Unit8-Worksheets (sample solutions)

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Unit 1 Activity 2 - Observation Stations At each Observation Station, you will be asked to perform one or more activities and respond to the following three prompts: I. Describe what you did or the procedure you followed at the station from beginning to end. Include an accurate description of the items involved as well as actions performed. II. Describe what you saw or observed during this event. Note: Do not confuse what you saw/observed with what you THINK is happening. III. Describe what changes you observed during this event, from beginning to end. Based on your observations and prior knowledge, why do you think these changes occurred? Observation Stations Station Name: Cheetos on Fire I. What did you do ? I took a piece of Cheeto and placed it in an aluminum pie pan, then I ignited it with a lighter wand, letting it burn. II. What did you observe ? The piece of Cheeto caught fire and produced a large flame, leaving behind a dark grey burned skeleton. It also became hot. III. What changes did you observe? Why do you think these changes occurred? The piece of Cheeto changed from its original red-orange state to a solid dark grey ash state as it burned. I think this is due to a chemical reaction with the oxygen of the air, ignited by the high temperature of the lighter flame. That reaction seems to have released heat in the process.

Observation Stations Station Name: Tumble Buggy I. What did you do ? Taking the buggy, I turned its power switch on, put it on the floor, let it go for a few seconds, took it back, and turned it off. II. What did you observe ? When I turned the power switch on, the wheels started to rotate. Put on the floor, the buggy moved in a straight line at what appeared as a relatively constant velocity. Wheels stopped once the buggy was turned off. III. What changes did you observe? Why do you think these changes occurred? The buggy changed from a rest state to a moving state (constant velocity). This was caused by the switch, and I believe it happened because it closed the circuit allowing the flow of electricity from the batteries. Station Name: Wind-Up Toys I. What did you do ? I slightly winded the toy, then let it go on a table. II. What did you observe ? After winding the toy and letting go of the dial, it started to move for a time, and then stopped. During the motion, the dial was rotating. III. What changes did you observe? Why do you think these changes occurred? The toy changed from a rest state to a moving state. I think this was due to the winding of the dial as the motion stopped when it was fully unwound.

Observation Stations Station Name: Popper Toy I. What did you do ? I turned the popper inside out, put on the table quickly, then released it. II. What did you observe ? As the popper regained its initial shape, it jumped into the air. III. What changes did you observe? Why do you think these changes occurred? Initially, the popper was deformed, and as it regained its form, it gained speed and moved up into the air, slowing down before falling back down. I think the reason is that the deformation stored elastic energy that was later transformed into kinetic energy (and some heat). In free fall, that kinetic energy transformed into gravitational potential energy, then back it kinetic energy (and heat when it hit the floor again). Station Name: Ball Drop I. What did you do ? I first dropped a small ball from my chest height. Then I did the same with a basketball. Finally, I did it again with the small ball on top of the basketball. II. What did you observe ? When I dropped the little ball, it bounced back lower that its initial height. When I dropped the basketball, it did the same, but not at the same height as the small ball. Using different kind of balls, even if of about the same size, won’t bounce all at the same height, although it will always be at less than the initial height (at least in my trials). Now, putting the small ball on top of the basketball, the small ball bounces back to an epic height while the basketball bounces back to an even lower height than before, when alone. III. What changes did you observe? Why do you think these changes occurred? Contrary to what happens when only one ball is dropped, when the small one (less massive) is on top of the big one (more massive), the big one bounces back lower and the small one bounces back at epic height. I think this happens because the basketball collides on its way up with the small ball on top. Thus, the basketball cannot go as high as it is hit down, but its energy has to go somewhere… in the small ball that is hit up?

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Observation Stations Station Name: Mass on a Spring I. What did you do ? I pulled down the mass attached to the vertical spring by about 3 inches, then I released the mass. II. What did you observe ? As I pulled down, the spring stretched even more than when it was at rest with the mast just hanging. The mass, after release, started to oscillate back and forth about an equilibrium position. After some time, the oscillations damped to rest. III. What changes did you observe? Why do you think these changes occurred? As I pulled down, the spring stretched. After release, the mass started to move up as the spring unstretched. Passed the equilibrium position, the spring compressed, and the mass came to rest. Then it did the same in reverse, back and forth, slowly going back to rest as in the initial state. I think the extension or compression of the spring puts the mass into oscillatory motion, trying to bring the mass back to equilibrium, but it doesn’t get there because the mass gains speed along the way. It’s only because of friction that ultimately equilibrium will be back. Station Name: Watching Ice Melt I. What did you do ? I placed an ice cube on a metal block, then an ice cube on the plastic block. II. What did you observe ? The two ice cubes melted, but it took longer on the plastic block. Furthermore, the metal block became colder, not so much for the plastic block. III. What changes did you observe? Why do you think these changes occurred? The ice melted at different rates. I think the ice cube on the metal block melted faster because heat could more easily enter the ice through thermal conduction.

Observation Stations Station Name: Airzooka I. What did you do ? Holding the Airzooka, I pulled back the elastic air launcher knob and aimed at a suspended piece of paper. Then, I released the air launcher. II. What did you observe ? When I released, the piece of paper moved as if hit by something invisible. III. What changes did you observe? Why do you think these changes occurred? The piece of paper was initially still, but after the release, it moved. I think this happened because air was pushed by the airzooka, and then travelled toward the target. Station Name: Pop, Pop, Fizz, Fizz, … I. What did you do ? I place 1⁄2 tablet of Alka Seltzer in a film canister. Then I poured 1 test tube of water into the film canister and quickly replaced the cap. Finally, I quickly placed the closed film canister, lid side down, into a clear jar. II. What did you observe ? After all was mixed and closed, it took a few seconds for the film canister to explode. We can notice that when we just put the Alka Seltzer in an open container, it dissolves and make bubbles of gas. III. What changes did you observe? Why do you think these changes occurred? The water mixed with Alka Seltzer changes to a bubbling solution. In a closed film canister, the thing just builds up and explodes. I think the gas builds up the pressure in the canister, and the whole thing pops up when too high.

