CK-12 People's Physics Book Version 2 Read online

  Key Applications

  In “roller coaster” problems, a cart's gravitational potential energy at the top of one hill is transformed into kinetic energy at the next valley. It turns back into potential energy as cart climbs the next hill, and so on. In reality a fraction of the energy is lost to the tracks and air as heat, which is why the second rise is rarely as big as the first in amusement parks.

  In “pole-vaulter” problems, the athlete’s body breaks down the food molecules to change some of the bonding energy into energy that is used to power the body. This energy is transformed into kinetic energy as the athlete gains speed. The kinetic energy can be changed into potential energy as the athlete gains height.

  In “pendulum” problems, the gravitational potential energy of the pendulum at its highest point changes to kinetic energy as it swings to the bottom and then back into potential energy as it swings up. At any in-between point there is a combination of kinetic energy and potential energy, but the total energy remains constant.

  Examples

  Example 1

  Question A roller coaster begins at rest above the ground at point A. Assume no energy is lost. The radius of the loop is 40

  a) Find the speed of the roller coaster at point B, D, F, and H.

  b) At point G the roller coaster's speed is . How high off the ground is point G?

  Figure 2.1

  Roller coaster for problem 8.

  Solution a) To solve for the speed at any point on the roller coaster, we use conservation of energy: the cart's total energy, equal to its initial energy (all potential) is split between kinetic and potential energy at all points during the trip. Therefore, at any point Solving for :

  Therefore

  B:

  D:

  F:

  H:

  b) As in part a), we start with the equation but this time we will solve for h. Now simply input the known variables to solve for h.

  Energy Conservation Problem Set

  A stationary bomb explodes into hundreds of pieces. Which of the following statements best describes the situation? The kinetic energy of the bomb was converted into heat.

  The chemical potential energy stored in the bomb was converted into heat and gravitational potential energy.

  The chemical potential energy stored in the bomb was converted into heat and kinetic energy.

  The chemical potential energy stored in the bomb was converted into heat, sound, kinetic energy, and gravitational potential energy.

  The kinetic and chemical potential energy stored in the bomb was converted into heat, sound, kinetic energy, and gravitational potential energy.

  You hike up to the top of Granite Peak in the Trinity Alps to think about physics. Do you have more potential or kinetic energy at the top of the mountain than you did at the bottom? Explain.

  Do you have more, less, or the same amount of energy at the top of the mountain than when you started? (Let’s assume you did not eat anything on the way up.) Explain.

  How has the total energy of the Solar System changed due to your hike up the mountain? Explain.

  If you push a rock off the top, will it end up with more, less, or the same amount of energy at the bottom? Explain.

  For each of the following types of energy, describe whether you gained it, you lost it, or it stayed the same during your hike: Gravitational potential energy

  Energy stored in the atomic nuclei in your body

  Heat energy

  Chemical potential energy stored in the fat cells in your body

  Sound energy from your footsteps

  Energy given to you by a wind blowing at your back

  Just before your mountain bike ride, you eat a Calorie exercise bar. (You can find the conversion between food Calories and Joules in the chapter.) The carbon bonds in the food are broken down in your stomach, releasing energy. About half of this energy is lost due to inefficiencies in your digestive system. Given the losses in your digestive system how much of the energy, in Joules, can you use from the exercise bar?

  After eating, you climb a hill on your bike. The combined mass of you and your bike is .

  How much gravitational potential energy has been gained by you and your bike?

  Where did this energy come from?

  If you ride quickly down the mountain without braking but losing half the energy to air resistance, how fast are you going when you get to the bottom?

  You find yourself on your bike at the top of Twin Peaks in San Francisco. You are facing a descent. The combined mass of you and your bicycle is .

  How much gravitational potential energy do you have before your descent?

  You descend. If all that potential energy is converted to kinetic energy, what will your speed be at the bottom?

  Name two other places to which your potential energy of gravity was transferred besides kinetic energy. How will this manifest itself in your speed at the bottom of the hill? (No numerical answer is needed here.)

  Before a run, you eat an apple with Joules of binding energy. Joules of binding energy are wasted during digestion. How much remains?

  Some % of the remaining energy is used for the basic processes in your body (which is why you can warm a bed at night!). How much is available for running?

  Let’s say that, when you run, you lose % of your energy overcoming friction and air resistance. How much is available for conversion to kinetic energy?

  Let’s say your mass is . What could be your top speed under these idealized circumstances?

