# 2.1 Energy

- Page ID
- 939

### 2.1.1 The energy concept

You'd probably like to be able to drive your car and light your apartment without having to pay money for gas and electricity, and if you do a little websurfing, you can easily find people who say they have the solution to your problem. This kind of scam has been around for centuries. It used to be known as a perpetual motion machine, but nowadays the con artists' preferred phrase is “free energy.”^{1} A typical “free-energy” machine would be a sealed box that heats your house without needing to be plugged into a wall socket or a gas pipe. Heat comes out, but nothing goes in, and this can go on indefinitely. But an interesting thing happens if you try to check on the advertised performance of the machine. Typically, you'll find out that either the device is still in development, or it's back-ordered because so many people have already taken advantage of this Fantastic Opportunity! In a few cases, the magic box exists, but the inventor is only willing to demonstrate very small levels of heat output for short periods of time, in which case there's probably a tiny hearing-aid battery hidden in there somewhere, or some other trick.

Since nobody has ever succeeded in building a device that creates heat out of nothing, we might also wonder whether any device exists that can do the opposite, turning heat into nothing. You might think that a refrigerator was such a device, but actually your refrigerator doesn't destroy the heat in the food. What it really does is to extract some of the heat and bring it out into the room. That's why it has big radiator coils on the back, which get hot when it's in operation.

If it's not possible to destroy or create heat outright, then you might start to suspect that heat was a conserved quantity. This would be a successful rule for explaining certain processes, such as the transfer of heat between a cold Martini and a room-temperature olive: if the olive loses a little heat, then the drink must gain the same amount. It would fail in general, however.

Sunlight can heat your skin, for example, and a hot lightbulb filament can cool off by emitting light. Based on these observations, we could revise our proposed conservation law, and say that there is something called heatpluslight, which is conserved. Even this, however, needs to be generalized in order to explain why you can get a painful burn playing baseball when you slide into a base. Now we could call it heatpluslightplusmotion. The word is getting pretty long, and we haven't even finished the list.

Rather than making the word longer and longer, physicists have hijacked the word “energy” from ordinary usage, and give it a new, specific technical meaning. Just as the Parisian platinum-iridium kilogram defines a specific unit of mass, we need to pick something that defines a definite unit of energy. The metric unit of energy is the joule (J), and we'll define it as the amount of energy required to heat 0.24 grams of water from 20 to 21 degrees Celsius. (Don't memorize the numbers.)^{2}

Example 1: Temperature of a mixture |
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\(\triangleright\) If 1.0 kg of water at 20°C is mixed with 4.0 kg of water at 30°C, what is the temperature of the mixture? \(\triangleright\) Let's assume as an approximation that each degree of temperature change corresponds to the same amount of energy. In other words, we assume \(\Delta E\text{=} mc\Delta T\), regardless of whether, as in the definition of the joule, we have \(\Delta T=\text(21°\text{C}\ - 20°\text{C})\) or, as in the present example, some other combination of initial and final temperatures. To be consistent with the definition of the joule, we must have \(c=(1°C)/(0.24g)/(1°C)=4.2\times 10^{3}J/kg\cdot °C\), which is referred to as the specific heat of water. Conservation of energy tells us \(\Delta{} E=0\), so \[\begin{equation*} m_1c\Delta{} T_1+ m_2c\Delta{} T_2 = 0 \end{equation*}\] or \[\begin{align*} \frac{\Delta{} T_1}{\Delta{} T_2} &= -\frac{ m_2}{ m_1} \\ & = - 4.0 {}. \end{align*}\] If \(T_1\) has to change four times as much as \(T_2\), and the two final temperatures are equal, then the final temperature must be 28°C . |

Note how only *differences* in temperature and energy appeared in the preceding example. In other words, we don't have to make any assumptions about whether there is a temperature at which all an object's heat energy is removed. Historically, the energy and temperature units were invented before it was shown that there is such a temperature, called absolute zero. There is a scale of temperature, the Kelvin scale, in which the unit of temperature is the same as the Celsius degree, but the zero point is defined as absolute zero. But as long as we only deal with temperature differences, it doesn't matter whether we use Kelvin or Celsius. Likewise, as long as we deal with differences in heat energy, we don't normally have to worry about the total amount of heat energy the object has. In standard physics terminology, “heat” is used only to refer to differences, while the total amount is called the object's “thermal energy.”This distinction is often ignored by scientists in casual speech, and in this book I'll usually use “heat” for either quantity.