Observation Stations Station Name: Rubbing Hands I. What did you do ? I put my hands together and rubbed them back and forth. II. What did you observe ? Over time, heat builds up in my hands as a continue rubbing them together. III. What changes did you observe? Why do you think these changes occurred? What changes was the heat building up in my hands. I think that happened because of the friction. Station Name: NA I. What did you do ? II. What did you observe ? III. What changes did you observe? Why do you think these changes occurred?

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©Modeling Instruction – AMTA 2013 1 U8 Energy - ws 1a v3.1 Name Date Pd Energy Storage and Transfer Model Worksheet 1a: Qualitative Analysis - Pie Charts Use pie charts to analyze the energy changes in each situation given. • Designate your choice of system with a dotted line. Choose your system so that the energies involved are internal (within the system). • Carefully label the pies to correspond with the positions of the objects given. (A, B,C, etc.) • The pies should be accurately divided and labeled with the energy storage mechanisms involved. • Remember the 3 energy questions in deciding about the energy changes: 1. Where does the energy come from? (What's the source of the energy?) 2. What does the energy do? 3. Where does the energy go? 1. A wind-up toy is fully wound and at rest. 2. A wind-up toy is wound up and moving across level ground. The toy is speeding up. The toy is wound up and is moving at a constant speed up an incline. 3. The toy is wound up and is moving at a constant speed up an incline. E el E k E el E el E k E g E th E el E k E el E k E th E el E k E th E el E k E g E th

©Modeling Instruction – AMTA 2013 2 U8 Energy - ws 1a v3.1 4. The toy is wound up and moving along at a constant speed. 5. The toy is wound up and slowing down as it moves up an incline. 6. The toy is wound up and speeding up as it moves up an incline. E k E el E th E k E el E el E k E el E k E el E k E th E th E g E el E k E g E el E k E th E g E el E k E th E th E g E el E k

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©Modeling Instruction – AMTA 2013 1 U8 Energy - ws 1b v3.1 Name Date Pd Energy Storage and Transfer Model Worksheet 1b: Qualitative Analysis - Pie Charts Use pie charts to analyze the energy changes in each situation given. Designate your choice of system with a dotted line. Choose your system so that the energies involved are internal (within the system). Carefully label the pies to correspond with the positions of the objects given. (A, B,C, etc.) The pies should be accurately divided and labeled with the energy storage mechanisms involved. Remember the 3 energy questions in deciding about the energy changes: 1. Where does the energy come from? (What's the source of the energy?) 2. What does the energy do? 3. Where does the energy go? 1. A ball is held above the ground, and then is dropped so it falls straight down. (Restrict your analysis to the ball being in the air, BEFORE it hits the ground.) 2. A wind-up toy is wound up, then "walks" across a table and comes to a stop. 3. A baseball is thrown up in the air and then falls back down. Place velocity vectors beside each corresponding baseball in the drawing, and draw an energy storage pie for each lettered position. E el E k E el E k E th E th E k E k E g E g E g E k E g E k E g E k E g E g E k

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©Modeling Instruction – AMTA 2013 2 U8 Energy - ws 1b v3.1 4. An object rests on a coiled spring, and is then launched upwards. Assuming that rest position on spring implies reference height (y = 0) 5. A piece of clay is dropped to the floor. 6. A ball rolls to a stop on the floor. 7. A truck being driven down the street. Assuming E ch is gas in tank of truck 8. A superball is dropped and bounces up and down. Draw a pie chart for each position of the ball shown. Why does the ball not bounce as high each time? Where does the energy "go"? Some of the energy is converted to thermal on each bounce so the ball’s bounce height is reduced. E el E g E k E g E g E g E k E th E k E th E k E th E th E k E ch E k E k E th E ch E ch E g E g E th E el E th E el E th E g E th

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©Modeling Instruction - AMTA 2013 1 Energy ws 2 v3.1 Name Date Pd Energy Storage and Transfer Model Worksheet 2: Hooke’s Law and Elastic Energy Suppose one lab group found that F = 1000 N/m (∆x). Construct a graphical representation of force vs. displacement. (Hint: make the maximum displacement 0.25 m. ) 1. Graphically determine the amount of energy stored while stretching the spring described above from x = 0 to x = 10. cm. The area marked with a squiggle is equal to the energy stored. 2. Graphically determine the amount of energy stored while stretching the spring described above from x = 15 to x = 25 cm. The area marked with stars is equal to the energy stored The graph below was made from data collected during an investigation of the relationship between the amounts two different springs stretched when different forces were applied. 3. Determine the spring constant for each spring. The spring constant is equal to the slope of the line. 4. For each spring, compare: a. the amount of force required to stretch the spring 3.0 m. By inspection: F 1 = 24 N F 2 = 15 N b. the E el stored in each spring when stretched 3.0m. ( )( ) ( )( ) 1 1 0.1 100 5 2 2 E area b h m N J = = = = ( )( ) ( )( ) ( )( ) ( )( ) 1 2 1 2 1 0.1 150 0.1 100 20 2 E area b h b h E m N m N J = = + = + = 1 2 24 8 / m 3 15 5 / 3 N k N m N k N m m = = = = E 1 = area = 1 2 b ( ) h ( ) = 1 2 3.0 m ( ) 24 N ( ) = 36 J E 2 = area = 1 2 b ( ) h ( ) = 1 2 3.0 m ( ) 15 N ( ) = 22.5 J 0 0.1m 0.2m 0 200N N 100N N

