Lab 5_ Newton's Second Law

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Florida International University *

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2048L

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Physics

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Dec 6, 2023

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pdf

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Uploaded by DrComputerNightingale32

Lab #5 Lab partner-2: Alexia Reyes ID: 6313144 Title: Newton’s Second Law Preliminary Question Answers: 1) As a small amount of force is applied to the ball with the use of a flat end of a ruler, the ball will not move at a constant speed, this is because we would need to apply a constant force in order for the ball to move at a constant acceleration. 2) Similar to the previous question, the ball will not move at a constant speed, but it will move at a constant acceleration if constant force is applied to the ball. 3) When a small force is applied compared to an applied larger force, the acceleration of the ball will be small. This means the ball will move at a slower speed. Analysis: Part 1: 1) Yes, the net force of an object and the acceleration of an object is directly proportional. This is illustrated by the force vs acceleration graph in the experiment. As the acceleration increases the force increases, which is supported by the positive slope of the graph. 2) Force is being measured in kg*m/s/s and acceleration is measured in m/s/s. The slope is in units of kg. 3) In table one the mass of the car is 0.399 kg while the slope of the regression line is 0.3666 N/m/s/s (kg). Between the two values there is an 8% difference. In table 2 the mass of the car is 0.880 kg, While the slope of the regression line is 0.8422 N/m/s/s (kg). Between the two values there is a 4% difference. The slope represents the Force on the car over the acceleration of the car which equals the mass of the car, and the relationship between force, acceleration, and mass. 4) Force= mass x acceleration Part 2
5) For the data retrieved for table 5.4, we can examine that the plot displays a linear relationship between acceleration and force resulting in a positive slope. It is known that the slope of this line is equal to the mass of the object (the cart). Because the mass of the object is greater than 0, the slope results in being significantly different from 0. What is significant about the slope is the linear relationship between force and acceleration. 6) The plot is linear with a positive slope, and the data fails to have any consistent/concerning outliers. The positive linear slope is important as it references the directly proportional relationship between acceleration and force. The slope also represents our mass. The value of the slope is 0.418 N/m/s/s and the value is calculated by measuring the Force on the object against the acceleration of the object. Data Tables: PART 1: Table 1 Mass of cart with sensors (kg) 0.399 Regression line for Force Vs. acceleration data (N/m/s/s)(kg) 0.366 Table 2 Mass of cart with sensors and additional mass (kg) 0.880 Regression Line for Force Vs. Acceleration data (N/m/s/s)(kg) 0.8422
PART 2: Table 3 Trial F/a (Ns 2 /m) Hanging mass (kg) 1 0.964 0.05 2 0.956 0.07 3 0.936 0.09 4 0.919 0.110 5 0.917 0.130 6 0.922 0.150 7 0.913 0.170 Mass of cart (kg): 0.932 Trial F/a (Ns 2 /m) Hanging mass (kg) Mass added to cart (kg) Total cart mass (kg) 1 0.418 0.1 0 0.400 2 0.535 0.1 0.125 0.525 3 0.686 0.1 0.250 0.650 4 0.781 0.1 0.375 0.775 5 0.840 0.1 0.500 0.900
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