  But only % of the available energy goes to KE, another % goes into heat exhaust from your body. Now you come upon a hill if the remaining energy is converted to gravitational potential energy. How high do you climb before running out of energy completely?

  A car goes from rest to a speed of v in a time t. Sketch a schematic graph of kinetic energy vs. time. You do not need to label the axes with numbers.

  A car traveling with a speed of drives horizontally off of a cliff. Sketch the situation.

  Calculate the potential energy, the kinetic energy, and the total energy of the car as it leaves the cliff.

  Make a graph displaying the kinetic, gravitational potential, and total energy of the car at each increment of height as it drops

  A roller coaster begins at rest above the ground at point , as shown above. Assume no energy is lost from the coaster to frictional heating, air resistance, sound, or any other process. The radius of the loop is .

  Figure 2.2

  Roller coaster for problem 8.

  Find the speed of the roller coaster at points and .

  At point the speed of the roller coaster is . How high off the ground is point ?

  A pendulum has a string with length . You hold it at an angle of degrees to the vertical and release it. The pendulum bob has a mass of .

  Figure 2.3

  Pendulum for problem 9.

  What is the potential energy of the bob before it is released? (Hint: use geometry to determine the height when released.)

  What is its speed when it passes through the midpoint of its swing?

  Now the pendulum is transported to Mars, where the acceleration of gravity g is . Answer parts (a) and (b) again, but this time using the acceleration on Mars.

  On an unknown airless planet an astronaut drops a ball from a ledge. The mass hits the bottom with a speed of . What is the acceleration of gravity on this planet?

  The planet has a twin moon with exactly the same acceleration of gravity. The difference is that this moon has an atmosphere. In this case, when dropped from a ledge with the same height, the ball hits bottom at the speed of . How much energy is lost to air resistance during the fall?

  A car starts at rest and speeds up to . What is the gain in kinetic energy?

  We define efficiency as the ratio of output energy (in this case kinetic energy) to input energy. If this car’s efficiency is , how much input energy was provided by the gasoline?

  If gallons were used up in the process, w
hat is the energy content of the gasoline in Joules per gallon?

  A pile driver’s motor expends Joules of energy to lift a mass. The motor is rated at an efficiency of (see 11b). How high is the mass lifted?

  Answers to Selected Problems

  d

  (discuss in class)

  a.

  Chemical bonds in the food.

  a.

  a.

  .

  b.

  a.

  a.

  a.

  a.

  12.

  Chapter 3: Energy Conservation Appendix

  Equivalence between [2] and [3]

  The formulas above look pretty different, but the main conceptual split is that the in the first formula above ([2]) varies from planet to planet, but the in [3] constant throughout the universe --- in fact, it's called the Gravitational Constant. You might think that this makes the second formula more fundamental than the first, and you would be right. The first is actually a "special case" of the second. That is, the [3] always holds, but [2] only holds when certain conditions are met: that is, you are at the surface of a spherical body. In this case they are equivalent, but [2] is obviously simpler.

  It is important to see the relationship between [2] and [3], since it is typical of the stuff of physics. If [3] is the more fundamental equation, we should be able to start with it and derive [2]. First, we will make a minor simplification: we will assume that the "object of interest" starts at the surface of the earth, and not some arbitrary height near it (now we don't have to deal with the deltas and hairier details without sacrificing any of the content). Then the question we want to answer is: if we raise this object from the surface to a height , what will its gravitational potential energy change by?

  Figure 3.1

  Illustration for the Equivalence Derivation

  Since equation [3] always holds exactly (as far as we're concerned), we can certainly use it. Before it was raised, the object was at the surface, so its potential energy from Earth's gravity was given by: This held because the in [3] is the distance between the two objects' centers of mass. The center of mass of the earth is, predictably, at its center, so we can replace above with for objects on its surface. After the object is raised, its distance from the center of the Earth increases by . In other words, its new energy is If we explore the quantities of the type , we will see that when , . This kind of adjustment is called a Taylor approximation. To prove this to yourself, use your calculator to try a bunch of different numbers and see what the error is.

  The ratio of , the distance the object was raised, to the radius of the earth is miniscule. So, we can use the theorem above, remembering that it is this ratio that plays the role of above. This ratio is only small for large objects, such as planets; and an object's center of mass may not always be at its geographic center. Both have to be true for our results to hold. Hence, special case.

  Then we find that [6] reduces to The last step is to finally calculate the change in potential energy due to the movement. This is straightforward, since This is the equation we have been looking for. Although it doesn't look like it, it is completely equivalent to the formula . The only variable in this formula is h, everything else --- the radius and mass of Earth and G --- are constants. If we rearrange the formula, we see that: The quantity labeled g, if calculated with the appropriate radius and mass, will give the effective acceleration due to gravity near the surface. When Earth's mass and radius are used. the result is 9.8 m/s, but feel free to check.