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©Modeling Instruction - AMTA 2013 2 Energy ws 2 v3.1 5. Determine the amount that spring 2 needs to be stretched in order to store 24 joules of energy. 6. The spring below has a spring constant of 10. N/m. If the block is pulled 0.30 m horizontally to the right, and held motionless, what force does the spring exert on the block? Sketch a force diagram for the mass as you hold it still. (Assume a frictionless surface.) h = b ( ) slope ( ) E el = area = 1 2 b ( ) b ( ) slope ( ) Þ b = 2 E el slope = 48 J 5.0 N m = 3.1 m F = k ( ) D x ( ) = 10 N m ( ) 0.30 m ( ) = 3.0 N F el F g F N F T

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©Modeling Instruction - AMTA 2013 3 Energy ws 2 v3.1 7. The spring below has a spring constant of 20. N/m. The µ s between the box and the surface is 0.40. The line in the center of the box lines up at the 0 when the spring is not stretched or compressed. a. The box is pushed to the right, then released. Draw a force diagram for the box above when the spring is stretched, yet the box is stationary b. What is the maximum distance that the spring can be stretched from equilibrium before the box begins to slide back? The maximum distance is reached when F f = F el . c. Do pie chart analysis for this situation, when the spring is stretched beyond its maximum (from part b above) so it slides back, and then the box oscillates back and forth until it comes to a stop. Assume your system includes the spring, box, and table top. A = position of max stretch… B = box passes through equilibrium position…C = box compresses spring and stops…D = box returns to equilibrium position… E= box at interim position… F = box has stopped Due to static friction, the box would come to rest with some E el in the spring, but you may choose to ignore this. F el = k D x F fk = μ mg = .4 ( ) 1 kg ( ) 10 N kg ( ) = 4.0 N 4.0 N = 20 N m ( ) D x ( ) D x = 0.20 m F el F g F N F fs E th E el E el E th E th E th E th E el E k E k E k A B C D E F x max 0

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Name Date Pd Energy Model Worksheet 3: Qualitative Energy Storage & Conservation with Bar Graphs For each situation shown below: 1. List objects in the system within the circle. ** Always include the earth’s gravitational field in your system. 2. On the physical diagram, indicate your choice of zero height for measuring gravitational energy. 3. Sketch the energy bar graph for position A, indicate any energy flow into or out of the system from position A to position B on the System/Flow diagram, and sketch the energy bar graph for position B. 4. Write a qualitative energy equation that indicates the initial, transferred, and final energy of your system. 1a. In the situation shown below, a spring launches a roller coaster cart from rest on a frictionless track into a vertical loop. Assume the system consists of the cart, the earth, the track, and the spring. 1b. Repeat problem 1a for a frictionless system that includes the cart, the earth, and the track, but not the spring. 1c. Use the same system as problem 1a, but assume that there is friction between the cart and the track. Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow Qualitative Energy Conservation Equation: ࠵? !" # = ࠵? $ % + ࠵? & % A B cart track spring earth Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow Qualitative Energy Conservation Equation: ࠵? = ࠵? $ % + ࠵? & % A B W Cart track earth spring Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow Qualitative Energy Conservation Equation: ࠵? !" # = ࠵? $ % + ࠵? & % + ࠵? ’( % A B cart, track spring, earth

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1d. This situation is the same as problem 1a except that the final position of the cart is lower on the track. Make sure your bars are scaled consistently between problem 1a and 1d. Assume the system consists of the cart, the earth, the track, and the spring. 2a. A moving car rolls up a hill until it stops. Do this problem for a system that consists of the car, the road, and the earth. Assume that the engine is turned off, the car is in neutral, and there is no friction. 2b. Repeat problem 2a for the same system with friction. 3a. A person pushes a car, with the parking brake on, up a hill. Assume a system that includes the car, the road, and the earth, but does not include the person. Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow Qualitative Energy Conservation Equation: ࠵? !" # = ࠵? $ % + ࠵? & % A B cart, track spring, earth y y A = 0 v A > 0 y B > 0 v B = 0 A B Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow Qualitative Energy Conservation Equation: ࠵? $ # = ࠵? & % Car, road earth y y A = 0 v A > 0 y B > 0 v B = 0 A B Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow Qualitative Energy Conservation Equation: ࠵? $ # = ࠵? & % + ࠵? ’( % car, road earth car, road earth W y h A = 0 v A = 0 h B > 0 v B = 0 A B Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow Qualitative Energy Conservation Equation: ࠵? = ࠵? & % + ࠵? ’( % person

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3b. Repeat problem 3a for a system that includes the person. 4a. A load of bricks rests on a tightly coiled spring and is then launched into the air. Assume a system that includes the spring, the bricks and the earth. Do this problem without friction. 4b. Repeat problem 4a with friction. 4c. Repeat problem 4a for a system that does not include the spring. y h A = 0 v A = 0 h B > 0 v B = 0 A B Position A Energy (J) 0 E k E g E el E ch Position B Energy (J) 0 E k E g E el E th System/Flow Qualitative Energy Conservation Equation: ࠵? )( # = ࠵? & % + ࠵? ’( % E ch is chemical energy car, road, earth person ࠵? !" # = ࠵? $ % + ࠵? & % Energy Equation: Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow y A B h A = 0 v A = 0 h B > 0 v B > 0 bricks, spring, earth ࠵? !" # = ࠵? $ % + ࠵? & % + ࠵? ’( % Energy Equation: Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow y A B h A = 0 v A = 0 h B > 0 v B > 0 bricks, spring, earth Energy Equation: y A B h A = 0 v A = 0 h B > 0 v B > 0 Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow ࠵? = ࠵? $ % + ࠵? & % W bricks, earth spring

