Directions 1. Place equal masses of 150 - 200 grams on each side of the apparatus. 2. Continue to add a small amount of
Posted: Mon May 23, 2022 11:52 am
The mass found to overcome friction 4.00 g Distance 139 cm Both sides of the Atwood Machine start with 200.00 g then 4.00 g were added to one side to overcome friction. Then additional mass was added to make the system accelerate. The photos below show the setup. Trial 1 time when 10 g added. Trial 2 time when 15g total were added. Then trial 3 when 25 g total were added. To the frictional mass.
1000359 000 ON/RESET FLINN SCIENTIFIC, INC START/STOP 100286 LAP ON/RESET FLINN SCIENTIFIC, INC START/STOP
0:0002 15 FLINN SCIENTIFIC, INC LAP ON/RESET START/STOP
Atwood Machine Results (kg) 1. Mass to Overcome Friction_ 2. Frictional Force 3. Using Motion Equations Average Time (s) Distance (m) Acceleration (m/s) 4. Using Newton's 2nd Law Small Mass (kg) Large Mass (kg) Net Force (N) Acceleration (m/s²) 5. Calculate the % Difference %Difference Trial 1 (N) Trial 2 Trial 3
Sample Atwood Catulations Intram MASS to overcome Frict. 3.00g Forces of flection Fring: 300g (ny) (9.81m) - 0.0294 ) ik Aug tome - 2.165 2.125 2.20, 215 + 3 Note I gave you Ang times. Distances 0.50m acad 2 (0.560m).0.240m (2.165) st MEASUD Starting out m = m2 200 g or 0.200kg Small mass aoog - 0.200 kg Lange MASS - 200g+ 3.00g + wog= 213 or 0.213 kg : MARS Abp to O smail faut Fret mig-mag-meg-(m, M2-M.) (0.213 Kg - 0.200ly -0.005x) 4.813.) - 0.0981~ - Note: equal 10g that's what makes the system accelert = -0.28m 3 a= Furt 0.0981. mume 0.21319 +0.2006 % D.fference = \Exp, EXT] X100 O % Diff - (0.240 -0.236) Torayon +0.23894 ) X 100- 0.837%