Fall2023 Atwood's Machine Lab Online-1

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Northwest Vista College *

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1101

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Mechanical Engineering

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Apr 3, 2024

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Analysis of Atwood’s Machine Lab Online Constant Total Mass Table (20 points) Run m 1 (kg) m 2 (kg) m 1 +m 2 (kg) a(m/s 2 ) F net (N) 1 1.1 kg 0.9 kg 2.0 kg 1.00 m/s 2 2.00 N 2 1.2 kg 0.8 kg 2.0 kg 2.00 m/s 2 4.00 N 3 1.3 kg 0.7 kg 2.0 kg 3.00 m/s 2 6.00 N 4 1.4 kg 0.6 kg 2.0 kg 4.00 m/s 2 8.00 N 5 1.5 kg 0.5 kg 2.0 kg 5.00 m/s 2 10.0 N 6 1.6 kg 0.4 kg 2.0 kg 6.00 m/s 2 12.0 N 7 1.7 kg 0.3 kg 2.0 kg 7.00 m/s 2 14.0 N 8 1.8 kg 0.2 kg 2.0 kg 8.00 m/s 2 16.0 N 9 1.9 kg 0.1 kg 2.0 kg 9.00 m/s 2 18.0 N 10 2.0 kg 0.0 kg 2.0 kg 10.00 m/s 2 20.0 N Complete the above chart. Use the acceleration and total mass to calculate F net = a ( m 1 + m 2 ) . Show some calculations to receive credit. F net = 2.00 ( 1.2 + 0.8 ) F net = 4.00 F net = 7.00 ( 1.7 + 0.3 ) F net = ¿ 14.0 1. What is a real-world application of an Atwood's Machine? (4 points) 1
A real world example of an Atwood machine is an elevator or any type of pulley system. 2. For the Constant Total Mass data (Table 1), using Excel, or some other graphing software, plot a graph of F net vs. a, with the trendline displayed on the graph. Make sure to turn this graph in with your lab worksheets. (15 points) 0 2 4 6 8 10 12 0 5 10 15 20 25 f(x) = 2 x Fnet vs. Acceleration Acceleration (m/s2) Fnet (N) 3. (a) What are the units of the slope? (4 points) N/ M*s -1 (b) What physical quantity does the slope of the best-fit line represent? (4 points) Constant Total Mass Constant Net Force Table (20 points) 2
Run m 1 (kg) m 2 (kg) m 1 +m 2 (kg) a(m/s 2 ) F net (N) 1 1.1 kg 0.4 kg 1.5 4.67 7.005 N 2 1.2 kg 0.5 kg 1.7 4.12 7.004 N 3 1.3 kg 0.6 kg 1.9 3.68 6.992 N 4 1.4 kg 0.7 kg 2.1 3.33 6.993 N 5 1.5 kg 0.8 kg 2.3 3.04 6.992 N 6 1.6 kg 0.9 kg 2.5 2.80 7.000 N 7 1.7 kg 1.0 kg 2.7 2.59 6.993 N 8 1.8 kg 1.1 kg 2.9 2.41 6.989 N 9 1.9 kg 1.2 kg 3.1 2.26 7.006 N 10 2.0 kg 1.3 kg 3.3 2.12 6.996 N Complete the above chart. Use the acceleration and total mass to calculate F net = a ( m 1 + m 2 ) . Show some calculations to receive credit. F net = 3.04 ( 1.5 + 0.8 ) F net = 6.992 F net = 2.12 ( 2.0 + 1.3 ) F net = ¿ 6.996 5. For the Constant Net Force data (Table 2), using Excel, or some other graphing software, plot a graph of, a vs 1/M tot , with the trendline displayed on the graph. Make sure to turn the graph in with your lab worksheets. (15 points) 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 f(x) = 6.86 x + 0.06 Acceleration vs. 1/Mtot 1/Mtot (kg) Acceleration (m/s2) 6. (a) What are the units of the slope? (4 points) 3
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M*s -1 /1/kg (b) What physical quantity does the slope of the best-fit line represent? (4 points) Constant Net Force 7. In this experiment, we made the assumption that the tension and the acceleration experienced by the two subsystems, the two different masses, were exactly the same. Why are these good and/or valid assumptions? (5 points) These assumptions are valid since the performance of the two subsystems performed the same despite different conditions. 8. Above, we derived an equation for the acceleration: a = g ( m 2 m 1 ) ( m 2 + m 1 ) . Briefly explain what the numerator and denominator are in a physical sense. (5 points) With “g” in the numerator, this acts as the gravitational pull on the masses while the denominator acts as the total mass. 4