Ave. Time Acceleration Acceleration Normal Force Objective: To examine our model of friction by investigating the sliding of a block a Uncertainty N Title: up a ramp. Ru (s) m/s^2) 8a (N) (m/s^2) Raw Data: 2.08 1.57 0.65 0.08 Date: Distance of travel s = 0.8 m 2.08 2 1.63 0.60 0.07 Ramp Block Hanging Times 3.85 Angle 6 Height h Mass M Mass mz t1, tz, t3 0.07 Name: Run (deg) (cm (kg) (kg (s) 3 1.68 0.57 1.88 25 42.3 0.2338 0.2 1.63, 1.53, 1.55 4 1.85 0.47 0.05 Partner(s): 1.73, 1.66, 1.50 1.62 .06 2 25 42.3 0.2338 0.2 5 1.71 0.55 1.70, 1.68, 1.66 3 25 42.3 .4338 0.35 1.82, 1.87, 1.86 4 35 57.4 0.2338 0.22 1.66, 1.75, 1.71 Friction Force Friction Force Coefficient 5 70.7 0.2338 0.25 Uncertainty Run (N 1 0.710 .035 0.341 + 0.017 Sample: h = (100 cm) sine = (100 cm) sin (25) = 42.3 cm 2 0.73 0.030 0.351+0.014 3 1.187 0.055 0.308+0.014 4 0.629 0.023 0.335+0.012 5 0.564 0.029 .348+0.018 Calculated Data:Plot: Sample Calculations: Plot 1: Average Time (Plot the calculated { vs. / values. Also, include the point (0,0). Calculate the slope of the line. According to our theoretical model, the slope is the coefficient of kinetic friction t = (t1 + t2 + 13) / 3 = (1.63s + 1.53s + 1.55s) / 3 = 1.57s since fx = uk N. Record the value of uk obtained from your line.) Acceleration Results: a = 2s/ f2 = (2 * 0.8m)/(1.57s^2) = 0.65 m/s^2 Plot 1: Acceleration Uncertainty (Report the slope of the line which is the coefficient of friction.) 6a = 2a (8t / t) = (2 *0.65m/s^2)(0.1s/1.57s) = 0.08m/s^2 Run 2: Normal Force (We now see if the area of contact between the ramp and block makes a difference. Are the two coefficient of friction values the same when their respective uncertainties N = Mg cos e = 0.2338kg * 9.8 * cos(25degrees) = 2.08N are considered? Is the observation made in 2-1 consistent with our model of friction?) Friction Force Run 3: fr = (m2 - M sin 0)g - (m2 + M)a = (0.2kg - 0.2338 * sin(25deg)9.8- (We now see how a change in mass affects the results. Is the normal force larger, (0.2kg+0.2338)0.65m/s^2 = 0.710N smaller or the same compared to when the block was lighter? Does this agree with theory? Is the kinetic friction force larger, smaller, or the same? Does this agree with Friction Force Uncertainty theory? Is the coefficient of kinetic friction larger, smaller, or the same? Consider the uncertainties of each value. Does this agree with our theoretical model?) of* = (m2 + M) ba = (0.2kg + 0.2338kg) (0.08m/s^2) = 0.035N Runs 4-6 Coefficient of Friction (We now see how the ramp angle affects the results. Is the normal force constant? Does it increase or decrease as the ramp angle increases? What happens to the UK = tk/N = (0.710N)4(2.08N)= 0.341 uk friction force as the ramp angle increases? What should happen to the coefficient of friction as the ramp angle increases? Do you observe this trend if you consider the Coefficient Uncertainty ranges of each coefficient?) 6uk = 8tx/ N = (0.035N/2.08N) = 0.017 8uk Uncertainties: (What were the sources of measurement uncertainty? Which source was the largest contributor to measurement uncertainty? Does the measurement uncertainty alone account for any differences between calculated values and accepted values?)