Physics Lab 4

Results: Trial 1 Trial 2

Trial 5

Force/N Work eqn 1 J Max Velocity/m·s -1 Work eqn 2 J Trial 1 -0.693 -.2079 -1.267 -.1605 Trial 2 -0.738 -.2214 -1.215 -.1276 Trial 3 -0.685 .2055 -1.271 -.1615 Trial 4 -0.792 .2376 -1.187 -1.408 Trial 5 -0.718 .2154 -1.165 -.1357 Trial 6 -0.750 .225 -1.129 -.1275 Equation 1: W net = F net * Δ x Equation 2 = W net = Δ KE , KE = ½ mv 2 Average force = (-0.693 - 0.748 - 0.685 - 0.792 - 0.718 - 0. 750)/6 = -0.731 N Average velocity = (-1.267 - 1.215 - 1.271 - 1.187 - 1.165 - 1.129)/6 = -1.206 m/s Eqn 1 W net = -0.731 N * 30 cm = -0.731 N * .30 m = 0.2193 J Eqn 2 W net = Δ KE = KE f - KE i = ½ (0.2 kg) (-1.206 2 ) - ½ (0.2)(0 2 ) = 0.1454 J T’ comparison test Standard Deviation Eqn 1 = (.225 - 0.2055) / 2 = .00975 Standard error of Eqn 1 = 0.0975 / √(6) = 0.00398 Standard Deviation of Eqn 2 = (.1605 - .1275) / 2 = 0.033 Standard error of Eqn2 = 0.033 / √(6) = 0.013472 T’ = |0.1454- 0.2193 |/√(0.033 2 - 0.00398 2 ) = 2.256 Conclusion: By using the mean and uncertainty of the force data during the IO Lab’s acceleration, the t’ value can be found. The t’ value of the data is 2.256. With the t’ prime value it can be concluded that the values likely happened if the parent distributions were the same. Therefore the variance in the value of work done is within the expected range and it can be concluded that the friction from the IOLab is negligible. The first method and second method yielded the same results, and the variance between the force was due to both the methods used and general experimental uncertainty. Both forces calculated were around .14 to .2 joules of work done. While we can be relatively confident that friction between the IOLab wheels is negligible, since the t’ value is somewhat high at 2.256, there's a chance that a small amount of friction is affecting trial 2’s results. However, this is again unlikely as the t’ prime is below 3. Ultimately the t’