7.4: Work-Energy Theorem
- Apply the work-energy theorem to find information about the motion of a particle, given the forces acting on it
- Use the work-energy theorem to find information about the forces acting on a particle, given information about its motion
We have discussed how to find the work done on a particle by the forces that act on it, but how is that work manifested in the motion of the particle? According to Newton’s second law of motion, the sum of all the forces acting on a particle, or the net force, determines the rate of change in the momentum of the particle, or its motion. Therefore, we should consider the work done by all the forces acting on a particle, or the net work , to see what effect it has on the particle’s motion.
Let’s start by looking at the net work done on a particle as it moves over an infinitesimal displacement, which is the dot product of the net force and the displacement:
\[dW_{net} = \vec{F}_{net} \cdotp d \vec{r}. \nonumber\]
Newton’s second law tells us that
\[\vec{F}_{net} = m \left(\dfrac{d \vec{v}}{dt}\right) \nonumber\]
so
\[dW_{net} = m \left(\dfrac{d \vec{v}}{dt}\right) \cdotp d \vec{r}. \nonumber\]
For the mathematical functions describing the motion of a physical particle, we can rearrange the differentials dt, etc., as algebraic quantities in this expression, that is,
\[\begin{align*} dW_{net} &= m \left(\dfrac{d \vec{v}}{dt}\right) \cdotp d \vec{r} \\[4pt] &= m\, d \vec{v}\; \cdotp \left(\dfrac{d \vec{r}}{dt}\right) \\[4pt] &= m \vec{v}\; \cdotp d \vec{v}, \end{align*}\]
where we substituted the velocity for the time derivative of the displacement and used the commutative property of the dot product. Since derivatives and integrals of scalars are probably more familiar to you at this point, we express the dot product in terms of Cartesian coordinates before we integrate between any two points A and B on the particle’s trajectory. This gives us the net work done on the particle:
\[\begin{align} W_{net,\; AB} & = \int_{A}^{B} (mv_{x} dv_{x} + mv_{y}dv_{y} + mv_{z}dv_{z} \\[4pt] & = \frac{1}{2} m \left| v_{x}^{2} + v_{y}^{2} + v_{z}^{2} \right|_{A}^{B} = \left|\frac{1}{2} mv^{2} \right|_{A}^{B} = K_{B} - K_{A} \ldotp \end{align} \label{7.8}\]
In the middle step, we used the fact that the square of the velocity is the sum of the squares of its Cartesian components, and in the last step, we used the definition of the particle’s kinetic energy. This important result is called the work-energy theorem.
The net work done on a particle equals the change in the particle’s kinetic energy:
\[W_{net} = K_{B} - K_{A} \ldotp \label{7.9}\]
According to this theorem, when an object slows down, its final kinetic energy is less than its initial kinetic energy, the change in its kinetic energy is negative, and so is the net work done on it. If an object speeds up, the net work done on it is positive. When calculating the net work, you must include all the forces that act on an object. If you leave out any forces that act on an object, or if you include any forces that do not act on it, you will get a wrong result.
The importance of the work-energy theorem, and the further generalizations to which it leads, is that it makes some types of calculations much simpler to accomplish than they would be by trying to solve Newton’s second law. For example, in the section on Newton’s Laws of Motion , we found the speed of an object sliding down a frictionless plane by solving Newton’s second law for the acceleration and using kinematic equations for constant acceleration, obtaining
\[v_{f}^{2} = v_{i}^{2} + 2g(s_{f} - s_{i}) \sin \theta,\]
where \(s\) is the displacement down the plane.
We can also get this result from the work-energy theorem (Equation \ref{7.9}). Since only two forces are acting on the object—gravity and the normal force—and the normal force does not do any work, the net work is just the work done by gravity. This only depends on the object’s weight and the difference in height, so
\[W_{net} = W_{grav} = -mg (y_{f} - y_{i}),\]
where \(y\) is positive up. The work-energy theorem says that this equals the change in kinetic energy:
\[-mg (y_{f} - y_{i}) = \frac{1}{2} (v_{f}^{2} - v_{i}^{2}) \ldotp\]
Using a right triangle, we can see that
\[(y_f − y_i) = (s_f − s-i)\sin \theta, \nonumber\]
so the result for the final speed is the same.
What is gained by using the work-energy theorem? The answer is that for a frictionless plane surface, not much. However, Newton’s second law is easy to solve only for this particular case, whereas the work-energy theorem gives the final speed for any shaped frictionless surface. For an arbitrary curved surface, the normal force is not constant, and Newton’s second law may be difficult or impossible to solve analytically. Constant or not, for motion along a surface, the normal force never does any work, because it’s perpendicular to the displacement. A calculation using the work-energy theorem avoids this difficulty and applies to more general situations.
- Draw a free-body diagram for each force on the object.
- Determine whether or not each force does work over the displacement in the diagram. Be sure to keep any positive or negative signs in the work done.
- Add up the total amount of work done by each force.
- Set this total work equal to the change in kinetic energy and solve for any unknown parameter.
- Check your answers. If the object is traveling at a constant speed or zero acceleration, the total work done should be zero and match the change in kinetic energy. If the total work is positive, the object must have sped up or increased kinetic energy. If the total work is negative, the object must have slowed down or decreased kinetic energy
The frictionless track for a toy car includes a loop-the-loop of radius \(R\). How high, measured from the bottom of the loop, must the car be placed to start from rest on the approaching section of track and go all the way around the loop?
Strategy
The free-body diagram at the final position of the object is drawn in Figure \(\PageIndex{2}\). The gravitational work is the only work done over the displacement that is not zero. Since the weight points in the same direction as the net vertical displacement, the total work done by the gravitational force is positive. From the work-energy theorem, the starting height determines the speed of the car at the top of the loop,
\[mg(y_{2} - y_{1}) = \dfrac{1}{2} mv_{2}^{2}, \nonumber\]
where the notation is shown in the accompanying figure. At the top of the loop, the normal force and gravity are both down and the acceleration is centripetal, so
\[a_{top} = \dfrac{F}{m} = \dfrac{N + mg}{m} = \frac{v_{2}^{2}}{R} \ldotp \nonumber\]
The condition for maintaining contact with the track is that there must be some normal force, however slight; that is, \(N > 0\). Substituting for \(v_{2}^{2}\) and \(N\), we can find the condition for \(y_1\).
Solution
Implement the steps in the strategy to arrive at the desired result:
\[N = -mg + \frac{mv_{2}^{2}}{R} = \frac{-mgR + 2mg(y_{1} - 2R)}{R} > 0\; or\; y_{1} > \frac{5R}{2} \ldotp \nonumber\]
Significance
On the surface of the loop, the normal component of gravity and the normal contact force must provide the centripetal acceleration of the car going around the loop. The tangential component of gravity slows down or speeds up the car. A child would find out how high to start the car by trial and error, but now that you know the work-energy theorem, you can predict the minimum height (as well as other more useful results) from physical principles. By using the work-energy theorem, you did not have to solve a differential equation to determine the height.
Suppose the radius of the loop-the-loop in Example \(\PageIndex{1}\) is 15 cm and the toy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop?
In situations where the motion of an object is known, but the values of one or more of the forces acting on it are not known, you may be able to use the work-energy theorem to get some information about the forces. Work depends on the force and the distance over which it acts, so the information is provided via their product.
A bullet has a mass of 40 grains (2.60 g) and a muzzle velocity of 1100 ft./s (335 m/s). It can penetrate eight 1-inch pine boards, each with thickness 0.75 inches. What is the average stopping force exerted by the wood, as shown in Figure \(\PageIndex{3}\)?
Strategy
We can assume that under the general conditions stated, the bullet loses all its kinetic energy penetrating the boards, so the work-energy theorem says its initial kinetic energy is equal to the average stopping force times the distance penetrated. The change in the bullet’s kinetic energy and the net work done stopping it are both negative, so when you write out the work-energy theorem, with the net work equal to the average force times the stopping distance, that’s what you get. The total thickness of eight 1-inch pine boards that the bullet penetrates is 8 x \(\frac{3}{4}\) in. = 6 in. = 15.2 cm.
Solution
Applying the work-energy theorem, we get
\[W_{net} = - F_{ave} \Delta s_{stop} = - K_{initial} , \nonumber\]
so
\[F_{ave} = \frac{\frac{1}{2} mv^{2}}{\Delta s_{stop}} = \frac{\frac{1}{2} (2.66 \times 10^{-3}\; kg)(335\; m/s)^{2}}{0.152\; m} = 960\; N \ldotp \nonumber\]
Significance
We could have used Newton’s second law and kinematics in this example, but the work-energy theorem also supplies an answer to less simple situations. The penetration of a bullet, fired vertically upward into a block of wood, is discussed in one section of Asif Shakur’s recent article [“Bullet-Block Science Video Puzzle.” The Physics Teacher (January 2015) 53(1): 15-16]. If the bullet is fired dead center into the block, it loses all its kinetic energy and penetrates slightly farther than if fired off-center. The reason is that if the bullet hits off-center, it has a little kinetic energy after it stops penetrating, because the block rotates. The work-energy theorem implies that a smaller change in kinetic energy results in a smaller penetration. You will understand more of the physics in this interesting article after you finish reading Angular Momentum .
Learn more about work and energy in this PhET simulation ( https://phet.colorado.edu/en/simulation/the-ramp ) called “the ramp.” Try changing the force pushing the box and the frictional force along the incline. The work and energy plots can be examined to note the total work done and change in kinetic energy of the box.


7.2: Laws of Conservation : A Review
Learning Objectives
- Correctly define a law as it pertains to science.
- State the law of conservation of matter.
In science, a law is a general statement that explains a large number of observations. Before being accepted, a law must be verified many times under many conditions. Laws are therefore considered the highest form of scientific knowledge and are generally thought to be inviolable. Scientific laws form the core of scientific knowledge. One scientific law that provides the foundation for understanding in chemistry is the law of conservation of matter. It states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant. A concise way of expressing this law is to say that the amount of matter in a system is conserved .
With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. These eventually formed the basis of Dalton's Atomic Theory of Matter.
Law of Conservation of Mass
According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.
\[ \overbrace{\underbrace{\ce{HgO (s)}}_{\text{100 g}}}^{\text{Mercuric oxide}} \rightarrow \underbrace{\overbrace{\ce{Hg (l) }}^{\text{Mercury}}}_{\text{92.6 g}} + \underbrace{\overbrace{\ce{O2 (g)}}^{\text{Oxygen}}}_{\text{7.4 g}} \]
Another way of stating this is, "In a chemical reaction, matter is neither created nor destroyed." The law of conservation of mass is also known as the "law of indestructibility of matter."
