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Newtons - Which is shown on the graph above The slope of...

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Adam J. Englert October 8, 2008 Newton’s 2 nd Law 8:30 – 10:20 Lab partners: Unsure of their names
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The purpose of this experiment is to test Newton’s Second Law of Motion, ∑F = ma. We will set up a system in which we hold m constant and vary F. The equipment used on this experiment are as follows: an air track glider, two pulleys, a set of weights, two weight hangers and a computerized data acquisition system. m1 m2 a (m2- m1)*g 160 170 0 98 Slope Intercept 155 175 0.057 196 Theory 308.5g 98 150 180 0.363 294 Linest 433.5962 140.5069 145 185 0.493 392 56.66726 26.73009 140 190 0.857 490 113.3345 53.46017 Agreement Yes Yes The plot is a straight line when I plotted the (m2-m1)*g versus a using Excel.
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Unformatted text preview: Which is shown on the graph above. The slope of the line would be 433.6, while the intercept is 140.5. So that concludes that it is not equal to the mass of the glider. But it is relatively close to the correct slope range. It is off by about twelve. The slop decreases dramatically if you use the ∑F= sqrt(2)*ma. So that definitely is not the correct form for Newton’s 2 nd Law. In conclusion, my experiment did indeed confirm Newton’s 2 nd Law of ∑F = ma....
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