  Chapter 4: One-Dimensional Motion Version 2

  The Big Idea

  One dimensional motion describes objects moving in straight lines. Speed is a scalar measure of how quickly an object is moving along this line: units of length per one unit of time. If an object's speed changes, it is said to be accelerating (or decelerating). As we will see, understanding an object's acceleration is the key to understanding its motion. At this point, we are not worried about where the acceleration is coming from --- we will deal with that question later.

  In general, position, displacement, velocity and acceleration have directions and are therefore vectors. It's important to note, however, that in one dimension there are only two possible directions for vectors to point in, and these are usually labeled and . We will therefore typically avoid calling one dimensional quantities vectors, since their direction can be represented with a sign. The quantities labeled vectors below can and will be treated as scalars with signs in one dimensional situations throughout the book.

  Key Definitions

  Deriving the Kinematics Equations

  The simplest case of one dimensional motion is an object at rest. A slightly more difficult problem is that of an object moving at a constant velocity. Such an object's position at time is given by the familiar formula, that is, distance equals rate times time. In our language, this would be:

  If an object is undergoing an acceleration that changes with time, it is in general quite difficult to find its position and velocity as a function of time. However, it's always true that over a period of time average velocity and average acceleration are given by: Therefore, finding an object's position or velocity can be reduced to finding the average velocity or average acceleration, respectively. Usually, this is just as difficult as the problem mentioned above, but in one very common and specific case --- constant acceleration --- these formulas are very useful. In this case, velocity changes at a linear rate with time, that is: You should realize this is just another version of equation [1], which in fact describes anything changing at a linear rate. Since the average of a linear function over some time is just the average of its endpoints (figure 1), we have: Now, we

  We have obtained the equations of uniformly, accelerated motion, also known as the

  Big Three Equations

  Hints for Problems

  When beginning a one dimensional problem, define a positive direction. The other direction is then taken to be negative. Traditionally, "positive" is taken to mean "to the right"; however, any definition of direction used consistency throughout the problem will yield the right answer.

  Be sure you understand the difference between average velocity (measured over a period of time) and instantaneous velocity (measured at a single moment in time).

  Gravity near the Earth pulls an object toward the surface of the Earth with an acceleration of . In the absence of air resistance, all objects will fall with the same acceleration. Air resistance can cause low-mass, large area objects to accelerate more slowly.

  Deceleration is the term used when an object’s speed is decreasing due to an acceleration in the opposite direction of its velocity.

  The Big Three equations define the graphs of position and velocity as a function of time. When there is no acceleration (constant velocity), position increases linearly with time -- distance equals rate times time. Under constant acceleration, velocity increases linearly with time but distance does so at a quadratic rate. The slopes of the position and velocity graphs will give instantaneous velocity and acceleration, respectively.

  At first, you might get frustrated trying to figure out which of the Big Three equations to use for a certain problem, but don’t worry, this comes with practice. Making a table that identifies the variables given in the problem and the variables you are looking for can sometimes help.

  One Dimensional Examples

  Example 1

  Question: If Bob walks 100 meters north, turns around, and walks 100 meters south, what is his total distance and displacement ?

  Solution: If Bob walked 100 meters in north, turned around, and walked back to his starting position his total distance traveled would be:

  However, his displacement would be:

  As you can see, if Bob returns to his original starting position his displacement (0 meters) and his total distance traveled (200 meters) are very different.

  Example 2

  Question: Joe throws a ball straight up with a initial velocity of 70. For this
problem ignore Joe's height.

  a) How high does the ball go?

  b) For how many seconds is the ball in the air?

  c) Joe is now standing underneath a ceiling that is 237m high. How fast will the ball be traveling when it strikes the ceiling?

  Solution:

  a) We know the initial velocity (70) and the acceleration (the only acceleration that acts upon the ball after it has left Joe's hand is the acceleration of gravity, which is . We can figure out the final velocity by realizing that when the ball is at the highest point, the velocity will be . To find the height of the ball we will use the equation and solve for x.

  We can now substitute in the known values to get x.

  b) To solve for the total time the ball is in the air we will for the time that the ball is traveling up and double it (the trip down will take the same time as the trip up). We know the initial velocity is and the acceleration due to gravity is . We also know that the final velocity is . We will use the equation and solve for t.