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Contents

Section 13.1 - Work: the transfer of mechanical energy

Section 13.2 - Work in three dimensions

Section 13.3 - Varying force

Section 13.4 - Applications of calculus (optional calculus-based section)

Section 13.5 - Work and potential energy

Section 13.6 - When does work equal force times distance? (optional)

Section 13.7 - The dot product (optional)

Section 13.8 - Summary

Section 13.1 - Work: the transfer of mechanical energy

Section 13.2 - Work in three dimensions

Section 13.3 - Varying force

Section 13.4 - Applications of calculus (optional calculus-based section)

Section 13.5 - Work and potential energy

Section 13.6 - When does work equal force times distance? (optional)

Section 13.7 - The dot product (optional)

Section 13.8 - Summary

The mass contained in a closed system is a conserved quantity, but if the system is not closed, we also have ways of measuring the amount of mass that goes in or out. The water company does this with a meter that records your water use.

Likewise, we often have a system that is not closed, and would like to know how much energy comes in or out. Energy, however, is not a physical substance like water, so energy transfer cannot be measured with the same kind of meter. How can we tell, for instance, how much useful energy a tractor can “put out” on one tank of gas?

The law of conservation of energy guarantees that all the chemical energy in the gasoline will reappear in some form, but not necessarily in a form that is useful for doing farm work. Tractors, like cars, are extremely inefficient, and typically 90% of the energy they consume is converted directly into heat, which is carried away by the exhaust and the air flowing over the radiator. We wish to distinguish the energy that comes out directly as heat from the energy that serves to accelerate a trailer or to plow a field, so we define a technical meaning of the ordinary word “work” to express the distinction:

Work is the amount of energy transferred into or out of a system, not counting energy transferred by heat conduction.

Based on this definition, is work a vector, or a scalar? What are its units?

(answer in the back of the PDF version of the book)The conduction of heat is to be distinguished from heating by friction. When a hot potato heats up your hands by conduction, the energy transfer occurs without any force, but when friction heats your car's brake shoes, there is a force involved. The transfer of energy with and without a force are measured by completely different methods, so we wish to include heat transfer by frictional heating under the definition of work, but not heat transfer by conduction. The definition of work could thus be restated as the amount of energy transferred by forces.

The examples in figures b-d show that there are many different ways in which energy can be transferred. Even so, all these examples have two things in common:

- A force is involved.
- The tractor travels some distance as it does the work.

In b, the increase in the height of the weight, \(\Delta y\), is the same as the distance the tractor travels, which we'll call \(d\). For simplicity, we discuss the case where the tractor raises the weight at constant speed, so that there is no change in the kinetic energy of the weight, and we assume that there is negligible friction in the pulley, so that the force the tractor applies to the rope is the same as the rope's upward force on the weight. By Newton's first law, these forces are also of the same magnitude as the earth's gravitational force on the weight. The increase in the weight's potential energy is given by \(F\Delta y\), so the work done by the tractor on the weight equals \(Fd\), the product of the force and the distance moved:

\[\begin{equation*}
W = Fd .
\end{equation*}\]

In example c, the tractor's force on the trailer accelerates it, increasing its kinetic energy. If frictional forces on the trailer are negligible, then the increase in the trailer's kinetic energy can be found using the same algebra that was used on page 307 to find the potential energy due to gravity. Just as in example b, we have

\[\begin{equation*}
W = Fd .
\end{equation*}\]

Does this equation always give the right answer? Well, sort of. In example d, there are two quantities of work you might want to calculate: the work done by the tractor on the plow and the work done by the plow on the dirt. These two quantities can't both equal \(Fd\). Most of the energy transmitted through the cable goes into frictional heating of the plow and the dirt. The work done by the plow on the dirt is less than the work done by the tractor on the plow, by an amount equal to the heat absorbed by the plow. It turns out that the equation \(W=Fd\) gives the work done by the tractor, not the work done by the plow. How are you supposed to know when the equation will work and when it won't? The somewhat complex answer is postponed until section 13.6. Until then, we will restrict ourselves to examples in which \(W=Fd\) gives the right answer; essentially the reason the ambiguities come up is that when one surface is slipping past another, \(d\) may be hard to define, because the two surfaces move different distances.

We have also been using examples in which the force is in the same direction as the motion, and the force is constant. (If the force was not constant, we would have to represent it with a function, not a symbol that stands for a number.) To summarize, we have:

The work done by a force can be calculated as

\[\begin{equation*}
W = Fd ,
\end{equation*}\]

if the force is constant and in the same direction as the motion. Some ambiguities are encountered in cases such as kinetic friction.

\(\triangleright\) Multiplying the force by the distance gives \(9\times10^{17}\ \text{J}\). For comparison, the Northridge earthquake of 1994, which killed 57 people and did 40 billion dollars of damage, released 22 times less energy.

Along her route, the climber has placed removable rock anchors (not shown) and carabiners attached to the anchors. She clips the rope into each carabiner so that it can travel but can't pop out. In both 1 and 2, she has ascended a certain distance above her last anchor, so that if she falls, she will drop through a height \(h\) that is about twice this distance, and this fall height is about the same in both cases. In fact, \(h\) is somewhat larger than twice her height above the last anchor, because the rope is intentionally designed to stretch under the big force of a falling climber who suddenly brings it taut.

To see why we want a stretchy rope, consider the equation \(F=W/d\) in the case where \(d\) is zero;
\(F\) would theoretically become infinite. In a fall, the climber loses a fixed amount of gravitational
energy \(mgh\). This is transformed into an equal amount of kinetic energy as she falls, and eventually
this kinetic energy has to be transferred out of her body when the rope comes up taut.
If the rope was not stretchy, then the distance traveled at the point where the rope attaches to
her harness would be zero, and the force exerted would theoretically be infinite.
Before the rope reached the theoretically infinite tension \(F\) it would break (or her back
would break, or her anchors would be pulled out of the rock).
We want the rope
to be stretchy enough to make \(d\) fairly big, so that dividing \(W\) by \(d\) gives a small
force.^{1}

In g/1 and g/2, the fall \(h\) is about the same.
What is different is the length \(L\) of rope that has been paid out. A longer rope can stretch
more, so the distance \(d\) traveled after the “catch” is proportional to \(L\).
Combining \(F=W/d\), \(W\propto h\), and \(d\propto L\), we have \(F\propto h/L\).
For these reasons, rock climbers define a *fall factor* \(f=h/L\). The larger fall factor
in g/1 is more
dangerous.

Figure h shows a pulley arrangement for doubling the force supplied by the tractor (book 1, section 5.6). The tension in the left-hand rope is equal throughout, assuming negligible friction, so there are two forces pulling the pulley to the left, each equal to the original force exerted by the tractor on the rope. This doubled force is transmitted through the right-hand rope to the stump.

It might seem as though this arrangement would also double the work done by the tractor, but look again. As the tractor moves forward 2 meters, 1 meter of rope comes around the pulley, and the pulley moves 1 m to the left. Although the pulley exerts double the force on the stump, the pulley and stump only move half as far, so the work done on the stump is no greater that it would have been without the pulley.

