Lab 7

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Columbus State Community College *

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1250

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Mechanical Engineering

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Apr 3, 2024

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5t?,e:7 LAB 7: WORK AND ENERGY Introduction The experiments in this lab investigate energy and how it's transformed from one type to another. We'll focus on just two types of energy: kinetic energy (K) and gravitational potential energy (U). Their definitions are and K = 1 mv 2 2 U=mgh 7.1 7.2 Work transforms energy from one type to another. The work done on an object by a constant force is calculated from this relationship: W=Fdcos¢ 7.3 When a force varies with time, the work it does can be calculated by making a graph of Fcos¢ vs. d; the area under the curve is equal to the work done. There are two ways to use energy to predict what a body subject to a set of forces will do. In general, when a net external force does work Won a body, the resulting change in the body's kinetic energy is found from the work-energy theorem: W=M 7.4 In words, the net work done on a body during any time interval is equal to the change in its kinetic energy L1K - K, K,. The key word here is net; equation Error! Reference source not found. doesn't apply to any single force acting on a body, unless all other forces balance out. A second predictive method uses the principle of energy conservation. When non- conservative forces (friction, air resis~ance, normal forces) do no work on a body, then its total mechanical energy remains constant. Total mechanical energy Emec is defined as Emec =K +U 7.5 There are two equivalent ways to formulate the conservation principle mathematically: 7.6 and /J..U=-M 7.7 The first version compares the net mechanical energy at two different instants and predicts that it's the same at both: E 1 = E 2 . The second version explicitly claims that, when mechanical energy is conserved, any gain in U is offset by a loss of K, and vice versa. 63

PRELAB QUESTIONS Lab 7: Work & energy Name: J)~/2Al1 :;3.orD 1. In Experiment I, suppose the average tension is 0.51 N over a displacement of 1.1 m. The work done by tension during that displacement is ~~56j) b. 1.7 J C. 2.4 J 2. In Experiment I, suppose the falling mass is m = 156 g, and the total mass of the sensor, carts, and load is M = 1239 g. During one trial, the initial and final speeds are 0.36 m/s and 1.29 m/s. The change in the carts' kinetic energy for this trial is a. 0.62 J C. 1.7 J 3. Before pressing the TARE button on the force sensor, you should be sure that the string attached to its hook is taut. ~ALSE 4. In Experiment II, suppose the final speed of the masses is 3.1 m/s, and mA = 455 g, m 8 . = 412 g. The system's final kinetic energy is a. 1.5 J b. 2. 7 J cg 5. In Experiment II, mass l's potential energy change is negative, and mass 2's potential energy change is positive. cg FALSE 64

Experiment I: Work & Kinetic Energy In this experiment, a mass hanging from a string running over a pulley accelerates two carts. According to equation Errorl Reference source not found., the work done by tension should equal the carts' change in kinetic energy. Does it? Equipment 2 carts with Velcro pads 2.2 m dynamics track with bumper and pulley 4 table rods 2 right-angle clamp assorted masses ( ~1 kg) string and scissors force sensor motion sensor mass hanger, assorted masses Setup Mount the track on four table rods, using track clamps to hold it in place o the end of the track should project over the end of the lab table o the track should be 1-2 m above the floor Install a bumper near the end of the track over the floor Attach a pulley to the same end of the track as the bumper Clip the motion sensor to the opposite end of the track o set its switch to the "narrow beam" position Level the track Screw in the hook to the force sensor, if necessary Recalibrate the force sensor for positive force readings, if desired o see Appendix C or the instructo~ Attach the force sensor to one cart Attach the two carts together by their Velcro pads Place a total load of 500-1000 g in the carts Measure and record M, the total mass of the carts, sensor, and load o for simplicity, M will be referred to as "the mass of the carts" from now on Place the carts on the track with the force sensor hook facing the pulley o hold or block the carts while getting ready Tie a string to the hook and run it over the pulley Tie a mass hanger to the free end of the string Adjust the string's length so the carts can roll roughly 1 meter (or more) before o they reach the bumper o the mass hanger hits the floor Check your setup to make sure that o the motion sensor is aimed at the carts and set to "narrow beam" o the force sensor's cord won't get hung up on anything A diagram of the equipment setup appears on the next page. 65

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Equipment setup: bumper force sensor & carts (M) motion sensor falling m mass table rods ------ 7 Computer setup Open DataStudio and choose Create Experiment Select Force Sensor and Motion Sensor from the Sensors menu Connect both sensors to the data interface as shown Set the sample rate to at least 20 Hz (same for both sensors) Open a single Graph, and use it to plot o force vs. position o velocity vs. position Lock the graph's horizontal axes with the ~ii button o if you use separate graph windows, the lock button won't work Procedure Roll the carts back until they're about 15 cm from the motion sensor Make the string slack, then press the force sensor's TARE button Holding the cart in place, put 100g on the mass hanger Get someone in position to catch the cart before it hits the bumper Make sure the string is horizontal Click Start Release the carts Click Stop and catch the carts before they hit the bumper If the data on either graph look bad, run the trial again If the graphs show good data, then o measure and record m, the total mass hanging from the string o add 50 g to the mass hanger and repeat Run a total of five trials with increasing m's o M doesn't change; it's the mass of the sensor, carts, and load NOTE: It's important to zero the force sensor before every trial. Be sure that no force acts on the force sensor when you press TARE. 66