Example \(\PageIndex{1}\)
If heating 10 grams of \(\ce{CaCO3}\) produces 4.4 g of \(\ce{CO2}\) and 5.6 g of \(\ce{CaO}\), show that these observations are in agreement with the law of conservation of mass.
A sample of calcium carbonate (CaCO 3 ). (Public Domain; Walkerma ).
Solution
- Mass of the reactants: \(10 \,g\)
- Mass of the products: \(4.4 \,g+ 5.6\, g = 10\, g\).
Because the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass.
What does this mean for chemistry? In any chemical change, one or more initial substances change into a different substance or substances. Both the initial and final substances are composed of atoms because all matter is composed of atoms. According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and type of atoms after the chemical change as were present before the chemical change.
Before looking at explicit examples of the law of conservation of matter, we need to examine the method chemists use to represent chemical changes.
Exercise \(\PageIndex{1}\)
- What is the law of conservation of matter?
- How does the law of conservation of matter apply to chemistry?
- Answer a:
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The law of conservation of matter states that in any given system that is closed to the transfer of matter, the amount of matter in the system stays constant
- Answer b:
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The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.
Key Takeaway
The amount of matter in a closed system is conserved.
Exercises
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Express the law of conservation of matter in your own words.
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Explain why the concept of conservation of matter is considered a scientific law.
Answer
1. Matter may not be created or destroyed.
2. The concept is a scientific law because it is based on experimentation.


21.3: Adding Physical Activity to Your Life
Overcoming Barriers to Physical Activity
Given the health benefits of regular physical activity, we might have to ask why two out of three (60%) Americans are not active at recommended levels.
Many technological advances and conveniences that have made our lives easier and less active, as well as many personal variables, including physiological, behavioral, and psychological factors, may affect our plans to become more physically active. In fact, the 10 most common reasons adults cite for not adopting more physically active lifestyles are (Sallis and Hovell, 1990; Sallis et al., 1992):
- Do not have enough time to exercise
- Find it inconvenient to exercise
- Lack self-motivation
- Do not find exercise enjoyable
- Find exercise boring
- Lack confidence in their ability to be physically active (low self-efficacy)
- Fear being injured or have been injured recently
- Lack self-management skills, such as the ability to set personal goals, monitor progress, or reward progress toward such goals
- Lack encouragement, support, or companionship from family and friends, and
- Do not have parks, sidewalks, bicycle trails, or safe and pleasant walking paths convenient to their homes or offices.
Understanding common barriers to physical activity and creating strategies to overcome them may help you make physical activity part of your daily life.
Suggestions for Overcoming Physical Activity Barriers |
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|---|---|
| Lack of time | Identify available time slots. Monitor your daily activities for one week. Identify at least three 30-minute time slots you could use for physical activity. |
| Add physical activity to your daily routine. For example, walk or ride your bike to work or shopping, organize school activities around physical activity, walk the dog, exercise while you watch TV, park farther away from your destination, etc. | |
| Select activities requiring minimal time, such as walking, jogging, or stairclimbing. | |
| Social influence | Explain your interest in physical activity to friends and family. Ask them to support your efforts. |
| Invite friends and family members to exercise with you. Plan social activities involving exercise. | |
| Develop new friendships with physically active people. Join a group, such as the YMCA or a hiking club. | |
| Lack of energy | Schedule physical activity for times in the day or week when you feel energetic. |
| Convince yourself that if you give it a chance, physical activity will increase your energy level; then, try it. | |
| Lack of motivation | Plan ahead. Make physical activity a regular part of your daily or weekly schedule and write it on your calendar. |
| Invite a friend to exercise with you on a regular basis and write it on both your calendars. | |
| Join an exercise group or class. | |
| Fear of injury | Learn how to warm up and cool down to prevent injury. |
| Learn how to exercise appropriately considering your age, fitness level, skill level, and health status. | |
| Choose activities involving minimum risk. | |
| Lack of skill | Select activities requiring no new skills, such as walking, climbing stairs, or jogging. |
| Take a class to develop new skills. | |
| Lack of resources | Select activities that require minimal facilities or equipment, such as walking, jogging, jumping rope, or calisthenics. |
| Identify inexpensive, convenient resources available in your community (community education programs, park and recreation programs, worksite programs, etc.). | |
| Weather conditions | Develop a set of regular activities that are always available regardless of weather (indoor cycling, aerobic dance, indoor swimming, calisthenics, stair climbing, rope skipping, mall walking, dancing, gymnasium games, etc.) |
| Travel | Put a jump rope in your suitcase and jump rope. |
| Walk the halls and climb the stairs in hotels. | |
| Stay in places with swimming pools or exercise facilities. | |
| Join the YMCA or YWCA (ask about reciprocal membership agreement). | |
| Visit the local shopping mall and walk for half an hour or more. | |
| Bring your mp3 player your favorite aerobic exercise music. | |
| Family obligations | Trade babysitting time with a friend, neighbor, or family member who also has small children. |
| Exercise with the kids-go for a walk together, play tag or other running games, get an aerobic dance or exercise tape for kids (there are several on the market) and exercise together. You can spend time together and still get your exercise. | |
| Jump rope, do calisthenics, ride a stationary bicycle, or use other home gymnasium equipment while the kids are busy playing or sleeping. | |
| Try to exercise when the kids are not around (e.g., during school hours or their nap time). | |
| Retirement years | Look upon your retirement as an opportunity to become more active instead of less. Spend more time gardening, walking the dog, and playing with your grandchildren. Children with short legs and grandparents with slower gaits are often great walking partners. |
| Learn a new skill you’ve always been interested in, such as ballroom dancing, square dancing, or swimming. | |
| Now that you have the time, make regular physical activity a part of every day.  Treat yourself to an exercycle and ride every day while reading a favorite book or magazine. |
Content in the “Personal Barriers” section was taken from Promoting Physical Activity: A Guide for Community Action (USDHHS, 1999).
- Adding Physical Activity to Your Life. Provided by : Centers for Disease Control and Prevention. Located at : http://www.cdc.gov/physicalactivity/basics/adding-pa/barriers.html . License : Public Domain: No Known Copyright


9: Cell Communication
While the necessity for cellular communication in larger organisms seems obvious, even single-celled organisms communicate with each other. Yeast cells signal each other to aid mating. Some forms of bacteria coordinate their actions in order to form large complexes called biofilms or to organize the production of toxins to remove competing organisms. The ability of cells to communicate through chemical signals originated in single cells and was essential for the evolution of multicellular organisms. The efficient and error-free function of communication systems is vital for all life as we know it.
Thumbnail image credit: "Cell-to-cell Communication" by NIH Image Gallery is licensed under CC BY-NC 4.0
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- 9.2: Signaling Molecules and Cellular Receptors
- Chemical signals are released by signaling cells in the form of small, usually volatile or soluble molecules called ligands. A ligand is a molecule that binds another specific molecule, in some cases, delivering a signal in the process. Ligands can thus be thought of as signaling molecules. Ligands interact with proteins in target cells, which are cells that are affected by chemical signals; these proteins are also called receptors.
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- 9.3: Propagation of the Signal
- Once a ligand binds to a receptor, the signal is transmitted through the membrane and into the cytoplasm. Continuation of a signal in this manner is called signal transduction. Signal transduction only occurs with cell-surface receptors because internal receptors are able to interact directly with DNA in the nucleus to initiate protein synthesis. When a ligand binds to its receptor, conformational changes occur that affect the receptor’s intracellular domain.
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- 9.4: Response to the Signal
- Inside the cell, ligands bind to their internal receptors, allowing them to directly affect the cell’s DNA and protein-producing machinery. Using signal transduction pathways, receptors in the plasma membrane produce a variety of effects on the cell. The results of signaling pathways are extremely varied and depend on the type of cell involved as well as the external and internal conditions. A small sampling of responses is described below.
Contributors and Attributions
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Connie Rye (East Mississippi Community College), Robert Wise (University of Wisconsin, Oshkosh), Vladimir Jurukovski (Suffolk County Community College), Jean DeSaix (University of North Carolina at Chapel Hill), Jung Choi (Georgia Institute of Technology), Yael Avissar (Rhode Island College) among other contributing authors. Original content by OpenStax (CC BY 4.0; Download for free at<EMAIL_ADDRESS>).


11.5: Butadiene is Stabilized by a Delocalization Energy
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Learning Objectives
- Apply Hückel theory to an extended \(\pi\)-bonding network
- Identify the origin of delocalization energy from Hückel theory and relate it to resonances structures in valence bond theory
1,3-Butadiene is a simple conjugated diene with the formula \(\ce{C_4H_6}\) and can be viewed structurally as two vinyl groups (\(\ce{CH_2=CH_2}\)) joined together with a single bond. Butadiene can occupy either a cis or trans conformers and at room temperature, 96% of butadiene exists as the trans conformer, which is 2.3 kcal/mole more stable than the cis structure.