We're defining energy by adding up things from a list, which we lengthen as needed: heat, light, motion, etc. One objection to this approach is aesthetic: physicists tend to regard complication as a synonym for ugliness. If we have to keep on adding more and more forms of energy to our laundry list, then it's starting to sound like energy is distressingly complicated. Luckily it turns out that energy is simpler than it seems. Many forms of energy that are apparently unrelated turn out to be manifestations of a small number of forms at the atomic level, and this is the topic of section 2.4.

##### Discussion Questions

◊ The ancient Greek philosopher Aristotle said that objects “naturally” tended to slow down, unless there was something pushing on them to keep them moving. What important insight was he missing?”

Another possible objection is that the open-ended approach to defining energy might seem like a kind of cheat, since we keep on inventing new forms whenever we need them. If a certain experiment seems to violate conservation of energy, can't we just invent a new form of invisible “mystery energy” that patches things up? This would be like balancing your checkbook by putting in a fake transaction that makes your calculation of the balance agree with your bank's. If we could fudge this way, then conservation of energy would be untestable --- impossible to prove or disprove.

Actually all scientific theories are unprovable. A theory can never be proved, because the experiments can only cover a finite number out of the infinitely many situations in which the theory is supposed to apply. Even a million experiments won't suffice to prove it in the same sense of the word “proof” that is used in mathematics. However, even one experiment that contradicts a theory is sufficient to show that the theory is wrong. A theory that is immune to disproof is a bad theory, because there is no way to test it. For instance, if I say that 23 is the maximum number of angels that can dance on the head of a pin, I haven't made a properly falsifiable scientific theory, since there's no method by which anyone could even attempt to prove me wrong based on observations or experiments.

Conservation of energy is testable because new forms of energy are expected to show regular mathematical behavior, and are supposed to be related in a measurable way to observable phenomena. As an example, let's see how to extend the energy concept to include motion.

### 2.1.3 Kinetic energy

Energy of motion is called *kinetic energy*. (The root of the word is the same as the word “cinema” -- in French, kinetic energy is “énergie cinétique.”) How does an object's kinetic energy depend on its mass and velocity? Joule attempted a conceptually simple experiment on his honeymoon in the French-Swiss Alps near Mt. Chamonix, in which he measured the difference in temperature between the top and bottom of a waterfall. The water at the top of the falls has some gravitational energy, which isn't our subject right now, but as it drops, that gravitational energy is converted into kinetic energy, and then into heat energy due to internal friction in the churning pool at the bottom:

\[\begin{equation*} \text{gravitational energy} \rightarrow \text{kinetic energy} \rightarrow \text{heat energy} \end{equation*}\]

In the logical framework of this book's presentation of energy, the significance of the experiment is that it provides a way to find out how an object's kinetic energy depends on its mass and velocity. The increase in heat energy should equal the kinetic energy of the water just before impact, so in principle we could measure the water's mass, velocity, and kinetic energy, and see how they relate to one another.^{4}

Although the story is picturesque and memorable, most books that mention the experiment fail to note that it was a failure! The problem was that heat wasn't the only form of energy being released. In reality, the situation was more like this:

\[\begin{align*} \text{gravitational energy} \rightarrow & \text{kinetic energy} \\ \rightarrow & \text{heat energy} \\ & + \text{sound energy} \\ & + \text{energy of partial evaporation} . \end{align*}\]

The successful version of the experiment, shown in figures d and f, used a paddlewheel spun by a dropping weight. As with the waterfall experiment, this one involves several types of energy, but the difference is that in this case, they can all be determined and taken into account. (Joule even took the precaution of putting a screen between himself and the can of water, so that the infrared light emitted by his warm body wouldn't warm it up at all!) The result^{5} is

Whenever you encounter an equation like this for the first time, you should get in the habit of interpreting it. First off, we can tell that by making the mass or velocity greater, we'd get more kinetic energy. That makes sense. Notice, however, that we have mass to the first power, but velocity to the second. Having the whole thing proportional to mass to the first power is necessary on theoretical grounds, since energy is supposed to be additive. The dependence on \(v^2\) couldn't have been predicted, but it is sensible. For instance, suppose we reverse the direction of motion. This would reverse the sign of \(v\), because in one dimension we use positive and negative signs to indicate the direction of motion. But since \(v^2\) is what appears in the equation, the resulting kinetic energy is unchanged.