The same is true for any mechanical arrangement that increases or decreases force, such as the gears on a ten-speed bike. You can't get out more work than you put in, because that would violate conservation of energy. If you shift gears so that your force on the pedals is amplified, the result is that you just have to spin the pedals more times.

It strikes most students as nonsensical when they are told that if they stand still and hold a heavy bag of cement, they are doing no work on the bag. Even if it makes sense mathematically that \(W=Fd\) gives zero when \(d\) is zero, it seems to violate common sense. You would certainly become tired! The solution is simple. Physicists have taken over the common word “work” and given it a new technical meaning, which is the transfer of energy. The energy of the bag of cement is not changing, and that is what the physicist means by saying no work is done on the bag.

There is a transformation of energy, but it is taking place entirely within your own muscles, which are converting chemical energy into heat. Physiologically, a human muscle is not like a tree limb, which can support a weight indefinitely without the expenditure of energy. Each muscle cell's contraction is generated by zillions of little molecular machines, which take turns supporting the tension. When a particular molecule goes on or off duty, it moves, and since it moves while exerting a force, it is doing work. There is work, but it is work done by one molecule in a muscle cell on another.

When object A transfers energy to object B, we say that A does positive work on B. B is said to do negative work on A. In other words, a machine like a tractor is defined as doing positive work. This use of the plus and minus signs relates in a logical and consistent way to their use in indicating the directions of force and motion in one dimension. In figure i, suppose we choose a coordinate system with the \(x\) axis pointing to the right. Then the force the spring exerts on the ball is always a positive number. The ball's motion, however, changes directions. The symbol \(d\) is really just a shorter way of writing the familiar quantity \(\Delta x\), whose positive and negative signs indicate direction.

While the ball is moving to the left, we use \(d\lt0\) to represent its direction of motion, and the work done by the spring, \(Fd\), comes out negative. This indicates that the spring is taking kinetic energy out of the ball, and accepting it in the form of its own potential energy.

As the ball is reaccelerated to the right, it has \(d>0\), \(Fd\) is positive, and the spring does positive work on the ball. Potential energy is transferred out of the spring and deposited in the ball as kinetic energy.

In summary:

The work done by a force can be calculated as

\[\begin{equation*}
W = Fd ,
\end{equation*}\]

if the force is constant and along the same line as the motion. The quantity \(d\) is to be interpreted as a synonym for \(\Delta x\), i.e., positive and negative signs are used to indicate the direction of motion. Some ambiguities are encountered in cases such as kinetic friction.

In figure
i, what about the work done by the ball on the spring?

[4]

There are many examples where the transfer of energy out of an object cancels out the transfer of energy in. When the tractor pulls the plow with a rope, the rope does negative work on the tractor and positive work on the plow. The total work done by the rope is zero, which makes sense, since it is not changing its energy.

It may seem that when your arms do negative work by lowering a bag of cement, the cement is not really transferring energy into your body. If your body was storing potential energy like a compressed spring, you would be able to raise and lower a weight all day, recycling the same energy. The bag of cement does transfer energy into your body, but your body accepts it as heat, not as potential energy. The tension in the muscles that control the speed of the motion also results in the conversion of chemical energy to heat, for the same physiological reasons discussed previously in the case where you just hold the bag still.

One of the advantages of electric cars over gasoline-powered cars is that it is just as easy to put energy back in a battery as it is to take energy out. When you step on the brakes in a gas car, the brake shoes do negative work on the rest of the car. The kinetic energy of the car is transmitted through the brakes and accepted by the brake shoes in the form of heat. The energy cannot be recovered. Electric cars, however, are designed to use regenerative braking. The brakes don't use friction at all. They are electrical, and when you step on the brake, the negative work done by the brakes means they accept the energy and put it in the battery for later use. This is one of the reasons why an electric car is far better for the environment than a gas car, even if the ultimate source of the electrical energy happens to be the burning of oil in the electric company's plant. The electric car recycles the same energy over and over, and only dissipates heat due to air friction and rolling resistance, not braking. (The electric company's power plant can also be fitted with expensive pollution-reduction equipment that would be prohibitively expensive or bulky for a passenger car.)

◊

Besides the presence of a force, what other things differentiate the processes of frictional heating and heat conduction?

◊

Criticize the following incorrect statement: “A force doesn't do any work unless it's causing the object to move.”

◊

To stop your car, you must first have time to react, and then it takes some time for the car to slow down. Both of these times contribute to the distance you will travel before you can stop. The figure shows how the average stopping distance increases with speed. Because the stopping distance increases more and more rapidly as you go faster, the rule of one car length per 10 m.p.h. of speed is not conservative enough at high speeds. In terms of work and kinetic energy, what is the reason for the more rapid increase at high speeds?

Suppose work is being done to change an object's kinetic energy. A force in the same direction as its motion will speed it up, and a force in the opposite direction will slow it down. As we have already seen, this is described as doing positive work or doing negative work on the object. All the examples discussed up until now have been of motion in one dimension, but in three dimensions the force can be at any angle \(\theta \) with respect to the direction of motion.

What if the force is perpendicular to the direction of motion? We have already seen that a force perpendicular to the motion results in circular motion at constant speed. The kinetic energy does not change, and we conclude that no work is done when the force is perpendicular to the motion.

So far we have been reasoning about the case of a single force acting on an object, and changing only its kinetic energy. The result is more generally true, however. For instance, imagine a hockey puck sliding across the ice. The ice makes an upward normal force, but does not transfer energy to or from the puck.

Suppose the force is at some other angle with respect to the motion, say \(\theta=45°\). Such a force could be broken down into two components, one along the direction of the motion and the other perpendicular to it. The force vector equals the vector sum of its two components, and the principle of vector addition of forces thus tells us that the work done by the total force cannot be any different than the sum of the works that would be done by the two forces by themselves. Since the component perpendicular to the motion does no work, the work done by the force must be

\[\begin{equation*}
W = F_{\parallel} |\mathbf{d}| , \text{[work done by
a constant force]}
\end{equation*}\]

where the vector \(\mathbf{d}\) is simply a less cumbersome version of the notation \(\Delta\mathbf{r}\). This result can be rewritten via trigonometry as

\[\begin{equation*}
W = |\mathbf{F}| |\mathbf{d}| \cos \theta .
\text{[work done by a constant force]}
\end{equation*}\]

Even though this equation has vectors in it, it depends only on their magnitudes, and the magnitude of a vector is a scalar. Work is therefore still a scalar quantity, which only makes sense if it is defined as the transfer of energy. Ten gallons of gasoline have the ability to do a certain amount of mechanical work, and when you pull in to a full-service gas station you don't have to say “Fill 'er up with 10 gallons of south-going gas.”

Students often wonder why this equation involves a cosine
rather than a sine, or ask if it would ever be a sine. In
vector addition, the treatment of sines and cosines seemed
more equal and democratic, so why is the cosine so special
now? The answer is that if we are going to describe, say, a
velocity vector, we must give both the component *parallel*
to the \(x\) axis and the component *perpendicular* to
the \(x\) axis (i.e., the \(y\) component). In calculating work,
however, the force component perpendicular to the motion is
irrelevant --- it changes the direction of motion without
increasing or decreasing the energy of the object on which
it acts. In this context, it is *only* the parallel
force component that matters, so only the cosine occurs.