For the simple application of applying Hückel theory for understanding the electronic structure of butadiene, we will ignore the energetic differences between the two conformers. As discussed previously, the molecular orbitals are linear combination of the four \(|p \rangle\) atomic orbitals on the carbon atoms that are not participating in the \(\sigma\) bonding network:
\[|\psi_i \rangle = \sum_j^4 c_{ij} |p_{i} \rangle \nonumber \]
or explicitly
\[|\psi_i \rangle =c_{i1} |p_1 \rangle+ c_ {i2} | p_2 \rangle + c_ {i3} | p_3 \rangle + c_ {i4} | p_4 \rangle \label{general} \]
for the i th molecular orbital \(|\psi_i \rangle \). The secular equations that need to be solved are
\[\begin{bmatrix} H_{11} - ES_{11} & H_{12} - ES_{12} & H_{13} - ES_{13} & H_{14} - ES_{14} \\ H_{12} - ES_{12} & H_{22} - ES_{22} & H_{23} - ES_{23} & H_{24} - ES_{24} \\ H_{13} - ES_{13} & H_{23} - ES_{23} & H_{33} - ES_{33} & H_{34} - ES_{34} \\ H_{14} - ES_{14} & H_{24} - ES_{24} & H_{34} - ES_{34} & H_{44} - ES_{44} \\ \end{bmatrix} \times\begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ \end{bmatrix}= 0 \label{complete} \]
If the standard Hückel theory approximations were used
\[ H_{ii}- ES_{ii} = \alpha\nonumber \]
and
\[ H_{ij}- ES_{ij} = \beta\nonumber \]
when \(i=j\pm 1\), otherwise
\[ H_{ij}- ES_{ij} = 0\nonumber \]
then the secular equations for butadiene in Equation \(\ref{complete}\) become
\[\begin{bmatrix} \alpha - E & \beta & 0 & 0 \\ \beta & \alpha - E & \beta & 0 \\ 0 & \beta & \alpha - E & \beta \\ 0 & 0 & \beta & \alpha - E \\ \end{bmatrix} \times\begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ \end{bmatrix}= 0 \label{Huckel1} \]
Solving Equation \(\ref{Huckel1}\) for \(\{c_i\}\) coefficients and energy secular equation requires extracting the roots of the secular determinant:
\[\left|\begin{array}{cccc}\alpha-E&\beta&0&0\\\beta&\alpha-E&\beta&0\\0&\beta&\alpha-E&\beta\\0&0&\beta&\alpha-E\end{array}\right|=0\label{24} \]
If both sides of Equation \(\ref{24}\) were divided by \(\beta^{4}\) and a new variable \(x\) is defined
\[x=\dfrac{\alpha-E}{\beta}\label{25} \]
then Equation \(\ref{24}\) simplifies further to
\[\left|\begin{array}{cccc}x&1&0&0\\1&x&1&0\\0&1&x&1\\0&0&1&x\end{array}\right|=0\label{26} \]
This is essentially the connection matrix for the butadiene molecule. Each pair of connected atoms is represented by 1, each non-connected pair by 0 and each diagonal element by \(x\). Expansion of the determinant in Equation \(\ref{26}\) gives the 4th order polynomial equation
\[x^{4}-3x^{2}+1=0 \label{27} \]
While solving 4th order equations typically require numerical estimation, Equation \(\ref{27}\) can be further simplified by recognizing that it is a quadratic equation in terms of \(x^{2}\). Therefore, the roots are
\[x^{2}= \dfrac{3\pm\sqrt{5}}{2} \nonumber \]
or \(x=\pm\; 0.618\) and \(x= \pm\; 1.618\). Since \(\alpha\) and \(\beta\) are negative, these molecular orbital energies can ordered in terms of energy (from lowest to highest):
\[E_1=\alpha+1.618\beta \label{E1} \]
\[E_2=\alpha+0.618\beta \label{E2} \]
\[E_3=\alpha-0.618\beta \label{E3} \]
\[E_4=\alpha-1.618\beta \label{E4} \]
This sequence of energies is displayed in the energy diagram of Figure 10.6.1 .
Each p atomic orbital of carbon contributes a single electron to the \(\pi\) manifold, so the ground-state occupation of the resulting four \(\pi\) electrons have a \(\pi_1^{2}\pi_2^{2}\) configuration (Figure 10.6.1 ). The the total \(\pi\)-electron energy is then determined by adding up the energies in Equations \(\ref{E1}\)-\(\ref{E4}\) and scaling by their occupations to get
\[\begin{align} E_{\pi} (\text{butadiene}) &= 2 \times E_1 + 2 \times E_2 + 0 \times E_3 + 0 \times E_4 \nonumber \\[4pt] &=2(\alpha+1.618\beta)+2(\alpha+0.618\beta) \nonumber \\[4pt] &=4\alpha + 4.472\beta \label{29} \end{align} \]
If the bonding of butadiene were described only as two localized double bond as in its dominant valence-bond structure (Figure 10.6.1 ), then its \(\pi\)-electron energy would be given by twice the \(E_{\pi}\) predicted for the ethlyene molecule:
\[\begin{align} E_{\pi} (\text{butadiene}) &= 2 \times E_{\pi} (\text{ethylene}) \nonumber \\[4pt] &=2 \times 2(\alpha+\beta) \nonumber \\[4pt] &= 4\alpha + 4\beta \label{ 30}\end{align} \]
Comparing Equation \(\ref{29}\) with Equation \(\ref{29}\) , the total \(\pi\) energy of butadiene lies lower than the total \(\pi\) energy of two double bonds by \(0.48\beta\) (the \(\sigma\) bond does not contribute). This difference is known as the delocalization energy ; a typical estimate of \(\beta\) is around -75 kJ/mol, which results in a delocalization energy for butadiene of -35 kJ/mol.
The delocalization energy is the extra stabilization resulting from the electrons extending over the whole molecule.
Delocalization Energy in Valence Bond Theory
Delocalization energy is intrinsic to molecular orbital theory, since it results from breaking the two-center bond concept with the molecular orbitals that spread over more that just one pair of atoms. However, within the two-center theory of valence bond theory, the delocalization energy results from a stabilization energy attributed to resonance . Several conventional valence bond resonance structures that can be written for 1,3-butadiene, four of which are shown in Figure 10.6.2 . However, while structure \(2a\) dominates, the other resonance structures also contribute to describing the total molecule and hence predict a corresponding stabilization energy akin to the delocalization energy in molecular orbital theory.
In general, the true description of the bonding within the valence bond theory is a superposition of resonance structures with amplitudes that are determined via a variational optimization to find the lowest possible energy for the valence bond wavefunctions.
The solving the secular equations (Equation \ref{complete}) gives the \(\{c_{ij}\}\) coefficients for the molecular orbitals in Equation \(\ref{general}\) (not demonstrated):
\[|\psi_1 \rangle =0.37 |p_1 \rangle + 0.60 | p_2 \rangle + 0.60 | p_3 \rangle + 0.37 | p_4 \rangle \label{MO1} \]
\[|\psi_3 \rangle =0.60 |p_1 \rangle + 0.37 | p_2 \rangle -0.37 | p_3 \rangle - 0.60 | p_4 \rangle \label{MO2} \]
\[|\psi_3 \rangle =0.60 |p_1 \rangle - 0.37 | p_2 \rangle -0.37 | p_3 \rangle + 0.60 | p_4 \rangle \label{MO3} \]
\[|\psi_4 \rangle =0.37 |p_1 \rangle - 0.60 | p_2 \rangle + 0.60 | p_3 \rangle - 0.37 | p_4 \rangle \label{MO4} \]
These are depicted in Figure 10.6.3 .
Note the correlation of the energy of the \(\pi\) molecular orbitals of butadiene to the number of nodes in the wavefunction; this is the general trend observed in previous systems like the particle in the box and atomic orbitals. The four 3-D calculated molecular orbitals are contrasted in Figure 10.6.1 .
Exercise 10.6.1
What would Hückel theory predict for the energy levels and \(\pi\) molecular orbitals of cis-butadiene? Do you believe this approach adequately describes the energy levels of the cis-trans isomerization reaction of butadiene?


9.7E: Mu- A Double-Stranded Transposable DNA Bacteriophage
- Outline the life cycle of Mu phages
Bacteriophage Mu, or phage Mu, is a temperate bacteriophage, a type of virus that infects bacteria. It belongs to the family Myoviridae, and consists of an icosahedral head, a contractile tail, and six tail fibers.
All of the known temperate phages employ one of only three different systems for their lysogenic cycle: lambda-like integration/excision, Mu-like transposition, or the plasmid-like partitioning of phage N15.
Mu bacteriophage uses DNA-based transposition to integrate its genome into the genome of the host cell that it is infecting. It can then use transposition to initiate its viral DNA replication. Once the viral DNA is inserted into the bacteria, the Mu’s transposase protein/enzyme in the cell recognizes the recombination sites at the ends of the viral DNA (gix-L and gix-R sites) and binds to them, allowing the process of replicating the viral DNA or embedding it into the host genome. A transposable element (TE) is a DNA sequence that can change its relative position (self-transpose) within the genome of a single cell. The mechanism of transposition can be either “copy and paste” or “cut and paste. ” Transposition can create phenotypically significant mutations and alter the cell’s genome size.
Mu phage transposition is the best known example of replicative transposition. Its transposition mechanism is somewhat similar to a homologous recombination. Replicative transposition is a mechanism of transposition in molecular biology, proposed by James A. Shapiro in 1979, in which the transposable element is duplicated during the reaction, so that the transposing entity is a copy of the original element. In this mechanism, the donor and receptor DNA sequences form a characteristic intermediate “theta” configuration, sometimes called a “Shapiro intermediate. ” Replicative transposition is characteristic to retrotransposons and occurs from time to time in class II transposons.
Key Points
- All of the known temperate phages employ one of only three different systems for their lysogenic cycle: lambda-like integration/excision, Mu-like transposition, or the plasmid-like partitioning of phage N15.
- Phage Mu uses DNA -based transposition to integrate its genome into the genome of the host cell that it is infecting. It can then use transposition to initiate its viral DNA replication.
- Mu phage transposition is the best-known example of replicative transposition. Its transposition mechanism is somewhat similar to a homologous recombination.
Key Terms
- replicative transposition : A mechanism of transposition in which the transposable element is duplicated during the reaction, so that the transposing entity is a copy of the original element.


22.13: Indonesian Art
Indonesian Architecture
Indonesian architecture has been shaped by interaction between indigenous customs and foreign influences, including Hinduism, Buddhism, and Islam.
Summarize the developments of the golden age of Indonesian architecture
Key Points
- Indonesia has a particularly rich tradition of Hindu–Buddhist sculpture and architecture, and it was strongly influenced by India from the 1st century CE onward.
- Buddhist art in Indonesia reached its golden era under the Sailendra dynasty of the Sri Vijaya Empire between the 8th and 13th centuries.
- Sculpture flourished between the 8th and 10th centuries AD in Java and Bali, taking the form of free-standing statues or relief sculptures incorporated into temples. They are characterized by their delicacy and serenity of expression.
- Indonesian art enjoyed another golden age under the Majapahit Empire (1293–1500), during which a large number of Hindu–Buddhist brick temples were built, characterized by tall, slender-roofed red brick gates and a strong geometrical quality.
- By the 15th century, Islam had become the dominant religion in Indonesia, and local mosques reflected both indigenous and Islamic influences. They lacked the Islamic dome and had tall timber -tiered roofs similar to the pagodas of Balinese Hindu temples.
Key Terms
- animism : A belief that spirits inhabit some or all classes of natural objects or phenomena.
- mandala : Any ritualistic geometric design, symbolic of the universe, used as an aid to meditation, particularly in Hinduism and Buddhism.
- syncretism : The reconciliation or fusion of different systems or beliefs (or the attempt at such fusion).
