VI 1 M glider 295 1 M weight hanger 5 3 M g M2 g a cm s2 0 0 0 0 0 5 5 5 5 5 10 10 10 10 10 15 15 15 15 15 20 20 20 20 20 25 25 25 25 25 5 5 5 5 5 10 10 10 10 10 15 15 15 15 15 20 20 20 20 20 25 25 25 25 25 30 30 30 30 30 18 8 18 8 19 6 21 9 22 1 25 9 33 6 33 4 33 5 33 1 47 1 47 3 47 3 47 46 7 60 61 7 61 7 61 7 61 5 76 76 76 9 76 9 76 3 90 5 91 1 90 3 91 2 90 8 a vs M2 f x 2 86 x 4 57 0 5 10 15 20 25 30 35 a cm s2 Average acceleratrion was calculated using excel Slope Slope uncertainty 2 86 77 4 573333 33 03 0 036243 0 705746 Intercept Intercept uncertainty M 1 M2 295 1 5 3 5 300 4 this value is a constant 300 4 2 86 300 4 300 4 cancels out on the right hand side g 300 4 100 90 80 70 60 50 40 30 20 10 0 g 2 M Solving for g a M 2 g M 1 M 2 s g M 1 M 2 s g 300 4 g 859 144 cm s2 Solving for g s g M 1 M 2 s2 s g g The accepted value of g 980 cm s2 does not lie within the range that was calculated g 859 144 g 859 144 cm s2 g 18 41265054 cm s2 g g 859 18cm s2 VI 2 Average for a1using values V 3of data sheet average a1 of recorded accelrations number of recorded accelerations average a1 61 4 61 2 61 2 61 2 61 1 61 22 cm s2 Average for a2using therecorded accelerations when M 15 average a2 of recorded accelrations number of recorded accelerations average a2 60 0 61 7 61 6 61 7 61 5 61 3cm s2 Finding the frictional force Frictional Force M 1 M2 a1 a2 5 5 2 Frictional Force 300 4 61 2 61 3 15 02 cm g 2 s2 15 02 N We take the positive value of this since we assume all calculations are positive The frictional force is less than the smallest value of M 2 g which is 49 98 N