What about the factor of 1/2 in front? It comes out to be exactly 1/2 by the design of the metric system. If we'd been using the old-fashioned British engineering system of units (which is no longer used in the U.K.), the equation would have been \(K=(7.44\times10^{-2}\ \text{Btu}\cdot\text{s}^2/\text{slug}\cdot\text{ft}^2)mv^2\). The version of the metric system called the SI,^{6} in which everything is based on units of kilograms, meters, and seconds, not only has the numerical constant equal to 1/2, but makes it unitless as well. In other words, we can think of the joule as simply an abbreviation, 1 J=1 \(\text{kg}\cdot\text{m}^2/\text{s}^2\). More familiar examples of this type of abbreviation are 1 minute=60 s, and the metric unit of land area, 1 hectare=\(10000\ \text{m}^2\).

Example 2: Ergs and joules |
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\(\triangleright\) There used to be two commonly used systems of metric units, referred to as mks and cgs. The mks system, now called the SI, is based on the meter, the kilogram, and the second. The cgs system, which is now obsolete, was based on the centimeter, the gram, and the second. In the cgs system, the unit of energy is not the joule but the erg, 1 erg=1 \(\text{g}\cdot\text{cm}^2/\text{s}^2\). How many ergs are in one joule? \(\triangleright\) The simplest approach is to treat the units as if they were algebra symbols. \[\begin{align*} \text{1 J} &= 1\ \frac{\text{kg}\cdot\text{m}^2}{\text{s}^2} \\ &= 1\ \frac{\text{kg}\cdot\text{m}^2}{\text{s}^2} \times \frac{\text{1000 g}}{\text{1 kg}} \times \left(\frac{\text{100 cm}}{\text{1 m}}\right)^2 \\ &= 10^7\ \frac{\text{g}\cdot\text{cm}^2}{\text{s}^2} \\ &= 10^7\ \text{erg} \end{align*}\] |

Example 3: Cabin air in a jet airplane |
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\(\triangleright\) A jet airplane typically cruises at a velocity of 270 m/s. Outside air is continuously pumped into the cabin, but must be cooled off first, both because (1) it heats up due to friction as it enters the engines, and (2) it is heated as a side-effect of being compressed to cabin pressure. Calculate the increase in temperature due to the first effect. The specific heat of dry air is about 1.0\(\times10^3\) \(\text{J/kg}\cdot^{\circ}\)\(\text{C}\). \(\triangleright\) This is easiest to understand in the frame of reference of the plane, in which the air rushing into the engine is stopped, and its kinetic energy converted into heat. \[\begin{align*} 0 &= \Delta{} E \\ &= \Delta{} K+\Delta{} E_{heat} . \end{align*}\] In the plane's frame of reference, the air's initial velocity is \(v_i\)=270 m/s, and its final velocity is zero, so the change in its kinetic energy is negative, \[\begin{align*} \Delta{} K &= K_{f} - K_{i} \\ &= 0-\text{(1/2)} m{ v_{i}}^2 \\ &= -\text{(1/2)} m{ v_i}^2 . \end{align*}\] Assuming that the specific heat of air is roughly independent of temperature (which is why the number was stated with the word “about”), we can substitute into \(0 = \Delta{} K+\Delta{} E_{heat}\), giving \[\begin{equation*} 0 = -\frac{1}{2} m{ v_{i}}^2+ mc\Delta{} T \end{equation*}\] Assuming that the specific heat of air is roughly independent of temperature (which is why the number was stated with the word “about”), we can substitute into \(0 = \Delta{} K+\Delta{} E_{heat}\), giving \[\begin{equation*} 0 = -\frac{1}{2} m{ v_{i}}^2+ mc\Delta{} T \end{equation*}\]
\[\begin{equation*} \frac{1}{2}{ v_{i}}^2 = c\Delta{} T . \end{equation*}\] Note how the mass cancels out. This is a big advantage of solving problems algebraically first, and waiting until the end to plug in numbers. With a purely numerical approach, we wouldn't even have known what value of \(m\) to pick, or if we'd guessed a value like 1 kg, we wouldn't have known whether our answer depended on that guess. Solving for \(\Delta T\), and writing \(v\) instead of \(v_i\) for simplicity, we find \[\begin{align*} \Delta{} T & = \frac{ v^2}{2 c} \\ &\approx 40^{\circ}\text{C} . \end{align*}\] The passengers would be boiled alive if not for the refrigeration. The first stage of cooling happens via heat exchangers in the engine struts, but a second stage, using a refrigerator under the floor of the cabin, is also necessary. Running this refrigerator uses up energy, cutting into the fuel efficiency of the airplane, which is why typically only 50% of the cabin's air is replaced in each pumping cycle of 2-3 minutes. Although the airlines prefer to emphasize that this is a much faster recirculation rate than in the ventilation systems of most buildings, people are packed more tightly in an airplane. |

