The wooden block was placed sideways with its short face touching the track. As before, the maximum static friction force and the kinetic friction force at constant velocity was determined. Results are reported on Table 3. Table 3 Static Kinetic Normal frictiun Friction Force Block sideway with 2 extra masses m 1 546 1 384 FROM TABLE 1 1.632 1.298 5.88 Rewrite the results for Trial 1 Block +2masses ooy [ Error analysis for Table 3 Report on Table 3 also the uncertainty on each column for Total mass, Normal force, Static Friction, and Kinetic Friction. = Uncertainty on the Kinetic and static friction Force: The Force sensor has a resolution of 0.02 N, use this number as uncertainty on the data for Force. * Uncertainty on the mass: The mass measured will have its uncertainty from the scale resolution. Use 0.1g. Do not forget to convert into kg. = Uncertainty on the acceleration of gravity: According to Wolfram Alfa at Washington D.C. g = 9.81454 +0.00001 5. This uncertainty is very small and it gives a 5 negligible contribution to the overall error for the normal force. Uncertainty on the Normal force: To find the uncertainty on the Normal force, you want simply use the uncertainty on the mass measurement. To confirm the hypothesis that the area of contact is not related to the coefficient of friction, compare the forces of friction in the two conditions reported on table 3. Determine the quantity \"difference\" below |Force . Force large area small area 2 2 (ermr on Force \") +(error on Force ] large are small area. Kinetic Normal Total mass Static Friction Force (N) (kg) friction (us) (uk) (FN) Trial 1 0.6 1.546 1.384 5.88 Trial 2 0.6 1.632 1.298 5.88 Uncertainity 0.0002 0 . 02 0.02 0.02 Differnu . Differnustatic = / Forulagraria - Forusmall areal revor on Forulangrava) + ( evior on Forus A ) 2 1 1 . 5 46 - 1. 632) V( 0 . 02 ) 2 + ( 0 . 02 ) 2 1 -0. 086) 0 . 02020 = 3 . 044 1 1.384 - 1.2981 & Difference kinetic S (0 02)2 + ( 0 . 02 ) 2 = 3. 041