does it make any sense?
Yes? Student: Because you're
going to have the same potential energy [Inaudible]
problem if you're oscillating [inaudible]
Professor Ramamurti Shankar: Or,
if a spring is going back and forth at a given point x,
you can have two velocities because you could be going away
from the center, or it can be going back towards
the center. So, this tells you the velocity
on those two will be related by a minus sign,
right? If you pull the spring this
way, when you go through the origin you say,
"What's my velocity?" I'll get two answers because I
could be going to the origin or I could go the other way and on
the way back again pass the origin.
So, every x you'll get two velocities.
And that agrees well with the fact that v^(2) is what
you will solve for. And because it's quadratic you
will get two answers and you should get two answers;
you should be expecting that. Okay, now, we are going to find
one bad boy, bad girl, bad person, which is going to
ruin this whole thing. There is--you know what I'm
talking about here, yes?
Student: Friction [inaudible]
Professor Ramamurti Shankar: So,
friction is a force acting on a body, so what's the thing I
should try to do? Let's take a body with only
friction acting on it. Let's not worry about anything
else, or maybe friction and a spring.
So, I try as usual, to get a Law of Conservation of
Energy when friction is acting, and you will see that I will
not succeed. And it looks like everything
here is water-tight, right?
Everything looks good. You tell me the force acting on
a body. I integrate the force from
start to finish and I call that the difference of potential
energies; then I've got my result.
Try to follow this logic: K_2 -
K_1 is integral to force,
integral to G(x_2) - G(x_1) or which
is U(x_1) - U(x_2);
you can rearrange it to get K_2 +
U_2 = K_1 +
U_1. Looks like you can always get
Law of Conservation of Energy. And you want to know why that
fails when there is friction. Okay, why does it fail when
there is friction? So, when there is friction--Let
me try to do it again. K_2 -
K_1 is the force due to the spring,
which is whatever –kx. Let me do that integral plus
the force of friction dx, from x_1 to
x_2. This one is from
x_1 to x_2 and we saw
what this is. Let me write it one more time,
½kx_1^(2) makes
½kx_2^(2); if you didn't have friction you
are done. Because you can get kinetic
plus potential for spring, is kinetic plus potential.