- candi : Hindu and Buddhist temples and sanctuaries of Indonesia, mostly built during the 8th to 15th centuries; however, ancient non-religious structures such as gates, urban ruins, and bathing places are often also called by this name.
Overview: Indonesia
The culture and art of Indonesia has been shaped by interaction between local indigenous customs and multiple foreign influences. Situated on the ancient maritime trading routes between the Near East and the Far East, Indonesia was exposed to a multitude of foreign cultural practices and religions, including Hinduism , Buddhism , and Islam. The result is a complex fusion of many different customs, expressed in Indian art forms.
Architecture in Indonesia before 1200 CE
The Sri Vijaya Empire
Indonesia has a particularly rich tradition of Hindu–Buddhist sculpture and architecture, which was strongly influenced by India from the 1st century CE onward. The earliest Buddhist structures in Indonesia to survive to the present day are the 4th century Batujaya plastered brick stupas in West Java. However, Buddhist art reached its golden era under the Sailendra dynasty of the Sri Vijaya Empire. The islands of Sumatra and Java in western Indonesia were the seat of the Sri Vijaya Empire (8th–13th centuries), which practiced Mahayana and Vajrayana Buddhism and were a major political and cultural influence in the Southeast Asian peninsula.
Stone and bronze sculpture flourished between the 8th and 10th century CE under the Sailendra dynasty in Java and Bali. These sculptures were either free-standing statues or relief sculptures and friezes incorporated into temples; they are characterized by their delicacy and serenity of expression.
Borobudur
The most outstanding example of this classical Hindu–Buddhist sculpture in Indonesia is found in the temple of Borobudur in central Java. Built in the 8th century, Borobudur is the largest Buddhist temple in the world and is supposed to represent a map of the Buddhist cosmos; it is a masterful combination of didactic narrative sculptures, spiritual symbolism , monumental design, and meditative serenity. The entire structure resembles a stupa, and when seen from above, looks like a mandala , a concentric diagram with spiritual significance in Buddhism. As part of its structure, it has 504 statues of the Buddha and 2,672 relief panels depicting the life of the Buddha. As a visitor ascends through the eight levels of the temple, the story unfolds from beginning to end. The last three levels simply contain stupas and statues of the Buddha.
Prambanan
Near Borobudur is the 9th century temple complex of Prambanan, one of the oldest and largest Hindu temples in Southeast Asia. The complex consists of eight main shrines, surrounded by 224 smaller ones. The Indian influence on the building is unmistakable, not only in the architectural style but also in the stone reliefs featuring scenes from the Hindu epic Ramayana, which adorn the outer walls of the main temples.
Architecture in Indonesia after 1200 CE
The Majapahit Empire
The decline of the Sri Vijaya Empire in the 11th century was accompanied by a corresponding decline in Buddhism and a shift of power to Eastern Java. The Majapahit Empire was established in 1293 and lasted until around 1500. The Majapahit rulers practiced a mix of Hinduism, Buddhism, and animism , and the religious architecture reflects this synthesis. Indonesian art enjoyed a golden age under Majapahit rule, and an enormous number of Hindu–Buddhist temples, or candi , were built during this period. The building material of choice was brick and mortar of vine sap and palm sugar. Majapahit architecture is characterized by tall and slender roofed red brick gates, a strong geometrical quality, and a sense of verticality, achieved through numerous horizontal lines . Majapahit influence can still be seen to the present day in Hindu temples in Bali.
Introduction of Islam
By the 15th century, Islam had become the dominant religion in Java and Sumatra, Indonesia’s most populous islands. Islamic cultural and artistic influences were absorbed and reinterpreted in the local landscape, resulting in mosques that reflected both Indonesian and Islamic background, with additional influences from Hindu, Buddhist, and Chinese architecture. Indonesian mosques lacked the ubiquitous Islamic dome until the 19th century, and they had tall timber tiered roofs similar to the pagodas of Balinese Hindu temples.
Indonesian Painting
Indonesian painting has been shaped by a myriad of cultural influences, including Islam, Hinduism, Buddhism, and colonial forces.
Examine the characteristics of Indonesian painting during its golden age
Key Points
- Indonesian art has been shaped by interaction between indigenous customs and foreign influences, including Hinduism , Buddhism , and Islam .
- Indonesian painting before the 19th century was mostly restricted to the decorative arts, considered to be a religious and spiritual activity.
- Notable artwork includes the mural paintings on the long houses of the Kenyah people of Borneo , which are based on endemic natural motifs such as ferns and hornbills.
- A tradition of Balinese painting uses narrative imagery to depict scenes from Balinese legends and religious scripts; these paintings can usually found in Indonesian palm-leaf manuscripts and on the ceilings of Balinese temples.
- Under the influence of the Dutch colonial power, a trend toward Western- style painting emerged in the 19th century.
- The most famous indigenous 19th century Indonesian painter is Raden Saleh (1807–1877), who was the first indigenous artist to study in Europe and was heavily influenced by Romanticism .
Key Terms
- Borneo : The third-largest island in the world and the largest island in Asia.
- Raden Saleh : A pioneering Indonesian Romantic painter of Arab-Javanese ethnicity; he was considered to be the first “modern” artist from Indonesia (then Dutch East Indies), and his paintings corresponded with 19th century romanticism, which was popular in Europe at the time.
Overview: Art in Indonesia
The art and culture of Indonesia have been shaped by interactions between original indigenous customs and multiple foreign influences. Indonesia is centrally located along ancient trading routes between the Far East, South Asia, and the Middle East, resulting in art and paintings that are strongly influenced by a multitude of religions, including Hinduism, Buddhism, Confucianism , Islam, and Christianity.
Indonesian Painting Before the 19th Century
Indonesian painting before the 19th century was mostly restricted to the decorative arts, considered to be a religious and spiritual activity. Artists’ names were often anonymous, as the individual human creator was seen as far less important than his creation to honor the deities or spirits. Notable artwork includes the mural paintings on the long houses of the Kenyah people of Borneo, which are based on endemic natural motifs such as ferns and hornbills. Other traditional art includes the geometric wood carvings of the Toraja people of South Sulawesi.
There is a tradition of Balinese painting that uses narrative imagery to depict scenes from Balinese legends and religious scripts. These classical paintings can usually be found in Indonesian lontar or palm-leaf manuscripts and on the ceilings of Balinese temples.
Indonesian Painting After the 19th Century
Under the influence of the Dutch colonial power, a trend toward Western style painting emerged in the 19th century. In the Netherlands, the term “Indonesian Painting” is often applied to the paintings produced by Dutch or other foreign artists who lived and worked in the former Netherlands-Indies. The most famous indigenous 19th century Indonesian painter is Raden Saleh (1807–1877), who was also the first indigenous artist to study in Europe. His art is heavily influenced by Romanticism.
The 1920s to 1940s was a time of growing nationalism in Indonesia. The previous period of romanticism was not seen as a purely Indonesian movement and began to wane, and painters began to turn to the natural world for inspiration. Some examples of Indonesian painters during this period are the Balinese Ida Bagus Made and the realist Basuki Abdullah. The Indonesian Painters Association (or PERSAGI, 1938–1942) was formed during this period and established a contemporary art philosophy that saw art as a reflection of the artist’s individual views, as well as an expression of national cultural thoughts. During the 1960s, new elements were added when abstract expressionism and Islamic art began to be absorbed by the art community. The national identity of Indonesia was stressed by painters through the use of a realistic, documentary style.
- Curation and Revision. Provided by : Boundless.com. License : CC BY-SA: Attribution-ShareAlike
- Stupa Borobudur. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Stupa_Borobudur.jpg. License : Public Domain: No Known Copyright
- Sri Vijaya. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Sri_Vijaya. License : Public Domain: No Known Copyright
- Buddhist art. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Buddhist_art. License : CC BY-SA: Attribution-ShareAlike
- Indonesian architecture. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Indonesian_architecture. License : CC BY-SA: Attribution-ShareAlike
- Culture of Indonesia. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Culture_of_Indonesia. License : CC BY-SA: Attribution-ShareAlike
- Majapahit empire. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Majapahit_empire. License : CC BY-SA: Attribution-ShareAlike
- mandala. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/mandala. License : CC BY-SA: Attribution-ShareAlike
- syncretism. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/syncretism. License : CC BY-SA: Attribution-ShareAlike
- animism. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/animism. License : CC BY-SA: Attribution-ShareAlike
- candi. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/candi. License : CC BY-SA: Attribution-ShareAlike
- Kenyah_mural_painting.jpg. Provided by : Wikimedia Commons. Located at : commons.wikimedia.org/wiki/File:Kenyah_mural_painting.jpg. License : Public Domain: No Known Copyright
- Raden_Saleh_-_Diponegoro_arrest.jpg. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Culture_of_Indonesia#/media/File:Raden_Saleh_-_Diponegoro_arrest.jpg. License : Public Domain: No Known Copyright
- The Arrest of Pangeran Diponegoro. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/The_Arrest_of_Pangeran_Diponegoro. License : CC BY-SA: Attribution-ShareAlike
- Culture of Indonesia. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Culture_of_Indonesia#Traditional_visual_arts. License : CC BY-SA: Attribution-ShareAlike
- Raden Saleh. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Raden_Saleh. License : CC BY-SA: Attribution-ShareAlike
- Borneo. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Borneo. License : CC BY-SA: Attribution-ShareAlike


7.1: California’s Counties
On January 4, 1850, a committee of California’s first constitutional convention, chaired by General Mariano Vallejo, recommended the creation of eighteen counties. They were Benicia, Butte, Fremont, Los Angeles, Mariposa, Monterey, Mt. Diablo, Oro, Redding, Sacramento, San Diego, San Francisco, San Joaquin, San Jose, San Luis Obispo, Santa Barbara, Sonoma, and Sutter.
Between January 4 and February 18, 1850, the California legislature added nine counties to the list recommended by General Vallejo’s committee, some of the changes based on additional recommendations by the committee. The nine added counties were Branciforte, Calaveras, Coloma, Colusi, Marin, Mendocino, Napa, Trinity, and Yuba. This brought the total number of counties to 27. The legislature also approved several name changes. Benicia was renamed El Dorado, Fremont was renamed Yola, Mt. Diablo was renamed Contra Costa, San Jose was renamed Santa Clara, Oro was renamed Tuolumne, and Redding was renamed Shasta.