### 2.1.4 Power

Power, \(P\), is defined as the rate of change of energy, \(d{}E/d{}t\). Power thus has units of joules per second, which are usually abbreviated as watts, 1 W=1 J/s. Since energy is conserved, we would have \(d{}E/d{}t=0\) if \(E\) was the total energy of a closed system, and that's not very interesting. What's usually more interesting to discuss is either the power flowing in or out of an open system, or the rate at which energy is being transformed from one form into another. The following is an example of energy flowing into an open system.

Example 4: Heating by a lightbulb |
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\(\triangleright\) The electric company bills you for energy in units of kilowatt-hours (kilowatts multiplied by hours) rather than in SI units of joules. How many joules is a kilowatt-hour? \(\triangleright\) 1 kilowatt-hour = (1 kW)(1 hour) = (1000 J/s)(3600 s) = 3.6 MJ. |

Now here's an example of energy being transformed from one form into another.

Example 5: Human wattage |
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\(\triangleright\) Food contains chemical energy (discussed in more detail in section 2.4), and for historical reasons, food energy is normally given in non-SI units of Calories. One Calorie with a capital “C” equals 1000 calories, and 1 calorie is defined as 4.18 J. A typical person consumes 2000 Calories of food in a day, and converts nearly all of that directly to body heat. Compare the person's heat production to the rate of energy consumption of a 100-watt lightbulb. \(\triangleright\) Strictly speaking, we can't really compute the derivative \(d{} E/d{} t\), since we don't know how the person's metabolism ebbs and flows over the course of a day. What we can really compute is \(\Delta E/\Delta t\), which is the power averaged over a one-day period. Converting to joules, we find \(\Delta E=8\times10^6\) J for the amount of energy transformed into heat within our bodies in one day. Converting the time interval likewise into SI units, \(\Delta t=9\times10^4\) s. Dividing, we find that our power is 90 J/s = 90 W, about the same as a light bulb. |

### 2.1.5 Gravitational energy

Gravitational energy, to which I've already alluded, is different from heat and kinetic energy in an important way. Heat and kinetic energy are properties of a single object, whereas gravitational energy describes an interaction between two objects. When the skater in figures g and h is at the top, his distance from the bulk of the planet earth is greater. Since we observe his kinetic energy decreasing on the way up, there must be some other form of energy that is increasing. We invent a new form of energy, called gravitational energy, and written \(U\) or \(U_g\), which depends on the distance between his body and the planet. Where is this energy? It's not in the skater's body, and it's not inside the earth, either, since it takes two to tango. If either object didn't exist, there wouldn't be any interaction or any way to measure a distance, so it wouldn't make sense to talk about a distance-dependent energy. Just as marriage is a relationship between two people, gravitational energy is a relationship between two objects.

There is no precise way to define the distance between the skater and the earth, since both are objects that have finite size. As discussed in more detail in section 2.3, gravity is one of the fundamental forces of nature, a universal attraction between any two particles that have mass. Each atom in the skater's body is at a definite distance from each atom in the earth, but each of these distances is different. An atom in his foot is only a few centimeters from some of the atoms in the plaster side of the pool, but most of the earth's atoms are thousands of kilometers away from him. In theory, we might have to add up the contribution to the gravitational energy for every interaction between an atom in the skater's body and an atom in the earth.

For our present purposes, however, there is a far simpler and more practical way to solve problems. In any region of the earth's surface, there is a direction called “down,” which we can establish by dropping a rock or hanging a plumb bob. In figure h, the skater is moving up and down in one dimension, and if we did measurements of his kinetic energy, like the made-up data in the figure, we could infer his gravitational energy. As long as we stay within a relatively small range of heights, we find that an object's gravitational energy increases at a steady rate with height. In other words, the strength of gravity doesn't change much if you only move up or down a few meters. We also find that the gravitational energy is proportional to the mass of the object we're testing. Writing \(y\) for the height, and \(g\) for the overall constant of proportionality, we have

\[\begin{equation*} \begin{array}{cr} U_g=mgy . & \text{[gravitational energy; $y$=height; only ac-} \\ & \text{curate within a small range of heights]} \end{array} \end{equation*}\]

The number \(g\), with units of joules per kilogram per meter, is called the *gravitational field*. It tells us the strength of gravity in a certain region of space. Near the surface of our planet, it has a value of about 9.8 J/kg·m, which is conveniently close to 10 J/kg·m for rough calculations.

