PHYS182A_195L-WorkEnergyTheorem_.docx

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PHYSICS 182A/195L LAB REPORT - LAB 9: Work Energy Theorem Lab 9: Work Energy Theorem San Diego State University Department of Physics Physics 182A/195L TA: Lab partner 1: Ariana Gomes Lab partner 2: Karylle Querimit, Sarah Valenzuela Date: 10/31/23 Score: Data has been entered in blue. Theory Work is the change in energy of a system (one or more objects with mass) as the result of a force applied to that system. How exactly is force related to energy? Let’s think of an example. Recall that if a mass is held at a height h above the earth, then it has a potential energy . In order to move an object from the ground ( ) to a height , then it must have at least force acting on it while it moves from to (otherwise gravity would pull it back down). If we simply replace with in the expression for energy, we get: That is, the potential energy of the object is equal to the force we had to apply to it, times the distance , that it moved while the force was applied. More generally, any constant force applied to an object over some distance will change the energy of that object: This change in energy could be potential energy or kinetic energy, depending on the circumstances. A change in energy of a system due to a force is called Work. 1 Department of Physics

What if the force is not constant? A situation with non-constant force requires calculus. For a force that changes with , the work done by that force is This is one way to write the work-energy theorem , which tells us exactly how to compute the work done on a system by a changing force . Constant force Let’s use the work-energy theorem to calculate the work done on a cart with mass by a constant force. To generate a constant force, we will use a hanging mass attached to the cart with a pulley system. This way, a known hanging mass can be used to calculate the force applied to the system. Note that the force acting on the cart is not , it is the tension in the string. Using the free-body diagram above, and the fact that both masses must have the same acceleration, we can determine that the tension acting on the cart is: (For a detailed derivation, see the appendix.) Therefore, the work done on the cart by the pulley system is : Here, is the distance that the cart has moved. (It is also the distance that the hanging mass has fallen. Why?). Linear force Next, we’ll use the work-energy theorem to calculate the work done on a cart with mass by a linear force . To generate a force which changes linearly with x, we use a simple spring, which obeys Hooke's law:

PHYSICS 182A/195L LAB REPORT - LAB 9: Work Energy Theorem Here, is the spring constant and is the distance the spring is stretched from equilibrium. Suppose that the cart starts from some initial displacement , where the spring is taut. The cart is then released, so that the spring pulls it to the right. The force applied by the spring is changing linearly, since as the cart changes it’s x-position, the length of the spring changes and so does the force it applies. The total work done on the cart by the spring from the carts initial position to its final position is given by the work-energy theorem: This integral can be computed with the power-rule from calculus: An integral of a function is nothing but the area under the curve. Therefore, the work done by the force can also be seen as the area under the curve. In this case, the curve is a linear function with slope . The area is just the area of a triangle with base length and height , so 3 Department of Physics

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Procedure Setup 1. Use the Balance Scale to measure the mass of the Smart Cart alone, the Smart Cart + the four 250g Stackable Masses. Record these values in Table A.1. 2. Connect the Wireless Smart Cart via Bluetooth under Hardware Setup in Pasco Capstone TM . 3. Lay the cart flat on the table and make sure only the hook is attached. Go back to hardware setup and click on the smart cart force sensor settings (blue gear). Click “zero sensor now” to zero the force sensor. Part A: Constant Force 1. Set up the equipment as shown in Figure A.1 , but place magnetic End-Stops on both ends of the track. There is a hole in the End-Stop that you can thread the string through during the experiment. Put the 4 stackable 250g masses on the cart. 2. Cut about 1 meter of string. Tie one end of the string to the hook of the cart. Thread the string through the End-Stop and tie the other end to a Mass Hanger. 3. Hang the mass hanger over the pulley and place 45g on the mass hanger. Record the total hanging mass (hanger + weights) in Table A.1. Figure A.1: The cart is placed in the center of a level track, attached to a pulley system . 4. Roll the cart as far back as you can before the mass hanger touches the pulley. Hold it in place. 5. Measure the displacement of the cart from the end of the track (pulley end) using the measuring tape on the track. Record this value in Table A.1. 6. Wait for the hanging mass to stop swinging, then hit START to begin recording. 7. Release the cart to roll freely down the track, and then catch the cart before it rolls into the End-Stops. Hit STOP to stop recording. 8. Use the “Highlight tool” to select only the data points recorded when the cart was in motion, and which have positive force values. 9. Record the area under the Force vs. Position curve. using the “Display Area tool” to calculate the area. Record the value in Table A.2. 10. Determine the average force using the “Mean tool”. Record the value in Table A.2. 11. Use the “Coordinate tool” to find the final velocity of the cart on the Velocity vs. Position graph. (This should be the peak of the velocity graph.) Record the value in Table A.2.

