Today, we will talk exclusively
about work and energy. First, let's do
a one-dimensional case. The work that a force
is doing, when that force is moving
from point A to point B-- one-dimensional, here's point A
and here is point B-- and the force is along
that direction or... either in this direction
or in this direction but it's completely
one-dimensional, that work is the integral
in going from A to B of that force dx,
if I call that the x-axis. The unit of work,
you can see, is newton-meters. So work is newton-meters, for which we...
we call that "Joule." If there's more than one force
in this direction, you have to add these forces
in this direction vectorially, and then this is the work
that the forces do together. Work is a scalar, so this
can be larger than zero, it can be zero,
or it can be smaller than zero. If the force and the direction
in which it moves are in opposite directions,
then it is smaller than zero. If they're
in the same direction, either this way or that way, then the work is larger
than zero. F = ma, so therefore, I can
also write for this m dv/dt. And I can write down for dx,
I can write down v dt. I substitute that in there, so
the work in going from A to B is the integral from A to B
times the force, which is m dv/dt, dx which is v dt. And look what I can do. I can eliminate time, and I can now go to a integral
over velocity-- velocity A to the velocity B, and I get m times v times dv. That's a very easy integral. That is 1/2 m v squared,
which I have to evaluate between vA and vB, and that is 1/2 m vB squared,
minus 1/2 m vA squared. 1/2 m v squared is what we call
in physics "kinetic energy." Sometimes we write just a K
for that. It's the energy of motion. And so the work that is done
when a force moves from A to B is the kinetic energy
in point B-- you see that here-- minus the kinetic energy
in point A, and this is called
the work-energy theorem. If the work is positive, then
the kinetic energy increases when you go from A to B. If the work is smaller
than zero, then the work ... then the kinetic energy
decreases. If the work is zero, then there
is no change in kinetic energy. Let's do a simple example. Applying this
work-energy theorem, I have an object that I want
to move from A to B. I let gravity do that. I give it a velocity. Here's the velocity v of A,
and let the separation be h, and this could be
my increasing y direction. The object has a mass m, and so there is a force,
gravitational force which is mg, and if I want to give it
a vector notation, it's mg y roof, because this
is my increasing value of Y. When it reaches point B,
it comes to a halt, and I'm going to ask you now
what is the value of h. We've done that in the past
in a different way. Now we will do it purely based
on the energy considerations. So I can write down that
the work that gravity is doing in going from A to B,
that work is clearly negative. The force is in this direction and the motion is
in this direction, so the work that gravity is
doing in going from A to B equals minus mgh. That must be the kinetic energy
at that point B, so that this kinetic energy
at point B minus the kinetic energy
at point A, this is zero, because it comes to a halt here, and so you find that
1/2 m vA squared equals mgh. m cancels, and so you'll find
that the height that you reach equals vA squared divided by 2g. And this is something
we've seen before. It was easy for us to derive it
in the past, but now we've done it purely
based on energy considerations. I'd like to do a second example. I lift an object from A to B--
I, Walter Lewin. I take it at A. It has no speed here;
vA is zero. It has no speed there. And I bring it
from here to here. There's a gravitational force mg
in this direction, so the force by Walter Lewin
must be in this direction, so the motion and my force
are in the same direction, so the work that I'm doing
is clearly plus mgh. So the work that Walter Lewin is
doing is plus mgh when the object goes
from A to B. The work that gravity was doing
was minus mgh-- we just saw that. So the net work that is done
is zero, and you see there is indeed
no change in kinetic energy. There was no kinetic energy
here to start with, and there was
no kinetic energy there. If I take my briefcase
and I bring it up here, I've done positive work. If I bring it down,
I've done negative work. If I bring it up,
I do again positive work. When I do positive work,
gravity does negative work. When I do negative work,
like I do now, gravity does positive work. And I can do that a whole day, and the net amount of work
that I have done is zero-- positive work, negative work,
positive work, negative work. I will get very tired. Don't confuse getting tired
with doing work. I would have done no work
and I would be very tired. I think we would all agree that if I stand here
24 hours like this that I would get very tired. I haven't done any work. I might as well put it here and let the table just
hold that briefcase for me. So it's clear that
you can get very tired without having done any work. So this is the way
we define work in physics. Now let's go from one
dimensions to three dimensions. It is not very much different,
as you will see. I go in three dimensions
from point A to point B, and I now have a force... which could be pointing not
just along the x direction, but in general,
in all directions. Now the work that the force is
doing in going from A to B is F dot dr r is the position in
three-dimensional space where the force is
at that moment, and dr is a small displacement. So if this is from A to B, then dr here,
if you're going this direction, this would be
the little vector dr. And here, that would be
a little vector dr. And the force itself
