Chapter 9: Basic Physics SE-Energy

Difficulty Level: At Grade Created by: CK-12

The Big Idea

Energy is a measure of the amount of or potential for movement in something. The total amount of energy in the universe is always the same. This symmetry is called a conservation law. Conservation of energy is one of five conservation laws that govern our universe. A group of things (we’ll use the word system) has a certain amount of energy. Energy can be added to a system; for instance, when chemical bonds in a burning log break, they release heat. Energy can be lost from a system; for instance, when a spacecraft “burns up” its energy of motion during re-entry, it loses energy and the surrounding atmosphere gains the lost energy. A closed system is one for which the total energy does not change, or is conserved. In this chapter, we will often consider closed systems, for which the total amount of energy stays the same, but transforms from one kind to another. The transfer of energy from one system to another is called Work. Work is equal to the amount of energy transferred in (positive work) or out (negative work) of a system. Work is equal to the distance the object is moved multiplied by the amount of force in the direction of its motion. The rate at which work is done is called Power.

Key Definitions

• m=\begin{align*}m =\end{align*} mass (in kilograms, kg)
• h=\begin{align*}h =\end{align*} height above the ground (in meters, m)
• v=\begin{align*}v =\end{align*} speed (in meters per second, m/s)
• g=\begin{align*}g =\end{align*} acceleration due to gravity (9.8 m/s210 m/s2)\begin{align*}(9.8 \ m/s^2 \approx 10 \ m/s^2)\end{align*}
• E=\begin{align*}E =\end{align*} energy (in Joules; 1 J=1 kgm2/s2\begin{align*}1 \ J = 1 \ kg \cdot m^2/ \text{s}^2\end{align*})
• W=\begin{align*}W =\end{align*} work (in Joules; work is just energy being transferred)
• P=\begin{align*}P =\end{align*} power (in Watts; 1 W=1 J/s\begin{align*}1 \ W = 1 \ J/s\end{align*})

Key Equations

• \begin{align*} \sum E_{\text{initial}} = \sum E_\text{final} \end{align*} ; The total energy does not change in closed systems
• \begin{align*}KE = \frac{1}{2} \ mv^2\end{align*} ; Kinetic energy
• \begin{align*}PE_g = mgh\end{align*} ; Potential energy of gravity
• \begin{align*}PE_{sp} = \frac{1}{2} \ kx^2\end{align*} ; Potential energy of a spring
• \begin{align*}W = F_{x} \Delta x = Fd \ \cos \circleddash\end{align*} ; Work is equal to the distance multiplied by the component of the force in the direction it is moving.
• \begin{align*}P = W/t\end{align*} ; Power is equal to the energy released per second
• \begin{align*}Eff = \frac{P_{out}}{P_{in}}\end{align*} ; The efficiency equals the output power divided by the input power

Key Concepts

• The energy of motion is kinetic energy, KE. Whenever an object is in motion it has kinetic energy. The faster it is going, the more energy it has.
• The energy due to gravity is called gravitational potential energy, \begin{align*}PE_g\end{align*}, which gets larger the farther off the ground you are.
• Molecules have chemical potential energy due to the bonds between the electrons; when these bonds are broken, energy is released which can be transferred into kinetic and/or potential energy. 1 food Calorie (1 kilocalorie) is equal to 4180 Joules of stored chemical potential energy.
• Energy can be transformed from one kind into the other; if the total energy at the end of the process appears to be less than at the beginning, the “lost” energy has been transferred to another system, often by heat or sound waves.
• Work is simply the word for the amount of energy transferred in or out of the system. Work can be computed by multiplying the distance with the component of force that is parallel to the distance
• Efficiency is equal to the output energy divided by the input energy.

Key Applications

• Work is defined as the product of a force times the distance over which the force acts. The force and the distance must act along the same line; if the angle between them is not \begin{align*}0^\circ\end{align*} (or \begin{align*}180^\circ\end{align*}) then you must use the component of force; hence the \begin{align*}\cos \circleddash \end{align*} in the equation.
• Mechanical Advantage is the ability to lift or move objects with great force while utilizing only a little force. The trade-off is that you must operate the smaller input force for a large distance. This is all seen through the work Equation. Work equals force times distance. Energy is conserved. Thus one can get a large force for a small distance equal to a small force for a large distance.
• Mechanical advantage equals the distance of effort divided by the distance the object moves. It is also equal to the output force divided by the input force.
• In “roller coaster” problems, the gravitational potential energy at the top of one hill turns into kinetic energy at the bottom of the next valley. It turns back into potential energy as you round the next hill, and so on. However, some of the energy is lost to the tracks and air as heat, which is why the second rise is often not as big as the first.
• When we say energy is lost as heat, often what we mean is that a friction force does negative work on an object. That negative work is lost energy - heat. Keep in mind that work is force times distance.
• In “pendulum” problems, the potential energy at the highest point in a pendulum’s swing changes to kinetic energy when it reaches the bottom and then back into potential energy when it reaches the top again. At any in-between point there is a combination of kinetic energy and potential energy, but the total energy remains the same.
• In everyday life, the human body breaks down the food molecules to change some of the bonding energy into energy that is used to power the body. This energy goes on to turn into kinetic energy as a person moves, or for example an athlete gains speed. The kinetic energy can be changed into potential energy as a person gains height above the ground.

