Physics123 Lab3 Report

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Apr 3, 2024

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Newton’s Second Law Amena S. Tajammul February 7, 2023 1 Introduction Intro and description of the procedure used Newton’s Second Law states that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. This can be expressed with the equation below: a = Fnet / m where a is acceleration of the object, Fnet is the sum of forces applied to, and m is its mass. It is measured in kg*m/s^2, or Newtons (N). When taking the magnitude of any acceleration, you must take into account the angle it makes. The force of gravity is always pointed downwards. When an object is at an angle, gravity is pointing directly down but angled in relation to the axes. Thus the new gravity in the y direction is g*cos(theta) and in the x direction is g*sin(theta). When combining this information with other formulas, we can solve for the acceleration with the following formula: g*sin(theta)=vf^2/(2*x) where g is gravity, theta is the angle at which the object rests, vf is the final velocity, and x is the distance it travels. Experiment 1: The first experiment was to determine whether the acceleration of an object is proportional to the force applied. To do this, we used a velocity sensor to measure the velocity as the glider went down the ramp and measured the distance it traveled. After taking 5 trials, we changed the height by adding blocks underneath it and completed more trials. Experiment 2: For this experiment, we kept the height the same and added mass to the glider. With each change in mass, we did three trials to see how the velocity changed. We attempted to prove that it is impossible to test whether acceleration is inversely proportional to mass using only this data. 2 Data Experiment 1 neat data table with units Mass: 202.2 kg Distance: 28.7 Length: 129.4 cm
Experiment 2 neat data table with units
3 Results + Graph Analysis - create a graph for experiment 1, based on the equation g*sin(theta) = vf^2/(2 x) plot Vf^2 on the y-axis and 2*x*sin(theta) on the x-axis so that the slope is equal to g. On the graph each set of 5 trials should be averaged to give a single point, so there will be 5 total points on the graph. using the standard deviation from each set add error bars to each point on the graph in the y direction. Find the standard deviation of velocity for each set of 5 trials, square it (because the plot has Vf^2 on the y axis) and add error bars of that size to the data point. If the error bars are too small to see, multiply it by a factor of 10 (or whatever makes them visible) and add the error bars at that value, and make a note saying error bars are 10x (or whatever you use) the standard deviation. Make sure the graph has units and labels on the axis. I'll send out a document with a more detailed explanation of uncertainty and types of error after the test, for now this will do. 2*x*sin(theta) Average vf^2 (m/s) Standard Deviation 0 0.0374812 0.0004898979486 3.3 0.6042 0.0004
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6.6 0.8328 0.0007483314774 10.1 1.0156 0.001019803903 13.6 1.1768 0.004166533331 Results - Calculate acceleration for each experiment and compare it to the accepted value quantitatively, using [((experimental) - (accepted))*100]/(accepted) = % error. Also answer the questions at the bottom of the procedure Calculated Acceleration Expected Acceleration Percent Error 0.06529784 0 - 0.63598892 0.560432767 13.48175151 1.208285436 1.128438949 7.075835788 1.796939652 1.726738794 4.06551693 2.412645017 2.309891808 4.448399213 1. Calculate the accelerations using Equations 2 and 4 for the two experiments. Compare your results. Experiment 1 Accelerations velocity angle g*sin(theta) vf^2/(2*x) 0.193 0 0 0.00014392967 54 0.193 0 0 0.00014392967 54 0.194 0 0 0.00014542503 86 0.194 0 0 0.00014542503 86 0.194 0 0 0.00014542503 86 0.604 3.3 -1.545907803 0.00140964451 3 0.604 3.3 -1.545907803 0.00140964451 3 0.604 3.3 -1.545907803 0.00140964451
3 0.604 3.3 -1.545907803 0.00140964451 3 0.605 3.3 -1.545907803 0.00141431607 4 0.832 6.6 3.053105362 0.00267474497 7 0.833 6.6 3.053105362 0.002681178516 0.832 6.6 3.053105362 0.00267474497 7 0.833 6.6 3.053105362 0.002681178516 0.834 6.6 3.053105362 0.00268761978 4 1.017 10.1 -6.125692359 0.00399647990 7 1.014 10.1 -6.125692359 0.00397293663 1 1.016 10.1 -6.125692359 0.00398862442 1.016 10.1 -6.125692359 0.00398862442 1.015 10.1 -6.125692359 0.00398077666 2 1.176 13.6 8.419785786 0.00534380216 4 1.17 13.6 8.419785786 0.00528941267 4 1.177 13.6 8.419785786 0.00535289412 7 1.178 13.6 8.419785786 0.00536199381 8 1.183 13.6 8.419785786 0.00540760819 2 Experiment 2 Accelerations velocity angle g*sing(theta) vf^2/(2*x) 0.604 3.3 -1.545907803 0.00140964451
3 0.605 3.3 -1.545907803 0.00141431607 4 0.601 3.3 -1.545907803 0.00139567619 8 0.606 3.3 -1.545907803 0.00141899536 3 0.605 3.3 -1.545907803 0.00141431607 4 0.605 3.3 -1.545907803 0.00141431607 4 0.606 3.3 -1.545907803 0.00141899536 3 0.605 3.3 -1.545907803 0.00141431607 4 0.605 3.3 -1.545907803 0.00141431607 4 For experiment 1, the accelerations changed when the angle changed, but for experiment two, the mass changing had no effect on the acceleration. 2. Derive an expression for the normal force. Do a sample calculation of the normal force for one of your data sets. What happens to the normal force if the incline angle is completely vertical (θ = 90◦)? How about when horizontal? Normal force(Fn) can be calculated using the FBD of an angled object. Fn=m*g*cos(theta) Where Fn is normal force, m is mass of the object, g is gravity, and theta is the angle. For experiment 1, when the angle is 0, the normal force is 1.98156. When the incline is completely vertical and there are no other vertical forces acting on the object besides gravity, Fn=m*g. When the incline angle is horizontal, for example when it is pushed against a wall, the Fn becomes a horizontal force instead of a vertical force. In this case, Fn is equal to the force of the push towards the wall which keeps the object stable. 3. Select one of the velocities you obtained from the second experiment. If the glider goes up the incline with that initial velocity, how far up the plane will it go? How much time would it take for the glider to return to the starting point? If the glider started with vo=0.605 m/s at the same angle, it would have gone up
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4 Conclusion Conclusion - explain the relationship between F and m, and why in the second experiment the acceleration didn't change as m was varied (it should not have changed much) and list any possible sources of error in the experiment. Force = acceleration * mass. The acceleration in the second experiment did not change as m changed because the mass changed, so did the force of gravity. Since they increased proportionally, the acceleration stayed the same. Incorrect measurements and calculations due to human error are possible errors in the experiment.