F=ma summer 2023

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Apr 3, 2024

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Lab: Newton’s Second Law of Motion Answer all the questions in the green-shaded tables below. [50] Remember to use the correct significant figures and precisions. Names of members in your group (of 2): (Only one group of 3 is allowed by the instructor whenever applicable) [0.2] 1) Iyanna Blount 2) Teresa Alvarez Purpose: The objective of this lab is - To learn about Newton’s second law of motion F net = ma Experimental Setup The experimental setup is shown below. Part I For part I of the experiment, the acceleration of a fixed mass m due to various forces is measured. According to the second law of motion F net = ma , we have F net / m . For a fixed mass m , the graph of acceleration (y axis) 𝑎 = vs force (x axis) will be a straight line whose slope is the reciprocal of the mass (slope = 1/ m .) Note that we can also plot force on the y axis and acceleration on the x axis. Then the slope of the graph is the mass. But since force is the independent variable in this part of the experiment, i.e., the variable under direct control of the experimenter, it should be plotted on the x axis. Procedure ● Measure the mass of the cart. ● Start the Capstone software. ● Initialize the motion sensor. ● Open a graph window to measure the position of the cart on the track.

● Make sure that the track is level by adjusting the screws on the legs of the track. The track is level when the cart does not move when placed at various positions on the track. ● Place the motion sensor at one end of the track and the pulley at the other end. ● Attach a long string to the cart. ● Add to the cart an extra 0.590 kg mass (two 0.200 kg masses, one 0.100 kg mass, one 0.050 kg mass and two 0.020 kg masses). ● Record the total mass of the cart and extra mass in Question 1. ● Hang 0.020 kg of the extra mass on the end of the string and hang it over the pulley. This will provide the force to pull the cart. ● Put the rest of the extra masses (0.590 kg - 0.020 kg = 0.570 kg) on the cart. ● Start taking data with Capstone. ● Hold the cart about 20 cm from the motion sensor for a few seconds before letting it go. The cart should move away from the motion sensor as the mass hanging over the pulley falls. The position graph should look like the picture below (without the highlight and curve fit.) ● Click on the highlight icon, , to get a highlight window. Resize the window and move it to highlight the data corresponding to the cart moving. Avoid the flat part (which means the cart is not moving) and the other end (open-down curve) where the falling mass hits the ground and stops pulling on the cart. ● Click on the curve-fit icon, , and choose a quadratic fit . You will get a display 𝑥 = ?? 2 + ?? + ? of the values of A, B and C. The curve should fit the highlighted data well. If not, resize the highlight window or retake the data. The acceleration of the cart is twice the value of A, i.e., . Record the 𝑎 = 2? acceleration in the table in Question 2. Repeat the measurement to complete the table in Question 2 with the following division of the extra masses: ● 0.040 kg hanging and 0.550 kg on the cart. ● 0.070 kg hanging and 0.520 kg on the cart. ● 0.090 kg hanging and 0.500 kg on the cart. ● 0.120 kg hanging and 0.470 kg on the cart. Question 1: [1] What is the total mass of the system (cart + extra masses)?

Answer: 1.090 kg. Question 2: [5] Complete the acceleration data table below. Mass hanging on string (kg) Mass on cart (kg) Acceleration of cart (m/s 2 ) 0.020 0.570 0.132 0.040 0.550 0.324 0.070 0.520 0.604 0.090 0.500 0.804 0.120 0.470 1.056 Analysis In Excel, graph the force pulling on the system vs the acceleration of the system. The force pulling on the system is the mass hanging on the string (column 1 of table in Question 2) times the acceleration due to gravity g (= 9.80 m/s 2 .) The acceleration of the system is column 3 of the table in Question 2. Remember that force is the independent variable, so it should be plotted on the x axis. Fit a straight line to the data. Remember to display the equation and the R-squared value of the fit. Label the graph properly: each axis title should have what is plotted with units in parentheses. Upload your graph in Question 6. Proportionality How do we answer the question whether y is proportional to x ? If a variable y is proportional to a variable x , it means that y = mx , where m is a constant of proportionality. That means that a graph of y versus x will be a straight line (linear) passing through 0 ( y intercept = 0.) Therefore, to prove the proportionality of a set of data, the graph of the data must meet two conditions: (a) It is a straight line. This is determined by the R-squared value of the best-fit line. Note: R 2 is the coefficient of determination, which is a measure of how good the model fits the data. In our experiment, we like to see R 2 > 0.95 for the model to be reliable enough. It means that the linear model (straight line) accounts for at least 95% of the variability of the data. See https://towardsdatascience.com/r%C2%B2-or-r%C2%B2-when-to-use-what-4968eee68ed3