could be like this here, and the force could be
like this there. The force can obviously
change along this path. So let the force be... F of x, x roof, plus F of y, y roof, plus F of z, z roof. I'll move this A
up a little, put it here. And let dr-- the general
notation for vector dr-- equals dx, x roof, plus dy, y roof, plus dz, z roof. It cannot be any more general. So the work that
this force is doing when it moves from A to B is the integral of this F dr. Let's first take a small
displacement over dr, then I get dw. That is simply Fx times dx--
it's a scalar-- because this is a dot product... plus Fy dy, plus Fz dz. That is little bit amount
of work if the force is displaced
over a distance dr. Now I have to do the integral
over the entire path to get W. From A to B, that's an integral
going from A to B, integral going from A to B. I don't need this anymore. Integral in going from A to B,
integral in going from A to B. Now we're home free,
because we already did this. This is a one-dimensional
problem, and a one-dimensional problem,
we already know the outcome. The integral F dx, we found that is
1/2 m vB squared minus m vA squared, which in this case
is obviously the velocity in the x direction, because this is
a one-dimensional problem. And the one-dimensional problem
indicates that the velocity
that I'm dealing with is the component
in this direction. So we have that
this is 1/2 m v B squared-- and this is the x component-- minus vA squared,
and that is the x component. This is also a one-dimensional
problem now, except that now I deal
with the component... with the y component
of the velocity, so I get 1/2 m
times vB y squared minus vA y squared, plus 1/2 m vB z squared
minus vA z squared. And now we're home free,
because what you see here is you see v squared
in the x direction, v squared y component,
v squared z component. And if you add those three up, you get exactly
the square of the velocity. You get the square of the speed. So if you add up
these three terms, you get up ... you get vB squared...
I lost my m. Let me put my m in there. 1/2 m times vB squared, and here you see Ax squared,
Ay squared, Az squared minus vA squared, and you get exactly the same
result that you had before, namely that the work done is the
difference in kinetic energy. You can always think
of these as speeds. Velocity squared is the speed. It's the magnitude squared
of the velocity. All right, I'd like
to return to gravity and work on
a three-dimensional situation. We have here, let this be x,
this be y and this be z. And there is here, this is
the increasing value of y. And there's here point A in
three dimensions like this. And there is here point B, so you get a rough idea
about the three dimensions. And y of B minus y of A
equals h. It's a given--
there is a height difference between A and between B. There is a gravitational force. The object moves from A to B. Suppose it moves
in some crazy way. Of course, gravity alone
could not do that. There has to be another force
if it goes in a strange way. But I'm only calculating now the work that's going
to be done by gravity. The other forces
I ignore for now. I only want to know the work
that gravity is doing. The object has a mass m,
and so there is a force mg, and I can write down the force
in vector notation. It's in this direction. So now I notice that there
is only a value for F of y, but there is no value
for F of x, and there is no value
for Fx; they are zero. And so F of y equals minus mg. And so if I calculate now
the work in going from A to B, which is the integral in going
from A to B of F dot dr, and the only term that I have is the one that deals
with the y direction. so it is the integral in
going from A to B of Fy dy. And that equals minus mg, because we have the minus mg,
times y of B minus y of h, so that is minus mg times h. And what you see here, that it
is completely independent of the path that I have chosen. It doesn't matter how I move. The only thing that matters is the difference in height
between point A and point B. h could be larger than zero,
if B is above A. It could be smaller than zero
if B is below A. It could be equal to zero
if B has the same height as A. Whenever the work
that is done by a force is independent of its path-- it's only determined by the starting point
and the end point-- that force is called
a "conservative force." It's a very important concept
in physics. I will repeat it. Whenever the work
that is done by a force in going from one point
to another is independent of the path-- it's only determined by the starting point
and the end point-- we call that
a conservative force. Gravity is a conservative force. It's very clear. Suppose that I do the work--
that I go from A to B in some very strange way. Then it is very clear that
the work that I would have done would be plus mgh,
because my force, of course, is exactly in the opposite
direction as gravity. So whenever gravity is
doing positive work, I would be doing negative work. If I hold it in my hand, when I'm doing positive work,
gravity is doing negative work. Again, I'm going
to concentrate now on a case where we deal
with gravity only. When there's only gravity,
then we have that minus mgh is the work done
in going from A to B equals minus mg, times y of B
minus y of A, and that now is the kinetic
energy at point B minus the kinetic energy
at point A. This is the work-energy theorem. Look closely here. I can rearrange this, and I can
bring the Bs in one side, I can bring the As on one side. I then get mg times y of B plus
the kinetic energy at point B equals mg times y of A plus the kinetic energy
at point A. And this is truly
an amazing result. We call mgy...
