Gravitational potential energy
Conservation of energy can be understood only if we have the formula for all of its forms. I wish to discuss the formula for gravitational energy near the surface of the Earth, and I wish to derive this formula in a way which has nothing to do with history but is simply a line of reasoning invented for this particular lecture to give you an illustration of the remarkable fact that a great deal about nature can be extracted from a few facts and close reasoning. It is an illustration of the kind of work theoretical physicists become involved in. It is patterned after a most excellent argument by Mr. Carnot on the efficiency of steam engines.1 Consider weight-lifting machines—machines which have the property that they lift one weight by lowering another. Let us also make a hypothesis: that there is no such thing as perpetual motion with these weight-lifting machines. (In fact, that there is no perpetual motion at all is a general statement of the law of conservation of energy.)  We must be careful to define perpetual motion. First, let us do it for weight-lifting machines. If, when we have lifted and lowered a lot of weights and restored the machine to the original condition, we find that the net result is to have lifted a weight, then we have a perpetual motion machine because we can use that lifted weight to run something else. That is, provided the machine which lifted the weight is brought back to its exact original condition, and furthermore that it is completely self-contained—that it has not received the energy to lift that weight from some external source—like Bruce’s blocks. A very simple weight-lifting machine is shown in Fig. 4–1. This machine lifts weights three units “strong.” We place three units on one balance pan, and one unit on the other. However, in order to get it actually to work, we must lift a little weight off the left pan. On the other hand, we could lift a one-unit weight by lowering the three-unit weight, if we cheat a little by lifting a little weight off the other pan. Of course, we realize that with any actual lifting machine, we must add a little extra to get it to run. This we disregard, temporarily. Ideal machines, although they do not exist, do not require anything extra. A machine that we actually use can be, in a sense, almost reversible: that is, if it will lift the weight of three by lowering a weight of one, then it will also lift nearly the weight of one the same amount by lowering the weight of three. We imagine that there are two classes of machines, those that are not reversible, which includes all real machines, and those that are reversible, which of course are actually not attainable no matter how careful we may be in our design of bearings, levers, etc. We suppose, however, that there is such a thing—a reversible machine—which lowers one unit of weight (a pound or any other unit) by one unit of distance, and at the same time lifts a three-unit weight. Call this reversible machine, Machine $A$. Suppose this particular reversible machine lifts the three-unit weight a distance $X$. Then suppose we have another machine, Machine $B$, which is not necessarily reversible, which also lowers a unit weight a unit distance, but which lifts three units a distance $Y$. We can now prove that $Y$ is not higher than $X$; that is, it is impossible to build a machine that will lift a weight any higher than it will be lifted by a reversible machine. Let us see why. Let us suppose that $Y$ were higher than $X$. We take a one-unit weight and lower it one unit height with Machine $B$, and that lifts the three-unit weight up a distance $Y$. Then we could lower the weight from $Y$ to $X$, obtaining free power, and use the reversible Machine $A$, running backwards, to lower the three-unit weight a distance $X$ and lift the one-unit weight by one unit height. This will put the one-unit weight back where it was before, and leave both machines ready to be used again! We would therefore have perpetual motion if $Y$ were higher than $X$, which we assumed was impossible. With those assumptions, we thus deduce that $Y$ is not higher than $X$, so that of all machines that can be designed, the reversible machine is the best. We can also see that all reversible machines must lift to exactly the same height. Suppose that $B$ were really reversible also. The argument that $Y$ is not higher than $X$ is, of course, just as good as it was before, but we can also make our argument the other way around, using the machines in the opposite order, and prove that $X$ is not higher than $Y$. This, then, is a very remarkable observation because it permits us to analyze the height to which different machines are going to lift something without looking at the interior mechanism. We know at once that if somebody makes an enormously elaborate series of levers that lift three units a certain distance by lowering one unit by one unit distance, and we compare it with a simple lever which does the same thing and is fundamentally reversible, his machine will lift it no higher, but perhaps less high. If his machine is reversible, we also know exactly how high it will lift. To summarize: every reversible machine, no matter how it operates, which drops one pound one foot and lifts a three-pound weight always lifts it the same distance, $X$. This is clearly a universal law of great utility. The next question is, of course, what is $X$? Suppose we have a reversible machine which is going to lift this distance $X$, three for one. We set up three balls in a rack which does not move, as shown in Fig. 4–2. One ball is held on a stage at a distance one foot above the ground. The machine can lift three balls, lowering one by a distance $1$. Now, we have arranged that the platform which holds three balls has a floor and two shelves, exactly spaced at distance $X$, and further, that the rack which holds the balls is spaced at distance $X$, (a). First we roll the balls horizontally from the rack to the shelves, (b), and we suppose that this takes no energy because we do not change the height. The reversible machine then operates: it lowers the single ball to the floor, and it lifts the rack a distance $X$, (c). Now we have ingeniously arranged the rack so that these balls are again even with the platforms. Thus we unload the balls onto the rack, (d); having unloaded the balls, we can restore the machine to its original condition. Now we have three balls on the upper three shelves and one at the bottom. But the strange thing is that, in a certain way of speaking, we have not lifted two of them at all because, after all, there were balls on shelves $2$ and $3$ before. The resulting effect has been to lift one ball a distance $3X$. Now, if $3X$ exceeds one foot, then we can lower the ball to return the machine to the initial condition, (f), and we can run the apparatus again. Therefore $3X$ cannot exceed one foot, for if $3X$ exceeds one foot we can make perpetual motion. Likewise, we can prove