(a) Work is the transfer of energy. According to this definition, is the horse in the picture doing work on the pack? (b) If you calculate work by the method described in this section, is the horse in figure o doing work on the pack?

(answer in the back of the PDF version of the book)\(\triangleright\) If you exert a force of 21 N on a push broom, at an angle 35 degrees below horizontal, and walk for 5.0 m, how much work do you do? What is the physical significance of this quantity of work?

\(\triangleright\) Using the second equation above, the work done equals

\[\begin{equation*}
(21\ \text{N})(5.0\ \text{m})(\cos 35°) = 86\ \text{J} .
\end{equation*}\]

The form of energy being transferred is heat in the floor and the broom's bristles. This comes from the chemical energy stored in your body. (The majority of the calories you burn are dissipated directly as heat inside your body rather than doing any work on the broom. The 86 J is only the amount of energy transferred through the broom's handle.)

As a violinist draws the bow across a string, the bow hairs exert both a normal force and a kinetic frictional force on the string. The normal force is perpendicular to the direction of motion, and does no work. However, the frictional force is in the same direction as the motion of the bow, so it does work: energy is transferred to the string, causing it to vibrate.

One way of playing a violin more loudly is to use longer strokes. Since \(W=Fd\), the greater distance results in more work.

A second way of getting a louder sound is to press the bow more firmly against the strings. This increases the normal force, and although the normal force itself does no work, an increase in the normal force has the side effect of increasing the frictional force, thereby increasing \(W=Fd\).

The violinist moves the bow back and forth, and sound is produced on both the “up-bow” (the stroke toward the player's left) and the “down-bow” (to the right). One may, for example, play a series of notes in alternation between up-bows and down-bows. However, if the notes are of unequal length, the up and down motions tend to be unequal, and if the player is not careful, she can run out of bow in the middle of a note! To keep this from happening, one can move the bow more quickly on the shorter notes, but the resulting increase in \(d\) will make the shorter notes louder than they should be. A skilled player compensates by reducing the force.

Up until now, we have not found any physically useful way to define the multiplication of two vectors. It would be possible, for instance, to multiply two vectors component by component to form a third vector, but there are no physical situations where such a multiplication would be useful.

The equation \(W= |\mathbf{F}| |\mathbf{d}| \cos \theta\) is an example of
a sort of multiplication of vectors that is useful. The
result is a scalar, not a vector, and this is therefore
often referred to as the *scalar product* of the
vectors \(\mathbf{F}\) and \(\mathbf{d}\). There is a standard shorthand
notation for this operation,

\[\begin{multline*}
\mathbf{A}\cdot\mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta ,
\shoveright{\text{[definition of the notation $\mathbf{A}\cdot\mathbf{B}$;}}\\
\text{$\theta$ is the angle between vectors $\mathbf{A}$ and $\mathbf{B}$]}
\end{multline*}\]

and because of this notation, a more common term for this
operation is the *dot product*. In dot product
notation, the equation for work is simply

\[\begin{equation*}
W = \mathbf{F}\cdot\mathbf{d} .
\end{equation*}\]

The dot product has the following geometric interpretation:

\[\begin{align*}
\mathbf{A}\cdot\mathbf{B} &= |\mathbf{A}| (\text{component of $\mathbf{B}$ parallel to $\mathbf{A}$}) \\
&= |\mathbf{B}| (\text{component of $\mathbf{A}$ parallel to $\mathbf{B}$})
\end{align*}\]

The dot product has some of the properties possessed by ordinary multiplication of numbers,

\[\begin{align*}
\mathbf{A}\cdot\mathbf{B} &= \mathbf{B}\cdot\mathbf{A} \\
\mathbf{A}\cdot(\mathbf{B}+\mathbf{C}) &= \mathbf{A}\cdot\mathbf{B}+\mathbf{A}\cdot\mathbf{C} \\
\left(c\mathbf{A}\right)\cdot\mathbf{B} &= c\left(\mathbf{A}\cdot\mathbf{B}\right) ,
\end{align*}\]

but it lacks one other: the ability to undo multiplication by dividing.

If you know the components of two vectors, you can easily calculate their dot product as follows:

\[\begin{equation*}
\mathbf{A}\cdot\mathbf{B} = A_xB_x+A_yB_y+A_zB_z .
\end{equation*}\]

(This can be proved by first analyzing the special case where each vector has only an \(x\) component, and the similar cases for \(y\) and \(z\). We can then use the rule \(\mathbf{A}\cdot(\mathbf{B}+\mathbf{C}) = \mathbf{A}\cdot\mathbf{B}+\mathbf{A}\cdot\mathbf{C}\) to make a generalization by writing each vector as the sum of its \(x\), \(y\), and \(z\) components. See homework problem 17.)

\[\begin{align*}
\mathbf{b}\cdot\mathbf{b} &=
\left( b_{x}\hat{\mathbf{x}}+ b_{y}\hat{\mathbf{y}}+ b_z\hat{\mathbf{z}}\right)
\cdot
\left( b_{x}\hat{\mathbf{x}}+ b_{y}\hat{\mathbf{y}}+ b_z\hat{\mathbf{z}}\right) \\
&= b_{x}^2+ b_{y}^2+ b_z^2 ,\\
\text{so its magnitude can be expressed as}
|\mathbf{b}| &= \sqrt{\mathbf{b}\cdot\mathbf{b}} .
\end{align*}\]

We will often write \(b^2\) to mean \(\mathbf{b}\cdot\mathbf{b}\), when the context makes it
clear what is intended. For example, we could express kinetic energy as
\(\text{(1/2)} m|\mathbf{v}|^2\), \(\text{(1/2)} m\mathbf{v}\cdot\mathbf{v}\),
or \(\text{(1/2)} m v^2\). In the third version, nothing but context tells
us that \(v\) really stands for the magnitude of some vector \(\mathbf{v}\).
\(\triangleright\) A mule pulls a barge with a force \(\mathbf{F}\text{=(1100 N)}\hat{\mathbf{x}}+\text{(400 N)}\hat{\mathbf{y}}\), and the total distance it travels is \(\text{(1000 m)}\hat{\mathbf{x}}\). How much work does it do?

\(\triangleright\) The dot product is \(1.1\times10^6\ \text{N}\!\cdot\!\text{m} = 1.1\times10^6\ \text{J}\).

Up until now we have done no actual calculations of work in cases where the force was not constant. The question of how to treat such cases is mathematically analogous to the issue of how to generalize the equation \((\text{distance})=(\text{velocity})(\text{time})\) to cases where the velocity was not constant. There, we found that the correct generalization was to find the area under the graph of velocity versus time. The equivalent thing can be done with work:

The work done by a force \(F\) equals the area under the curve on a graph of \(F_{\parallel}\) versus \(x\). (Some ambiguities are encountered in cases such as kinetic friction.)

The examples in this section are ones in which the force is varying, but is always along the same line as the motion, so \(F\) is the same as \(F_{\parallel}\).

In which of the following examples would it be OK to calculate work using \(Fd\), and in which ones would you have to use the area under the \(F-x\) graph?

(a) A fishing boat cruises with a net dragging behind it.

(b) A magnet leaps onto a refrigerator from a distance.

