Lecture 11: Work, Energy, and Universal Gravitation

Walter Lewin, 8.01 Physics I: Classical Mechanics, Fall 1999.
(Massachusetts Institute of Technology: MIT OpenCourseWare).
http://ocw.mit.edu (accessed 04 08, 2013).
License: Creative Commons Attribution-Noncommercial-Share Alike.

1D Work and Kinetic Energy

The work that a force is doing, when that force is moving from point A to point B--

$W_{AB} = \int_A^B F \mathrm{d}x$

The unit of work is N m, for which we...call that "joule", J.

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 it can be > 0, = 0, or < 0.

F = ma, so therefore, F = m dv/dt.

And I can write down for dx = v dt.

I substitute that in there, so the work in going from A to B is

$W_{AB} = \int_{A}^{B} m \frac{dv}{dt} v dt = \int_{v_A}^{v_B} m v dv$

That's a very easy integral.

$W_{AB} = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2$

kinetic energy

$\frac{1}{2} m v^2 = KE = K$

work-energy theorem

$W_{AB} = \mathrm{KE}_B - \mathrm{KE}_A$

work is positive ==> KE increases when going from A to B
work is negative ==> KE decreases when going from A to B
work is zero ==> KE does not change when going from A to B

Example

I have an object that I want to move from A to B.
I give it a velocity at point A
when it reaches point B, it comes to a halt

What is the value of h?

$W_{AB} = \int_A^B -mg dy = -mgh = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2 = - \frac{1}{2} m v_A^2$

So you find that

$mgh = \frac{1}{2} m v_A^2$

m cancels, and so you'll find that

$h = \frac{v_A^2}{2g}$

Example:

Professor Lewin lifts an object from A to B
at point A and B, the velocities are all zero

The work that Walter Lewin is doing is +mgh when the object goes from A to B.

The work that gravity was doing was -mgh.

So the net work that is done is zero, and you see there is indeed no change in kinetic energy.

Example:

bring a briefcase up and down
bring up => positive work
bring down => negative work
gravity does opposite amount of work
net work is zero

Don't confuse getting tired with doing work.

I would have done no work and I would be very tired.

Work Calculated in 3-Dimensions

Now the work that the force is doing in going from A to B is

$W_{AB} = \int_A^B \vec{F} \cdot d \vec{r}$

r is the position in three-dimensional space where the force is at that moment, and dr is a small displacement.

So let the force be...

$\vec{F} = F_x \hat{x} + F_y \hat{y} + F_z \hat{z}$

And let dr--

$d\vec{r} = dx \hat{x} + dy \hat{y} + dz \hat{z}$

So the work that this force is doing when it moves from A to B is

$W_{AB} = \int_A^B \vec{F} \cdot d\vec{r} = \int_A^B F_x dx + \int_A^B F_y dy + \int_A^B F_z dz$

So we have that this is

$W_{AB} = \frac{1}{2} m \left( v_{B_x}^2 - v_{A_x}^2 \right) + \frac{1}{2} m \left( v_{B_y}^2 - v_{A_y}^2 \right) + \frac{1}{2} m \left( v_{B_z}^2 - v_{A_z}^2 \right)$

If you add up these three terms, you get

$W_{AB} = \frac{1}{2} m \left( v_{B}^2 - v_{A}^2 \right)$

and you get exactly the same result that you had before, namely that the work done is the difference in kinetic energy.

Gravity is a Conservative Force

And so if I calculate now the work in going from A to B, this is

$W_{AB} = \int_A^B \vec{F} \cdot d\vec{r} = \int_A^B F_y dy = -mg(y_B - y_A) = -mgh$

And what you see here, that it is completely independent of the path that I have chosen.

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."

Gravity is a conservative force.

When Gravity is the only Force

$-mgh = -mg(y_B - y_A) = \mathrm{KE}_B - \mathrm{KE}_A$

I can rearrange this, and I then get

$mgy_B + \mathrm{KE}_B = mgy_A + \mathrm{KE}_A$

We call mgy..."gravitational potential energy." Often we write for that PE, or we write for that a U.

And what you're seeing here is that

$\mathrm{U}_B + \mathrm{KE}_B = \mathrm{U}_A + \mathrm{KE}_A$

One can be converted into the other and it can be converted back.

Sum of kinetic energy and potential energy--called "mechanical energy"-- is conserved.

And mechanical energy is only conserved if the force is a conservative force.

Spring forces are also conservative
Friction is not a conservative force

What Matters is the Difference in Potential Energy

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.

But where is y a zero?

Well, you are completely free to choose where you put U = 0.

