Lab 7 - Energy_updated (1)

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University of North Carolina, Chapel Hill *

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2101

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Apr 3, 2024

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PHYS 2101 – Fall 2022 Fall 2022 ENERGY Fig. 1: Investigating Energy Exchanges - Kinetic Energy and Gravitational Potential Energy

PHYS 2101 – Fall 2022 Fall 2022 Objectives: At the end of this lab, students should be able to 1) calculate different forms of mechanical energy, and 2) verify the law of conservation of energy. 1. Introduction: The law of conservation of energy states that the total amount of energy in an isolated system remains constant. In other words, Energy can neither be created nor destroyed but can change its forms. The total energy E of a system (the sum of its mechanical energy and its internal energies, including thermal energy) can change only by the amounts of energy that are transferred to or from the system. If work W is done on the system, then W = ΔE = ΔE mech + ΔE th + ΔE int (7.1) If the system is isolated (i.e., W = 0): ΔE mech + ΔE th + ΔE int = 0 (7.2) The skate-park is an excellent example of the conservation of energy. For the isolated skate-track- Earth system, the law of conservation of energy equation has the form ΔE mech + ΔE th = 0 (7.3) The unit of Energy is Joules (J ) in the SI system. 1.1 Mechanical Energy : The mechanical energy E mec h of a system is the sum of its kinetic energy K and its potential energy, U; E mech = K + U (7.4) The conservation of mechanical energy can be written as; ΔE mech = ΔK + ΔU = 0. (7.5) It can also be rewritten as K 1 + U 1 = K 2 + U 2 (7.6) where the subscript refers to different instants during an energy transfer process. 1.2 Gravitational Potential Energy : The potential energy associated with a system consisting of Earth and a nearby particle is gravitational potential energy. If the particle moves from y 1 to height y 2 , the change in gravitational potential energy of the particle-Earth system is ΔU = mg (y 2 – y 1 ) = mgΔy. (7.7) 1.3 Kinetic Energy : The kinetic energy is associated with the state of motion of an object. If an object changes its speed from v 1 to v 2 , the change in kinetic energy is ΔK = K 2 – K 1 = ½ mv 2 2 - ½ mv 1 2 (7.8)

PHYS 2101 – Fall 2022 Fall 2022 2. Procedure Please click here ( https://excelschools.net/en/simulation/energy-skate-park.html ) to download the PhET simulation link for Energy Skate Park . Take some time and play with the skater and his track . It helps to practice with the following features and controls. Track selector : click on ‘ Tracks’ located on the top right (Next to ‘File’) and select from the drop- down menu. For example, the “ Double well (Roller Coaster) ” shown above. Reset (located on the top right): This rests the simulation to default values and sets the track to friction parabola track. Skater selector : clicking on ‘ Choose skater…’ will allow you to choose a skateboarder with a different mass option. Measuring Tape: Check the ‘ Measuring Tape Bo x ’ when you want to make measurements. Drag the left end of the tape measure to where you start your measurement and then drag the right end to the final location. To make a reference horizontal line to your measurement, check the potential energy reference box and drag the blue line you see on the screen to the initial position. Graph Selector: If you would like to observe graphs that depict the relationships among potential, kinetic, and thermal energy of the simulation, click the buttons under the Energy Graphs. The types of graphs are shown above. You can also add pie graphs by checking the show pie chart box. These graphs can be shown with or without Thermal energy . Gravity: you may change the gravitational force by changing the location or the sliding bar underneath the Gravity box. Additional Features : Clicking the C lear Heat makes the track frictionless. You can also edit the track friction and the skater mass using the Track friction and Edit Skater buttons. You can also control the speed of the skater using the slide bar under the screen. Whenever you are done playing make sure you reset the simulation and start the steps below: Part I: Parabolic Track 1. Click Reset and click on ‘Bar Graph’ , ‘Energy vs. Position’ , and ‘Energy vs. Time’ . Observe the energy bars as the skater moves back and forth. As the skater descends, his kinetic energy (green) Increases and his potential energy (blue) Decreases . The ‘Total’ energy bar Stays the same . [3 Points] 2. Check the ‘ Measuring Tape box’ (please refer procedure section about how to use measuring tape). Considering the bottom of the parabola as a reference line, measure the maximum height ( h ) the skater climbs h = 3.93m [1 Point]