(c) Earth's gravity does work on an outward-bound space probe.

(answer in the back of the PDF version of the book)An important and straightforward example is the calculation of the work done by a spring that obeys Hooke's law,

\[\begin{equation*}
F \approx -k\left(x-x_\text{o}\right) .
\end{equation*}\]

The minus sign is because this is the force being exerted by the spring, not the force that would have to act on the spring to keep it at this position. That is, if the position of the cart in figure p is to the right of equilibrium, the spring pulls back to the left, and vice-versa.

We calculate the work done when the spring is initially at equilibrium and then decelerates the car as the car moves to the right. The work done by the spring on the cart equals the minus area of the shaded triangle, because the triangle hangs below the \(x\) axis. The area of a triangle is half its base multiplied by its height, so

\[\begin{equation*}
W = -\frac{1}{2}k\left(x-x_\text{o}\right)^2 .
\end{equation*}\]

This is the amount of kinetic energy lost by the cart as the spring decelerates it.

It was straightforward to calculate the work done by the spring in this case because the graph of \(F\) versus \(x\) was a straight line, giving a triangular area. But if the curve had not been so geometrically simple, it might not have been possible to find a simple equation for the work done, or an equation might have been derivable only using calculus. Optional section 13.4 gives an important example of such an application of calculus.

The graph shows the force between the carbon nucleus and the proton as the proton is on its way in, with the distance in units of femtometers (1 fm=\(10^{-15}\) m). Amusingly, the force turns out to be a few newtons: on the same order of magnitude as the forces we encounter ordinarily on the human scale. Keep in mind, however, that a force this big exerted on a single subatomic particle such as a proton will produce a truly fantastic acceleration (on the order of \(10^{27}\ \text{m}/\text{s}^2\)!).

Why does the force have a peak around \(x=3\) fm, and become smaller once the proton has actually merged with the nucleus? At \(x=3\) fm, the proton is at the edge of the crowd of protons and neutrons. It feels many attractive forces from the left, and none from the right. The forces add up to a large value. However if it later finds itself at the center of the nucleus, \(x=0\), there are forces pulling it from all directions, and these force vectors cancel out.

We can now calculate the energy released in this reaction by using the area under the graph to determine the amount of mechanical work done by the carbon nucleus on the proton. (For simplicity, we assume that the proton came in “aimed” at the center of the nucleus, and we ignore the fact that it has to shove some neutrons and protons out of the way in order to get there.) The area under the curve is about 17 squares, and the work represented by each square is

\[\begin{equation*}
(1\ \text{N})(10^{-15}\ \text{m}) = 10^{-15}\ \text{J} ,
\end{equation*}\]

so the total energy released is about

\[\begin{equation*}
(10^{-15}\ \text{J}/\text{square})(17\ \text{squares}) = 1.7\times10^{-14}\ \text{J} .
\end{equation*}\]

This may not seem like much, but remember that this is only
a reaction between the nuclei of two out of the zillions of
atoms in the sun. For comparison, a typical *chemical*
reaction between two atoms might transform on the order of
\(10^{-19}\) J of electrical potential energy into heat ---
100,000 times less energy!

As a final note, you may wonder why reactions such as these only occur in the sun. The reason is that there is a repulsive electrical force between nuclei. When two nuclei are close together, the electrical forces are typically about a million times weaker than the nuclear forces, but the nuclear forces fall off much more quickly with distance than the electrical forces, so the electrical force is the dominant one at longer ranges. The sun is a very hot gas, so the random motion of its atoms is extremely rapid, and a collision between two atoms is sometimes violent enough to overcome this initial electrical repulsion.

The student who has studied integral calculus will recognize that the graphical rule given in the previous section can be reexpressed as an integral,

\[\begin{equation*}
W = \int_{x_1}^{x_2}F dx .
\end{equation*}\]

We can then immediately find by the fundamental theorem of calculus that force is the derivative of work with respect to position,

\[\begin{equation*}
F = \frac{dW}{dx} .
\end{equation*}\]

For example, a crane raising a one-ton block on the moon would be transferring potential energy into the block at only one sixth the rate that would be required on Earth, and this corresponds to one sixth the force.

Although the work done by the spring could be calculated without calculus using the area of a triangle, there are many cases where the methods of calculus are needed in order to find an answer in closed form. The most important example is the work done by gravity when the change in height is not small enough to assume a constant force. Newton's law of gravity is

\[\begin{equation*}
F = \frac{GMm}{r^2} ,
\end{equation*}\]

which can be integrated to give

\[\begin{align*}
W &= \int_{r_1}^{r_2} \frac{GMm}{r^2} dr\\
&= -GMm\left(\frac{1}{r_2}-\frac{1}{r_1}\right) .
\end{align*}\]

The techniques for calculating work can also be applied to the calculation of potential energy. If a certain force depends only on the distance between the two participating objects, then the energy released by changing the distance between them is defined as the potential energy, and the amount of potential energy lost equals minus the work done by the force,

\[\begin{equation*}
\Delta PE = -W .
\end{equation*}\]

The minus sign occurs because positive work indicates that the potential energy is being expended and converted to some other form.

It is sometimes convenient to pick some arbitrary position as a reference position, and derive an equation for once and for all that gives the potential energy relative to this position

\[\begin{equation*}
PE_x = -W_{\text{ref}\rightarrow x} .
\text{[potential energy at a point $x$]}
\end{equation*}\]

To find the energy transferred into or out of potential energy, one then subtracts two different values of this equation.

These equations might almost make it look as though work and
energy were the same thing, but they are not. First,
potential energy measures the energy that a system
*has* stored in it, while work measures how much energy
is *transferred* in or out. Second, the techniques for
calculating work can be used to find the amount of energy
transferred in many situations where there is no potential
energy involved, as when we calculate the amount of kinetic
energy transformed into heat by a car's brake shoes.

\(\triangleright\) A toy gun uses a spring with a spring constant of 10 N/m to shoot a ping-pong ball of mass 5 g. The spring is compressed to 10 cm shorter than its equilibrium length when the gun is loaded. At what speed is the ball released?

\(\triangleright\) The equilibrium point is the natural choice for a reference point. Using the equation found previously for the work, we have

\[\begin{equation*}
PE_x = \frac{1}{2}k\left(x-x_\text{o}\right)^2 .
\end{equation*}\]

The spring loses contact with the ball at the equilibrium point, so the final potential energy is

\[\begin{equation*}
PE_f = 0 .
\end{equation*}\]

The initial potential energy is

\[\begin{align*}
PE_i &= \frac{1}{2}(10\ \text{N}/\text{m})(0.10\ \text{m})^2 . \\
&= 0.05\ \text{J}.
\end{align*}\]

The loss in potential energy of 0.05 J means an increase in kinetic energy of the same amount. The velocity of the ball is found by solving the equation \(KE=(1/2)mv^2\) for \(v\),

\[\begin{align*}
v &= \sqrt{\frac{2KE}{m}}\\
&= \sqrt{\frac{(2)(0.05\ \text{J})}{0.005\ \text{kg}}}\\
&= 4\ \text{m}/\text{s} .
\end{align*}\]

\(\triangleright\) The potential energy equals minus the work that would have to be done to bring the object from \(r_1=\infty\) to \(r= r_2\), which is

\[\begin{equation*}
PE = -\frac{GMm}{r} .
\end{equation*}\]

This is simpler than the equation for the work, which is an example of why it is advantageous to record an equation for potential energy relative to some reference point, rather than an equation for work.