we give that a name, and we call that "gravitational
potential energy." Often we write for that PE,
or we write for that a U. And what you're seeing here is that the sum of potential energy
at point B and the kinetic energy
at point B is the same as
the potential energy at A and the kinetic energy
at point A. One can be converted
into the other and it can be converted back. Kinetic energy can be converted
back to potential energy, and potential energy
can be converted back, but the sum of them-- which we
call "mechanical energy"-- is conserved. And mechanical energy
is only conserved if the force is
a conservative force. It's extremely useful. We will use it many times, but
you have to be very careful. It's a dangerous tool
because it's only true when the force is conservative. Spring forces are
also conservative, but, for instance, friction
is not a conservative force. If I move an object
from here to here... Let's suppose
I move this object, and I go along a straight line, then the friction is
doing negative work, I am doing positive work. But now suppose I go from here
to here through this routing. You can see that the work
I have to do is much more. Friction is not
a conservative force. The frictional force
remains constant, dependent on the friction, the kinetic
friction coefficient, is always the same... the frictional force, which
I have to overcome as I move, and so if I go all the way here and then all the way back to
this point where I wanted to be, then I have done a lot more work than if I go along
the shortest distance. So friction is a classic example of a force that is
not conservative. If I look at this result-- the sum of gravitational
potential energy and kinetic energy is conserved
for gravitational force-- then it is immediately obvious where we put the zero
of kinetic energy. The zero of kinetic energy is when the object
has no velocity, because kinetic energy
equals 1/2 m v squared. So if the object has
no velocity, then there is no kinetic energy. How about potential energy? Well, you will say, sure, potential energy must be zero
when y is zero, because that's the way
that we defined it. You see? mgy is gravitational
potential energy. So you would think that
u is zero when y is zero. Not an unreasonable thing
to think. But where is y a zero? Is y zero at the surface
of the Earth? Or is y zero
at the floor of 26.100? Or is y zero here,
or is y zero the roof? Well, you are completely free
to choose where you put u equals zero. It doesn't matter as long as point A and point B
are close enough together that the gravitational
acceleration, g, is very closely the same
for both points. The only thing that matters then is how far they are
separated vertically. The only thing that matters is
that uB minus uA... uB minus uA would be mgh. It is only the h that matters, and so you can then
simply choose your zero anywhere you want to. It's easy to see. Suppose I have here point A
and I have here point B. And suppose
this separation was h. Well, if you prefer to call
zero potential energy at A, I have no problem with that. So we can call this
u equals zero here. Then you would have
to call this u... you have to call it plus mgh. If you say, "No, I don't want to do that;
I want to call this zero"... that's fine. Then this becomes minus mgh. If you prefer to call
this zero, that's fine, too. Then this will have a positive
gravitational potential energy, and this will have one
that is higher than this one by this amount. If you say, really "I'd like
to call this zero," of course the same holds. What matters is what the difference between
potential energy is. That is what we need when we apply the conservation
of mechanical energy. That is what we need
in order to evaluate how the object changes
its kinetic energy. So where you choose your zero
is completely up to you. As long as A and B
are close enough so that there is
no noticeable difference in the gravitational
acceleration g. Eh...Before the end of this hour,
I will also evaluate the situation that g
is changing. When you go far way from
the Earth, g is changing. So let us first do...
look at a consequence of the conservation
of mechanical energy. Very powerful concept, and as
long as we deal with gravity, you can always use it. You see here on the desk something that looks
like a roller coaster, and I'm going to slide
an object from this direction. Let's clean it
a little bit better. Here is that roller coaster. This is a circle,
and then it goes up again. So let the circle have
a radius R. This point be A. I release it with zero speed. I assume that there is
no friction for now. This point be B. And I define here y equals zero, or what is even more important,
I define that u equals zero. And this is the direction,
positive direction, of y. At A, the object has
no velocity, no speed. At B, of course, it does. It has converted some potential
energy to kinetic energy. At this point C,
it has reached a maximum velocity
that it can ever have because all the potential energy
has been converted to kinetic energy. And at this point D,
if it ever reaches that point, that will be the velocity, C Okay, I start off, point A is at
a distance h above this level, and so I apply now the conservation
of mechanical energy. So I know that u at A plus
the kinetic energy at A-- which is zero-- must be u at B
plus kinetic energy at B, must be u at C
plus kinetic energy at C, must be u at D
plus kinetic energy at D. If there is no friction, if there are no other forces,
only gravity. So we lose no... no energy goes
lost in terms of friction. We know that this
height difference is 2R. And so now I can write this
in general terms of y... Point... Take this point B. Think of that being at a
location y above the zero line. Then I can write down now
that uA, which is mgh... That was a given when I started. That was all the energy I had. That was my total
mechanical energy. If I call this u zero,
which is free choice I have, equals u of B, which is
now mgy, plus 1/2 m v squared at that position y. This should hold...