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5a. A crate is propelled up a hill by a tightly coiled spring. Analyze this situation for a frictionless system that includes the spring, the hill, the crate, and the earth. 5b. Repeat problem 5a for a system that does not include the spring and does have friction. 6a. A bungee jumper falls off the platform and reaches the limit of stretch of the cord. Analyze this situation for a frictionless system that consists of the jumper, the earth, and the cord. 6b. Repeat problem 6a if the cord is not part of the system. y A B h A = 0 v A = 0 h B > 0 v B > 0 0 Energy Equation: Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow ࠵? !" # = ࠵? $ % + ࠵? & % spring crate earth y A B h A = 0 v A = 0 h B > 0 v B > 0 0 Energy Equation: ࠵? = ࠵? $ % + ࠵? & % + ࠵? ’( % W Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow crate, hill, earth spring y A h A > 0 v A = 0 h B > 0 v B = 0 0 y 0 B Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow Energy Equation: ࠵? & # = ࠵? & % + ࠵? !" % + ࠵? ’( % jumper, cord, earth jumper earth cord ࠵? & # − |࠵?| = ࠵? & % W B Energy Equation: y A h A > 0 v A = 0 h B > 0 v B = 0 0 y 0 System/Flow Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th

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©Modeling Instruction - AMTA 2013 1 U8 Energy - ws 4 v3.1 Name Date Pd Energy Storage and Transfer Model Worksheet 4: Quantitative Energy Calculations & Energy Conservation Be careful with units and unit conversions! 1. How much kinetic energy does a 2000 kg SUV traveling 70 mph have? (1 mile = 1600 meters) 2. How much energy does a 180 Calorie, half-pint carton of chocolate milk store? (One food Calorie = 4186 Joules) 3. Consider your 3 kg physics binder resting on the table in the classroom. Determine the gravitational energy of the earth- book system if the zero-reference level is chosen to be: a) the table Because h = 0, E g = 0 b) the floor, 0.68 meters below the book c) the ceiling, 2.5 meters above the book E g is (-) because the object is below the zero reference level. 4. A bungee cord stretches 25 meters and has a spring constant of 140 N/m. How much energy is stored in the bungee? 5. How fast does a 50-gram arrow need to travel to have 40 joules of kinetic energy? 6. How much energy is stored when a railroad car spring is compressed 10 cm? (The spring requires about 10,000 N to be compressed 3.0 cm.) 70 miles hr ´ 1600 m 1 mi ´ 1 hr 3600 s = 31.1 m s E k = 1 2 mv 2 = 1 2 2000 kg ( ) 31.1 m s ( ) 2 = 967,000 J or 967 kJ 180 Cal ´ 4186 J 1 Cal = 753 kJ E g = mgh E g = 3.0 kg ( ) 10 N kg ( ) 0.68 m ( ) = 20.4 J ( ) ( ) ( ) 3.0 10 2.50 75 N g kg E kg m J = - ( )( ) 2 2 1 1 2 2 140 25 43,800 N el m E k x m J = = = E k = 1 2 mv 2 Þ v = 2 40 J ( ) 0.050 kg = 40 m s 10,000 .030 333,000 N N m m k = = E el = 1 2 k x 2 = 1 2 333,000 N m ( ) 0.10 m ( ) 2 = 1670 J

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©Modeling Instruction - AMTA 2013 2 U8 Energy - ws 4 v3.1 7. A cart moving at 5.0 m/s collides with a spring. At the instant the cart is motionless, what is the largest amount that the spring could be compressed? Assume no friction. a. Define the system with the energy flow diagram, then complete the energy bar graphs qualitatively. b. Quantitative Energy Conservation Equation: c. Determine the maximum compression of the spring. 8. A rock is shot straight up into the air with a slingshot that had been stretched 0.30 m. Assume no air resistance. a. Qualitatively complete the energy flow diagram and the energy bar graphs. b. Quantitative Energy Conservation Equation: c. Determine the greatest height the rock could reach. E k = E el Þ 1 2 mv 2 = 1 2 k x 2 1 2 mv 2 = 1 2 k x 2 Þ x = mv 2 k = 8.0 kg 25 m 2 s 2 ( ) 50 N m = 2.0 m E el = E g Þ 1 2 k x 2 = mgh 1 2 k x 2 = mgh Þ h = k x 2 2 mg = 100 N m 0.30 m ( ) 2 2 0.50 kg ( ) 10 N kg ( ) = 9.0 Nm 10 N = 0.90 m Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow 0 A B k = 100 N/m D x = 0.30 m m = 500 g v = 0 Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow A B m = 8.0 kg v = 5.0 m/s k = 50 N/m v = 0 cart spring rock slingshot

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©Modeling Instruction - AMTA 2013 3 U8 Energy - ws 4 v3.1 9. Determine final velocity of the rollercoaster, assuming a 10% loss to friction. ° 10. The moon could be an ideal spaceport for exploring the solar system. A moon launching system could consist of a magnetic rail gun that shoots items into moon orbit. How much energy would be needed from the rail gun to get a 10,000 kg capsule into an orbit 100 km above the moon surface? The moon’s gravitational field strength is 1.6 N/kg and the orbital velocity for this altitude is 1700 m/s. Hint: Put the rail gun outside of the system. ࠵? = ∆࠵? = ࠵?࠵?∆ℎ + 1 2 ࠵?∆(࠵? ! ) ࠵? = (10 000 ࠵?࠵?) 11.6 ࠵? ࠵?࠵? 5 (10 " ࠵?) + 1 2 (10 000 ࠵?࠵?) 61700 ࠵? ࠵? 9 ! ࠵? = 1.60 × 10 # ࠵? + 1.45 × 10 $% ࠵? = 1.61 × 10 $% ࠵? Original qualitative EBC showed roughly equal amounts of E k and E g . After these were calculated, the diagram was adjusted. Note: as the height increases, the gravitational field strength decreases (1.45 N/kg at 100 km), so the value of E g calculated above is slightly too large. E g = E k + E th Þ E g - 1 10 E g = E k 0.90 mgh = 1 2 mv 2 Þ 2 0.90 gh ( ) = v 2 v = 1.8 10 N kg ( ) 5.0 m = 90 m 2 s 2 = 9.5 m s Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow A B m = 40 kg v = 0 5.0 m 0 Position A Energy (J) 0 E k E g E el Position B Energy (J) 0 E k E g E el E th System/Flow W coaster Earth track capsule Moon rail gun