Although the equations derived in the previous two examples may seem arcane and not particularly useful except for toy designers and rocket scientists, their usefulness is actually greater than it appears. The equation for the potential energy of a spring can be adapted to any other case in which an object is compressed, stretched, twisted, or bent. While you are not likely to use the equation for gravitational potential energy for anything practical, it is directly analogous to an equation that is extremely useful in chemistry, which is the equation for the potential energy of an electron at a distance \(r\) from the nucleus of its atom. As discussed in more detail later in the course, the electrical force between the electron and the nucleus is proportional to \(1/r^2\), just like the gravitational force between two masses. Since the equation for the force is of the same form, so is the equation for the potential energy.

◊

What does the graph of \(PE=(1/2)k\left(x-x_\text{o}\right)^2\) look like as a function of \(x\)? Discuss the physical significance of its features.

◊

What does the graph of \(PE=-GMm/r\) look like as a function of \(r?\) Discuss the physical significance of its features. How would the equation and graph change if some other reference point was chosen rather than \(r=\infty\)?

◊

Starting at a distance \(r\) from a planet of mass \(M\), how fast must an object be moving in order to have a hyperbolic orbit, i.e., one that never comes back to the planet? This velocity is called the escape velocity. Interpreting the result, does it matter in what direction the velocity is? Does it matter what mass the object has? Does the object escape because it is moving too fast for gravity to act on it?

◊

Does a spring have an “escape velocity?”

◊

Calculus-based question: If the form of energy being transferred is potential energy, then the equations \(F=dW/dx\) and \(W=\int F dx\) become \(F=-dPE/dx\) and \(PE=-\int F dx\). How would you then apply the following calculus concepts: zero derivative at minima and maxima, and the second derivative test for concavity up or down.

In the example of the tractor pulling the plow discussed on page 319, the work did not equal \(Fd\). The purpose of this section is to explain more fully how the quantity \(Fd\) can and cannot be used. To simplify things, I write \(Fd\) throughout this section, but more generally everything said here would be true for the area under the graph of \(F_{\parallel}\) versus \(d\).

The following two theorems allow most of the ambiguity to be cleared up.

The change in kinetic energy associated with the motion of an object's center of mass is related to the total force acting on it and to the distance traveled by its center of mass according to the equation \(\Delta KE_{cm}=F_{total}d_{cm}\).

This can be proved based on Newton's second law and the equation \(KE=(1/2)mv^2\). Note that despite the traditional name, it does not necessarily tell the amount of work done, since the forces acting on the object could be changing other types of energy besides the KE associated with its center of mass motion.

The second theorem does relate directly to work:

When a contact force acts between two objects and the two surfaces do not slip past each other, the work done equals \(Fd\), where \(d\) is the distance traveled by the point of contact.

This one has no generally accepted name, so we refer to it simply as the second theorem.

A great number of physical situations can be analyzed with these two theorems, and often it is advantageous to apply both of them to the same situation.

The work-kinetic energy theorem tells us how to calculate the skater's kinetic energy if we know the amount of force and the distance her center of mass travels while she is pushing off.

The second theorem tells us that the wall does no work on the skater. This makes sense, since the wall does not have any source of energy.

\(\triangleright\) Is it possible to absorb an impact without recoiling? For instance, would a brick wall “give” at all if hit by a ping-pong ball?

\(\triangleright\) There will always be a recoil. In the example proposed, the wall will surely have some energy transferred to it in the form of heat and vibration. The second theorem tells us that we can only have nonzero work if the distance traveled by the point of contact is nonzero.

Newton's first law tells us that the total force on the refrigerator must be zero: your force is canceling the floor's kinetic frictional force. The work-kinetic energy theorem is therefore true but useless. It tells us that there is zero total force on the refrigerator, and that the refrigerator's kinetic energy doesn't change.

The second theorem tells us that the work you do equals your hand's force on the refrigerator multiplied by the distance traveled. Since we know the floor has no source of energy, the only way for the floor and refrigerator to gain energy is from the work you do. We can thus calculate the total heat dissipated by friction in the refrigerator and the floor.

Note that there is no way to find how much of the heat is dissipated in the floor and how much in the refrigerator.

If you push on a cart and accelerate it, there are two forces acting on the cart: your hand's force, and the static frictional force of the ground pushing on the wheels in the opposite direction.

Applying the second theorem to your force tells us how to calculate the work you do.

Applying the second theorem to the floor's force tells us that the floor does no work on the cart. There is no motion at the point of contact, because the atoms in the floor are not moving. (The atoms in the surface of the wheel are also momentarily at rest when they touch the floor.) This makes sense, since the floor does not have any source of energy.

The work-kinetic energy theorem refers to the total force, and because the floor's backward force cancels part of your force, the total force is less than your force. This tells us that only part of your work goes into the kinetic energy associated with the forward motion of the cart's center of mass. The rest goes into rotation of the wheels.

Up until now, we have not found any physically useful way to define the multiplication of two vectors. It would be possible, for instance, to multiply two vectors component by component to form a third vector, but there are no physical situations where such a multiplication would be useful.

The equation \(W= |\mathbf{F}| |\mathbf{d}| \cos \theta\) is an example of
a sort of multiplication of vectors that is useful. The
result is a scalar, not a vector, and this is therefore
often referred to as the *scalar product* of the
vectors \(\mathbf{F}\) and \(\mathbf{d}\). There is a standard shorthand
notation for this operation,

\[\begin{multline*}
\mathbf{A}\cdot\mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta ,
\shoveright{\text{[definition of the notation $\mathbf{A}\cdot\mathbf{B}$;}}\\
\text{$\theta$ is the angle between vectors $\mathbf{A}$ and $\mathbf{B}$]}
\end{multline*}\]

and because of this notation, a more common term for this
operation is the *dot product*. In dot product
notation, the equation for work is simply

\[\begin{equation*}
W = \mathbf{F}\cdot\mathbf{d} .
\end{equation*}\]

The dot product has the following geometric interpretation:

\[\begin{align*}
\mathbf{A}\cdot\mathbf{B} &= |\mathbf{A}| (\text{component of $\mathbf{B}$ parallel to $\mathbf{A}$}) \\
&= |\mathbf{B}| (\text{component of $\mathbf{A}$ parallel to $\mathbf{B}$})
\end{align*}\]

The dot product has some of the properties possessed by ordinary multiplication of numbers,

\[\begin{align*}
\mathbf{A}\cdot\mathbf{B} &= \mathbf{B}\cdot\mathbf{A} \\
\mathbf{A}\cdot(\mathbf{B}+\mathbf{C}) &= \mathbf{A}\cdot\mathbf{B}+\mathbf{A}\cdot\mathbf{C} \\
\left(c\mathbf{A}\right)\cdot\mathbf{B} &= c\left(\mathbf{A}\cdot\mathbf{B}\right) ,
\end{align*}\]

but it lacks one other: the ability to undo multiplication by dividing.