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PHYS 2101 – Fall 2022 Fall 2022 3. Observe the ‘Energy vs. time graph’ as the skater moves back and forth. The gravitational potential energy at the maximum height to which the skater reaches is equal to 3950.32 Joules. [1 Point] 4. Apply the principle of conservation of mechanical energy to calculate the skater’s speed at the lowest point of the parabola. Show your work to get full credit. [4 Points] Ke=1/2mv^2 2843.57*2=5687.14 5687.14/75kg= 75.8285 Sqrt(758285)= 8.707 8.707m/s 5. Replace the skateboarder with a Bulldog and repeat steps 2, 3, and 4. [6 Points] a. h = 4 m b. gravitational potential energy = 1053.92 Joules c. speed = 8.658 m/s 6. Is the law of conservation of energy affected by the mass of the skater? Yes/no? [1 Point] no 7. Now click Reset . Check the ‘potential Energy Reference’ box. Drag the reference line at the lowest point of the parabola and observe the ‘ Energy versus Position graph ’ as the skater moves back and forth. a. Pause the simulation at the bottom of the parabola and record the following values: [3 Points] i. Kinetic energy = 3000 Joules ii. Potential energy = 0 Joules iii. Total energy= 3000 Joules b. Play the simulation and then pause it at the maximum height. Record the following: [3 Points] i. Kinetic energy = 0 Joules ii. Potential energy = 3000 Joules iii. Total energy= 3000 Joules

PHYS 2101 – Fall 2022 Fall 2022 What do you conclude about the values of different types of mechanical energies at top and bottom of its path? [2 Points] They are inversely related so as one goes up the other one goes down. 8. Apply the following settings: a. Stop the simulation. b. Click ‘ Reset ’ then ‘ Return to skater ’ buttons. c. Click the ‘Track friction’ and adjust the coefficient of friction to the one-eighth mark on the slide. d. Open the ‘ Energy versus Time ’ graph and click on the ‘Stop’ button located in the graph. e. Drag the skater to the top of the parabola and release him. Press ‘ Go ’. f. Run the simulation for 20 seconds 9. What are the energies at 12 seconds and 17 seconds [3 Points] a. at 12 seconds: K = 0.01 J U = 2315.72 J E th = 1638.71 J b. at 17 seconds: K = 0.43 J U = 1878.48 J E th = 2073.53 J 10. Calculate the change and total energies between the 12 th and 17 th second. [5 Points] a. 𝛥 K = .42 J 𝛥 U = - 437.24 J 𝛥 E th = 434.82 J b. Total change in Energy: 𝛥 E = 𝛥 K + 𝛥 U + 𝛥 E th = -2 J 11. What are the two times when the KE and PE of the skater are the same? Where do you think the skater is along its path when this happens? [4 Points] Roughly 3 and 5 seconds The skater is halfway up or down to his maximum height or the bottom of the parabola. Part II: Double Well (Roller Coaster)

PHYS 2101 – Fall 2022 Fall 2022 Fig. 2 1. Click the ‘ Reset’ and ‘ Return Skater’ buttons. From ‘ Tracks ’ select ‘ Double Well (Roller Coaster) ’ . Check the ‘Potential Energy Reference’ box and drag the reference line to the position as shown in Figure 2 above. Measure the height of each control point (1,2,3,4 and 5 in the figure above) from the reference line. Calculate the potential (U), kinetic (K), and total (E) energy of the skater at these points (assuming the track is frictionless). Show your calculation for (a) and (c) only . [6 Points] a. At point 1: h 1 = 6.03 m. U 1 = 4432.05 J K 1 = 0 J E 1 = 4432.05 J U= 6.03*75*9.8= 4432.05J K= ½*75*0 = 0J b. At point 2: h 2 = 0 m. U 2 = 0 J K 2 = 4432.05 J E 2 = 4432.05 J c. At point 3: h 3 = 2.95 m. U 3 = 2168.25 J K 3 = 2263.8 J E 3 = 4432.05 J d. At point 4: h 4 = 1.17 m. U 4 = 859.95 J K 4 = 3572.1 J E 4 = 4432.05 J U=1.17*75*9.8 = 859.95J K = 4432.05-859.95= 3572.1J 2. Calculate the speeds at control points 3 and 4 using the kinetic energy you calculated in the previous step. Show your work. [4 Points] (2263.8*2)/75=60.368

