M5 Newton s 2nd Law M glider 296 5 M weight hanger 5 1 M2 M M hanger g 5 1 5 1 5 1 5 1 5 1 10 1 10 1 10 1 10 1 10 1 15 1 15 1 15 1 15 1 15 1 20 1 20 1 20 1 20 1 20 1 25 1 25 1 25 1 25 1 25 1 30 1 30 1 30 1 30 1 30 1 a cm s2 14 8 14 6 15 1 15 1 15 3 29 1 30 1 28 9 29 5 30 5 43 4 43 9 44 8 43 1 43 1 57 5 60 6 60 6 61 6 57 7 72 1 73 4 72 0 77 4 73 7 87 6 87 1 90 8 90 6 90 4 VI 1 M g 0 0 0 0 0 5 0 5 0 5 0 5 0 5 0 10 0 10 0 10 0 10 0 10 0 15 0 15 0 15 0 15 0 15 0 20 0 20 0 20 0 20 0 20 0 25 0 25 0 25 0 25 0 25 0 Graph Using LINEST on excel I found the slope s the intercept b the uncertainty of the slope s and the uncertainty of the intercept b s 2 970514 b 0 46772 s 0 030073 b 0 588288 VI 2 Calculations for 1 2 g 2 97 5 1 296 5 895 752 cm s 2 s M1 2 0 030073 296 5 5 1 9 0700168 895 752 9 0700168 cm s 2 The accepted value of g 980 cm s 2 does not lie in the range of my calculated value of g but they are not too far from one another The frictional force calculated is f 649 35 N while the minimum value of M2g 4331 88 N which helps illustrate the influence of friction on the acceleration The frictional force f acts in the opposite direction to the velocity of the glider Nevertheless since at the minimum force the 2 frictional force is such that consequently the glider will keep moving even though the frictional force is countering its motion 2 4331 88 N and f 649 35 N it follows that