If you know the components of two vectors, you can easily calculate their dot product as follows:

\[\begin{equation*}
\mathbf{A}\cdot\mathbf{B} = A_xB_x+A_yB_y+A_zB_z .
\end{equation*}\]

(This can be proved by first analyzing the special case where each vector has only an \(x\) component, and the similar cases for \(y\) and \(z\). We can then use the rule \(\mathbf{A}\cdot(\mathbf{B}+\mathbf{C}) = \mathbf{A}\cdot\mathbf{B}+\mathbf{A}\cdot\mathbf{C}\) to make a generalization by writing each vector as the sum of its \(x\), \(y\), and \(z\) components. See homework problem 17.)

\[\begin{align*}
\mathbf{b}\cdot\mathbf{b} &=
\left( b_{x}\hat{\mathbf{x}}+ b_{y}\hat{\mathbf{y}}+ b_z\hat{\mathbf{z}}\right)
\cdot
\left( b_{x}\hat{\mathbf{x}}+ b_{y}\hat{\mathbf{y}}+ b_z\hat{\mathbf{z}}\right) \\
&= b_{x}^2+ b_{y}^2+ b_z^2 ,\\
\text{so its magnitude can be expressed as}
|\mathbf{b}| &= \sqrt{\mathbf{b}\cdot\mathbf{b}} .
\end{align*}\]

We will often write \(b^2\) to mean \(\mathbf{b}\cdot\mathbf{b}\), when the context makes it
clear what is intended. For example, we could express kinetic energy as
\(\text{(1/2)} m|\mathbf{v}|^2\), \(\text{(1/2)} m\mathbf{v}\cdot\mathbf{v}\),
or \(\text{(1/2)} m v^2\). In the third version, nothing but context tells
us that \(v\) really stands for the magnitude of some vector \(\mathbf{v}\).
\(\triangleright\) A mule pulls a barge with a force \(\mathbf{F}\text{=(1100 N)}\hat{\mathbf{x}}+\text{(400 N)}\hat{\mathbf{y}}\), and the total distance it travels is \(\text{(1000 m)}\hat{\mathbf{x}}\). How much work does it do?

\(\triangleright\) The dot product is \(1.1\times10^6\ \text{N}\!\cdot\!\text{m} = 1.1\times10^6\ \text{J}\).

*work* — the amount of energy transferred into or out of a
system, excluding energy transferred by heat conduction

\(W\) — work

{}

Work is a measure of the transfer of mechanical energy, i.e., the transfer of energy by a force rather than by heat conduction. When the force is constant, work can usually be calculated as

\[\begin{equation*}
W = F_{\parallel} |\mathbf{d}| , \text{[only if the
force is constant]}
\end{equation*}\]

where \(\mathbf{d}\) is simply a less cumbersome notation for \(\Delta \mathbf{r}\), the vector from the initial position to the final position. Thus,

- A force in the same direction as the motion does positive work, i.e., transfers energy into the object on which it acts.
- A force in the opposite direction compared to the motion does negative work, i.e., transfers energy out of the object on which it acts.
- When there is no motion, no mechanical work is done. The human body burns calories when it exerts a force without moving, but this is an internal energy transfer of energy within the body, and thus does not fall within the scientific definition of work.
- A force perpendicular to the motion does no work.

When the force is not constant, the above equation should be generalized as the area under the graph of \(F_{\parallel}\) versus \(d\).

Machines such as pulleys, levers, and gears may increase or decrease a force, but they can never increase or decrease the amount of work done. That would violate conservation of energy unless the machine had some source of stored energy or some way to accept and store up energy.

There are some situations in which the equation \(W=F_{\parallel}\) \(|\mathbf{d}|\) is ambiguous or not true, and these issues are discussed rigorously in section 13.6. However, problems can usually be avoided by analyzing the types of energy being transferred before plunging into the math. In any case there is no substitute for a physical understanding of the processes involved.

The techniques developed for calculating work can also be applied to the calculation of potential energy. We fix some position as a reference position, and calculate the potential energy for some other position, \(x\), as

\[\begin{equation*}
PE_x = -W_{\text{ref}\rightarrow x} .
\end{equation*}\]

The following two equations for potential energy have broader significance than might be suspected based on the limited situations in which they were derived:

\[\begin{equation*}
PE = \frac{1}{2}k\left(x-x_\text{o}\right)^2 .
\end{equation*}\]

[potential energy of a spring having spring constant \(k\), when stretched or compressed from the equilibrium position \(x_o;\) analogous equations apply for the twisting, bending, compression, or stretching of any object.]

\[\begin{equation*}
PE = -\frac{GMm}{r}
\end{equation*}\]

[gravitational potential energy of objects of masses \(M\) and \(m\), separated by a distance \(r\); an analogous equation applies to the electrical potential energy of an electron in an atom.]

\begin{homeworkforcelabel}{stoppingdistances}{1}{}{1} Two speedboats are identical, but one has more people aboard than the other. Although the total masses of the two boats are unequal, suppose that they happen to have the same kinetic energy. In a boat, as in a car, it's important to be able to stop in time to avoid hitting things. (a) If the frictional force from the water is the same in both cases, how will the boats' stopping distances compare? Explain. (b) Compare the times required for the boats to stop. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{posnegwork}{1}{}{2}In each of the following situations, is the work being done positive, negative, or zero? (a) a bull paws the ground; (b) a fishing boat pulls a net through the water behind it; (c) the water resists the motion of the net through it; (d) you stand behind a pickup truck and lower a bale of hay from the truck's bed to the ground. Explain. [Based on a problem by Serway and Faughn.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{heliumescape}{1}{}{3}In the earth's atmosphere, the molecules are constantly moving around. Because temperature is a measure of kinetic energy per molecule, the average kinetic energy of each type of molecule is the same, e.g., the average KE of the \(\text{O}_2\) molecules is the same as the average KE of the \(\text{N}_2\) molecules. (a) If the mass of an \(\text{O}_2\) molecule is eight times greater than that of a He atom, what is the ratio of their average speeds? Which way is the ratio, i.e., which is typically moving faster? (b) Use your result from part a to explain why any helium occurring naturally in the atmosphere has long since escaped into outer space, never to return. (Helium is obtained commercially by extracting it from rocks.) You may want to do problem 21 first, for insight. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{bubba}{1}{}{4}Weiping lifts a rock with a weight of 1.0 N through a height of 1.0 m, and then lowers it back down to the starting point. Bubba pushes a table 1.0 m across the floor at constant speed, requiring a force of 1.0 N, and then pushes it back to where it started. (a) Compare the total work done by Weiping and Bubba. (b) Check that your answers to part a make sense, using the definition of work: work is the transfer of energy. In your answer, you'll need to discuss what specific type of energy is involved in each case. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{ant}{1}{}{5}(answer check available at lightandmatter.com) In one of his more flamboyant moments, Galileo wrote “Who does not know that a horse falling from a height of three or four cubits will break his bones, while a dog falling from the same height or a cat from a height of eight or ten cubits will suffer no injury? Equally harmless would be the fall of a grasshopper from a tower or the fall of an ant from the distance of the moon.” Find the speed of an ant that falls to earth from the distance of the moon at the moment when it is about to enter the atmosphere. Assume it is released from a point that is not actually near the moon, so the moon's gravity is negligible. You will need the result of example 9 on p. 333. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{}{1}{}{6} [Problem 6 has been deleted.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{anchor}{1}{}{7}(a) The crew of an 18th century warship is raising the
anchor. The anchor has a mass of 5000 kg. The water is 30
m deep. The chain to which the anchor is attached has a
mass per unit length of 150 kg/m. Before they start raising
the anchor, what is the total weight of the anchor plus the
portion of the chain hanging out of the ship? (Assume that
the buoyancy of the anchor and is negligible.)