Energy Conservation Problem Set

1. A stationary bomb explodes into hundreds of pieces. Which of the following statements best describes the situation?
1. The kinetic energy of the bomb was converted into heat.
2. The chemical potential energy stored in the bomb was converted into heat and gravitational potential energy.
3. The chemical potential energy stored in the bomb was converted into heat and kinetic energy.
4. The chemical potential energy stored in the bomb was converted into heat, sound, kinetic energy, and gravitational potential energy.
5. The kinetic and chemical potential energy stored in the bomb was converted into heat, sound, kinetic energy, and gravitational potential energy.
2. You hike up to the top of Mt. Everest to think about physics.
1. Do you have more potential or kinetic energy at the top of the mountain than you did at the bottom? Explain.
2. 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.
3. How has the total energy of the Solar System changed due to your hike up the mountain? Explain.
4. 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.
5. For each of the following types of energy, describe whether you gained it, you lost it, or it stayed the same during your hike:
1. Gravitational potential energy
2. Energy stored in the atomic nuclei in your body
3. Heat energy
4. Chemical potential energy stored in the fat cells in your body
5. Sound energy from your footsteps
6. Energy given to you by a wind blowing at your back
3. One type of Olympic weightlifting is called the snatch, where the lifter must lift the weight over his or her head in one continuous movement. Hossein Rezazadeh of Iran snatched 467.5 pounds in the 2000 Sydney games.
1. Convert that weight to kg.
2. Suppose Rezazadeh lifted that weight 1.8 m. How much work did he do against gravity?
3. If Rezazadeh lifted the weight in 2.8 s, what power did he exert?
4. A shopper in Whole Foods pushes their cart with a force of 40 N directed at an angle of \begin{align*}30^\circ\end{align*} downward from the horizontal.
1. Find the work done by the shopper on the cart as he moves down a 15-m aisle to the tofu section.
2. If the mass of the cart is 24 kg and we neglect friction, how fast will the cart be moving when it reaches the tofu section if it started from rest?
3. The shopper brings the cart to rest is 2.7 s when he reaches the tofu. What power does he exert in stopping the cart?
5. Just before your mountain bike ride, you eat a 240 Calorie exercise bar. (You can find the conversion between food Calories and Joules in the ‘Key Concepts’.) 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. (a) 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 500 m hill on your bike. The combined mass of you and your bike is 75 kg. (b) How much gravitational potential energy has been gained by you and your bike? (c) Where did this energy come from? (d) 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?
6. You take the bus with your bike to the top of Twin Peaks in San Francisco. You are facing a long descent; the top of Twin Peaks is 600 m higher than its base. The combined mass of you and your bicycle is 85 kg.
1. How much gravitational potential energy do you have before your descent?
2. You descend. If all that potential energy is converted to kinetic energy, what will your speed be at the bottom?
3. Name two other places to which your potential energy of gravity was transferred besides kinetic energy. How will this change your speed at the bottom of the hill? (No numerical answer is needed here.)
7. A mover loads a 100-kg box into the back of a moving truck by pushing it up a ramp. The ramp is 4 m long and the back of the truck is 1.5 m high.
1. Calculate the potential energy gained by the box when it’s loaded into the truck.
2. Calculate the mechanical advantage of the ramp.
3. Calculate the force required to push the box up the ramp in the absence of friction.
4. If instead the force used to push the box is 500 N, calculate
1. the work done by the mover
2. the energy lost to friction.
3. the force of friction between the box and the ramp
4. the efficiency of the ramp.
8. A car goes from rest to a speed of \begin{align*}v\end{align*} in a time \begin{align*}t\end{align*}, with a constant acceleration. (a) Sketch a schematic graph of speed vs. time. You do not need to label the axes with numbers, but the shape of the graph should accurately describe what is happening. (b) Sketch a schematic graph of KE vs time. You do not need to label the axes with numbers, but the shape of the graph should accurately describe what is happening.
9. A 1200 kg car traveling with a speed of 29 m/s drives horizontally off of a 90 m cliff. (a) Sketch the situation. (b) Calculate the potential energy, the kinetic energy, and the total energy of the car as it leaves the cliff, using the base of the cliff as your zero point for PE. (c) Make a graph displaying the kinetic, gravitational potential, and total energy (draw each with a different color or number the curves) of the car at each 10 m increment of height as it drops
10. A roller coaster begins at rest 120 m above the ground at point \begin{align*}A\end{align*}, 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 40 m.
1. Find the speed of the roller coaster at the following points:
1. \begin{align*}B\end{align*}
2. \begin{align*}C\end{align*}
3. \begin{align*}D\end{align*}
4. \begin{align*}E\end{align*}
5. \begin{align*}F\end{align*}
6. \begin{align*}H\end{align*}.
2. At point \begin{align*}G\end{align*} the speed of the roller coaster is 22 m/s. How high off the ground is point \begin{align*}G\end{align*} ?
11. A pendulum has a string with length 1.2 m. You hold it 8.7 cm above its lowest point and release it. The pendulum bob has a mass of 2.0 kg.
1. What is the potential energy of the bob before it is released?
2. What is its speed when it passes through the midpoint of its swing?
3. Now the pendulum is transported to Mars, where the acceleration of gravity \begin{align*}g\end{align*} is \begin{align*}3.7 \ m/s^2\end{align*}. Answer parts (a) and (b) again, but this time using the acceleration on Mars.
12. On an unknown airless planet an astronaut drops a 4.0 kg ball from a 60 m ledge. The mass hits the bottom with a speed of 12 m/s.
1. What is the acceleration of gravity \begin{align*}g\end{align*} on this planet?
2. The planet has a twin in an alternate universe with exactly the same acceleration of gravity. The difference is that this planet has an atmosphere. In this case, when dropped from a ledge with the same height, the 4.0 kg ball hits bottom at the speed of 9 m/s.
1. How much energy is lost to air resistance during the fall?
2. What average force did air resistance exert on the ball?
13. A skateboarder, mass 65 kg, is traveling at 8 m/s down the sidewalk when he hits a rough patch 5.3 m long. The force of friction acting on him is 100 N.
1. How much work does friction do on the skateboarder?
2. How fast will he be traveling when he leaves the rough patch?
3. He continues down the sidewalk and runs head on into another skateboarder, mass 40 kg, at rest. They stick together and roll down the sidewalk. Calculate their velocity after the collision.
4. Is energy conserved in this collision? Show calculations to support your answer.
14. Jaya is flying her airplane of mass 20,000 kg at 60 m/s. She then revs up her engine so that the forward thrust of the propeller becomes 75,000 N. If the force of air resistance is 40,000 N and the plane stays in level flight, calculate the final speed of the plane after it travels 500 m. HINT: Be sure to add up ALL the places energy can go here - work done by engine goes to more than just plane KE!
15. A 1500 kg car starts at rest and speeds up to 3.0 m/s with a constant acceleration.
1. If the car reaches its final speed in 1.2 s, what is its acceleration?
2. How far does the car travel in that time?
3. What is the car’s gain in kinetic energy?
4. What power is exerted by the engine?
5. We define efficiency as the ratio of output energy (in this case kinetic energy) to input energy. If this car’s efficiency is 0.30, how much input energy was provided by the gasoline?
6. If 0.00015 gallons were used up in the process, what is the energy content of the gasoline in Joules per gallon?
7. Compare that energy to the food energy in a gallon of Coke, if a 12-oz can contains 150 Calories (food calories) and one gallon is 128 ounces.
16. A pile driver’s motor expends 310,000 Joules of energy to lift a 5400 kg mass. The motor is rated at an efficiency of 0.13 (see #15e).
1. How high is the mass lifted?
2. The mass is dropped on a 1000-kg concrete pile and drives it 12 cm into the ground. What force does the mass exert on the pile?
17. A crowbar is a very handy tool to get out tough nails, demolition of a wall or breaking into or out of jail. It can be used from either end. The man shown is prying out a board. He is applying 30 N of force, study the picture to see how he is using it and which dimensions should be used.
1. What is the Mechanical Advantage of this crowbar, shown above with its dimensions?
2. How much force is being applied to the wood plank?