Effective February 18, 1850, twenty-seven counties were created in California. The new counties were Branciforte, Butte, Calaveras, Colusi, Contra Costa, El Dorado, Los Angeles, Marin, Mariposa, Mendocino, Monterey, Napa, Sacramento, San Diego, San Francisco, San Joaquin, San Luis Obispo, Santa Barbara, Santa Clara, Solano, Sonoma, Sutter, Trinity, Tuolumne, Yola, and Yuba.
In early 1850, not long after the legislature adopted its first statute creating counties, new statutes were adopted changing some county names. Branciforte was changed to Santa Cruz, Colusi was changed to Colusa, and Yola was changed to Yolo. By 1960 these were further sub-divided to form sixteen additional counties, making 43 in total. Another fourteen counties were formed through further subdivision from 1861 to 1893, and most recent county to form was Imperial County, in 1907. California is currently divided into 58 counties.
California is home to San Bernardino County, the largest county in the contiguous United States, as well as Los Angeles County, the most populous county in the United States.
California counties are general law counties which means that they operate within the framework of general laws passed by the state legislature and signed by the governor. These laws dictate many aspects of county government, but may be chartered as provided in Article XI, Section 3 of the California Constitution. A charter county is granted limited home rule powers. Of the 58 counties in California, 14 are governed under a charter. They are Alameda, Butte, El Dorado, Fresno, Los Angeles, Orange, Placer, Sacramento, San Bernardino, San Diego, San Francisco, San Mateo, Santa Clara, and Tehama.
A fun fact; nine counties in California are named for Saints, tied with Louisiana for the largest number.
Defunct Counties
The current county names in California have historical and cultural meaning, lest we do not forget the defunct counties that have historical importance as well. Here is a list of forgotten counties that either were merged, cut, or renamed.
- Branciforte County was the original name of Santa Cruz County in 1850. The county was given the name of "Branciforte" after the Spanish pueblo founded there in 1797. A major watercourse in the county, Branciforte Creek, still bears this name. Less than two months later, on April 5, 1850, the name was changed to Santa Cruz, meaning "Holy Cross".
- Klamath County was created in 1851 from the northern half of Trinity County. Part of the county's territory went to Del Norte County in 1857, and in 1874 the remainder was divided between Humboldt and Siskiyou counties.
- Pautah County, California was created in 1852 out of territory which, the state of California assumed, was to be ceded to it by the United States Congress from territory in what is now the state of Nevada. When the cession never occurred, the California State Legislature officially abolished the never-organized county in 1859.
- Buena Vista County was created in 1855 by the California State Legislature out of the southeastern territory of Tulare County on the west of the Sierra Nevada but was never officially organized. The south of Tulare County was later organized as Kern County in 1866, with additions from Los Angeles and San Bernardino counties.
- Coso County was created in 1864 by the California State Legislature out of territory of Mono County and Tulare County on the east slope of the Sierra Nevada but was never officially organized. The region was later organized in 1866 as Inyo County with additions from Los Angeles and San Bernardino counties.
- Roop County, Nevada encompassed much of Lassen County, including the Honey Lake Valley and the community of Susanville, California; ambiguous organic legislation of Nevada Territory led to confusion about the geographic extent of Nevada's western border. This was later clarified, with the portions of Roop County in California being assigned to Lassen County; the remaining, sparsely portions of Roop County were eventually combined with Washoe County, Nevada.


1.5: Amateur Radio Equipment Basics
By the end of this section, you will be able to:
- Define transceiver, and its basic operation in an radio system.
- Define repeater, and explain why repeaters are used.
We will begin our discussion of radio equipment with a basic overview of the topic. The AM or FM broadcast radios that are common in everyday life are receivers only. In other words, they only receive the electromagnetic waves that are broadcast from the remote AM or FM radio station and then convert the received signal to sound. While this setup is reasonable for music or news, it lacks the interactivity that most Amateur Radio operators desire to make two-way contacts.
Transceivers
In contrast, radios that are able to both receive and transmit are called transceivers . The basic components of a transceiver are illustrated in Fig. \(\PageIndex{1}\). The operator communicating by voice can switch between receiving and transmitting, usually through a "push-to-talk" (PTT) button on a microphone. This button enables a transmit/receive switch so that the transmitter sends its electrical signals through a feed line out to the antenna. The antenna then converts the electrical signals in the radio to the electromagnetic waves that travel outwards into the air. When the button is released, the transmit-receive switch connects the antenna to the receiver, and any electromagnetic waves detected by the antenna are then converted into an electrical signal that can be turned into audio. The transceiver operates using simplex (also half-duplex ) communication as it can simply receive or transmit using the same frequency but not both simultaneously.
Figure \(\PageIndex{2}\) shows photos of actual transceivers useable for Amateur Radio. In both models, a "push-to-talk" button enables the operator to speak into the microphone and transmit the signal. When the button is released, the radios will return to receiving the signals from the connected antenna.
Figure \(\PageIndex{2}\): (left to right) Examples of an Amateur Radio base-station transceiver [2] and handheld transceiver [3]. For the base-station transceiver, the feed line connects to the back of the unit and is not shown. For the handheld transceiver, the feed line is just the connector from the antenna to the unit.
Repeaters
In the VHF and UHF range (above 30 MHz), the signals from the radio's antenna travel will usually travel primarily along their line of sight [4], which is as far as the wave can travel in a straight line away from the antenna and still directly interact with another antenna, as illustrated in Figure \(\PageIndex{3}\).
Figure \(\PageIndex{3}\): Illustration of the idealized range of "line of sight" propagation of electromagnetic waves. (Note that the antenna size is exaggerated to show the effect.) [5]
To extend the ability of a station to communicate beyond its "line of sight" range, repeaters are employed. As the name suggests, repeaters are special radios which have antennas that receive signals transmitted on a channel of a particular frequency, and then re-transmit (repeat) the incoming signal on a different channel of a second frequency that is offset slightly from the receiving frequency. Figure \(\PageIndex{4}\) shows a block diagram of the operation of a repeater.
Figure \(\PageIndex{4}\): Block diagram of a repeater. Signals received by the antenna on a channel of one frequency are re-transmited on a second channel of a different frequency. [6]
A second station that is not within the range of the first station can then communicate with the first station using the repeater as an intermediary, as illustrated in Figure \(\PageIndex{5}\)(B). In some cases, the communication may even be further extended using a network of multiple repeaters. As suggested by Fig. \(\PageIndex{5}\)., it is advantageous to place repeaters in high locations like towers, tall building, hills, or mountains to maximize their range.
Figure \(\PageIndex{5}\): (A) Simplex communication occurs on the same frequency between the stations with each station either transmitting or receiving but not both at the same time. (B) Duplex communication uses a repeater as intermediate station with transmitting and receiving on different frequencies, possibly at the same time. [7]
Other examples of communication systems which use duplexing include wired and cellular telephones, although the method for the duplexing may be different than described above [8].
References
- Kumon RE. Block diagram of a transceiver. (CC BY-SA 4.0)
- Wikimedia Commons contributors. File: Yaesu FT991.jpg [Internet]. Wikimedia Commons. (Image: CC0 1.0; mick m43, All mode HF-6m-2m-70cm-AM-SSB-FM-C4FM Fusion Yaesu FT-991 transceiver, Annotation: CC BY-SA 4.0, Ronald E. Kumon)
- Wikimedia Commons contributors. File: Baofeng UV-5R transceiver 5.jpg [Internet]. Wikimedia Commons. (Image: CC BY-SA 4.0; Barbara Kaminskaya, Dual band amateur VHF/UHF transceiver Baofeng UV-5R, Annotation: CC BY-SA 4.0, Ronald E. Kumon)
- Wikipedia contributors. Line-of-sight propagation [Internet]. Wikipedia, The Free Encyclopedia.
- Wikimedia Commons contributors. File: VHF propagation.png [Internet]. Wikimedia Commons. (CC BY-SA 4.0; Rudolf Davis Strazds, VHF radio wave propagation)
- Kumon RE. Block diagram of a repeater. (CC BY-SA 4.0)
- Wikimedia Commons contributors. File: Repeater Talk around channel.png [Internet]. Wikimedia Commons. (Original image: CC0 1.0; Goodtiming8871, Repeater Talk around channel diagram, Modifications: CC BY-SA 4.0; Ronald Kumon)
- Wikipedia contributors. Duplex (telecommunications) [Internet]. Wikipedia, The Free Encyclopedia.
Technician Exam Questions
Relevant exam questions include: T7A02, T1F09.


7.2: Mass, Weight and Gravitation Near a Planet's Surface
- Explain the difference between mass and weight
- Explain why falling objects on Earth are never truly in free fall
- Describe the concept of weightlessness
- Explain the connection between the constants G and g
- Determine the mass of an astronomical body from free-fall acceleration at its surface
- Describe how the value of g varies due to location and Earth’s rotation
Mass and weight are often used interchangeably in everyday conversation. For example, our medical records often show our weight in kilograms but never in the correct units of newtons. In physics, however, there is an important distinction. Weight is the pull of Earth on an object. It depends on the distance from the center of Earth. Unlike weight, mass does not vary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon, it depends on the amount of matter not on its location.
Units of Force
The equation F net = ma is used to define net force in terms of mass, length, and time. As explained earlier, the SI unit of force is the newton. Since F net = ma,
\[1\; N = 1\; kg \cdotp m/s^{2} \ldotp \nonumber\]
Although almost the entire world uses the newton for the unit of force, in the United States, the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb. Thus, a 225-lb person weighs 1000 N.
Weight and Gravitational Force
When an object is dropped, it accelerates toward the center of Earth. Newton’s second law says that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight \(\vec{w}\), or its force due to gravity acting on an object of mass m. Weight can be denoted as a vector because it has a direction; down is, by definition, the direction of gravity, and hence, weight is a downward force. The magnitude of weight is denoted as w. Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration g. Using Galileo’s result and Newton’s second law, we can derive an equation for weight.
Consider an object with mass m falling toward Earth. It experiences only the downward force of gravity, which is the weight \(\vec{w}\). Newton’s second law says that the magnitude of the net external force on an object is \(\vec{F}_{net} = m \vec{a}\). We know that the acceleration of an object due to gravity is \(\vec{g}\), or \(\vec{a} = \vec{g}\). Substituting these into Newton’s second law gives us the following equations.