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

$\mathrm{U}_B - \mathrm{U}_A = mgh$

A Roller Coaster, Upside-down

Here is that roller coaster.

$\mathrm{U}_A + \mathrm{KE}_A = \mathrm{U}_B + \mathrm{KE}_B = \mathrm{U}_C + \mathrm{KE}_C = \mathrm{U}_D + \mathrm{KE}_D$

If there is no friction, if there are no other forces, only gravity.

So no energy goes lost in terms of friction.

And so now I can write this in general terms of y...

$mgh = mgy + \frac{1}{2} mv^2$

This should hold...for every point that I have here.

and we summarize it as

Therefore, it should also hold for point D.

However, at point D, it requires a centripetal acceleration.

$a_c = \frac{v^2}{R} \ge g$

So I have

$2g(h-y) = v^2 \ge gR$

But y for that point D is 2R,

$2g(h-2R) \ge gR$

Hence,

$h \ge \frac{5}{2}R$

Newton's Law of Universal Gravitation

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.

The formal definition of 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, or equivalently, the negative work the gravity does.

Newton's Universal Law of Gravity

$F = \frac{mMG}{r^2}$

Gravity is always attractive.

The gravitational constant G is an extremely low number--

$G = 6.67 \times 10^{-11}$ N m^2/kg^2

That's an extremely low number.

e.g. 2 objects, each 1 kg, are 1 m apart, they would only experience 6.67 x 10^-11 newtons.

$F = \frac{mMG}{r^2} = ma$

$F = \frac{mM_{\bigoplus}G}{R_{\bigoplus}^2} = mg$

You substitute in there

$M_{\bigoplus} = 6 \times 10^{24}$ kg

$G = 6.67 \times 10^{-11}$

$R_{\bigoplus} = 6.4 \times 10^6$ m

out pops our well-known number of 9.8 m/s^2.

Okay, my goal was to evaluate the gravitational potential energy that is defined in general, not in a special case when we are near the Earth.

$W_{WL} = \int_\infty^R \frac{mMG}{r^2} dr = \left -\frac{mMG}{r} \right|_\infty^R = -\frac{mMG}{R}$

And this is the gravitational potential energy at any distance capital R away from this object.

At infinity, it's now always zero.

Earlier, you had a choice where you chose your zero.

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.

If you move an object from A to B and this separation is h, and if A and B are very apart, the difference in potential energy is no longer mgh, but the difference in potential energy is the difference between U_A and U_B.

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 -mMG/r result and the mgh result that we found there?

The answer is NO.

I invite you to go through the following exercise.

$r_A = R_{\bigoplus}$

$r_B = R_{\bigoplus} + h$

And h << the radius of the Earth.

So I can calculate now what the difference in potential energy is between point B and point A.

And when I use the general equation and apply the first order of Taylor's expansion, you will immediately see that the result collapses into mgh because the g at the two points is so close that you will find then that it is approximately mgh.

We will, many, many times in the future, use the 1/r relationship for gravitational potential energy.

We will get very used to the idea that gravitational potential energy is negative everywhere, 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 use the simplification that the difference in gravitational potential energy is given by mgh.

Conservation of Mechanical Energy and a Wrecking Ball

I have here a pendulum.

15 kg, lifted 1m higher
potential energy is increased by about 150 J
150 J is enough to kill a person

They use these devices--it's called a wrecker ball--

Demo: release the pendulum to break a glass pane

Demo: Conservation of Mechanical Energy

Professor Lewin standing with his back against a wall, holding a 15-kg pendulum under his chin, and let it go
The pendulum swings back and barely touches Professor Lewin's chin

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11. Work, Energy, and Universal Gravitation.md

11. Work, Energy, and Universal Gravitation.md

Lecture 11: Work, Energy, and Universal Gravitation

1D Work and Kinetic Energy

Example

Example:

Example:

Work Calculated in 3-Dimensions

Gravity is a Conservative Force

When Gravity is the only Force

What Matters is the Difference in Potential Energy

A Roller Coaster, Upside-down

Newton's Law of Universal Gravitation

Conservation of Mechanical Energy and a Wrecking Ball

Demo: Conservation of Mechanical Energy

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11. Work, Energy, and Universal Gravitation.md

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Lecture 11: Work, Energy, and Universal Gravitation

1D Work and Kinetic Energy

Example

Example:

Example:

Work Calculated in 3-Dimensions

Gravity is a Conservative Force

When Gravity is the only Force

What Matters is the Difference in Potential Energy

A Roller Coaster, Upside-down

Newton's Law of Universal Gravitation

Conservation of Mechanical Energy and a Wrecking Ball

Demo: Conservation of Mechanical Energy