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PHYS 2101 – Fall 2022 Fall 2022 Sqrt(60.368) = 7.76968m/s (3572.1*2)/75= 95.256 Sqrt(95.256) = 9.7599m/s a. The skaters speed at point 3: v 3 = 7.76968 m/s b. The skaters speed at point 4: v 4 = 9.7599 m/s 3. Now open the Energy vs Position Graph and read the potential (U), kinetic(K), and total (E) energies at the control points [4 Points] a. At point 1: U 1 = 4100 J K 1 = 0 J E 1 = 4100 J b. At point 2: U 2 = 0 J K 2 = 4100 J E 2 = 4100 J c. At point 3: U 3 = 2200 J K 3 = 1900 J E 3 = 4100 J d. At point 4: U 4 = 3250 J K 4 = 850 J E 4 = 4100 J 4. Calculate the heights at each of the control points using the information from step 3. Show your work. [5 Points] Mgh= U U/mg = h 4100/75*9.8 = 5.578m 0/75*9.8 = 0m 2200/75*9.8=2.993 3250/75*9.8 = 4.422 a. h 1 = 5.578 m, h 2 = 0 m, h 3 = 2.993 m, h 4 = 4.422 m, h 5 = 5.578 m.

PHYS 2101 – Fall 2022 Fall 2022 5. How does the shape of potential and kinetic energies relate to the shape of the track? [2 Points] a. Potential energy to the track: Potential energy follows the shape of the track b. Kinetic energy to the track: Is the inverse of the track 6. If you change the location to the Moon instead of Earth , will the shape of the energies change? If not, what will change if any? [2 Points] Yes it flattens the shape of the graph because the drecrease in gravity leads to a decrease in potential and kinetic energy. 7. Apply the following settings for the simulation to answer the proceeded questions. a. Stop the simulation. b. Click ‘Reset’ followed by the ‘ Return skater ’ button. Choose the ‘ Double Well Roller Coaster’ track. c. Adjust the coefficient of friction to the one-eighth mark on the slide d. Open the Energy versus Time Graph . Click on “ Stop ’ and then “ Clear ” button on the graph. e. Drag the skater to the top of the roller coaster and then press on ‘ Go !’. f. Run the simulation for 20 seconds 8. What are the energies at 9 seconds and 16 seconds? [4 Points] a. at the 9 th second: K = 905.69 J U = 1895.62 J E th = 1956.01 J b. at the 16 th second: K = 611.65 J U = 1130.96 J E th = 3014.71 J 9. Calculate the change and total energies between the 9 th and 16 th -second timestamp. E th is thermal energy [4 Points] a. 𝛥 K = -294.04 J 𝛥 U = -764.66 J 𝛥 E th = 1058.7 J b. Total change in Energy: 𝛥 E = 𝛥 K + 𝛥 U + 𝛥 E th = 0 J 10. Pick any time between 1 and 9 seconds and record the values of K, U, E th , and Total E . Describe how they are related. [2 Points] K 2.39J U 3821.19 Eth 933.74 E 4757.32 The relation I see is that as potential energy increases kinetic energy decreases and vice versa. I also see that the total energy minus kinetic plus potential gives you friction. 11. Pick any time other than that you chose for Step 10 and repeat the process. Do the relations you found in 10 still hold good? Why/why not does this relation hold/ not hold? [2 Points] Yes because U and K are still inverse and the total minus K and U still equals frictiom 12. Why do you think thermal energy just keeps on rising? [1 Point]

PHYS 2101 – Fall 2022 Fall 2022 Be cause friction causes the rider to lose energy continuously till they stop moving Part III: Create your own Track [where a skater can make a loop (a sample figure below) and then flies off] (8 Points) You can keep adding extra tracks by dragging “Tracks” located on the top right inside the blue rectangle to the screen. Connect them by connecting the red circles. Take a screenshot of your track and paste it below. For the track, you created, calculate the velocity at the top and bottom of the loop and the speed by which the skater flies off from the edge. Show all the measurements and calculations you performed to get full credit.

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PHYS 2101 – Fall 2022 Fall 2022

Lab 7 - Energy_updated (1)

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