Lab 7 - Energy_updated

PHYS 1 – Spring 2023 Spring 2023 Lab 7: ENERGY Fig. 1: Investigating Energy Exchanges - Kinetic Energy and Gravitational Potential Energy

PHYS 1 – Spring 2023 Spring 2023 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 1 – Spring 2023 Spring 2023 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 3465.64 J 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] 5. 6.71 m/s 6. Replace the skateboarder with a Bulldog and repeat steps 2, 3, and 4. [6 Points] a. h = 4.60 m b. gravitational potential energy = 3465.64 Joules c. speed = 6.71 m/s 7. Is the law of conservation of energy affected by the mass of the skater? Yes/ no ? [1 Point] 8. 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 = 3,000 Joules ii. Potential energy = 0 Joules iii. Total energy= 3,000 Joules

PHYS 1 – Spring 2023 Spring 2023 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 What do you conclude about the values of different types of mechanical energies at top and bottom of its path? [2 Points] The total energy value remains the same despite the position however, at the maximum height and the lowest point the kinetic and potential energy values flip. 9. 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 10. What are the energies at 12 seconds and 17 seconds [3 Points] a. at 12 seconds: K = 4000 J U = 1000 J E th =0 J b. at 17 seconds: K = 2500 J U = 2500 J E th = 0 J 11. Calculate the change and total energies between the 12 th and 17 th second. [5 Points] a. ࠵? K = 1500 J ࠵? U = 1500 J ࠵? E th = 0 J b. Total change in Energy: ࠵? E = ࠵? K + ࠵? U + ࠵? E th = 3000 J 12. 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] When the skater is ascending/descending from the two highest points. This would be about mid-way between the highest points of the parabola and the lowest.

PHYS 1 – Spring 2023 Spring 2023 b. The skaters speed at point 4: v 4 = 9.31 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 = 4125 J K 1 = 0 J E 1 = 4100 J b. At point 2: U 2 = 0 J K 2 = 4125 J E 2 = 4100 J c. At point 3: U 3 = 2200 J K 3 = 1825 J E 3 = 4100 J d. At point 4: U 4 = 775 J K 4 = 3250 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] a. h 1 = 5.61 m, h 2 = 0 m, h 3 = 3.00 m, h 4 = 1.05 m, h 5 = 5.95 m. 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 same trend of the shape of the track, at its highest point potential energy is also at its highest and vice versa. b. Kinetic energy to the track: Kinetic energy was opposite to the shape of the track. At its highest point, kinetic energy was at its lowest and vice versa. 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]

PHYS 1 – Spring 2023 Spring 2023 The shape of the energies will not change but the total energy will because of the significant decrease in the gravitational pull on the moon. 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 = 243.90 J U = 3050 J E th = 1631.42 J b. at the 16 th second: K = 1316.50 J U = 1124 J E th = 2477 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. ࠵? K = 1072.6 J ࠵? U = -1926 J ࠵? E th = 845.58J b. Total change in Energy: ࠵? E = ࠵? K + ࠵? U + ࠵? E th = -7.82 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] 4 seconds: K=13.13J, U=3973.03J, Eth=927.91J Total energy: 4914.07J When potential energy as highest, kinetic energy is at its lowest so with a much higher potential energy value, the kinetic energy will be significantly smaller. The total energy supports the fact that energy cannot be created nor destroyed, only transferred. 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] At 6 seconds these relations continue to hold true. This is because of the law of conservation of energy which states that energy cannot be created nor destroyed, only transferred. Even though the values have changed, the total energy remains the same. 12. Why do you think thermal energy just keeps on rising? [1 Point] As the skater continues, friction also builds which would result in an increase of thermal energy.

PHYS 1 – Spring 2023 Spring 2023 Velocities: Top of loop = 8.45 m/s Bottom of loop = 53.59 m/s End of track = 52.13 m/s