Physics Laboratory Report - Lab_ 6a1

Physics Laboratory Report Lab 6a1: Work and Energy Name: John Hanna Group ID: C Date of Experiment: 10/18/2023 Date of Report Submission: 10/26/2023 Course & Section Number: PHYS111A105 Instructor’s Name: Subodh Dahal Partners’ Names: Jeremiah, David 1. INTRODUCTION a. OBJECTIVES i. To demonstrate the work-energy theorem by measuring the work done on an object by a constant force and the kinetic energy of the object. ii. To understand how the work done on the object changes the object’s energy. b. THEORETICAL BACKGROUND i. Work is the measure of energy transfer that occurs when an object is moved over a distance by an external force which is, fully or partly,applied in the direction of the displacement. Work is a scalar quantity defined as W = F * s = Fscos(θ) and has a SI unit of ‘joule’ (J). Work can also be expressed in terms of kinetic energy using the work energy theorem,which states that the work done by the force on an object is equal to the change in the kinetic energy of the object. Therefore W = ∆KE = KE 2 − KE 1 . Kinetic energy is equal to 1/2mv 2 , ∴ W=1/2m(v 2 2 −v 1 2 ). ii. This lab will use a frictionless air-track to determine the work done on a glider while being pulled by a constant force on horizontal and inclined tracks and to demonstrate the work energy theorem. The change in the kinetic energy of the glider will also be determined. While using the horizontal track the net force acts in the same direction as the motion of the glider, so the work done is equal to the work done by the tension. Also, while using the inclined air track it is derived that work is equal

to the sum of the change in kinetic energy and the change in potential energy; meaning W T = ∆KE + ∆PE. 2. EXPERIMENTAL PROCEDURE I. Equipment List Computer with Capstone software installed, 850 Universal interface, Force sensor, Rotary motion sensor, Air-track, Air supply with a hose, Glider set (a glider, 50-g weights (x4), a hook, and a string), Lab jack (or wooden block), Single sheave pulley, Short Rod, Right-angle clamp, L-shape aluminum rod, Protractor, 1-m stick in front of a lab counter, and an electronic balance on a lab counter. II. Setup Part I: Work Energy in a Horizontal Air-Track System with a Constant Force 1. Set up the force and rotary motion sensors as shown in Figure 5 using straight and L-shaped rods and a right-angle clamp. 2. Insert a hook to the glider as shown in the below figure and put the glider on the air track. Connect one end of the string to the hook attached to the glider. 3. Connect the other end of string to the hook of the force sensor after inserting the single sheave pulley with the weights as shown in Figure 5. Make sure that the string is on the rims of the hanging pulley and the largest pulley of the rotary motion sensor. 4. Level the air-track to make the horizontal condition. 5. Adjust the vertical position of the rotary motion sensor so that the line of the string should be parallel to their track. This is how the angle between the applied force (tension) by the string and the displacement of the glider is zero.

Part II. Work and Energy in an Inclined Air-Track System with a Constant Force 1. Keep the glider with the same mass as in Part I. FInd the inclined angle using a protractor or a meter stick (Pythagorean Theorem). Input the values of mass and angle in the software and record in Table 2-1 in the Data Table section. 2. Follow steps 1 to 13 in the procedure of Part I. 3. RESULTS a. EXPERIMENTAL DATA Table 1-1 Mass = 0.2799 Angle = 0 Position x 1 (m) V i (m/s) Tension, T i (N) Net Force, F i (N) 1 0.182 0.671 0.47 0.341 2 0.46 1.071 0.382 0.328 3 1.081 1.633 0.411 0.349 4 1.433 1.877 0.352 0.339