The gravitational force on a mass is its weight. We can write this in vector form, where \(\vec{w}\) is weight and m is mass, as
\[\vec{w} = m \vec{g} \ldotp \label{5.8}\]
In scalar form, we can write
\[w = mg \ldotp \label{5.9}\]
Since g = 9.80 m/s 2 on Earth, the weight of a 1.00-kg object on Earth is 9.80 N:
\[w = mg = (1.00\; kg)(9.80 m/s^{2}) = 9.80\; N \ldotp\]
When the net external force on an object is its weight, we say that it is in free fall , that is, the only force acting on the object is gravity. However, when objects on Earth fall downward, they are never truly in free fall because there is always some upward resistance force from the air acting on the object.
Acceleration due to gravity g varies slightly over the surface of Earth, so the weight of an object depends on its location and is not an intrinsic property of the object. Weight varies dramatically if we leave Earth’s surface. On the Moon, for example, acceleration due to gravity is only 1.67 m/s 2 . A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon.
The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body, such as Earth, the Moon, or the Sun. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of “weightlessness” and “microgravity,” they are referring to the phenomenon we call “free fall” in physics. We use the preceding definition of weight, force \(\vec{w}\) due to gravity acting on an object of mass m, and we make careful distinctions between free fall and actual weightlessness.
Be aware that weight and mass are different physical quantities, although they are closely related. Mass is an intrinsic property of an object: It is a quantity of matter. The quantity or amount of matter of an object is determined by the numbers of atoms and molecules of various types it contains. Because these numbers do not vary, in Newtonian physics, mass does not vary; therefore, its response to an applied force does not vary. In contrast, weight is the gravitational force acting on an object, so it does vary depending on gravity. For example, a person closer to the center of Earth, at a low elevation such as New Orleans, weighs slightly more than a person who is located in the higher elevation of Denver, even though they may have the same mass.
It is tempting to equate mass to weight, because most of our examples take place on Earth, where the weight of an object varies only a little with the location of the object. In addition, it is difficult to count and identify all of the atoms and molecules in an object, so mass is rarely determined in this manner. If we consider situations in which \(\vec{g}\) is a constant on Earth, we see that weight \(\vec{w}\) is directly proportional to mass m, since \(\vec{w} = m \vec{g}\), that is, the more massive an object is, the more it weighs. Operationally, the masses of objects are determined by comparison with the standard kilogram, as we discussed in Units and Measurement . But by comparing an object on Earth with one on the Moon, we can easily see a variation in weight but not in mass. For instance, on Earth, a 5.0-kg object weighs 49 N; on the Moon, where g is 1.67 m/s 2 , the object weighs 8.4 N. However, the mass of the object is still 5.0 kg on the Moon.
A farmer is lifting some moderately heavy rocks from a field to plant crops. He lifts a stone that weighs 40.0 lb. (about 180 N). What force does he apply if the stone accelerates at a rate of 1.5 m/s 2 ?
Strategy
We were given the weight of the stone, which we use in finding the net force on the stone. However, we also need to know its mass to apply Newton’s second law, so we must apply the equation for weight, w = mg, to determine the mass.
Solution
No forces act in the horizontal direction, so we can concentrate on vertical forces, as shown in the following free-body diagram. We label the acceleration to the side; technically, it is not part of the free-body diagram, but it helps to remind us that the object accelerates upward (so the net force is upward).
\[w = mg \nonumber \]
\[m = \frac{w}{g} = \frac{180\; N}{9.8\; m/s^{2}} = 18\; kg \nonumber\]
\[\sum F = ma \nonumber\]
\[F - w = ma \nonumber\]
\[F - 180\; N = (18\; kg)(1.5\; m/s^{2}) \nonumber\]
\[F - 180\; N = 27\; N \nonumber\]
\[F = 207\; N = 210\; N\; \text{ to two significant figures} \nonumber\]
Significance
To apply Newton’s second law as the primary equation in solving a problem, we sometimes have to rely on other equations, such as the one for weight or one of the kinematic equations, to complete the solution.
For \(\PageIndex{1}\), find the acceleration when the farmer’s applied force is 230.0 N
Weight on Any Planet
Recall that the acceleration of a free-falling object near Earth’s surface is approximately g = 9.80 m/s 2 . The force causing this acceleration is called the weight of the object, and from Newton’s second law, it has the value mg. This weight is present regardless of whether the object is in free fall. We now know that this force is the gravitational force between the object and Earth. If we substitute mg for the magnitude of \(\vec{F}_{12}\) in Newton’s law of universal gravitation, m for m 1 , and M E for m 2 , we obtain the scalar equation
\[mg = G \frac{m M_{E}}{r^{2}} \label{13.1}\]
where r is the distance between the centers of mass of the object and Earth. The average radius of Earth is about 6370 km. Hence, for objects within a few kilometers of Earth’s surface, we can take \(r = R_E\) (Figure \(\PageIndex{1}\)). The mass m of the object cancels, leaving
\[g = G \frac{M_{E}}{r^{2}} \ldotp \label{13.2}\]
This explains why all masses free fall with the same acceleration. We have ignored the fact that Earth also accelerates toward the falling object, but that is acceptable as long as the mass of Earth is much larger than that of the object.
Have you ever wondered how we know the mass of Earth? We certainly can’t place it on a scale. The values of g and the radius of Earth were measured with reasonable accuracy centuries ago.
- Use the standard values of g, R E , and Equation \ref{13.2} to find the mass of Earth.
- Estimate the value of g on the Moon. Use the fact that the Moon has a radius of about 1700 km (a value of this accuracy was determined many centuries ago) and assume it has the same average density as Earth, 5500 kg/m 3 .
Strategy
With the known values of g and R E , we can use Equation \ref{13.2} to find M E . For the Moon, we use the assumption of equal average density to determine the mass from a ratio of the volumes of Earth and the Moon.
Solution
- Rearranging Equation \ref{13.2}, we have $$M_{E} = \frac{g R_{E}^{2}}{G} = \frac{(9.80\; m/s^{2})(6.37 \times 10^{6}\; m)^{2}}{6.67 \times 10^{-11}\; N\; \cdotp m^{2}/kg^{2}} = 5.95 \times 10^{24}\; kg \ldotp$$
- The volume of a sphere is proportional to the radius cubed, so a simple ratio gives us $$\frac{M_{M}}{M_{E}} = \frac{R_{M}^{3}}{R_{E}^{3}} \rightarrow M_{M} = \left(\dfrac{(1.7 \times 10^{6}\; m)^{3}}{(6.37 \times 10^{6}\; m)^{3}}\right) (5.95 \times 10^{24}\; kg) = 1.1 \times 10^{23}\; kg \ldotp$$We now use Equation \ref{13.2}. $$g_{M} = G \frac{M_{M}}{r_{M}^{2}} = (6.67 \times 10^{-11}\; N\; \cdotp m^{2}/kg^{2}) \left(\dfrac{1.1 \times 10^{23}\; kg}{(1.7 \times 10^{6}\; m)^{2}}\right) = 2.5\; m/s^{2}$$
Significance
As soon as Cavendish determined the value of G in 1798, the mass of Earth could be calculated. (In fact, that was the ultimate purpose of Cavendish’s experiment in the first place.) The value we calculated for g of the Moon is incorrect. The average density of the Moon is actually only 3340 kg/m 3 and g = 1.6 m/s 2 at the surface. Newton attempted to measure the mass of the Moon by comparing the effect of the Sun on Earth’s ocean tides compared to that of the Moon. His value was a factor of two too small. The most accurate values for g and the mass of the Moon come from tracking the motion of spacecraft that have orbited the Moon. But the mass of the Moon can actually be determined accurately without going to the Moon. Earth and the Moon orbit about a common center of mass, and careful astronomical measurements can determine that location. The ratio of the Moon’s mass to Earth’s is the ratio of [the distance from the common center of mass to the Moon’s center] to [the distance from the common center of mass to Earth’s center].
Later in this section, we will see that the mass of other astronomical bodies also can be determined by the period of small satellites orbiting them. But until Cavendish determined the value of G, the masses of all these bodies were unknown.
What is the value of g 400 km above Earth’s surface, where the International Space Station is in orbit?
Solution
Using the value of M E and noting the radius is r = R E + 400 km, we use Equation \ref{13.2} to find g. From Equation \ref{13.2} we have
\[g = G \frac{M_{E}}{r^{2}} = (6.67 \times 10^{-11}\; N\; \cdotp m^{2}/kg^{2}) \left(\dfrac{5.96 \times 10^{24}\; kg}{(6.37 \times 10^{6} + 400 \times 10^{3}\; m)^{2}}\right) = 8.67\; m/s^{2} \ldotp\]
Significance
We often see video of astronauts in space stations, apparently weightless. But clearly, the force of gravity is acting on them. Comparing the value of g we just calculated to that on Earth (9.80 m/s 2 ) , we see that the astronauts in the International Space Station still have 88% of their weight. They only appear to be weightless because they are in free fall. We will come back to this in Satellite Orbits and Energy .
How does your weight at the top of a tall building compare with that on the first floor? Do you think engineers need to take into account the change in the value of g when designing structural support for a very tall building?
The Gravitational Field
Equation \ref{13.2} is a scalar equation, giving the magnitude of the gravitational acceleration as a function of the distance from the center of the mass that causes the acceleration. But we could have retained the vector form for the force of gravity in Equation \ref{13.1}, and written the acceleration in vector form as
\[\vec{g} = G \frac{M}{r^{2}} \hat{r} \ldotp\]
We identify the vector field represented by \(\vec{g}\) as the gravitational field caused by mass M. We can picture the field as shown Figure \(\PageIndex{2}\). The lines are directed radially inward and are symmetrically distributed about the mass.
As is true for any vector field, the direction of \(\vec{g}\) is parallel to the field lines at any point. The strength of \(\vec{g}\) at any point is inversely proportional to the line spacing. Another way to state this is that the magnitude of the field in any region is proportional to the number of lines that pass through a unit surface area, effectively a density of lines. Since the lines are equally spaced in all directions, the number of lines per unit surface area at a distance r from the mass is the total number of lines divided by the surface area of a sphere of radius r, which is proportional to r 2 . Hence, this picture perfectly represents the inverse square law, in addition to indicating the direction of the field. In the field picture, we say that a mass m interacts with the gravitational field of mass M. We will use the concept of fields to great advantage in the later sections on electromagnetism.
Apparent Weight: Accounting for Earth’s Rotation
As we saw in Applications of Newton’s Laws , objects moving at constant speed in a circle have a centripetal acceleration directed toward the center of the circle, which means that there must be a net force directed toward the center of that circle. Since all objects on the surface of Earth move through a circle every 24 hours, there must be a net centripetal force on each object directed toward the center of that circle.