what you see there should hold for every point
that I have here. It should for A, for B, for C,
for D, for any point. I lose my m,
and so you find here that... We summarize it at v squared
equals 2g, times h minus y. So this should hold
for all these points. Therefore, it should
also hold for point D. However, at point D, there
is something very important. There is a requirement. There is a requirement that there is a centripetal
acceleration, which is in this direction,
a centripetal. And that centripetal
acceleration is a must for this object
to reach that point D. And that centripetal
acceleration, as we remember from when we played
with the bucket of water, that is v squared divided by R. And this must be larger
or equal to the gravitational
acceleration g. If it is not larger, the bucket
of water would not have made it to that point D. So this is my second equation
that I'm going to use, so look very carefully. So v squared must be larger
or equal than gR, so I have here v squared,
which is 2g times h minus y. But y for that point D is 2R, so I put in a 2R,
must be larger or equal to gR. I lose my g, so 2h minus 4R
must be larger or equal to R, so h must be larger
or equal to 2½R. This is a classic result that almost every person who has
taken physics will remember. It is by no means intuitive. It means that if I have
this ball here-- and I will show you
that shortly-- and I let the ball go
into this roller coaster, that it will not make this point
unless I release it from a point that is at least 2½ times
the radius of this circle above the zero level. If I do it any lower,
it will not make it. So think about this. That is something that you could
not have just easily predicted. It's a very strong result,
but it is not something that you say intuitively,
"Oh, yes, of course." It follows immediately from the conservation
of mechanical energy. So if I release it... That, that 2½ radius point,
by the way, is somewhere here. So if I release this object
way below that, it will not make this point. Let's do that. You see, it didn't make it. I go a little higher,
didn't make it. I go a little higher,
didn't make it. Go a little higher,
still didn't make it. Now I go to the 2½ mark... and now it makes it. 2½ times the radius, conservation of mechanical
energy tells you that that is the minimum
it takes to just go through that point. Of course, if there were
no loss of energy at all, if there were no mechanical
energy lost-- that means if there
were no friction-- then if I were to release it
at this point, it would have to make it back
to this point again, with zero kinetic energy. But that won't ..that's not the case. There is always
a little bit of friction with the track, for one thing,
and also, of course, with air. So if I release it
all the way here, you will not expect it that it will bounce up
all the way to here. It will probably stop
somewhere there. It may not even
make it to the end. We can try that. Oh, it made it
somewhere to here-- a little lower than that level. Of course there is some
friction, that is unavoidable. All right, this is
a classic one. There are many exams where
this problem has been given. I won't give it to you this
time, but it's a classic one. You see it on the general exams
for physics, and it's simply a matter of conservation
of mechanical energy. Let's now look at the situation
whereby A and B are so far apart that the gravitational
acceleration is no longer constant, and so you can
no longer simply say that the difference
in potential energy between point B and point A
is simply mgh. So now we are dealing with
a very important concept, and that is
the gravitational force. You can think of the Earth
acting on a mass or you can think of the sun
acting on a planet, whichever you prefer, but
that's what I want to deal with when the distances
are now very large. Let me first give you
the formal definition of gravitational
potential energy. The formal definition is that the gravitational
potential energy at a point P is the work that I,
Walter Lewin, have to do to bring that mass from
infinity to that point P. Now, you may say
that's very strange that in physics, there are
definitions which... where Walter Lewin comes in. Well, we can change it
to gravity, because my force is always the same force as gravity
with a minus sign, so it's also minus the work
that gravity does when the object moves from
infinity to that point P. I just like to think of it, it's
easier for me to think of it, as the work that I do. So if we apply that concept, then we first have to know what
is the gravitational force. If this is an object,
capital M-- and you can think of this as being the Earth,
if you want to-- and there is here an object
little m, then I have to know what the forces are
between the two. And this now is Newton's
Universal Law of Gravity, which he postulated... Universal Law of Gravity. He says the force
that little m experiences, this force equals-- I'll put a little m here
and a capital M here-- so it is little m
experiences that force due to the presence
of capital M-- equals little m times capital M
times a constant, which Newton, in his days,
didn't know yet what that value was, divided by r squared, if r
is the distance between the two. This object, since Newton's