(b) After they have raised the anchor by 1 m, what is the
weight they are raising?

(c) Define \(y=0\) when the anchor is resting on the bottom,
and \(y=+30\) m when it has been raised up to the ship. Draw a
graph of the force the crew has to exert to raise the anchor
and chain, as a function of \(y\). (Assume that they are
raising it slowly, so water resistance is negligible.) It
will not be a constant! Now find the area under the graph,
and determine the work done by the crew in raising
the anchor, in joules.

(d) Convert your answer from (c) into units of kcal. (answer check available at lightandmatter.com)
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{horsepower}{1}{}{8}In the power stroke of a car's gasoline engine, the
fuel-air mixture is ignited by the spark plug, explodes, and
pushes the piston out. The exploding mixture's force on the
piston head is greatest at the beginning of the explosion,
and decreases as the mixture expands. It can be approximated
by \(F=a/x\), where \(x\) is the distance from the cylinder to
the piston head, and \(a\) is a constant with units of \(\text{N}\!\cdot\!\text{m}\).
(Actually \(a/x^{1.4}\) would be more accurate, but the problem
works out more nicely with \(a/x\)!) The piston begins its
stroke at \(x=x_1\), and ends at \(x=x_2\). The 1965 Rambler had
six cylinders, each with \(a=220\ \text{N}\!\cdot\!\text{m}\), \(x_1=1.2\) cm, and \(x_2=10.2\) cm.

(a) Draw a neat, accurate graph of \(F\) vs \(x\), on graph paper.

(b) From the area under the curve, derive the amount of
work done in one stroke by one cylinder.(answer check available at lightandmatter.com)

(c) Assume the engine is running at 4800 r.p.m., so that (answer check available at lightandmatter.com)
during one minute, each of the six cylinders performs 2400
power strokes. (Power strokes only happen every other
revolution.) Find the engine's power, in units of horsepower (1 hp=746 W).

(d) The compression ratio of an engine is defined as
\(x_2/x_1\). Explain in words why the car's power would be
exactly the same if \(x_1\) and \(x_2\) were, say, halved or
tripled, maintaining the same compression ratio of 8.5.
Explain why this would *not* quite be true with the
more realistic force equation \(F=a/x^{1.4}\).
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{magnetpe}{1}{1}{9}The magnitude of the force between two magnets separated by a distance \(r\) can be approximated as \(kr^{-3}\) for large values of \(r\). The constant \(k\) depends on the strengths of the magnets and the relative orientations of their north and south poles. Two magnets are released on a slippery surface at an initial distance \(r_i\), and begin sliding towards each other. What will be the total kinetic energy of the two magnets when they reach a final distance \(r_f?\) (Ignore friction.) (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{keconstantaccel}{1}{1}{10}A car starts from rest at \(t=0\), and starts speeding up with constant acceleration. (a) Find the car's kinetic energy in terms of its mass, \(m\), acceleration, \(a\), and the time, \(t\). (b) Your answer in the previous part also equals the amount of work, \(W\), done from \(t=0\) until time \(t\). Take the derivative of the previous expression to find the power expended by the car at time \(t\). (c) Suppose two cars with the same mass both start from rest at the same time, but one has twice as much acceleration as the other. At any moment, how many times more power is being dissipated by the more quickly accelerating car? (The answer is not 2.) (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{space-probe}{2}{1}{11}A space probe of mass \(m\) is dropped into a previously unexplored spherical cloud of gas and dust, and accelerates toward the center of the cloud under the influence of the cloud's gravity. Measurements of its velocity allow its potential energy, \(PE\), to be determined as a function of the distance \(r\) from the cloud's center. The mass in the cloud is distributed in a spherically symmetric way, so its density, \(\rho(r)\), depends only on \(r\) and not on the angular coordinates. Show that by finding \(PE\), one can infer \(\rho(r)\) as follows:

\[\begin{equation*}
\rho(r) = \frac{1}{4\pi Gmr^2}\frac{d}{dr}\left(r^2\frac{dPE}{dr}\right) .
\end{equation*}\]

\end{homeworkforcelabel}

\begin{homeworkforcelabel}{railgun-friction}{1}{1}{12}A rail gun is a device like a train on a track, with the train propelled by a powerful electrical pulse. Very high speeds have been demonstrated in test models, and rail guns have been proposed as an alternative to rockets for sending into outer space any object that would be strong enough to survive the extreme accelerations. Suppose that the rail gun capsule is launched straight up, and that the force of air friction acting on it is given by \(F=be^{-cx}\), where \(x\) is the altitude, \(b\) and \(c\) are constants, and \(e\) is the base of natural logarithms. The exponential decay occurs because the atmosphere gets thinner with increasing altitude. (In reality, the force would probably drop off even faster than an exponential, because the capsule would be slowing down somewhat.) Find the amount of kinetic energy lost by the capsule due to air friction between when it is launched and when it is completely beyond the atmosphere. (Gravity is negligible, since the air friction force is much greater than the gravitational force.) (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{comet-and-binary-star}{1}{}{13}A certain binary star system consists of two stars with masses \(m_1\) and \(m_2\), separated by a distance \(b\). A comet, originally nearly at rest in deep space, drops into the system and at a certain point in time arrives at the midpoint between the two stars. For that moment in time, find its velocity, \(v\), symbolically in terms of \(b\), \(m_1\), \(m_2\), and fundamental constants. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{crosswind}{1}{1}{14}An airplane flies in the positive direction along the \(x\) axis, through crosswinds that exert a force \(\mathbf{F}=(a+bx)\hat{\mathbf{x}}+(c+dx)\hat{\mathbf{y}}\). Find the work done by the wind on the plane, and by the plane on the wind, in traveling from the origin to position \(x\). \end{homeworkforcelabel}

\begin{homeworkforcelabel}{yukawa}{1}{1}{15}In 1935, Yukawa proposed an early theory of the force that held the neutrons and protons together in the nucleus. His equation for the potential energy of two such particles, at a center-to-center distance \(r\), was \(PE(r)=gr^{-1}e^{-r/a}\), where \(g\) parametrizes the strength of the interaction, \(e\) is the base of natural logarithms, and \(a\) is about \(10^{-15}\) m. Find the force between two nucleons that would be consistent with this equation for the potential energy. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{dot-product-invariant}{1}{}{16}Prove that the dot product defined in section 13.7 is rotationally invariant in the sense of section 7.5. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{dot-product-by-components}{1}{}{17}Fill in the details of the proof of \(\mathbf{A}\cdot\mathbf{B} = A_xB_x+A_yB_y+A_zB_z\) on page 337. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{is-work-conserved}{1}{}{18}(solution in the pdf version of the book) Does it make sense to say that work is conserved? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{workreverseaxis}{1}{}{19}(a) Suppose work is done in one-dimensional motion. What happens to the work if you reverse the direction of the positive coordinate axis? Base your answer directly on the definition of work. (b) Now answer the question based on the \(W=Fd\) rule. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{microwaveice}{1}{}{20}A microwave oven works by twisting molecules one way and then the other, counterclockwise and then clockwise about their own centers, millions of times a second. If you put an ice cube or a stick of butter in a microwave, you'll observe that the solid doesn't heat very quickly, although eventually melting begins in one small spot. Once this spot forms, it grows rapidly, while the rest of the solid remains solid; it appears that a microwave oven heats a liquid much more rapidly than a solid. Explain why this should happen, based on the atomic-level description of heat, solids, and liquids. (See, e.g., figure b on page 305.)