Analysis For each trial, identify an interval of constant acceleration o force relatively constant (jiggling ok, trending up or down not ok) o velocity increasing, not leveled off or decreasing o the interval should be as long as the data permit On that interval, identify the carts' initial and final positions (x1, x2) Take measurements on the identified interval 'o force graph • use the Smart Tool to measure the displacement d = x2 - X1 • measure the average (Mean) tension T by selecting the interval • click the Statistics button to display Mean • don't take the average force for the entire trial o velocity graph: measure initial & final velocity ( V1, v2) • v1 = velocity at position x 1 • V2 = velocity at position x 2 Repeat for the remaining trials, choosing a new interval for each one From your measurements, calculate o the work Wr done by tension: equation Error! Reference source not found. o initial and final kinetic energy of the carts o the carts' kinetic energy change o the percent difference between Wr and llK 67

Questions for Lab 7, Experiment I 1. Although tension isn't the only force acting on the carts, the work it does on theni is theoretically equal to their change in Kif friction is negligible. a. Explain why theory predicts that WT = !lK in this experiment. Illustrate with a force diagram. b. In your own data, is WT generally greater than, less than, or equal to ~K? What could account for the difference? Explain. 2. In your trials, is the average tension more than, less than, or equal to the weight of the falling mass? Use a force diagram to illustrate your explanation. 68

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Data for Lab 7, Experiment I Total mass of force sensor, carts, and load: M = l( / 4' f.J f m Tavg d .1K diff. f((J P~i 14 If •2 C~l tJ-32 C,}(J t7.Jt. /,),j- I, 3 ,-._ I) tJ 1~4 14 l~t6 (), 67 a364 tJ, !~ y::,,. 7f 11-15 J f/(l "1 4 ,i I ; ~,3 ~.i, (J. f, I p. 43 "4·2 l'2 1, -~1o. •v :2J-tl ~,,l J,2 (vJ,J (),'ll ,, ()J IA ii ; o.s~ 7L.4- J_ 3&1l 4 'I .21 IQf./ tJ, q3 f,)J c.31 o~ 6J I &i)~ !> I, 7 Note: the mass m of the falling body is included only to distinguish among the five trials. It doesn't enter into any of the calculations in this first table. Kinetic energies should be calculated for the force sensor, carts, and load, using their total mass M. Weight of falling mass and average tension m mg Tavg /()(} I( 9; 0 o.i ISO 14 'lt:J I· f o&fl () I f6P pt., 4- oLJ-,J 24so 3 ,-t 3ofJ 1 ti'4il 4 ,/ 69

Experiment II: Energy Conservation In this experiment, you'll set up an Atwood's machine (two masses hanging on opposite sides of a pulley) and time how long it takes mass A to fall to the ground from rest. If the system's mechanical energy is conserved, then the change in its potential energy !).LJ = 11UA + 11Ua should, by equation Error! Reference source not found., have the same absolute value as the change in its kinetic energy !).K = !).KA + !).Ka, Equipment 2 table rods pulley 1-2 right-angle clamps string, scissors 2 mass hangers assorted masses ms stopwatch meter stick Setup Attach a pulley to a table rod Atwood's machine The pulley should extend over the edge of the table Cut a length of string and tie a loop in each end Run the string over the pulley and attach mass hangers to each loop Put 400 g on each mass hanger Procedure Put an additional 20 g on one of the mass hangers o this will be mass A Move the masses until they're at the same height Measure the height h of both masses Simultaneously release both masses and start a stopwatch Stop the stopwatch when mA hits the floor If the masses hit each other or were swinging as they fell, try again If the trial is good o record the elapsed time o measure & record mA and ma (including the masses of the hangers) Gather data for a total of five good runs o increase the mass of mA by 20 g each time o keep the initial height of the two masses the same in every trial Analysis Calculate the change in ma's potential energy o use equation Error! Reference source not found. o value should be the same for all 5 trials For each trial, calculate o the change in the system's total potential energy: 11U = 11UA + 11Ua o the masses' final speed: v2 = 2h/t o the change in the system's total kinetic energy: !).K = !).KA + 11Ka o percent difference between 11).Ul and /)..K 70

Questions for Lab 7, Experiment II 1. The Analysis section claims that the final speed of mass A is v 2 = 2h/t. a. Prove this claim, assuming mass 1 falls from rest with constant acceleration. b. Explain why this is also the final speed of mass B. 2. Choose the trial with the best agreement between l.1UI and .1K, and check the prediction of equation Error! Reference source not found .. a. Calculate the system's initial mechanical energy: Emec,1 = K1 + U1 b. Calculate the system's final mechanical energy: Emec,2 = K2 + U2 c. Calculate the percent difference between Emec, 1 and Emec,2 d. Which aspect of the experiment do you think is primarily responsible for the discrepancy, and why? How could the discrepancy be reduced? 71

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Data for Lab 7, Experiment II t1ha = 1~u1 t ~K Diff. •4;_; -(J,J; -1,48 J,81 J 0 44: (J, ,3 (). /lllf I• 67 ,44~ --- - ,:;, :;j -/•j_) ?, 'ft /,/~ (}tt// v• c,z,2 /, 7 ,:--- . . -¥~~ - ,?. 36> -,~,2 ~ttw'3 Q,, l( J (JI? ,7t7 v,,·J& '. C 41v ,P ? ' - '_,;>, ... - I, 7 3 • t I (/•7Jb fJ•-f4 (),, 111 , , 7; '-5it:? -e?~3; - I, ';'J 3, If o, ?J- o, f/6 . () • 9 "" ,.; '..I OL'? f ()/ Best trial Initial: KA1 = 'i) Ka1 = 0 K1 = 0 UA1 = - I. 't$ Ua1 = f2•ifj U1 = I .t,3~'> Emec,1 = I • zj ? Final: Ua2 = '7tJI Emec,2 = fJ• 'ltr- Difference: U % 72