Example 6: Velocity at the bottom of a drop |
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If the skater in figure g drops 3 meters from rest, what is his velocity at the bottom of the pool?
Starting from conservation of energy, we have \[\begin{align*} 0 &= \Delta E \\ &= \Delta K +\Delta U \\ &= K_f - K_i + U_f - U_i \\ &= \frac{1}{2} m { v_f }^2 + mgy_f - mgy_i & \text{(because $K_i$=0)} \\ &= \frac{1}{2} m { v_f }^2 + mg \Delta y , & \text{($\Delta y\lt$0)} \text{so} v &= \sqrt{-2 g \Delta y} \\ &= \sqrt{-\text{(2)(10\ J/kg}\!\cdot\!{}\text{m)(}-\text{3 m)}} \\ &= \text{8 m/s} & \text{(rounded to one sig. fig.)} \end{align*}\] There are a couple of important things to note about this example. First, we were able to massage the equation so that it only involved \(\Delta{}y\), rather than \(y\) itself. In other words, we don't need to worry about where \(y=0\) is; any coordinate system will work, as long as the positive \(y\) axis points up, not down. This is no accident. Gravitational energy can always be changed by adding a constant onto it, with no effect on the final result, as long as you're consistent within a given problem. The other interesting thing is that the mass canceled out: even if the skater gained weight or strapped lead weights to himself, his velocity at the bottom would still be 8 m/s. This isn't an accident either. This is the same conclusion we reached in section 1.2, based on the equivalence of gravitational and inertial mass. The kinetic energy depends on the inertial mass, while gravitational energy is related to gravitational mass, but since these two masses are equal, we were able to use a single symbol, \(m\), for them, and cancel them out. |

We can see from the equation \(v=\sqrt{-2g\Delta{}y}\) that a falling object's velocity isn't constant. It increases as the object drops farther and farther. What about its acceleration? If we assume that air friction is negligible, the arguments in section 1.2 show that the acceleration can't depend on the object's mass, so there isn't much else the acceleration *can* depend on besides \(g\). In fact, the acceleration of a falling object equals \(-g\) (in a coordinate system where the positive \(y\) axis points up), as we can easily show using the chain rule:

\[\begin{align*} \left(\frac{d{}v}{d{}t}\right)&= \left(\frac{d{}v}{d{}K}\right)\left(\frac{d{}K}{d{}U}\right)\left(\frac{d{}U}{d{}y}\right)\left(\frac{d{}y}{d{}t}\right) \\ &= \left(\frac{1}{mv}\right)(-1)(mg)(v) \\ &= -g , \end{align*}\]

where I've calculated \(d{}v/d{}K\) as \(1/(d{}K/d{}v)\), and \(d{}K/d{}U=-1\) can be found by differentiating \(K+U=\text{(constant)}\) to give \(d{}K+d{}U=0\).^{7}

We can also check that the units of \(g\), \(J/kg\cdot m\), are equivalent to the units of acceleration,

\[\begin{equation*} \frac{\text{J}}{\text{kg}\!\cdot\!\text{m}} = \frac{\text{kg}\!\cdot\!\text{m}^2/\text{s}^2} {\text{kg}\!\cdot\!\text{m}} \\ = \frac{\text{m}}{\text{s}^2} , \end{equation*}\]

and therefore the strength of the gravitational field near the earth's surface can just as well be stated as \(10\ \text{m}/\text{s}^2\).