Why don't we just integrate the force of friction and then turn
that into a U and then we're done?
What's going to be the problem? Let me just give some--anybody
else want--yes? Student: It's always
against the direction of motion so it would be always changing
direction force [inaudible] Professor Ramamurti
Shankar: Okay what was your answer?
Student: You can't associate a potential function
with friction because it depends on the path,
not only the initial position and final position.
The anti-derivative only considers the final and the
initial position. Professor Ramamurti
Shankar: Right, but let's suppose I'm somewhere
and I'm pulling a block and there is some friction,
there is some force F at the location,
there's my F. Why don't I just use the value
of F? Yes?
Student: Energy is lost due to heat.
Professor Ramamurti Shankar: I know but see,
you are saying something else which is speaking,
I had into the book, but right now what's the
problem with integrating the force of friction?
Student: I was just going to say the energy of the
force of friction is no longer within the system [inaudible]
Professor Ramamurti Shankar: No,
but you're all telling me why there is a loss of mechanical
energy but I'm just a mathematician.
I'm saying, give me a force, find its integral.
What's the problem? Yes.
Student: Is it because the function to the force of
friction depends on the velocity?
Professor Ramamurti Shankar: Yes,
I think somebody else said that earlier also.
The force of friction is not a function of x.
You might say, "What do you mean?"
I'm pushing this guy, I know how hard it's pushing me
back, but I'm telling you now, "Push it the opposite way!"
Then, you'll find the force of friction at the very same
location is pointing in a different direction whereas the
force of a spring is the force of the spring.
It's -kx, whether you're going towards
the spring or you're going away from the spring;
try to think about that. Same with gravity.
Gravity is pulling you down with a force -mg and it
doesn't care whether the chalk is going up or coming down.
On the way up it's pulling it down;
on the way down it's also pulling it down,
because at a given location there is a fixed force.
So, the problem is not the force varies with x or
y, but the force is a function only of x.
But if the function of x and velocity then you also have
to know the velocity, and no one's giving you then a
definite value for the velocity because in the vibrating mass
and spring, velocity can have either sign.
Therefore, we do this integral but we don't do this.
Even though I write it this way, mathematically it's wrong
because the force of friction--Is it really a
function of x? I'll tell you in what limited
sense it is a function and what technical sense it is not.
Let's defer that for now. Take all these guys to the
left-hand side, so this is
K_2, this'll become
U_2, so you'll get (K_2
+ U_2) - (K_1+U
_1) is equal to this last thing,
work done by friction from x_1 to
x_2 and the question is,
"Can I do the integral?" The answer is,
if you just tell me x_1 and
x_2, then I know that's the point
you're making; I cannot just do an integral
because I will have to know the whole story, but in limited
cases I can do this. Suppose I pulled a mass by an
amount A and I let it go, but now there is friction.
I think we all know that if it came to the center it won't be
going quite as fast. And we also know,
when it overshoots the other side, it will not go back to
–A; normally, it will go back to
-A. Everyone knows why it goes back
to -A, right?
When you go the other extremity without friction,
your energy is all potential. It was ½kA^(2) to begin
with, it's going to be ½kA^(2) at the end,
except A can go to -A without changing the
answer; that's why it'll go from
A to -A. But when you've got friction,
I cannot say the energies are equal because I've got to do
this integral. For this part of the journey
from here to here, during that part of the
journey, the force is a well-defined function.
Namely, it is a constant value -µ times mg and
it is acting this way. So, for this part of the trip,
I can do the integral. It is very easy to find out
what it is; it's -mg times µ
kinetic, times the distance traveled.
Of course, it's a little more complicated now.
You have to tell me where you want your velocities.
You might say, find the velocity at x =
0. If you want me to do that,
the distances traveled is the original displacement.
That's what is the stuff on that side.
And here, K_2 = ½mv^(2) which is what
I'm trying to find. I am at the origin so it's
½k times 0^(2), no potential energy.
Initially, there is no kinetic energy but there was potential
energy ½kA^(2). You follow that logic now?
This is the difference in energies due to the conserved
forces, the conserved energies, but the difference
E_2 - E_1 is no longer 0
but is given by this number. So, I can calculate it for this
part. I can even ask,
-- this is one of the homework problems, I think -- you're
asked, "How far does it go to the left?"
Well, what do you do? Let's call the place where it
goes to. The left most extremity is
A prime. Then we don't know,
so there when you go to K_2 + U_2
- K_1+ U_1,
you will go back to this formula.
First of all, the distance that the force of
friction acts on is how much--is the distance A to come
here and another A prime to go to the left.
A prime is a magnitude of how far you go to the left.
And likewise, on the left-hand side you've
got to have the total kinetic plus potential minus kinetic
plus potential. At the end of the day,
you only have potential energy ½k A prime squared,
no kinetic and initially, again, you only have potential
energy ½kA^(2). You will see that A prime now
satisfies a quadratic equation; you can solve it.

But what is the point? The point is you cannot do this
integral once and for all and say the answer depends on the
end points. In this particular problem,
if you told me you are doing this part of the journey,
during that part of the journey because for the entire trip,
the velocity had a particular sign,
namely to the left, force of friction had a
definite magnitude and sign, namely to the right,
and then of course it's as good as any legitimate function and
you can do the integral. In fact, the forces are
constant; you can pull it out of the
integral, and the integral dx is just the distance
traveled. Okay, so work done by
friction--it will always come out negative and you should keep
track of that.

Okay, so the final bottom line for Law of Conservation of
Energy that you guys should remember is the following.
You take all the forces acting on a body.
So, lets do it one more time in our heads, so you know where
everything comes from. Everything comes from this
result K_2 - K_1,
which is the work done, is integral of all the forces.
Divide the forces into forces that depend on location only and
not on anything else, like velocity.
Let me just divide them into two forces, one is a good force,
F_g dx, and one is a force of friction,
dx. This guy will turn into a
potential energy difference. Let me just take one good force
and the force of friction. If you had two,
each will generate its own potential when you do the
integral. Then, we may then say
E_2 - E_1;
that's the same as (K_2 +
U_2) - (K_1 +U_1) is equal to
the work done by friction. Work done by friction is the
shorthand for that integral which you can do only;
you must divide the motion involved into pieces.
For example, if you have a mass that swung
to the left then turns around and goes back to the right,
of course, again, it won't come back to the
original, it'll come to a smaller number.
That's because there's more loss of energy on the return
trip. For the return trip,
you must do the problem separately.
On the return trip, you start from this A
prime that you found, then start moving to the right,
the friction will act to the left.
So, you must divide the motion into segments.
During each segment the force should have a definite direction
and magnitude so you can do the integral.
The point is, there is no universal formula
for F(x) due to friction. So, in one dimension that's all
it is, bringing the good and bad forces, those which lead to
potential, those which don't; the difference is the good
forces are functions of x, bad forces are
functions of x and something else,
so that at a given x it's possible to have two values
for that function. Mathematically,
we don't call it a function, that's why you cannot do the
integral once and for all. So, this is the Law of
Conservation of Energy. Okay?
Alright now, what I have to do here for the
remaining time is a little more of some mathematical preparation
for what I'm going to do next. What I'm going to do next is to
go to higher dimensions and we've got to make sure we all
know some of the mathematical ideas connected to the higher
dimensions. So, I'm going to start off with
some things and this is like a mathematical interlude.
Again, I told you, if you're not strong on some of
these things you should read this math book I mentioned to
you on Basic Training; it'll certainly cover these
topics. But you don't have to have
that, I'm going to tell you now how to do it.
So, the math idea that I'm going to use,
see this is really a mathematical digression to make
sure everybody knows, they are very simple things,
but we do it all the time in physics so I want to make sure
you guys can also do that. So, let's take some function
F(x) that looks like this.
This is F(x), this is x [pointing to