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©Modeling Instruction - AMTA 2013 1 U8 Energy supplement v3.1 Name Date Pd Energy Storage and Transfer Model Energy Transferred by Non-Parallel Forces 1. Below are 4 situations where a tension force of 12 N is acting on a box and the box moves 5.0 meters while the force acts on it. Rank the following (from the least energy to the greatest) for the energy transferred to the box by the applied force. If situations transfer the same amount of energy place an equals sign between the letters. least A = B D C greatest Defend your sequence. In A there is no energy transferred to the box as the tension is perpendicular to the motion of the box. With no friction no energy is needed to keep the speed constant. In B we don't care where the energy went (it is most likely thermal account in the carrier), no energy has been transferred to the box. In D, only a portion of the force (12N ´ cos30° = 10.4N) acts in the same direction as the displacement, whereas in D all 12 N is in the same direction as D x. Viewed from above, the box moves 5.0 m around a circle due to the tension force acting on it. There is no friction between the box and the surface it is sliding on. The string is holding the 12 N box up with 12 N of force as the box is carried horizontally 5.0 m to the right at a constant speed. The string is pulling the box 5.0 m to the right on a frictionless surface with 12 N of force. The string at an angle is pulling the box 5.0 m to the right on a frictionless surface with 12 N of force. . A . B . C . 30° D .

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©Modeling Instruction - AMTA 2013 1 U8 Energy - ws 5 v3.1 Name Date Pd Energy Storage and Transfer Model Worksheet 5: Energy Transfer and Power 1. A student eats a tasty school lunch containing 700 Calories. (One food Calorie = 4186 joules.) Due to basal metabolism, the student radiates about 100 joules per second into the environment. a. How long would the student have to sit on a couch to radiate away all of the energy from lunch? ∆࠵? = ! " = ∆$ !" " = %(’(( *+,)./012 # $%& 3 %0(( # ’ = 4 56( 4(( 7 0(( # ’ = 29 302 ࠵? * 0 8 62(( 9 + = 8.14 ℎ b. If all of the energy from the student’s lunch did something useful, like lifting pianos weighing 5000 newtons to the top of a 10.0-meter tall apartment building, how many pianos could be lifted with the energy from lunch? (Ignore the energy radiated by the student.) Complete the energy bar graph below to aid your solution. Note: E ch category added Energy Conservation Equation: ࠵? :8 ; = ࠵? < = To lift one piano: −∆࠵? :8 = ∆࠵? < = ࠵? < ℎ = (5000 ࠵?)(10 ࠵?) = 50 000 ࠵? ࠵? >?+@A9 = ∆$ ()( ∆$ * ,-%.) = 4 56( 4(( 7 B( ((( # ,-%.) = 58.604 ࠵?࠵?࠵?࠵?࠵?࠵? , i.e., 58 pianos with 0.604(50 000 ࠵?) = 30 200 ࠵? leftover. 2. Jill pulls on a rope to lift a 12.0-kg pail out of a well, while the clumsy Jack watches. For a 10.0- meter segment of the lift, she lifts the bucket straight up at constant speed. How much power is required to complete this task in 5.00 seconds? Complete the energy bar graph as part of your solution. Energy Conservation Equation: ࠵? :8 ; = ࠵? < = ࠵? = ∆࠵? < ∆࠵? = ࠵?࠵?∆ℎ ∆࠵? = (12.0 ࠵?࠵?) C9.81 ࠵? ࠵?࠵? D (10.0 ࠵?) 5.00 ࠵? = 1177.2 ࠵? 5.00 ࠵? = 235 ࠵? Position A Energy (J) 0 Position B Energy (J) 0 E k E g E e E th System/Flow E k E g E el E ch Position A Energy (J) 0 Position B Energy (J) 0 E k E g E e E th System/Flow E k E g E el E ch student pianos Earth Jill pail Earth

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©Modeling Instruction - AMTA 2013 2 U8 Energy - ws 5 v3.1 3. Hulky and Bulky are two workers being considered for a job at the UPS loading dock. Hulky boasts that he can lift a 100 kg box 2.00 meters vertically, in 3.00 seconds. Bulky counters with his claim of lifting a 200 kg box 5.00 meters vertically, in 20.0 seconds. Which worker is more powerful? Hulky ࠵? C = ∆$ / ∆D = E<∆8 ∆D = (0(( F<).5.10 0 1/ 3(4.(( E) 6.(( 9 = 0524 7 6.(( 9 = 654 ࠵? Bulky ࠵? = = ∆$ / ∆D = E<∆8 ∆D = (4(( F<).5.10 0 1/ 3(B.(( E) 4(.( 9 = 510( 7 4(.( 9 = 491 ࠵? Hulky is more powerful. 4. The trains on the Boss rollercoaster are raised from 10.0 m above ground at the loading platform to a height of 60.0 m at the top of the first hill in 45.0 s. Assume that the train (including passengers) has a mass of 2500 kg. Ignoring frictional losses, how powerful should the motor be to accomplish this task? Complete the energy bar graphs below. Energy Conservation Equation: ࠵? = ∆࠵? < ࠵? EADAH = I ∆D = ∆$ / ∆D = E<∆8 ∆D = (4B(( F<).5.10 0 1/ 3(2(.(%0(.()J /B.( 9 = 0 442 4B( 7 /B.( 9 = 27 250 ࠵? = 27.3 kW Position A Energy (J) 0 Position B Energy (J) 0 E k E g E e E th System/Flow E k E g E el E ch W trains Earth motor