Let’s first consider an object of mass m located at the equator, suspended from a scale (Figure \(\PageIndex{3}\)). The scale exerts an upward force \(\vec{F}_{s}\) away from Earth’s center. This is the reading on the scale, and hence it is the apparent weight of the object. The weight (mg) points toward Earth’s center. If Earth were not rotating, the acceleration would be zero and, consequently, the net force would be zero, resulting in F s = mg . This would be the true reading of the weight.
With rotation, the sum of these forces must provide the centripetal acceleration , \(a_c\). Using Newton’s second law, we have
\[\sum F = F_{s} - mg = ma_{c} \quad where\; a_{c} = - \frac{v^{2}}{r} \ldotp \label{13.3}\]
Note that a c points in the same direction as the weight; hence, it is negative. The tangential speed v is the speed at the equator and r is R E . We can calculate the speed simply by noting that objects on the equator travel the circumference of Earth in 24 hours. Instead, let’s use the alternative expression for a c from Motion in Two and Three Dimensions . Recall that the tangential speed is related to the angular speed (\(\omega\)) by v = r\(\omega\). Hence, we have a c = −r\(\omega\) 2 . By rearranging Equation 13.3 and substituting r = R E , the apparent weight at the equator is
\[F_{s} = m (g - R_{E} \omega^{2}) \ldotp\]
The angular speed of Earth everywhere is
\[\omega = \frac{2 \pi\; rad}{24\; hr \times 3600\; s/hr} = 7.27 \times 10^{-5}\; rad/s \ldotp\]
Substituting for the values of R E and \(\omega\), we have R E \(\omega\) 2 = 0.0337 m/s 2 . This is only 0.34% of the value of gravity, so it is clearly a small correction.
How fast would Earth need to spin for those at the equator to have zero apparent weight? How long would the length of the day be?
Strategy
Using Equation \ref{13.3}, we can set the apparent weight (F s ) to zero and determine the centripetal acceleration required. From that, we can find the speed at the equator. The length of day is the time required for one complete rotation.
Solution
From Equation \ref{13.2}, we have \(\sum\)F = F s − mg = ma c , so setting F s = 0, we get g = a c . Using the expression for a c , substituting for Earth’s radius and the standard value of gravity, we get
\[\begin{split} a_{c} & = \frac{v^{2}}{r} = g \\ v & = \sqrt{gr} = \sqrt{(9.80\; m/s^{2})(6.37 \times 10^{6}\; m)} = 7.91 \times 10^{3}\; m/s \ldotp \end{split}\]
The period T is the time for one complete rotation. Therefore, the tangential speed is the circumference divided by T, so we have
\[\begin{split} v & = \frac{2 \pi r}{T} \\ T & = \frac{2 \pi r}{v} = \frac{2 \pi (6.37 \times 10^{6}\; m)}{7.91 \times 10^{3}\; m/s} = 5.06 \times 10^{3}\; s \ldotp \end{split}\]
This is about 84 minutes.
Significance
We will see later in this section that this speed and length of day would also be the orbital speed and period of a satellite in orbit at Earth’s surface. While such an orbit would not be possible near Earth’s surface due to air resistance, it certainly is possible only a few hundred miles above Earth.
Results Away from the Equator
At the poles, a c → 0 and Fs = mg , just as is the case without rotation. At any other latitude \(\lambda\), the situation is more complicated. The centripetal acceleration is directed toward point P in the figure, and the radius becomes \(r = R_E \cos \lambda\). The vector sum of the weight and \(\vec{F}_{s}\) must point toward point P, hence \(\vec{F}_{s}\) no longer points away from the center of Earth. (The difference is small and exaggerated in the figure.) A plumb bob will always point along this deviated direction. All buildings are built aligned along this deviated direction, not along a radius through the center of Earth. For the tallest buildings, this represents a deviation of a few feet at the top.
It is also worth noting that Earth is not a perfect sphere. The interior is partially liquid, and this enhances Earth bulging at the equator due to its rotation. The radius of Earth is about 30 km greater at the equator compared to the poles. It is left as an exercise to compare the strength of gravity at the poles to that at the equator using Equation \ref{13.2}. The difference is comparable to the difference due to rotation and is in the same direction. Apparently, you really can lose “weight” by moving to the tropics.
Gravity Away from the Surface
Earlier we stated without proof that the law of gravitation applies to spherically symmetrical objects, where the mass of each body acts as if it were at the center of the body. Since Equation \ref{13.2} is derived from Equation \ref{13.1}, it is also valid for symmetrical mass distributions, but both equations are valid only for values of \(r ≥ R_E\). As we saw in Example 13.4, at 400 km above Earth’s surface, where the International Space Station orbits, the value of \(g\) is 8.67 m/s 2 . (We will see later that this is also the centripetal acceleration of the ISS.)
For \(r < R_E\), Equation \ref{13.1} and Equation \ref{13.2} are not valid. However, we can determine g for these cases using a principle that comes from Gauss’s law , which is a powerful mathematical tool that we study in more detail later in the course. A consequence of Gauss’s law, applied to gravitation, is that only the mass within r contributes to the gravitational force. Also, that mass, just as before, can be considered to be located at the center. The gravitational effect of the mass outside r has zero net effect.
Two very interesting special cases occur. For a spherical planet with constant density, the mass within r is the density times the volume within r. This mass can be considered located at the center. Replacing M E with only the mass within r, M = \(\rho\) x (volume of a sphere), and R E with r, Equation \ref{13.2} becomes
\[g = G \frac{M_{E}}{R_{E}^{2}} = G \frac{\rho \left(\dfrac{4}{3} \pi r^{3}\right)}{r^{2}} = \frac{4}{3} G \rho \pi r \ldotp\]
The value of g, and hence your weight, decreases linearly as you descend down a hole to the center of the spherical planet. At the center, you are weightless, as the mass of the planet pulls equally in all directions. Actually, Earth’s density is not constant, nor is Earth solid throughout. Figure \(\PageIndex{4}\) shows the profile of \(g\) if Earth had constant density and the more likely profile based upon estimates of density derived from seismic data.
The second interesting case concerns living on a spherical shell planet. This scenario has been proposed in many science fiction stories. Ignoring significant engineering issues, the shell could be constructed with a desired radius and total mass, such that g at the surface is the same as Earth’s. Can you guess what happens once you descend in an elevator to the inside of the shell, where there is no mass between you and the center? What benefits would this provide for traveling great distances from one point on the sphere to another? And finally, what effect would there be if the planet was spinning?


14.2: Conservative Ascendance
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The Reagan Revolution marked the culmination of a long process of political mobilization on the American right. In the first two decades after World War II the New Deal seemed firmly embedded in American electoral politics and public policy. Even two-term Republican president Dwight D. Eisenhower declined to roll back the welfare state. To be sure, William F. Buckley tapped into a deep vein of elite conservatism in 1955 by announcing in the first issue of National Review that his magazine “stands athwart history yelling Stop.” 4 Senator Joseph McCarthy and John Birch Society founder Robert Welch stirred anticommunist fervor. But in general, the far right lacked organizational cohesion. Following Lyndon Johnson’s resounding defeat of Republican Barry Goldwater—“Mr. Conservative”—in the 1964 presidential election, many observers declared American conservatism finished. New York Times columnist James Reston wrote that Goldwater had “wrecked his party for a long time to come.” 5
Despite these dire predictions, conservatism not only persisted, it prospered. Its growing appeal had several causes. The expansive social and economic agenda of Johnson’s Great Society reminded anticommunists of Soviet-style central planning and deficits alarmed fiscal conservatives. Race also drove the creation of the New Right. The civil rights movement, along with the Civil Rights Act and the Voting Rights Act, challenged the racial hierarchy of the Jim Crow South. All of these occurred under Democratic leadership, pushing white southerners toward the Republican Party. In the late 1960s and early 1970s, Black Power, affirmative action, and court-ordered busing of children between schools to achieve racial balance brought “white backlash” in the North, often in cities previously known for political liberalism. To many white Americans, the urban rebellions, antiwar protests, and student uprisings of the late 1960s signaled social chaos. At the same time, slowing wage growth, rising prices, and growing tax burdens threatened many working- and middle-class citizens who long formed the core of the New Deal coalition. Liberalism no longer seemed to offer the great mass of white Americans a road map to prosperity, so they searched for new political solutions.
Former Alabama governor and conservative Democrat George Wallace masterfully exploited the racial, cultural, and economic resentments of working-class whites during his presidential runs in 1968 and 1972. Wallace’s record as a staunch segregationist made him a hero in the Deep South, where he won five states as a third-party candidate in the 1968 general election. Wallace’s populist message also resonated with blue-collar voters in the industrial North who felt left behind by the rights revolution. On the campaign stump, the fiery candidate lambasted hippies, antiwar protesters, and government bureaucrats. He assailed female welfare recipients for “breeding children as a cash crop” and ridiculed “over-educated, ivory-tower” intellectuals who “don’t know how to park a bicycle straight.” 6 Wallace also advanced progressive proposals for federal job training programs, a minimum wage hike, and legal protections for collective bargaining. Running as a Democrat in 1972, Wallace captured the Michigan primary and polled second in the industrial heartland of Wisconsin, Pennsylvania, and Indiana. In May 1972, an assassin’s bullet left Wallace paralyzed and ended his campaign. Nevertheless, his amalgamation of older, New Deal–style proposals and conservative populism represented the rapid reordering of party loyalties in the late 1960s and early 1970s. Richard Nixon similarly harnessed the New Right’s sense of grievance through his rhetoric about “law and order” and the “silent majority.” 7 But Nixon and his Republican successor, Gerald Ford, continued to accommodate the politics of the New Deal order. The New Right remained without a major public champion.
Christian conservatives also felt themselves under siege from liberalism. In the early 1960s, Supreme Court decisions prohibiting teacher-led prayer ( Engel v. Vitale ) and Bible reading in public schools ( Abington v. Schempp ) led some on the right to conclude that a liberal judicial system threatened Christian values.  Evangelical Protestants—Christians who professed a personal relationship with Jesus Christ, upheld the Bible as an infallible source of truth, and felt a duty to convert, or evangelize, nonbelievers—composed the core of the so-called religious right.