Answers (use \begin{align*}10 \ m/s^2\end{align*} for acceleration of gravity):

1. d
2. (discuss in class)
1. 212.5 kg
2. 3825 J
3. 1370 W
1. 520 J
2. 6.6 m/s
3. 193 W
1. \begin{align*}5.0 \times 10^5 \ J\end{align*}
2. \begin{align*}3.75 \times 10^5 \ J\end{align*}
3. Chemical bonds in the food.
4. 71 m/s
1. \begin{align*}5.0 \times 10^5 \ J\end{align*}
2. 110 m/s
1. 1500 J
2. 2.7
3. 375 N
1. 2000 J
2. 500 J
3. 125 N
4. 75%
1. (b) \begin{align*}KE = 504,600 \ J; \ U_g = 1,080,000 \ J; \ E_{total} = 1,584,600 \ J\end{align*}
1. 34 m/s at B; 28 m/s at D, 40 m/s at E, 49 m/s at C and F; 0 m/s at H
2. 96 m
1. 1.7 J
2. 1.3 m/s
3. 0.64 J, 0.80 m/s
1. \begin{align*}1.2 \ m/s^2\end{align*}
2. 130 J, 2.2 N
1. -530 J
2. 6.9 m/s
3. 4.3 m/s
2. 73 m/s
1. \begin{align*}2.5 \ m/s^2\end{align*}
2. 1.8 m
3. 6750 J
4. 5.6 kW
5. 22.5 kJ
6. 150 MJ/gallon of gas
7. 6.7 MJ/gallon Coke
1. 0.76 m
2. 336,000 N
1. M.A. = 12
2. Force out = 360 N

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