Third Law holds-- action equals minus reaction-- this force,
which is I will indicate it as capital M, little m-- it is the force that
this one experiences due to the presence
of this one-- is exactly the same
in magnitude but opposite in direction, and that is the Universal Law
of Gravity. Gravity is always attractive. Gravity sucks--
that's the way to think of it. It always attracts. There is no such thing
as repelling forces. The gravitational constant G
is an extremely low number-- 6.67 times 10 to the minus 11-- in our... SI units, which is newtons,
gram-meters per kilogram or something like that. That's an extremely low number. It means that
if I have two objects which are each one kilogram,
which are about one meter apart, which I have now here
about one meter, that the force which
they attract each other is only 6.67 times 10
to the minus 11 newtons. That is an extremely
small force. If this were the Earth, and I am
here and this is my mass, then I experience a force which
is given by this equation. This would be, then,
the mass of the Earth. Now, F equals ma. So if I'm here, I experience
a gravitational acceleration, and the gravitational
acceleration that I experience is therefore given
by MG divided by r squared. And so you see that the gravitational
acceleration that I experience at different distances
from the Earth, or, for that matter, at different distances
from the sun, is inversely proportional
with r squared. We have discussed that earlier
when we dealt with the planets, and we dealt with uniform
circular motions, and we evaluated
the centripetal acceleration. We came exactly
to that conclusion-- that the gravitational
acceleration falls off as one over r squared. Ten times further away, the gravitational acceleration
is down by a factor of 100. If you are standing near
the surface of the Earth, then, of course, the force
that I will experience is my mass
times the mass of the Earth times the gravitational constant divided by the radius
of the Earth squared-- just like we are
here in 26.100-- and so this must be mg. That's the gravitational
acceleration if we drop an object here. And so you see that
this now is our famous g, and that is the famous 9.8. If you substitute in there
the mass of the Earth, which is six times
10 to the 24 kilograms. You put in here the
gravitational constant, and you put in the radius
of the Earth, which is 6,400 kilometers, out pops our well-known number of 9.8 meters
per second squared. Okay, my goal was
to evaluate for you the gravitational
potential energy the way that it is defined
in general, not in a special case
when we are near the Earth. So we now have to move an object
from infinity to a point P, and we calculate the work
that I have to do. So here is capital M,
and here is that point P, and infinity is somewhere there. It's very, very far away,
and I come in from infinity with that object with mass m,
and I finally land at point P. Since gravity is
a conservative force, and since my force is always
the same in magnitude except in opposite direction, it doesn't matter how I move in; it will always come up
with the same answer. So we might as well do it
in a civilized way and simply move that object
in from infinity along a straight line. It should make no difference because gravity is
a conservative force. So infinity is somewhere there. The force that
I will experience, that I will have to produce,
is this force. The force of gravity
is this one. The two are identical except that mine is
in this direction-- this is increasing value of r-- so mine would be plus m MG
divided by r squared if I'm here at location r. And let this be at a distance
capital R from this object. You can already see that the
gravitational potential energy, when I come from infinity
with a force in this direction and I move inward, you can already see that
gravitational potential energy will always be negative
for all points anywhere. It doesn't matter where I am,
it will always be negative. You may say, gee, that's
sort of a strange thing-- negative potential energy. Well, that is not a problem. Remember that depending upon how
you define your zero level here, you also end up with negative
values for potential energy. So there's nothing sacred
about that. What is important, of course,
if we get the right answer for the gravitational
potential energy, that when we move away
from this object that the gravitational
potential energy increases. That's all that matters. But whether it is negative
or positive is irrelevant. So we already know
it's going to be negative, and so we can now evaluate
the work that I have to do when I go from infinity
to that position, capital R. So here comes the work
that Walter Lewin has to do when we go from infinity
to that point, which is capital R,
radius, from this object. Think of it as the sun or
the Earth; either one is fine. So that is the integral in going
from infinity to R of my force-- which is plus, because it's
an increasing value of R-- m MG divided by R squared dr. That's a very easy integral. This is minus one over r, so I get m MG
over r with a minus sign, and it has to be evaluated
between infinity and capital R. When I substitute for R,
infinity, I get a zero, and so the answer is
minus m MG over capital R. And this is the potential...