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©Modeling Instruction - AMTA 2013 3 U8 Energy - ws 5 v3.1 5. a. An aerodynamic 1000-kg car takes about 270 newtons of force to maintain a speed of 25.0 m/s. How much horsepower is required from the engine to maintain this speed? (1 hp = 746 W) The units of power are W = K L = N J L … ࠵? = ࠵? ∆࠵? = ࠵? M ∆࠵? ∆࠵? = ࠵? M ࠵?̅ M = (270 N) *25.0 m s + = 6750 J s = 6750 W C 1 hp 746 W D = 9.05 hp b. How much horsepower is required for the same car to accelerate from 0-25 m/s in 6 seconds? ࠵? = ࠵? ∆࠵? = ∆࠵? F ∆࠵? = 1 2 ࠵?R࠵? N 4 − ࠵? ? 4 S ∆࠵? = 1 2 (1000 kg) U*25 m s + 4 − *0 m s + 4 V 6.00 s = 312 500 J 6.00 s = 52 083 W ࠵? = 52 083 W C 1 hp 746 W D = 69.8 hp 6. Your electric utility company sends you a monthly bill informing you of the number of kilowatt- hours of energy you have used that month. a. What is a kilowatt-hour (kilowatt x hour, or kWh)? Determine how many Joules equal one kilowatt-hour. 1 ࠵?࠵?ℎ = (1000 ࠵?)(1 ℎ) = *1000 7 9 + (3600 ࠵?) = 3 600 000 ࠵? b. A frost free, 17 cu. ft. refrigerator-freezer uses energy at a rate of 500 watts when you hear the compressor running. If the fridge runs for 200 hours per month, how many kilowatt-hours of energy does the refrigerator use each month? ∆࠵? = ࠵?∆࠵? = (500 ࠵?)(200 h) = (100 000 Wh) = 100 kWh per month c. In the Phoenix area, electricity rates range from 8.0 cents per kilowatt-hour (winter) to 11.5 cents per kWh (summer). How much does the energy cost each month to run the refrigerator? Cost = rate ´ quantity Winter: ࠵?࠵?࠵?࠵? = *8.0 ¢ FI8 + *100 FI8 EA@D8 + = 800 ¢ EA@D8 = 8.00 $ EA@D8 Summer: ࠵?࠵?࠵?࠵? = *11.5 ¢ FI8 + *100 FI8 EA@D8 + = 1150 ¢ EA@D8 = 11.50 $ EA@D8

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©Modeling Instruction – AMTA 2013 1 U8 Energy – review v3.1 Name Date Pd Energy Storage and Transfer Model: Review Sheet 1. Three balls are rolled down three tracks starting from rest at the point marked “start.” a. Describe the acceleration of the ball traveling on track A. The acceleration of the ball is low at first, and gradually increases as it moves to the end of the track. b. Describe the acceleration of the ball traveling on track B. The ball undergoes constant acceleration. c. Describe the acceleration of the ball traveling on track C. At first, the acceleration of the ball is greater than that for the other two balls, but it gradually decreases as it moves to the end of the track. d. Describe the velocity of the ball traveling on track A. The velocity of the ball increases at an increasing rate. e. Describe the velocity of the ball traveling on track B. The velocity of the ball increases at a constant rate. f. Describe the velocity of the ball traveling on track C. The velocity of the ball increases rapidly at first, then at a decreasing rate. g. Rank the time needed for the balls to travel from start to finish. Explain your ranking. shortest: C B A :longest The displacement of each ball is the area under a v-t graph. Because of its greater initial acceleration, the curve for C is above that of B until the end. The area under C reaches D x sooner than for B. A similar argument holds for A . start finish A B C mid-height mid-range v t B C A same final v

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©Modeling Instruction – AMTA 2013 2 U8 Energy – review v3.1 h. Rank the instantaneous velocities of the balls at the mid-height line. Explain your ranking. fastest: A = B = C _ :slowest Each ball has the same decrease in E g which, in turns means that have the same increase in E k . They will have the same velocity at the mid-height line. i. Rank the instantaneous velocities of the balls at the mid-range line. Explain your ranking. fastest: C B A_ : slowest Due to differences in the D h for each track at the mid-range, the D E g is greatest for C and least for A. Hence the velocity would be greatest for C and least for A. j. Rank the instantaneous velocities of the balls at the finish point. Explain your ranking. fastest: _ A = B = C __:slowest Each ball has the same decrease in E g which, in turns means that have the same increase in E k . They will have the same velocity at the end. k. If the start is 1.0 m higher than the finish, determine the heights at which A, B, and C will have half of their final kinetic energy. The change in kinetic energy is a function of the initial and final positions, not the path. All three balls have half of their final kinetic energy at 0.5 m above the finish. l. If the start is 1.0 m higher than the finish, determine the heights at which A, B, and C will have half of their final velocity. Since E k µ v 2 , the balls have ¼ of their final E k when they are moving at ½ of their final velocity. They will have dropped ¼ of the total D h or 0.25m. They will be 0.75 m above the finish. 2. A baseball (m = 140 g) traveling at 30 m/s moves a fielder's glove backward 35 cm when the ball is caught. a. Construct an energy bar graph of the situation, with only the ball and Earth in the system. b. How large is the average force exerted by the ball on the glove? The average F (180N) acts in the opposite direction as the displacement. E i + W = E f Þ 1 2 mv 2 + F D x = 0 1 2 0.14 kg ( ) 900 m 2 s 2 = - F 0.35 m ( ) - 63 J 0.35 m = F = - 180 N Energy (J) 0 Energy (J) 0 System/Flow Iniital Final E k E g E el E k E g E el E th ball Earth glove

