A group of students performed the same \"Newton\'s Second Law\" experiment that

Question

A group of students performed the same "Newton's Second Law" experiment that you did in class. For this lab, assume g 9.81 m/s2. They obtained the following results m1(kg) t1 (s) Vi(m/s) t2(s) v2(m/s) 0.050 1.2000 0.2500 1.7998 0.3969 0.100 1.2300 0.3240 1.6369 0.6690 0.150 1.1500 0.3820 1.4793 0.8626 0.200 1.1100 0.4240 1.3965 1.0022 where m1 is the value of the hanging mass (including the mass of the hanger), v1 is the average velocity and ti is the tim at which v1 is the instantaneous velocity for the first photogate, and v2 is the average velocity and t2 is the time at which v2 is the instantaneous velocity for the second photogate (a) Use Excel to construct a spreadsheet to do the following. (You will not submit this spreadsheet. However, the results will be needed later in this problem.) (i) Enter the above data HINT (ii) Compute the acceleration, a, for each trial. HINT I (iii) Create a graph of the hanging weight mig vs. the acceleration. (iv) Use the trendline option to draw the best fit line for the above data and determine the slope and y- HINT intercept from it. (v) report your results below slope 11.863 kg

Explanation / Answer

(a) multiply all m1 values by g (9.8 m/s^2). find acceleration by finding the change in velocity by the change in time. (v2-v1)/(t2-t1) for each.

Essentially, the first acceleration value would be (0.3969-0.2500) / (1.7998-1.200).

Then type in acceleration values as x and m1g as your y values in excel.

-make a scatter plot, right click on one of the points, add trendline, and set it to where it displays chart on worksheet. your slope and y intercepts are right there! slope will be in kg and y intercept will be in N.

(b) is the same answer you got for slope.

(c) I am not sure how to do c correctly or explain it But there are a and b!

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