Example 7: Speed after a given time |
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An object falls from rest. How fast is it moving after two seconds? Assume that the amount of energy converted to heat by air friction is negligible.
Under the stated assumption, we have \(a=- g\), which can be integrated to give \(v=- gt+\text{constant}\). If we let \(t=0\) be the beginning of the fall, then the constant of integration is zero, so at \(t=\text{2 s}\) we have \(v=- gt=-\text{(10 m/s}^2\text{)}\times\text{(2 s)}=\text{20 m/s}\). |

Example 8: The Vomit Comet |
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The U.S. Air Force has an airplane, affectionately known as the Vomit Comet, in which astronaut trainees can experience simulated weightlessness. Oversimplifying a little, imagine that the plane climbs up high, and then drops straight down like a rock. (It actually flies along a parabola.) Since the people are falling with the same acceleration as the plane, the sensation is just like what you'd experience if you went out of the earth's gravitational field. If the plane can start from 10 km up, what is the maximum amount of time for which the dive can last?
Based on data about acceleration and distance, we want to find time. Acceleration is the second derivative of distance, so if we integrate the acceleration twice with respect to time, we can find how position relates to time. For convenience, let's pick a coordinate system in which the positive \(y\) axis is down, so \(a\)=\(g\) instead of \(- g\). \[\begin{align*} a &= g \\ v &= gt + \text{constant} & \text{(integrating)} \\ &= gt & \text{(starts from rest)} \\ y &= \frac{1}{2} gt^2 +\text{constant} & \text{(integrating again)} \end{align*}\] Choosing our coordinate system to have \(y=0\) at \(t=0\), we can make the second constant of integration equal zero as well, so \[\begin{align*} t &= \sqrt{\frac{2 y}{ g}} \\ &= \sqrt{\frac{2\cdot\text{10000 m}}{\text{10 m/s}^2}} \\ &= \sqrt{\text{2000\ s}^2} \\ &= \text{40\ s} & \text{(to one sig. fig.)} \end{align*}\] Note that if we hadn't converted the altitude to units of meters, we would have gotten the wrong answer, but we would have been alerted to the problem because the units inside the square root wouldn't have come out to be \(\text{s}^2\). In general, it's a good idea to convert all your data into SI (meter-kilogram-second) units before you do anything with them. |

Example 9: High road, low road |
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In figure i, what can you say based on conservation of energy about the speeds of the balls when the reach point B? What does conservation of energy tell you about which ball will get there first? Assume friction doesn't convert any mechanical energy to heat or sound energy.
Since friction is assumed to be negligible, there are only two forms of energy involved: kinetic and gravitational. Since both balls start from rest, and both lose the same amount of gravitational energy, they must have the same kinetic energy at the end, and therefore they're rolling at the same speed when they reach B. (A subtle point is that the balls have kinetic energy both because they're moving through space and because they're spinning as they roll. These two types of energy must be in fixed proportion to one another, so this has no effect on the conclusion.) Conservation of energy does not, however, tell us anything obvious about which ball gets there first. This is a general problem with applying conservation laws: conservation laws don't refer directly to time, since they are statements that something stays the same at all moments in time. We expect on intuitive grounds that the ball that goes by the lower ramp gets to B first, since it builds up speed early on. |

Example 10: Buoyancy |
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\(\triangleright\) A cubical box with mass \(m\) and volume \(V= b^3\) is submerged in a fluid of density \(\rho\). How much energy is required to raise it through a height \(\Delta y\)?
As the box moves up, it invades a volume \(V'= b^2\Delta y\) previously occupied by some of the fluid, and fluid flows into an equal volume that it has vacated on the bottom. Lowering this amount of fluid by a height \(b\) reduces the fluid's gravitational energy by \(\rho V' gb=\rho g b^3\Delta y\), so the net change in energy is \[\begin{align*} \Delta E &= mg\Delta y-\rho g b^3\Delta y \\ &= ( m-\rho V) g\Delta y . \end{align*}\] In other words, it's as if the mass of the box had been reduced by an amount equal to the fluid that otherwise would have occupied that volume. This is known as Archimedes' principle, and it is true even if the box is not a cube, although we'll defer the more general proof until page 203 in chapter 3. If the box is less dense than the fluid, then it will float. |

Example 11: A simple machine |
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\(\triangleright\) If the father and son on the seesaw in figure k start from rest, what will happen? \(\triangleright\) Note that although the father is twice as massive, he is at half the distance from the fulcrum. If the seesaw was going to start rotating, it would have to be losing gravitational energy in order to gain some kinetic energy. However, there is no way for it to gain or lose gravitational energy by rotating in either direction. The change in gravitational energy would be \[\begin{align*} \Delta U &= \Delta U_{1} + \Delta U_{2} \\ &= g( m_1\Delta y_{1} + m_{2}\Delta y_2 ) , \end{align*}\] but \(\Delta y_1\) and \(\Delta y_2\) have opposite signs and are in the proportion of two to one, since the son moves along a circular arc that covers the same angle as the father's but has half the radius. Therefore \(\Delta U=0\), and there is no way for the seesaw to trade gravitational energy for kinetic. |

The seesaw example demonstrates the principle of the lever, which is one of the basic mechanical building blocks known as simple machines. As discussed in more detail in chapters 3 and 4, the principle applies even when the interactions involved aren't gravitational.