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©Modeling Instruction – AMTA 2013 3 U8 Energy – review v3.1 3. A spring whose spring constant is 850 N/m is compressed 0.40 m. What is the maximum speed it can give to a 500 g ball? 4. A bullet with a mass of ten grams is fired from a rifle with a barrel that is 85 cm long. a. Do an energy bar graph analysis of the situation. ° b. Assuming that the force exerted by the expanding gas to be a constant 5500 N, what speed would the bullet reach? E i + W = E f 1 2 k x 2 = 1 2 mv 2 Þ v = k x 2 m = 850 N m 0.40 m ( ) 2 0.500 kg v = 272 m 2 s 2 = 16.5 m s E i + W = E f 0 + F D x = 1 2 mv 2 Þ v = 2 F D x m = 2 5500 N ( ) 0.85 m ( ) 0.010 kg = 140 m s Energy (J) 0 Energy (J) 0 System/Flow Iniital Final E k E g E el E k E g E el E th bullet expanding gases

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©Modeling Instruction – AMTA 2013 4 U8 Energy – review v3.1 5. A 24 kg child descends a 5.0 m high slide and reaches the ground with a speed of 2.8 m/s. ° a. Do a bar graph analysis for this situation. b. How much energy was transferred to the thermal account due to friction in the process? The bar graph is not to scale as less than 10% of the initial E g ends up as E k . 6. A 1500 kg car is traveling at 20 m/s. a. Calculate the E k of the car relative to the road. b. If the average braking force applied to the car is 6000 N, how far would it travel before it came to a stop? (Draw an energy bar graph of the situation.) ° c. If this same average braking force were applied to the car moving at twice the speed, what would be the stopping distance? If the initial E k is 4 ´ as great, the stopping distance will also be 4 ´ as far. E i + W = E f Þ mgh + 0 = 1 2 mv 2 + E th 24 kg 10 N kg ( ) 5.0 m ( ) - 12 kg 2.8 m s ( ) 2 = E th 1200 J - 94 J = E th = 1110 J E k = 1 2 mv 2 = 750 kg 400 m 2 s 2 ( ) = 300 kJ E i + W = E f Þ 300 kJ + 0 = 6,000 N ( ) D x D x = 300,000 J 6000 N = 50 m E k µ v 2 Þ 4 E k µ 2 v ( ) 2 Energy (J) 0 Energy (J) 0 System/Flow Iniital Final E k E g E el E k E g E el E th Energy (J) 0 Energy (J) 0 System/Flow Iniital Final E k E g E el E k E g E el E th

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©Modeling Instruction – AMTA 2013 1 U8 Energy - ws 6 v3.1 Name Date Pd Energy Storage and Transfer Model Worksheet 6: U.S. Energy Consumption and Supply 1. Looking at the table from the U.S. Energy Information Administration website, a. Which energy sources could be increased to meet our energy needs? b. Which energy sources will be forced to decline during your lifetime? Why? c. Conventional hydroelectric and geothermal energy sources are essentially “maxed out” in the U.S. Why? d. How does energy conservation (using energy efficiently) impact these numbers?

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©Modeling Instruction – AMTA 2013 2 U8 Energy - ws 6 v3.1 2. Coal will be an essential source in the years to come. The U.S. has huge, but finite, coal reserves. Name several of the major drawbacks to extensive coal use: 3. Petroleum Humans have really figured out how to use petroleum, but it won’t last long at the rate we’re using it. Here’s a dramatic way of diagramming petroleum use. The total area of the rectangles represents the estimated total amount of oil past and present on earth. Each square represents the amount of oil used during the indicated time period. In the 1970’s, it was predicted that we would exhaust our oil supplies by 2000 because of the previous two decades of 7% annual increases in oil consumption. Oil use pre-1953 Oil use 1963-73 (7.8% growth rate) Oil use 1985-95 (had the 7% growth rate continued.) Oil use 1953-63 (7% growth rate) Oil use 1975-85 (had the 7% growth rate continued.) In fact, the oil crisis of the 1970’s helped to curb the growth rate in oil consumption: Oil use pre-1953 Oil use 1963-73 (7.8% growth rate) Oil use 2000-2035 (growth rate ~ 2.2%) Oil use 1953-63 (7% growth rate) Oil use 1973-2000 (3% growth rate) Oil won’t run out in 2035, however, because consumption will not remain high and then suddenly stop when oil runs out. Instead, consumption will be forced to decline as oil becomes scarce (and therefore expensive). The 2006 growth rate in petroleum consumption declined to 0.7%. 1 A graph of the quantity of a resource vs. time is a bell-shaped graph called a Hubbert curve. It is estimated that the world has consumed 1 trillion barrels of oil in our entire history of use. If 1 trillion barrels of oil remain (as most researchers 2 estimate), where are we on the Hubbert curve? (Sketch a graph below.) 1 BP Statistical Review of World Energy 2007 2 for example: Colin Campbell and Jean Laherrere, 1998 Scientific American article; Ken Deffeyes, Hubbert’s Peak, 2001 now oil usage time