that one foot cannot exceed $3X$, by making the whole machine run the opposite way, since it is a reversible machine. Therefore $3X$ is neither greater nor less than a foot, and we discover then, by argument alone, the law that $X=\tfrac{1}{3}$ foot. The generalization is clear: one pound falls a certain distance in operating a reversible machine; then the machine can lift $p$ pounds this distance divided by $p$. Another way of putting the result is that three pounds times the height lifted, which in our problem was $X$, is equal to one pound times the distance lowered, which is one foot in this case. If we take all the weights and multiply them by the heights at which they are now, above the floor, let the machine operate, and then multiply all the weights by all the heights again, there will be no change. (We have to generalize the example where we moved only one weight to the case where when we lower one we lift several different ones—but that is easy.) We call the sum of the weights times the heights gravitational potential energy—the energy which an object has because of its relationship in space, relative to the earth. The formula for gravitational energy, then, so long as we are not too far from the earth (the force weakens as we go higher) is \begin{equation} \label{Eq:I:4:3} \begin{pmatrix} \text{gravitational}\\ \text{potential energy}\\ \text{for one object} \end{pmatrix}= (\text{weight})\times(\text{height}). \end{equation} It is a very beautiful line of reasoning. The only problem is that perhaps it is not true. (After all, nature does not have to go along with our reasoning.) For example, perhaps perpetual motion is, in fact, possible. Some of the assumptions may be wrong, or we may have made a mistake in reasoning, so it is always necessary to check. It turns out experimentally, in fact, to be true. The general name of energy which has to do with location relative to something else is called potential energy. In this particular case, of course, we call it gravitational potential energy. If it is a question of electrical forces against which we are working, instead of gravitational forces, if we are “lifting” charges away from other charges with a lot of levers, then the energy content is called electrical potential energy. The general principle is that the change in the energy is the force times the distance that the force is pushed, and that this is a change in energy in general: \begin{equation} \label{Eq:I:4:4}  \begin{pmatrix} \text{change in}\\ \text{energy} \end{pmatrix}= (\text{force})\times  \begin{pmatrix} \text{distance force}\\ \text{acts through} \end{pmatrix}. \end{equation} We will return to many of these other kinds of energy as we continue the course. The principle of the conservation of energy is very useful for deducing what will happen in a number of circumstances. In high school we learned a lot of laws about pulleys and levers used in different ways. We can now see that these “laws” are all the same thing, and that we did not have to memorize $75$ rules to figure it out. A simple example is a smooth inclined plane which is, happily, a three-four-five triangle (Fig. 4–3). We hang a one-pound weight on the inclined plane with a pulley, and on the other side of the pulley, a weight $W$. We want to know how heavy $W$ must be to balance the one pound on the plane. How can we figure that out? If we say it is just balanced, it is reversible and so can move up and down, and we can consider the following situation. In the initial circumstance, (a), the one pound weight is at the bottom and weight $W$ is at the top. When $W$ has slipped down in a reversible way, (b), we have a one-pound weight at the top and the weight $W$ the slant distance, or five feet, from the plane in which it was before. We lifted the one-pound weight only three feet and we lowered $W$ pounds by five feet. Therefore $W=\tfrac{3}{5}$ of a pound. Note that we deduced this from the conservation of energy, and not from force components. Cleverness, however, is relative. It can be deduced in a way which is even more brilliant, discovered by Stevinus and inscribed on his tombstone.2  Figure 4–4 explains that it has to be $\tfrac{3}{5}$ of a pound, because the chain does not go around. It is evident that the lower part of the chain is balanced by itself, so that the pull of the five weights on one side must balance the pull of three weights on the other, or whatever the ratio of the legs. You see, by looking at this diagram, that $W$ must be $\tfrac{3}{5}$ of a pound. (If you get an epitaph like that on your gravestone, you are doing fine.) Let us now illustrate the energy principle with a more complicated problem, the screw jack shown in Fig. 4–5. A handle $20$ inches long is used to turn the screw, which has $10$ threads to the inch. We would like to know how much force would be needed at the handle to lift one ton ($2000$ pounds). If we want to lift the ton one inch, say, then we must turn the handle around ten times. When it goes around once it goes approximately $126$ inches. The handle must thus travel $1260$ inches, and if we used various pulleys, etc., we would be lifting our one ton with an unknown smaller weight $W$ applied to the end of the handle. So we find out that $W$ is about $1.6$ pounds. This is a result of the conservation of energy.  Take now the somewhat more complicated example shown in Fig. 4–6. A rod or bar, $8$ feet long, is supported at one end. In the middle of the bar is a weight of $60$ pounds, and at a distance of two feet from the support there is a weight of $100$ pounds. How hard do we have to lift the end of the bar in order to keep it balanced, disregarding the weight of the bar? Suppose we put a pulley at one end and hang a weight on the pulley. How big would the weight $W$ have to be in order for it to balance? We imagine that the weight falls any arbitrary distance—to make it easy for ourselves suppose it goes down $4$ inches—how high would the two load weights rise? The center rises $2$ inches, and the point a quarter of the way from the fixed end lifts $1$ inch. Therefore, the principle that the sum of the heights times the weights does not change tells us that the weight $W$ times $4$ inches down, plus $60$ pounds times $2$ inches up, plus $100$ pounds times $1$ inch has to add up to nothing:  \begin{equation} \label{Eq:I:4:5} -4W+(2)(60)+(1)(100)=0,\quad W=\text{$55$ lb}. \end{equation} 
 \begin{equation} \begin{gathered} -4W+(2)(60)+(1)(100)=0,\\[.5ex] W=\text{$55$ lb}. \end{gathered} \label{Eq:I:4:5} \end{equation}  Thus we must have a $55$-pound weight to balance the bar. In this way we can work out the laws of “balance”—the statics of complicated bridge arrangements, and so on. This approach is called the principle of virtual work, because in order to apply this argument we had to imagine that the structure moves a little—even though it is not really moving or even movable. We use the very small imagined motion to apply the principle of conservation of energy.