Table 1-2 Table 2-1 Mass = 0.2799 Angle = 5 Position x 1 (m) V i Tension, T i (N) Net Force, F i (N) 1 0.52 0.226 0.382 0.138 2 0.247 0.489 0.411 0.132 3 0.559 0.745 0.382 0.146 4 1.075 1.075 0.411 0.131 Table 2-2 Position change ( i to f ) s = x f - x i (m) W = T * s (J) KE i = 1/2Mv i 2 (J) KE f = 1/2Mv f 2 (J) ΔKE = KE i - KE f (J) % Difference 1 to 2 2.78 0.9591 0.63 0.1605 0.0975 1.64 2 to 3 0.621 0.214245 0.1605 0.373 0.2125 0.817 3 to 4 0.352 0.12144 0.373 0.493 0.12 0.193 1 to 4 1.251 0.431595 0.63 0.493 0.43 0.32 Position change ( i to f ) s = x f - x i (m) W = T * s (J) ΔKE = KE i - KE f (J) ΔPE = MgΔh (Δh = s * sin(θ)) (J) ΔE = KE + PE (J) % Difference 1 to 2 0.195 0.07917 0.026317 0.046618 0.0729356 8.197 2 to 3 0.312 0.126672 0.0442108 0.0745898 0.1188006 6.41326 3 to 4 0.516 0.209496 0.084054 0.12336 0.207414 0.9988 1 to 4 1.023 0.415338 0.154582 0.244569 0.399151 3.97476

Table 2-2 s = x f - x i s = x f - x i = 0.195 m s = x f - x i = 0.312 m s = x f - x i = 0.516 m s = x f - x i = 1.023 m W = T * s ; T = 0.406 W = 0.406 * 0.195 = 0.07917 J W = 0.406 * 0.312 = 0.126672 J W = 0.406 * 0.516 = 0.209496 J W = 0.406 * 1.023 = 0.415338 J ΔKE = KE i - KE f ΔKE = KE i - KE f = 0.026317 J ΔKE = KE i - KE f = 0.0442108 J ΔKE = KE i - KE f = 0.084054 J ΔKE = KE i - KE f = 0.154582 J ΔPE = Mgs * sin(θ) ; θ = 5 M = 0.2799 ΔPE = 0.2799 * 9.8 * 0.195 * sin(5) = 0.046618 J ΔPE = 0.2799 * 9.8 * 0.312 * sin(5) = 0.0745898 J ΔPE = 0.2799 * 9.8 * 0.516 * sin(5) = 0.12336 J ΔPE = 0.2799 * 9.8 * 1.023 * sin(5) = 0.244569 J ΔE = KE + PE ΔE = 0.026317 + 0.046618 = 0.0729356 J ΔE = 0.0442108 + 0.0745898 = 0.1188006 J ΔE = 0.084054 + 0.12336 = 0.207414 J ΔE = 0.154582 + 0.244569 = 0.399151 J 4. ANALYSIS and DISCUSSION The work energy theorem asserts that the work done on an object is equal to the change in that object's kinetic energy for the conditions employed in part 1. The work energy theorem is a sound theory, as shown by the fact that the percentage

difference for each position change examined in part 1 was less than 1%. In the second section, work is defined as the product of the kinetic and potential energy changes. The fact that there was no percentage difference more than 7% demonstrated the validity of this application of the work energy theorem. Because more forces are acting on the object when the track is inclined, increasing the chance of errors, it may be claimed that the percentage differences in parts 2 are predicted to be higher than those in parts 1. This is because all of these forces will not be taken into account in the calculations. The studies produced accurate and exact results since there were only minor inaccuracies that might be attributed to friction because both sections had minimal percentage differences. Additionally, the percentage differences for each component of the experiment fell within the same range of numbers, demonstrating the accuracy of the values. According to the equation of work, when an object undergoes work, a net force acting on it moves over a distance. The aforementioned net force causes the object to move with an acceleration equal to the force divided by the mass, according to Newton's second law. The equations of motion demonstrate how the object's velocity will change over time since it is accelerating. Since kinetic energy, also known as motion energy, is defined as 1/2mv 2 , when the velocity changes, the energy will likewise vary. As a result, the energy of an object is altered by the total work done on it. 5. CONCLUSIONS The data of both parts were consistent, and they support the work energy theorem, and the experiment helped us understand how the work affects the energy, and different inclined points. The results from both parts of the experiment were consistent with the work energy theorem, proving that the theorem works for both flat and inclined systems.The experiment also showed in detail how and why the work done on an object affects and changes its energy.