And let's say you start here, at some point x.
And I'm going to go to a nearby point, x + Δx.
Δx is exaggerated so you can see it.
So, actually in the end it's going to be a very small number.

The question is, "What is the change in the
function in going from here to here?"
So, we all know what it is; that is the change in the

We are going to need approximations to the change in
the function. The useful approximation is to
say, let us pretend the function is a straight line and keeps
going with the slope it had to begin with.
You can see that if the function is not curving up or
down too much, you almost got the right change
in F. And what is the change in
F? That change in F is a
derivative of the function at the point x times the
change in x. But it is not an equality;
it is not an equality because there is this little guy on the
top that you have missed. Because the function is not
following the same slope that you have to begin with,
it's curving up. So, let's take a concrete
example. Suppose F(x) = x^(2),
then F(x + Δx ) = x^(2) + 2xΔx + Δx^(2).
Therefore, the change in F -- you got to take it
to the other side -- change in F = 2xΔx
+Δx^(2). 2xΔx,
2x is, of course, what you realize is
that the derivative of the function at that point.
And that's giving you this part but there's a tiny portion on
the top, which is Δx^(2) that you're not getting.
If the function is x^(3),
then x to the fourth, then hyperbolic cos (x)
there'll be more errors. But the point is,
the change in the function has a leading piece that's
proportional Δx, and then the corrections that
we see mathematically are of higher order in Δx.

So, if Δx is made very small but not zero,
then it's a very useful formula to know that the change in
function and going from one point to another point is this
rate of change times the change. I don't think anything could be
more natural than that. But we should be comfortable by
saying it's naively people say, "Well, you should do that
because in calculus you're canceling this Δx with
that dx." That's not what we're really
doing. We all know that this stands
for a certain limit, but the limit stands literally
for the slope which is at that point.
So, if functions had only one slope and they never changed
their slope and you knew that slope,
then if you know the value to begin with, you can find the
value a mile away from where you are because it's going at the
same clip. But most functions of course
will change the slope, and for that reason there will
be an error in the formula you get if you use this.
But if Δx is very, very small and it's your
intention in the end to make very,
very tiny Δxs then we say we will use this

That's very simple but it's very useful to know.
So, let's take an example of this function,
of this rule, because I'm going to use it all
the time.

So, one example is F(x) = (1 + x)^(n).

Well, F(1) is just one. Suppose you go to a nearby
point which is x away from the .1;
so, this x is really like your Δx,
then the change in the function will be df/dx at the
point 1 times the change in x.
Yes? Student: [inaudible]
Professor Ramamurti Shankar: I'm sorry,
ah, ok, that's right. F(0) if you like;
if x = 0, the function has a value of one.
Then, if you go to a neighboring point--;So,
here is what I'm saying, the function does something
which value here is one and I want to go to a nearby point
which I'm calling x. Then, df/dx you know is
n times 1 + x^(n) -1 times x.
But you have evaluated it at x = 0,
so this goes away, then you get nx.
So, the change in the function, this function as you move off
x = 0 is nx. So, what we are saying then is
1 + x raised to the n is 1 + nx + tiny
numbers, if x is small.

That is a very useful result. If something looks like 1 + x
raised to a power, and if you're not going too far
from the point 1, namely if your x is very
small, then don't bother with doing the whole 1 + x times 1 +
xn times, it's approximately equal to 1 +

So, let me give you one example; you will use this later on.