gravitational potential energy at any distance capital R
that you please away from this object. At infinity,
it's now always zero. Earlier, you had a choice
where you chose your zero. When you're near Earth
and when g doesn't change, you had a choice. Now you no longer have a choice. Now the gravitational potential
energy at infinity is fixed at zero. So let's look at this function, and let us make a plot
of this function as a function of distance. The one over r relationship of the gravitational
potential energy... the force, gravitational force,
falls off as one over r squared. Here's zero. This is the gravitational
potential energy. All these values
here are negative, and here I plot it
as a function. I use the symbol little r now
instead of capital R. And so the curve would
be something like this. This is proportional
to one over r. If you move an object from A
to B and this separation is h, and if A and B are very far apart, the difference in potential
energy is no longer mgh, but the difference
in potential energy is the difference between
this value and this value. And you have to use that
equation to evaluate that. But you can clearly see
that if I go from here to here-- if I take an object
and go from here to here-- that the potential energy
will increase, and that's all that matters. So it increases when you go
further away from the Earth if you look at the Earth, or from the sun
if you look at the sun. Is there any disagreement between this result
that we have here and the result
that we found there? And the answer is no. I invite you to go through
the following exercise. Take a point A in space, which is at a distance r of A from the center
of the Earth, say, and I do that... I start at
the surface of the Earth itself, so the radius is
the radius of the Earth. And I go to point B, which is a little bit
further away from the center of the Earth,
only a distance h. And h is way, way, way smaller
than the radius of the Earth. So I can calculate now what the difference
in potential energy is between point B and point A, and I can use, and I should use,
this equation. And when I use that equation and
you use the Taylor's expansion, the first order
of Taylor's expansion, you will immediately see
that the result that you find collapses into this result because the g
at the two points is so close that you will see
that you will find then that it is approximately mgh,
even though it is the difference between these two
rather clumsy terms. We will,
many, many times in the future, use the one over r relationship for gravitational
potential energy. We will get
very used to the idea that gravitational potential
energy is negative everywhere the way it's defined, and we will get used
to the idea that at infinity, the gravitational
potential energy is zero. But whenever we deal with near-
Earth situations like in 26.100, then, of course,
it is way more convenient to deal with the simplification that the difference in
gravitational potential energy is given by mgh. I always remember that-- mgh,
Massachusetts General Hospital. That's the best way that you can
remember these simple things. Now I want to return to the conservation
of mechanical energy. I have here a pendulum. I have an object
that weighs 15 kilograms, and I can lift it up one meter,
which I have done now. That means I've done work--
mgh is the work I have done. Believe me, I've increased the
potential energy of this object 15 times 10,
so about 150 joules. If I let it fall, then that will
be converted to kinetic energy. If I would let it swing
from one meter height, and you would be there and it
would hit you, you'd be dead. 150 joules is
enough to kill you. They use these devices--
it's called a wrecker ball-- they use them
to demolish buildings. You lift up a very heavy object,
even heavier than this, and then you let it go,
you swing it, thereby converting gravitational
potential energy into kinetic energy, and that way,
you can demolish a building. You just let it hit... (glass shattering ) and it breaks a building. And that's the whole idea
of wrecking. (laughter ) So you're using, then, the conversion of gravitational
potential energy to kinetic energy. Now, I am such a strong believer of the conservation
of mechanical energy that I am willing to put
my life on the line. If I release that bob
from a certain height, then that bob
can never come back to a point where the height
is any larger. If I release it from this height
and it swings, then when it reaches here,
it could not be higher. There is a conversion from gravitational potential
energy to kinetic energy back to gravitational
potential energy, and it will come to a stop here. And when it swings back, it should not be able
to reach any higher, provided that I do not give
this object an initial speed where I stand here. I trust the conservation
of mechanical energy for 100%. I may not trust myself. I'm going to release
this object, and I hope I will be able
to do it at zero speed so that when it comes back
it may touch my chin, but it may not crush my chin. I want you to be extremely
quiet, because this is no joke. If I don't succeed
in giving it zero speed, then this will be
my last lecture. (laughter ) I will close my eyes. I don't want to see this. So please be very quiet. I almost didn't sleep all night. Three, two, one, zero. (little scream) (class laughs with relief ) Physics works
and I'm still alive! (applause ) See you Wednesday. (applause continues )