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(b) The equation of the best-fit line has a y intercept of 0. Question 3 : [2] Is the acceleration proportional to the force? Justify your answer. Yes it is proportional, any net force applied to an object causes that object to accelerate. Acceleration is a vector quantity, which means it has both a magnitude and a direction. Force and acceleration are directly proportional, so force is also a vector quantity. Mass is a scalar, and whenever you multiply or divide a vector by a scalar, the answer is always another vector. Acceleration occurs whenever there is a change in either the magnitude or the direction of the velocity. So a net force may cause the object to speed up or slow down (change in magnitude) or to change directions both. Question 4 : [2] a) What is the slope of the best-fit line (numerical value)? .949 with unit -(m/s^2)/N-(m/s^2)(kg*m/s^2)- 1k/g. So the slope - .949/kg b) Using the slope of the best-fit line, calculate the total mass of the system, slope = 1/mass or mass = 1/slope. i) Total mass =1/.949 kg =1.05kg Question 5 : [2] What is the percent error between the values for the total mass obtained in question 4 and the exact mass in Question 1? Show your math. The percent error is % error = = =4% ? 𝑒𝑥𝑝𝑒?𝑖?𝑒?? −? 𝑒𝑥𝑎𝑐? | | ? 𝑒𝑥𝑎𝑐? × 100% = .1.05 −1.090 | | 1.090 × 100% .04 1.090 × 100% Question 6 : [2] Insert your graph here.

Part II Procedure If , then a graph of acceleration vs the reciprocal of the mass for a fixed force will be a straight line whose 𝑎 = 𝐹 ?𝑒? ? slope is the force net force F . For this part of the experiment, hang a 0.050 kg mass on the string and measure the acceleration of the cart with the following added masses: ● No added mass ● 0.100 kg added mass ● 0.300 kg added mass ● 0.600 kg added mass ● 1.000 kg added mass Record the acceleration of the system in the table in Question 7. Question 7: [5] Complete the acceleration data table below. Mass added to the cart (kg) Acceleration of cart (m/s 2 ) 0 0.892 0.100 0.766 0.300 0.594 0.600 0.402 1.000 0.306 Analysis The total mass of the system is the sum of the mass of the cart and the mass added to the cart (column 1 of the table in Question 7) and the mass hanging on the string (0.050 kg.) Compute the total mass and fill in the table in Question 8. In Excel, graph the acceleration (column 2 of table in Question 7) vs the RECIPROCAL of the total mass (Question 8) with acceleration on the y axis. Fit a straight line through the data and display the equation and R-squared value. Label the graph properly. Upload the graph in Question 9.

Question 8 : [5] Complete the total mass table below. Total system mass (kg) = mass of cart + added mass + mass on string 1/(Total mass) (1 / kg) 0.557 1.195 0.657 1.522 0.857 1.167 1.160 .862 1.560 .641 Question 9 : [5] Insert your Excel graph here. Question 10 : [2] Proportionality. According to the graph, is acceleration proportional to the reciprocal of mass (if yes then it is said that acceleration is inversely proportional to the mass)? Justify your answer. Yes, acceleration is inversely proportional to the reciprocal of mass because if you increase the mass at a given force as the graph shows, the rate of acceleration slows down as illustrated in the tables above. Question 11 : [2] a) What is the slope of the best-fit line (numerical value)? .518 =.518kg*m/s^2=.518N ?/? 2 1/𝑘𝑔

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b) From the slope of the best-fit line, what is the force acting on the system? .518 N is the weight of the hanging mass Question 12 : [2] What is the percent error between the values for the force in Question 11b and the true force which is the weight of the 0.050 kg mass hanging on the string, F exact = (0.050 kg)(9.80 m/s 2 )? 0.49N % error = = 𝐹 𝑒𝑥𝑝𝑒?𝑖?𝑒?? −𝐹 𝑒𝑥𝑎𝑐? | | 𝐹 𝑒𝑥𝑎𝑐? × 100% = 0.518−0.49 | | .049 × 100% = =5.7%= 6% 0.028 | | .049 × 100% .03 | | .049 × 100% Question 13 : [4] Do you think that these two experiments support Newton’s second law of motion? Justify your answer. I think that these two experiments support Newton’s second law of motion because according to this law an object will accelerate in the direction of the net force and because in this case the force of friction is opposite to the direction of travel, this acceleration causes the object to slow its forward motion and eventually stop.

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