With increasing assertiveness in the 1960s and 1970s, Christian conservatives mobilized to protect the “traditional” family. Women composed a striking number of the religious right’s foot soldiers. In 1968 and 1969 a group of newly politicized mothers in Anaheim, California, led a sustained protest against sex education in public schools. 8 Catholic activist Phyllis Schlafly marshaled opposition to the ERA, while evangelical pop singer Anita Bryant drew national headlines for her successful fight to repeal Miami’s gay rights ordinance in 1977. In 1979, Beverly LaHaye (whose husband, Tim—an evangelical pastor in San Diego—later coauthored the wildly popular Left Behind Christian book series) founded Concerned Women for America, which linked small groups of local activists opposed to the ERA, abortion, homosexuality, and no-fault divorce.
Activists like Schlafly and LaHaye valorized motherhood as women’s highest calling. Abortion therefore struck at the core of their female identity. More than perhaps any other issue, abortion drew different segments of the religious right—Catholics and Protestants, women and men—together. The Supreme Court’s 1973 Roe v. Wade ruling outraged many devout Catholics and evangelicals (who had been less universally opposed to the procedure than their Catholic counterparts). Christian author Francis Schaeffer cultivated evangelical opposition to abortion through the 1979 documentary film Whatever Happened to the Human Race? , arguing that the “fate of the unborn is the fate of the human race.” 9 With abortion framed in stark, existential terms, many evangelicals felt compelled to combat the procedure through political action.
Grassroots passion drove anti-abortion activism, but a set of religious and secular institutions turned the various strands of the New Right into a sophisticated movement. In 1979 Jerry Falwell—a Baptist minister and religious broadcaster from Lynchburg, Virginia—founded the Moral Majority, an explicitly political organization dedicated to advancing a “pro-life, pro-family, pro-morality, and pro-American” agenda. The Moral Majority skillfully wove together social and economic appeals to make itself a force in Republican politics. Secular, business-oriented institutions also joined the attack on liberalism, fueled by stagflation and by the federal government’s creation of new regulatory agencies like the Environmental Protection Agency and the Occupational Safety and Health Administration. Conservative business leaders bankrolled new “think tanks” like the Heritage Foundation and the Cato Institute. These organizations provided grassroots activists with ready-made policy prescriptions. Other business leaders took a more direct approach by hiring Washington lobbyists and creating political action committees (PACs) to press their agendas in the halls of Congress and federal agencies. Between 1976 and 1980 the number of corporate PACs rose from under three hundred to over twelve hundred.
Grassroots activists and business leaders received unlikely support from a circle of neoconservatives—disillusioned intellectuals who had rejected liberalism and the Left and become Republicans. Irving Kristol, a former Marxist who went on to champion free-market capitalism as a Wall Street Journal columnist, defined a neoconservative as a “liberal who has been mugged by reality.” 10 Neoconservative journals like Commentary and Public Interest argued that the Great Society had proven counterproductive, perpetuating the poverty and racial segregation that it aimed to cure. By the middle of the 1970s, neoconservatives felt mugged by foreign affairs as well. As ardent Cold Warriors, they argued that Nixon’s policy of détente left the United States vulnerable to the Soviet Union.
In sum, several streams of conservative political mobilization converged in the late 1970s. Each wing of the burgeoning New Right—disaffected northern blue-collar workers, white southerners, evangelicals and devout Catholics, business leaders, disillusioned intellectuals, and Cold War hawks—turned to the Republican Party as the most effective vehicle for their political counterassault on liberalism and the New Deal political order. After years of mobilization, the domestic and foreign policy catastrophes of the Carter administration provided the headwinds that brought the conservative movement to shore.


3.2: Motion with Constant Acceleration
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Until now, we have considered motion where the velocity is a constant (i.e. where velocity does not change with time). Suppose that we wish to describe the position of a falling object that we released from rest at time \(t=0\text{s}\) . The object will start with a velocity of 0 m/s and it will accelerate as it falls. We say that an object is “accelerating” if its velocity is not constant. As we will see in later chapters, objects that fall near the surface of the Earth experience a constant acceleration (their velocity changes at a constant rate).
Formally, we define acceleration as the rate of change of velocity. Recall that velocity is the rate of change of position, so acceleration is to velocity what velocity is to position. In particular, we saw that if the velocity, \(v_{x}\) , is constant, then position as a function of time is given by:
\[x(t) = x_{0} + v_{x}t\]
In analogy, if the acceleration is constant, then the velocity as a function of time is given by:
\[v_{x}(t) = v_{0x} + a_{x}t\]
where \(a_{x}\) is the “acceleration” and \(v_{0x}\) is the velocity of the object at \(t=0\) . We can work out the dimensions of acceleration for this equation to make sense. Since we are adding \(v_{0x}\) and \(a_{x}t\) , we need the dimensions of \(a_{x}t\) to be velocity:
\(\begin{aligned} \left[a_{x}t\right]&=\frac{L}{T} \\[4pt] \left[a_{x}\right]&=\frac{L}{T^{2}} \end{aligned}\)
Acceleration thus has dimensions of length over time squared, with corresponding S.I. units of \(\text{m/s} ^{2}\) (meters per second squared or meters per second per second). In order to describe the position of an object that is accelerating, we cannot use Equation 3.2.1 , since it is only correct if the velocity is constant.
In Section 3.3 Using calculus to obtain acceleration from position , we will show that the position as a function of time, \(x(t)\), of an object with constant acceleration , \(a_{x}\), is given by:
\[x(t)=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^{2}\]
where, at \(t = 0\), the object was at position \(x = x_{0}\) and had a velocity \(v_{0x}\).
Example \(\PageIndex{1}\)
A ball is thrown upwards with a velocity of \(10\text{m/s}\). After what distance will the ball stop before falling back down? Assume that gravity causes a constant downwards acceleration of \(9.8\text{m/s}^{2}\).
Solution
We will solve this problem in the following steps:
- Setup a coordinate system (define the \(x\)-axis).
- Identify the condition that corresponds to the ball stopping its upwards motion and falling back down.
- Determine the distance at which the ball stopped.
Since we throw the ball upwards with an initial velocity upwards, it makes sense to choose an \(x\)-axis that points up and has the origin at the point where we release the ball. With this choice, referring to the variables in Equation 3.2.3 , we have:
\(\begin{aligned} x_{0}&=0 \\[4pt] v_{0x}&=+10\text{m/s} \\[4pt] a_{x}&=-9.8\text{m/s}^{2} \end{aligned}\)
where the initial velocity is in the positive x-direction, and the acceleration, \(a_{x}\), is in the negative direction (the velocity will be getting smaller and smaller, so its rate of change is negative).
The condition for the ball to stop at the top of the trajectory is that its velocity will be zero (that is what it means to stop). We can use Equation 3.2.2 to find what time that corresponds to:
\(\begin{aligned} v(t)&=v_{0x}+a_{x}t \\[4pt] 0&=(10\text{m/s})+(-9.8\text{m/s}^{2})t \\[4pt] \therefore t&=\frac{(10\text{m/s})}{(9.8\text{m/s}^{2})}=1.02\text{s} \end{aligned}\)
Now that we know that it took \(1.02\text{s}\) to reach the top of the trajectory, we can find how much distance was covered:
\(\begin{aligned} x(t)&=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^{2} \\[4pt] x&=(0\text{m})+(10\text{m/s})(1.02\text{s})+\frac{1}{2}(-9.8\text{m/s}^{2})(1.02\text{s})^{2}=5.10\text{m} \end{aligned}\)
and we find that the ball will rise by \(5.10\text{m}\) before falling back down.
Visualizing motion with constant acceleration
When an object has a constant acceleration, its velocity and position as a function of time are described by the two following equations:
\(\begin{aligned} v(t)&=v_{0x}+a_{x}t \\[4pt] x(t)&=x_{0}+v_{0x}t + \frac{1}{2}a_{x}t^{2} \end{aligned}\)
where the velocity changes linearly with time, and the position changes quadratically with time (it goes as \(t^{2}\) ). Figure \(\PageIndex{1}\) shows the position and the speed as a function of time for the ball from Example 3.2.1 for the first three seconds of the motion.
We can divide the motion into three parts (shown by the vertical dashed lines in Figure \(\PageIndex{1}\)):
1) Between \(t=0\text{s}\) and \(t=1.02\text{s}\)
At time \(t=0\text{s}\) , the ball starts at a position of \(x=0\text{m}\) (left panel) and has a velocity of \(v_{0x}=10\text{m/s}\) (right panel). During the first second of motion, the position, \((t)\) , increases (the ball is moving up), until the position stops increasing at \(t=1.02\text{s}\) , as found in Example 3.2.1 . During that time, the velocity decreases linearly from \(10\text{m/s}\) to \(0\text{m/s}\) due to the constant negative acceleration from gravity. At \(t=1.02\text{s}\) , the velocity is instantaneously \(0\text{m/s}\) and the ball is momentarily at rest (as it reaches the top of the trajectory before falling back down).
2) Between \(t=1.02\text{s}\) and \(t=2.04\text{s}\)
At \(t=1.02\text{s}\) , the velocity continues to decrease linearly (it becomes more and more negative) as the ball start to fall back down faster and faster. The position also starts decreasing just after \(t=1.02\text{s}\) , as the ball returns back down to the point of release. At \(t=2.04\text{s}\) , the ball returns to the point from which it was thrown, and the ball is going with the same velocity \((10\text{m/s})\) as when it was released, but the velocity is negative (downwards motion).
3) After \(t=2.04\text{s}\)
If nothing is there to stop the ball, it continues to move downwards with ever increasing velocity. The position continues to become more negative and the velocity continues to become larger in magnitude and more negative.
Exercise \(\PageIndex{1}\)
Make a sketch of the acceleration as a function of time corresponding to the position and velocity shown in Figure \(\PageIndex{1}\).
- Answer
- The slope (or derivative) of the velocity vs. time graph is negative and constant. These graphs correspond to the example above where \(a = -9.8 \rm{m/s^2}\).
Speed versus velocity
In the previous example, our language was not quite as precise as it should be when conducting science. Specifically, we need a way to distinguish the situation when the velocity is decreasing (becoming more negative), while the object is actually going faster and faster (after \(t=1.02\text{s}\) ). We will use the term speed to refer to how fast an object is moving (how much distance it covers per unit time), and we will use the term velocity to also indicate the direction of the motion. In other words, the speed is the absolute value of the velocity. The speed is thus always positive, whereas the velocity can also be negative.
With this vocabulary, the speed of the ball decreases between \(t=0\text{s}\) and \(t=1.02\text{s}\) , and increases thereafter. On the other hand, the velocity continuously decreases (it is always becoming more and more negative). Velocity is thus the more general term since it tells us both the speed and the direction of the motion.