PHYSICS 182A/195L LAB REPORT - LAB 9: Work Energy Theorem Part B: Linear Force 1. Starting from the setup from Part A , remove the pulley, mass hanger, and the string. 2. Take a stiff long spring, and attach one end to the End-Stop and the other end to the hook of the cart. 3. Note the position of the cart when the spring is in equilibrium. 4. Now pull the cart backwards, allowing the spring to stretch. Move the cart away about 20 cm from equilibrium. 5. Measure the exact displacement of the cart from equilibrium using the measuring tape on the track. Record this value in Table B.1. 6. Hit START to begin recording. 7. Release the cart to roll freely down the track, and then catch the cart before it changes directions. Hit STOP to stop recording. 8. Use the “Highlight tool” to select only the data points recorded when the cart was in motion, and which have positive force values. 9. Record the area under the Force vs. Position curve, using the “Display Area tool” to calculate the area. Record the value in Table B.2. 10. Determine the average force using the “Mean tool”. Record the value in Table B.2. 11. Use the linear curve fit tool to find the slope of the force curve. Record the value in Table B.2. 12. Use the “Coordinate tool” to find the final velocity of the cart on the Velocity vs. Position graph. (This should be the peak of the velocity graph.) Record the value in Table B.2. Data Table A.1: Masses and Displacement Smart Cart Mass (kg) 0.252 kg Smart Cart + Four Stackable Masses (kg) 1.252 kg Hanging Mass (kg) 0.045 kg Displacement ( ) (m) 0.82 m Table A.2: Constant Force Variables Area under constant force curve (N*m) 0.32 N* m Average Force (N) 0.57 N Final Velocity (m/s) 0.640 m/s 5 Department of Physics

Table B.1: Cart Displacement Displacement ( )(m) -0.20 m Table B.2: Linear Force Variables Area under linear force curve (N*m) 0.16 N* m Average Force (N) 1.56 N Slope (N/m) -9.48 N/m Final Velocity (m/s) 0.472 m/s Analysis Part A: Constant Force 1. Use your area measurement in Table A.2 to find the work done on the cart by the pulley system. 0.32 N*m 2. What do you think would happen if the cart was being pulled on both sides by equal masses? If the cart was being pulled on both sides by equal masses the cart would stay in equilibrium. 3. Calculate the work done on the Smart Cart by the pulley system using the equation derived in the Theory section (show your work): W = ((0.045 kg)(1.252 kg)) / (0.045 kg + 1.252 kg) x (9.8 m/s^2)(0.82m) = 0.349 J 4. Calculate the percent error of the value of work you found in question 1 compared to the value predicted by the theory, which you found in question 3: % error = (0.32 N*m - 0.349 N*m) / (0.349 N*m) x 100 = 8.31 % 5. Using the final velocity given in Table A.2, calculate the final kinetic energy of the Smart Cart in Part A (show your work):

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PHYSICS 182A/195L LAB REPORT - LAB 9: Work Energy Theorem KE = 1/2 (1.252 kg + 0.045 kg) (0.640 m/s)^2 = 0.266 J 6. Compare the value of the final kinetic energy you calculated in question 5 to the work done on the cart calculated in question 3. Explain your findings. The work done on the cart in question 3 is larger in energy than the final kinetic energy in question 5. We came to this conclusion because the cart was slowing down so the final kinetic energy was less than the work done (given that work done is equal to the change in kinetic energy). Part B: Linear Force 1. Given the fact that the force from the spring is linearly proportional to the displacement of the spring, which of the measured values in Table B.2 represents the spring constant? Explain your answer. The measured value that represents the spring constant is the slope because it obeys hooke's law, the slope is force / displacement. 2. What is the value of the spring constant ? -9.48 N/m 3. Calculate the work done on the Smart Cart by the spring using the equation derived in the Theory section (show your work): W = ½(-9.48 N/m) (-0.20)^2 = -0.190 N*m 4. Calculate the percent error of the “Area under linear force curve” from Table B.2 compared to the value predicted by theory in the previous question. % error = (0.16 N*m -0.190 N*m / (0.190 N*m) x 100 = 15.8 % 5. Using the final velocity given in Table B.2, calculate the final kinetic energy of the Smart Cart in Part B (show your work): KE = 1/2 (1.252 kg ) (0.472 m/s)^2 = 0.139 J 6. Compare the value of the final kinetic energy you calculated in question 5 to the work done on the Smart Cart calculated in question 3. Explain your findings. The work done on the cart in question 3 is smaller in energy than the final kinetic energy in question 5. We came to this conclusion because as the cart passes equilibrium the change in final kinetic energy became larger and therefore the change in kinetic energy is larger than the work done. 7 Department of Physics

Questions 1. Sometimes, in physics, we appear to be equating two different quantities. Unit analysis is a useful sanity check when learning a new concept. Show that the base SI units for work and kinetic energy are the same: 1 Joule = kg x m^2 / s^2 & 1 N*m = kg x m/s^2 x m Appendix In the following, we will derive the tension pulling on the cart in the cart-pulley-hanging-mass system shown in the drawing below. There are two free-body diagrams to consider. First, the forces acting on the hanging mass are the tension from the string (pulling in the negative direction), and the force due to gravity. Thus, Newton’s second law tells us: . Second, there is only one force acting on the cart, and that is the tension. Again from Newton’s second law: . What can we say about the relationship between and ? Since both masses are connected by a rigid string, we know that . We can thus substitute into the first equation: . If we isolate in this equation, we obtain the result stated in the theory section: .

PHYS182A_195L-WorkEnergyTheorem_.docx

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