Mass of 15 0.06 Average Mass 0.004 Washers (kg) of Washer (kg) Procedure 1 Motion Data Mass of M1 (7 washers): 0.028 Mass of M2 (8 washers): 0.032 Height (m): 0.457m Trial Time (s) 1.33 2 1.28 3 1.31 4 1.35 5 1.37 Average 1.33 Average Acceleration (m/s?) 0.26m/s24. Set the two resulting expressions for the force of Tension from Part 1 equal to one another (as long as the string does not stretch, the magnitude of the acceleration in each equation is the same). Replace F91 and ng with M1 and M2, respectively. Rearrange the resulting expression for the acceleration, a, by isolating a to the left side of the equation. You should now have an expression for the acceleration of a_in terms of both masses. ThenI go back to Question 3 and solve for the FT by substituting your expression for the acceleration a, into one of the expressions for the Tension gFTz. This should result in an expression for the Tension (F1) in terms of the masses. Again, no numbers to plug in here yet. a = (M2 M119 (M1 +M2) 2M M FT: 1 2 5. Calculate the acceleration for the two sets of data you recorded using the expression for acceleration you determined in Post-Lab Question 4, and compare these values to those obtained by measuring distance and time using percent error. Cite two factors that ma).f cause discrepancies between the two values. 8. Calculate the Tension in the string for the falling washers for both procedures and H the one where the masses were equal. Use the expression for the Tension derived in Post-Lab Question 4. Show all calculations. From these two values: and the one where the masses were egualI what trend do you observe in the Tension in the string as the acceleration increases