Note that although a lever makes it easier to lift a heavy weight, it also decreases the distance traveled by the load. By reversing the lever, we can make the load travel a greater distance, at the expense of increasing the amount of force required. The human muscular-skeletal system uses reversed levers of this kind, which allows us to move more rapidly, and also makes our bodies more compact, at the expense of brute strength. A piano uses reversed levers so that a small amount of motion of the key produces a longer swing of the hammer. Another interesting example is the hydraulic jack shown in figure a. The analysis in terms of gravitational energy is exactly the same as for the seesaw, except that the relationship between \(\Delta{}y_1\) and \(\Delta{}y_2\) is now determined not by geometry but by conservation of mass: since water is highly incompressible, conservation of mass is approximately the same as a requirement of constant volume, which can only be satisfied if the distance traveled by each piston is in inverse proportion to its cross-sectional area.

##### Discussion Questions

◊ Hydroelectric power (water flowing over a dam to spin turbines) appears to be completely free. Does this violate conservation of energy? If not, then what is the ultimate source of the electrical energy produced by a hydroelectric plant?”

◊ You throw a steel ball up in the air. How can you prove based on conservation of energy that it has the same speed when it falls back into your hand? What if you threw a feather up? Is energy not conserved in this case?”

◊

Figure m shows a pendulum that is released at A and caught by a peg as it passes through the vertical, B. To what height will the bob rise on the right?”

◊ What is wrong with the following definitions of \(g\)?

- “\(g\) is gravity.”
- “\(g\) is the speed of a falling object.”
- “\(g\) is how hard gravity pulls on things.”

”2.1.6 Equilibrium and stability

The seesaw in figure k is in equilibrium, meaning that if it starts out being at rest, it will stay put. This is known as a neutral equilibrium, since the seesaw has no preferred position to which it will return if we disturb it. If we move it to a different position and release it, it will stay at rest there as well. If we put it in motion, it will simply continue in motion until one person's feet hit the ground.

Most objects around you are in stable equilibria, like the black block in figure o/3. Even if the block is moved or set in motion, it will oscillate about the equilibrium position. The pictures are like graphs of \(y\) versus \(x\), but since the gravitational energy \(U=mgy\) is proportional to \(y\), we can just as well think of them as graphs of \(U\) versus \(x\). The block's stable equilibrium position is where the function \(U(x)\) has a local minimum. The book you're reading right now is in equilibrium, but gravitational energy isn't the only form of energy involved. To move it upward, we'd have to supply gravitational energy, but downward motion would require a different kind of energy, in order to compress the table more. (As we'll see in section 2.4, this is electrical energy due to interactions between atoms within the table.)

A differentiable function's local extrema occur where its derivative is zero. A position where \(d{}U/d{}x\) is zero can be a stable (3), neutral (2), or unstable equilibrium, (4). An unstable equilibrium is like a pencil balanced on its tip. Although it could theoretically remain balanced there forever, in reality it will topple due to any tiny perturbation, such as an air current or a vibration from a passing truck. This is a technical, mathematical definition of instability, which is more restrictive than the way the word is used in ordinary speech. Most people would describe a domino standing upright as being unstable, but in technical usage it would be considered stable, because a certain finite amount of energy is required to tip it over, and perturbations smaller than that would only cause it to oscillate around its equilibrium position.

The domino is also an interesting example because it has two local minima, one in which it is upright, and another in which it is lying flat. A local minimum that is not the global minimum, as in figure o/5, is referred to as a metastable equilibrium.