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©Modeling Instruction – AMTA 2013 3 U8 Energy - ws 6 v3.1 4. Biomass is both an energy source and a food source. The fundamental conflict is that as population increases, more of the earth’s surface must be directed toward food production, and to efficiently produce food, modern agriculture requires 80 gallons of gasoline or its equivalent [in the form of fertilizers and pesticides] to produce each acre of corn 3 . (1 gallon of gasoline = 1.3 x 10 8 Joules) a. Calculate the fossil fuel energy input to produce an acre of corn: ࠵? !"#$% = #80 &’(()*+ )- &’+)(.*/ ’01/ )- 0)1* & # 2.4×26 ! 7 2 &’(()* )- 0)1* & = 1.04 × 10 26 J American corn production averages 130 bushels/acre. There are 25 kg of corn per bushel and 85 Calories per 77 grams of corn. (1 Calorie = 4186 J) b. Calculate the food energy output from one acre of corn. ࠵? 8$%#$% = #130 9:+;/(+ ’01/ )- 0)1* & # <= >& 9:+;/( & # 2666 & 2 >& & # ?= @’( AA & & # B2?C 7 2 @’( & = 1.50 × 10 26 J c. Therefore, the quip is made that modern agriculture is the process of turning petroleum into food. The problem is much worse when people eat beef rather than cornbread. Each pound of beef requires sixteen pounds of corn feed. Burroughs always has a variety of meats available at lunch. How can this be reconciled with our school philosophy? Eating less meet would be a way to do our part and reduce fossil fuel consumption. Energy use is less efficient in the production of meet than in the production of vegetables, fruits, plants, and herbs. 5. Windpower 4 Wind energy produced 19.5 billion kWh of electricity in 2008. However, that is still less than 1% of U.S. electricity generation. By contrast, the total amount of electricity that could potentially be generated from wind in the United States has been estimated at 10,777 billion kWh annually—three times the electricity generated in the U.S. today. So you want to build a wind farm . . . would you make money? A 50-MegaWatt wind farm has the capability to produce 50 million Joules per second, but would average 35% of full production capacity because of wind variations. Construction would cost about $50 million ($1 million per MW). 3 David Pimintel et al ., “Food Production and the Energy Crisis,” Science 182 , 448 (Nov. 2, 1973) 4 all data from the American Wind Energy Association, www.awea.org Installed US wind capacity as of Jan 09 (in Megawatts)

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©Modeling Instruction – AMTA 2013 4 U8 Energy - ws 6 v3.1 a. Running at 35% of full capacity, how much energy (in kWh) would the wind farm produce each year? ࠵? 8$%#$% = ࠵?࠵?∆࠵? = 35% 5(50 × 10 C W) # D 2666 &9 5(1 year) # 4C=.<= E’F+ 2 F/’1 & # <B ; 2 E’F &9 = 1.53 × 10 ? kWh b. If you can sell the energy for 4 cents per kWh, what is your annual gross revenue? ࠵?࠵?࠵? = # $ 6.6B >H; & (1.53 × 10 ? kWh) = $ 6.14 × 10 C = $ 6.14 million c. Which does wind energy provide a limited supply of, energy or power? What implication does this have for our energy consumption? The energy is renewable, but the power (average energy output per second) is limited. This means that the annual consumption of energy must remain below the annual energy output so that demand doesn’t exceed capacity. 6. Solar BP Solar advertises that “a 600 sq. ft. array [of photovoltaic cells] … rated about 3.3 kW . . . would generate about 3640 kWh per year. Our highest efficiency technology (also the most expensive) would generate about double this. The typical cost for a single residential system is about $10,000 per kW.” 5 a. Using the solar array advertised, determine the cost per kilowatt-hour over a 30-year period of use. ࠵?࠵?࠵?࠵? ࠵?࠵?࠵? ࠵?࠵?ℎ = I8%JK L8M% N "#$%#$ (46 PQJRM) = T26 666 $ ’( U(4.4 >H) T4CB6 ’() *+,- U(46 F/’1+) = $ 44 666 26V <66 >H; = 0.30 $ >H; b. Compare the energy output of this solar array to the energy needed by a water heater using 400 kWh per month. The solar array provides ࠵? = N ./ N 01 = W ./ W 01 = T4CB6 ’() *+,- UT 2 *+,- 24 5678)9 U B66 ’() 5678) = 464 B66 = 75.8% of what is needed by the water heater. 5 http://www.bpsolar.com/ContentDetails.cfm?page=135

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©Modeling Instruction – AMTA 2013 5 U8 Energy - ws 6 v3.1 It could be argued that using passive solar energy might decrease our energy needs by more than what might be gained by solar electrical generation. A water heater, for example, uses 375 – 525 kWh per month. 6 “Solar water heating systems typically reduce water heating costs by 50% to 80% over minimum efficiency electric resistance or gas-fired water heaters.” 7 Common passive solar water heating systems cost $2000. c. Suppose your $2000 solar water preheating system saved you from buying 200 kWh per month over 30 years. What is the effective cost per kWh for the solar-heated water? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵?࠵?࠵? ࠵?࠵?ℎ = ࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵? 8$%#$% (30 ࠵?࠵?࠵?࠵?࠵?) = 2000 $ #200 kWh month & ](30 years) # 12 months 1 year &_ = $ 2000 72 000 kWh = 0.028 $ kWh More on Solar: Despite the abundance of solar energy, solar is tricky to harness for electrical generation. Typical single crystal silicon cells usually average about 14 percent [efficiency]. 8 Solar cells are typically difficult, expensive, and energy intensive to make. I have found very little quantitative information about how much energy it takes to make a photovoltaic cell. (Find this information, and you will be rewarded handsomely.) It is possible that many solar cells never collect as much solar energy as the energy required to make the cell. Non-conventional Hydropower: Tidal energy – Coastal dams could trap water at high tide and then let the water out through a generator at low tide. Undersea turbines could generate electricity as the tide goes in and out. Thermal gradient energy – Uses the temperature difference between the warm surface of the ocean and cold depths of the ocean. Nuclear energy: Fission – energy is released as large atoms decay into smaller, more tightly bound atoms. The decay is induced by the addition of neutrons, which make the atom unstable. The decaying atom releases several neutrons, therefore creating a self-sustaining nuclear reaction. Fusion – Energy is released as small atoms fuse together to make bigger atoms. This is the Sun’s energy source. Although much research has been done with significant success, we are at least 25 to 50 years away from commercial fusion power. Hydrogen is often cited as an energy alternative, but a source of hydrogen is needed. Currently, fossil fuel or solar energy is used to separate water into hydrogen and oxygen. There is hope that through bioengineering a microorganism could be developed to do the job. 6 Ames, Iowa city government, http://www.city.ames.ia.us/ElectricWeb/energyguy/appliances.htm 7 www.energystar.gov 8 Steve Hester, Solar Electric Power Association Technical Director www.solarelectricpower.org

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