Please write out each step CLEARLY....Thank you!!! Part C Data TANETRALTRAIL05TC
ID: 3281215 • Letter: P
Question
Please write out each step CLEARLY....Thank you!!!
Explanation / Answer
#1. for all the cases average time to move the distance S is as under
Case 1 : (1.77 + 2.24 + 1.71)/3 = 1.9067 s
Case 2 : (1.78 + 1.51 + 1.78)/3 = 1.69 s
Case 3 : (1.38 + 1.26 + 1.58)/3 = 1.406 s
Case 4 : (1.25 + 1.25 + 1.13)/3 = 1.21 s
Case 5 : (2.69 + 2.23 + 3.32)/3 = 2.7466 s
#2 aceeleration is given by
a = 2s/T^2
SO FOR ALL THE CASES, s = 26 m
Case 1 : a1 = 2*26/1.9067^2 = 14.303377 m/s/s
Case 2 : a2 = 2*26/1.69^2 = 18.20664 m/s/s
Case 3 : a3 = 2*26/1.406^2 = 26.30466 m/s/s
Case 4 : a4 = 2*26/1.21^2 = 35.5166996 m/s/s
Case 5 : a5 = 2*26/2.7466^2 = 6.893 m/s/s
#3 from equation 8, da/a = ds/S + 2dt/t
so dt is the stnadard deviation ( found from excel ) in t, while dS = 0
case 1: da/14.30337 = 2*0.290229794/1.9067
da = 4.35439 m/s/s
Case 2: da/18.20664 = 2*0.155884573/1.69
da = 3.358738 m/s/s
Case 3: da/26.30466 = 2*0.161658075/1.406
da = 6.048877 m/s/s
Case 4: da/35.516699 = 2*0.069282032/1.21
da = 4.0672102 m/s/s
Case 5: da/6.839 = 2*0.547205019/2.7466
da = 2.725067 m/s/s
#4 from equation 6
coefficient of kinetic friciton k =[ml(g - a) - (ml + mb)a ]/(mb + ml)g
as the acceleration is greater than g ( which is not possible theoretically) the value obtained for k is -ve in all cases
case 1: -267.5454545
case 2 : -430.3524375
case 3 : -534.520969
case 4 : -633.7720009
case 5 : -246.3925898