Example 12: A neutral equilibrium |
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Figure p shows a special-purpose one-block funicular railroad near Hill and Fourth Streets in Los Angeles, California, used for getting passengers up and down a very steep hill. It has two cars attached to a single loop of cable, arranged so that while one car goes up, the other comes down. They pass each other in the middle. Since one car's gravitational energy is increasing while the other's is decreasing, the system is in neutral equilibrium. If there were no frictional heating, exactly zero energy would be required in order to operate the system. A similar counterweighting principle is used in aerial tramways in mountain resorts, and in elevators (with a solid weight, rather than a second car, as counterweight). |

Example 13: Water in a U-shaped tube |
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\(\triangleright\) The U-shaped tube in figure q has cross-sectional area \(A\), and the density of the water inside is \(\rho\). Find the gravitational energy as a function of the quantity \(y\) shown in the figure, and show that there is an equilibrium at \(y\)=0. \(\triangleright\) The question is a little ambiguous, since gravitational energy is only well defined up to an additive constant. To fix this constant, let's define \(U\) to be zero when \(y\)=0. The difference between \(U( y)\) and \(U(0)\) is the energy that would be required to lift a water column of height \(y\) out of the right side, and place it above the dashed line, on the left side, raising it through a height \(y\). This water column has height \(y\) and cross-sectional area \(A\), so its volume is \(Ay\), its mass is \(\rho Ay\), and the energy required is \(mgy\)=\((\rho Ay) gy\)=\(\rho gAy^2\). We then have \(U( y)= U(0)+\rho gAy^2=\rho gAy^2\). To find equilibria, we look for places where the derivative \(\text{d} U/\text{d} y=2\rho gAy\) equals 0. As we'd expect intuitively, the only equilibrium occurs at \(y\)=0. The second derivative test shows that this is a local minimum (not a maximum or a point of inflection), so this is a stable equilibrium. |

### 2.1.7 Predicting the direction of motion

Kinetic energy doesn't depend on the direction of motion. Sometimes this is helpful, as in the high road-low road example (p. 84, example 9), where we were able to predict that the balls would have the same final speeds, even though they followed different paths and were moving in different directions at the end. In general, however, the two conservation laws we've encountered so far aren't enough to predict an object's path through space, for which we need conservation of momentum (chapter 3), and the mathematical technique of vectors. Before we develop those ideas in their full generality, however, it will be helpful to do a couple of simple examples, including one that we'll get a lot of mileage out of in section 2.3.

Suppose we observe an air hockey puck gliding frictionlessly to the right at a velocity \(v\), and we want to predict its future motion. Since there is no friction, no kinetic energy is converted to heat. The only form of energy involved is kinetic energy, so conservation of energy, \(\Delta{}E=0\), becomes simply \(\Delta{}K=0\). There's no particular reason for the puck to do anything but continue moving to the right at constant speed, but it would be equally consistent with conservation of energy if it spontaneously decided to reverse its direction of motion, changing its velocity to \(-v\). Either way, we'd have \(\Delta{}K=0\). There is, however, a way to tell which motion is physical and which is unphysical. Suppose we consider the whole thing again in the frame of reference that is initially moving right along with the puck. In this frame, the puck starts out with \(K=0\). What we originally described as a reversal of its velocity from \(v\) to \(-v\) is, in this new frame of reference, a change from zero velocity to \(-2v\), which would violate conservation of energy. In other words, the physically possible motion conserves energy in all frames of reference, but the unphysical motion only conserves energy in one special frame of reference.

For our second example, we consider a car driving off the edge of a cliff (r). For simplicity, we assume that air friction is negligible, so only kinetic and gravitational energy are involved. Does the car follow trajectory 1, familiar from Road Runner cartoons, trajectory 2, a parabola, or 3, a diagonal line? All three could be consistent with conservation of energy, in the ground's frame of reference. For instance, the car would have constant gravitational energy along the initial horizontal segment of trajectory 1, so during that time it would have to maintain constant kinetic energy as well. Only a parabola, however, is consistent with conservation of energy combined with Galilean relativity. Consider the frame of reference that is moving horizontally at the same speed as that with which the car went over the edge. In this frame of reference, the cliff slides out from under the initially motionless car. The car can't just hover for a while, so trajectory 1 is out. Repeating the same math as in example 8 on p. 83, we have

in this frame of reference, where the stars indicate coordinates measured in the moving frame of reference. These coordinates are related to the ground-fixed coordinates \((x,y)\) by the equations

where \(v\) is the velocity of one frame with respect to the other. We therefore have

in our original frame of reference. Eliminating \(t\), we can see that this has the form of a parabola:

*self-check:*

What would the car's motion be like in the * frame of reference if it followed trajectory 3?

(answer in the back of the PDF version of the book)

### Contributors

Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.