According to relativity, a mass of a particle is given
by this [writes formula on the board].
It's a body sitting still, it has certain mass.
In Newtonian mechanics we say the mass is the mass and that's
the end of it. Relativity, you don't have to
know the origin of this; that'll be taught to you in due
course. But it just says particle to
the velocity have a mass different from when they are at
rest. And this is the velocity and
c is the velocity of light.
This is the exact formula. Now, most of the time we don't
care about it because v/c maybe 10^(-10) and the square of
that is 10^(-20) and one minus such a number is negligible and
we just put it as one. Suppose you want an answer
that's good to some accuracy. I will write this as follows,
as m_0 times 1 - v^(2) over c^(2)
to the power of -½ , do you agree?
That's what a square root in the bottom is,
it's this power upstairs. Now, we can write it as
follows, m_0, this is (1 + x)^(n),
okay? So the x that we use is
-v^(2)/c^(2) and the n that we use is -½.

That's what I'm using in this formula;
x is -v^(2)/c^(2) and
n, which is the power to which I'm raising,
is -½; therefore, this is equal to 1 +
v^(2)/2c^(2) plus corrections,
which are negligible if v^(2)/c^(2) is very
small. This is a very useful result.
It tells you the mass variation is that is the mass at rest plus
m_0v^(2)/ 2c^(2),
it's a very useful formula. Sometimes this formula is much
more friendly to use than the exact formula.
You cannot beat the exact formula;
it's the exact truth but if you're interested in that exact
function for small values of v/c, it's much more
useful to use this approximation.
And you will have a lot of problems like this in your
homework, now, later, everywhere else.
So, if I'm going to use it all the time I should tell you what
it is. So, let me summarize what I've
told you so far. First result is,
if a function is known at a certain point and you want to
know its value at a nearby point, what are you supposed to
do? That's not so hard to guess
right? For example,
your tuition costs are so much this year at Yale.
And we know how much it's going up every year;
you can make a guess on what it'll be a year from now.
Of course in reality, it'll be higher than that
because there is other terms, but that's the best guess you
would make. That's how we predict the
future by looking at past rates of change;
that's very normal. The point is that you can
predict the tuition in a year from now maybe,
but maybe not five years from now or ten years from now
because there's inflation on top of inflation.
That's like saying the rate of change itself has a rate of
change, the slope here is not the same as the slope there,
that's what the higher derivatives of the function do
for you. So, I will conclude by telling
you one last result which I'm not going to prove in this
course. And you guys should learn from
the math department the following result.
There is some crazy function, okay, this is f(x),
I want to know what it is at the point x.
I know what it is at the origin. If I know nothing else,
how do I predict the future value of x,
meaning for other x I don't know.
My best guess is, well, this is the number I knew
that's what it's going to do. But if I knew its rate of
change here, you know, its slope here,
the next guess is the derivative of f at the
origin times x. That's just what I did earlier;
instead of Δx I'm calling it x.
If x is small enough it'll work.
But of course, x need not be small;
if it gets bigger, it turns out you can do a
little better by adding this [writes formula on board].
This is the second derivative of the function at the origin
times x^(2) and you can go on adding more and more
terms. I'll just add one more term,
which is 1/3 times d^(3)x,
f/dx^(3) times x^(3).
This goes on forever. And it says that if you knew
every derivative of the function at the origin you can
reconstruct the function. Not just infinitesimally close
to it, but as far as you want to go, there are some mathematical
restrictions on this because there are infinite number of
terms in this thing. When you add them all up you
may get infinity which would be nonsense.
So, you should really sum up this infinite number of terms if
they add up to a finite number, and if they do that will give
you the function here. What we have done is to just
use the first two terms as an approximation.
Or what I'm telling you is that there is something called The
Taylor Expansion of a function that allows you to know the
function even here, macroscopically,
far from here if you knew all the derivatives.
And if you put those derivatives here and if the sum
added up to something finite, that finite thing is the value
of the function at the new point.
Here is an example you guys can take, f(x) = e^(x).
Why don't you go home and work out the series for f(x)
at all points? You have got to remember that
e^(x) has a beauty, that the derivative is
e^(x) itself. And its value at the origin of
every derivative is simply one. So, this just becomes 1 +
x + x^(2)/2. And that is,
in fact, the expression for e^(x) and it turns out
that no matter how big x is,
the sum in fact adds up to a finite number that we call
e^(x). And in particular,
what is e itself, e itself e to the
1; e is e to the 1
so put x = 1 you get 1 + 1 + 1^(2) over 2 factorial plus
1^(3) over 3 factorial. You do all that you get 2 point
something, something. Alright, that's e.
So, this is a very powerful notion.
If you knew all the derivatives of a function,
you can predict its value not only the tiny neighborhood,
but over big distance. If you knew the first
derivative and you were not very ambitious.
Look, the logic is very simple. If x is a tiny number
x^(2) is even tinier, x^(3) is even tinier.
We've got no respect for these terms.
We just drop them and we stop here.
Okay, I'll have something more to say about this later on.
But for now, these ideas will be useful for
what we're doing next time.