Don't repeat the following common mistakes:

*In a solid, the atoms are packed more tightly and have less
space between them.* Not true. Ice floats because it's *less* dense
than water.

*In a liquid, the atoms are moving much faster.* No, the difference
in average speed between ice at \(-1°\text{C}\) and water at \(1°\text{C}\) is
only 0.4%.

\end{homeworkforcelabel}

\begin{homeworkforcelabel}{escape-velocity}{1}{}{21}(answer check available at lightandmatter.com) Starting at a distance \(r\) from a planet of mass \(M\), how fast must an object be moving in order to have a hyperbolic orbit, i.e., one that never comes back to the planet? This velocity is called the escape velocity. Interpreting the result, does it matter in what direction the velocity is? Does it matter what mass the object has? Does the object escape because it is moving too fast for gravity to act on it? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{contracting-muscle}{1}{}{22}The figure, redrawn from *Gray's Anatomy*, shows the tension of which a muscle is capable.
The variable \(x\) is defined as the contraction of the muscle from its maximum length \(L\), so that at
\(x=0\) the muscle has length \(L\), and at \(x=L\) the muscle would theoretically have zero length.
In reality, the muscle can only contract to \(x=cL\), where \(c\) is less than 1.
When the muscle is extended to its maximum length, at \(x=0\), it is capable of the greatest tension, \(T_\text{o}\).
As the muscle contracts, however, it becomes weaker. Gray suggests approximating this function as a linear
decrease, which would theoretically extrapolate to zero at \(x=L\).
(a) Find the maximum work the muscle can do in one contraction, in terms of \(c\), \(L\), and \(T_\text{o}\).(answer check available at lightandmatter.com)

(b) Show that your answer to part a has the right units.

(c) Show that your answer to part a has the right behavior when \(c=0\) and when \(c=1\).

(d) Gray also states that the absolute maximum tension \(T_\text{o}\) has been found to be approximately
proportional to the muscle's cross-sectional area \(A\) (which is presumably measured at \(x=0\)), with proportionality constant
\(k\).
Approximating the muscle as a cylinder,
show that your answer from part a can be reexpressed in terms of the volume, \(V\), eliminating \(L\) and \(A\).(answer check available at lightandmatter.com)

(e) Evaluate your result numerically for a biceps muscle with a volume of 200 \(\text{cm}^3\), with \(c=0.8\) and
\(k=100\ \text{N}/\text{cm}^2\) as estimated by Gray.(answer check available at lightandmatter.com)

\end{homeworkforcelabel}

\begin{homeworkforcelabel}{maserati}{1}{}{23}A car accelerates from rest. At low speeds, its acceleration is limited
by static friction, so that if we press too hard on the gas, we will
“burn rubber” (or, for many newer cars, a computerized traction-control
system will override the gas pedal). At higher speeds, the limit on
acceleration comes from the power of the engine, which puts a limit on
how fast kinetic energy can be developed.

(a) Show that if a force \(F\) is applied to an object moving at
speed \(v\), the power required is given by \(P=vF\).

(b) Find the speed \(v\) at which we cross over from the first regime
described above to the second. At speeds higher than this, the engine
does not have enough power to burn rubber. Express your result in terms of the
car's power \(P\), its mass \(m\), the coefficient of static friction \(\mu_s\),
and \(g\).(answer check available at lightandmatter.com)

(c) Show that your answer to part b has units that make sense.

(d) Show that the dependence of your answer on each of the four variables
makes sense physically.

(e) The 2010 Maserati Gran Turismo Convertible has a maximum power of
\(3.23\times10^5\ \text{W}\) (433 horsepower) and a mass (including a 50-kg driver)
of \(2.03\times10^3\ \text{kg}\). (This power is the maximum the engine can supply
at its optimum frequency of 7600 r.p.m. Presumably the automatic transmission is designed so a gear is available
in which the engine will be running at very nearly this frequency when
the car is at moving at \(v\).) Rubber on asphalt has \(\mu_s\approx0.9\).
Find \(v\) for this car. Answer: \(18\ \text{m}/\text{s}\), or about 40 miles per hour.

(f) Our analysis has neglected air friction, which can probably be approximated
as a force proportional to \(v^2\). The existence of this force is the reason that the
car has a maximum speed, which is 176 miles per hour. To get a feeling for how good an approximation
it is to ignore air friction, find what fraction of the engine's
maximum power is being used to overcome air resistance when the car is moving at
the speed \(v\) found in part e. Answer: 1%
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{compound-bow}{1}{}{24}Most modern bow hunters in the U.S. use a fancy mechanical bow called a compound bow, which looks nothing like what most people imagine when they think of a bow and arrow. It has a system of pulleys designed to produce the force curve shown in the figure, where \(F\) is the force required to pull the string back, and \(x\) is the distance between the string and the center of the bow's body. It is not a linear Hooke's-law graph, as it would be for an old-fashioned bow. The big advantage of the design is that relatively little force is required to hold the bow stretched to point B on the graph. This is the force required from the hunter in order to hold the bow ready while waiting for a shot. Since it may be necessary to wait a long time, this force can't be too big. An old-fashioned bow, designed to require the same amount of force when fully drawn, would shoot arrows at much lower speeds, since its graph would be a straight line from A to B. For the graph shown in the figure (taken from realistic data), find the speed at which a 26 g arrow is released, assuming that 70% of the mechanical work done by the hand is actually transmitted to the arrow. (The other 30% is lost to frictional heating inside the bow and kinetic energy of the recoiling and vibrating bow.) (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{sliding-oscillator}{1}{}{25}A mass moving in one dimension is attached to a horizontal spring. It slides on the surface below it, with equal coefficients of static and kinetic friction, \(\mu_k=\mu_s\). The equilibrium position is \(x=0\). If the mass is pulled to some initial position and released from rest, it will complete some number of oscillations before friction brings it to a stop. When released from \(x=a\) (\(a>0\)), it completes exactly 1/4 of an oscillation, i.e., it stops precisely at \(x=0\). Similarly, define \(b>0\) as the greatest \(x\) from which it could be released and comlete 1/2 of an oscillation, stopping on the far side and not coming back toward equilibrium. Find \(b/a\). Hint: To keep the algebra simple, set every fixed parameter of the system equal to 1.(answer check available at lightandmatter.com) \end{homeworkforcelabel}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.