Class Notes, Fall 2001

Week of	Aug 22	Sep 5	Oct 1	Nov 5	Phy 205 page
	Aug 27	Sep 10	Oct 8	Nov 12
		Sep 17	Oct 15	Nov 19
		Sep 24	Oct 22	Nov 26
			Oct 29		Notes for Fall, 2000

Phy 205, Week 9

15 Oct More on the work-energy theorem. Reprise the one-dimensional constant force case to arrive at

What if the force isn't constant (even in one dimension)? As an example the Earth's gravitational force drops off with distance from its center.

Here R is the Earth's radius, about 6400 km, and x is the distance along the x-axis as measured from the center of the Earth. If you are at the Earth's surface then x = R and this reduces to the standard force at the surface. If you are above the surface by a distance of 6400 km, this reduces the force by a factor of four.

You can use the simple constant-force version of the work-energy theorem to get a crude approximation to this case. Just approximate the value of the force by some plausible constant value. I did say crude didn't I?

You can improve this rough approximation by approximating the graph of F_x by a set of steps: 2 steps or 3 steps or 100 steps or ... For each step, the force is a constant and I can apply the first form of the theorem to that smaller interval.

Start at initial coordinate x_i. The final coordinate is x_f. Pick some intermediate points x₁, x₂,... and in each of these intervals you can use the first and simplest form of the work-energy theorem.

and keep going. If you add all of these equations, notice the cancellation on the right-hand side. All the intermediate terms cancel; you're left with only the initial and final kinetic energies.

On the left you have an approximate value for the work done by this force. You can improve the approximation by putting in more intermediate points and doing more arithmetic. I did a detailed calculation in class in order to find the escape speed from the Earth. In order to do this efficiently I pulled a bit of a trick in the way that I chose the approximate values of the constants F_x1, F_x2, etc. For example, in the interval between x₂ and x₃, I noticed that of two plausible estimates

one is an overestimate and one is an underestimate of the values of F_x in this interval x₂ to x₃. I picked an intermediate estimate

Now when I substituted this value into the term with ( x₃ - x₂ ), I could simplify it and combine it with the other terms in a neat way. The overall result was

From this I showed how to get the escape speed from the Earth.

17 Oct Again state the one-dimensional work-energy theorem. Constant force, then to approximate a non-constant force by a set of steps, then to take a limit as this approximation becomes exact.

Example: Suppose that the force function is given in the MKS system to have the numerical value

In order to do the arithmetic I have to choose specific numbers here. A graph of this starts at 4 at the origin and by x = 1 has dropped to 2. What is the work done by this in the interval x = 0 to x = 1? I can make a crude approximation to the work done if I replace this by a constant equal to the value at the midpoint, x = 1/2. The result is

To make this a bit less crude, divide the interval into two pieces, from 0 to 1/2 and from 1/2 to 1. In each of these intervals, approximate the force by a constant value and for that, choose the value in the middle of the interval. The two terms for the work are then

this is 32/17 + 32/25 = 3.162 and is an improved approximation to whatever the exact answer is.
For homework, try this with four intervals.

When you let these distance intervals shrink to zero in order to get an exact answer, the resulting work-energy theorem is denoted

ó õ xf xi  Fx dx = Δ K = Kf - Ki = m vf2 / 2 - m vi2 / 2

This integral symbol is the abbreviation for the limit of that sum as the interval size shrinks to zero.

How do you evaluate these things? The fundamental theorem of calculus shows how. It relates this integral to the anti-derivative that we've been using all along.

The anti-derivative is the function whose derivative is what you've got. The fundamental theorem of calculus relates the integral to the anti-derivative. If dF/dx = f then

ó õ b a  f( x )  dx = F( b ) - F( a )

For the work energy theorem, the relationship is

Fx( x ) = - dU dx ,        so       W = ó õ xf xi  Fx dx = - U( x ) ]xixf = - U( xf ) + U( xi )

Use this in the work-energy theorem and you rearrange it to

This is in the form of a conservation law. Something stays the same, and in this case it is the combination of kinetic energy and potential energy, U.

I had actually worked this out for the special case of gravity, F_x = -m g, and talked about it before this generalization.

The units of energy. From mv²/2, the units are kg m / s², the Joule. Others are the kilowatt-hour, the calorie, and the kilocalorie (dietician's calorie). Some examples of each.

Power, dW / dt, the rate of doing work or the rate of expending energy. This has units Joule/second, or Watt. To see how this compares to something you live with every day, note that you probably consume in the neighborhood of 2000 kcal per day. Some people more, some less. Translate this into Watts, and it equals about 100 Watts.