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Roberts 1 Force, Mass, and Acceleration Physics 124 Section 004 Megan Roberts Samantha Hoehner Sanquetta Speiehts Abstract: The purpose of this lab is to show the relationship between force, mass, and acceleration. This is representative of Newton’s Second Law. A computer program is used to measure a velocity over a period of time. The acceleration can be derived from this. Newton’s second law is divided into its x and y components to give the net force on an object. From this law, a formula, g/a=(m1/m2) + 1 is derived to determine acceleration. Using just the x components, the acceleration due to gravity can be calculated with ax=gsin(Ө). The acceleration due to gravity was calculated to be 9.98 m/s^2 and the actual value is 9.8m/s^2. The experimental error in this portion of the experiment was 1.83%. This could be due to experimental or tools error. The y-intercept for the graph was -.006 rather than 0. For the second portion of the lab, m1 was calculated to be .347kg. The actual value was .33015kg. The percent difference was 8.87%. This could be due to tools error in the computer program or experimental error. The y-intercept for this graph was 1.03 and the expected y-intercept was 1.00. The Percent difference in this case was 2.96%. This could be due to experimental or tools error also. Roberts 2 Data: Exercise 1 H (cm) Ө SinӨ A (m/s^2) Vo (m/s) 2.57 1.49 0.026 0.250 0.052 2.88 1.68 0.029 .0290 0.081 3.24 1.89 0.032 0.310 0.088 4.22 2.45 0.043 0.423 0.123 4.57 2.66 0.046 0.453 0.125 Experimental Value of g: 9.98 Percent Error: 1.83% Experimental Value of g Acceleration vs. Sin(x) 0.5 a = 9.9801sinx - 0.0061 0.45 R2 = 0.9979 0.4 0.35 acceleration (m/s^2) 0.3 0.25 0.2 0.15 0.1 0.05 0 0.02 0.025 0.03 0.035 0.04 0.045 0.05 sin(x) Roberts 3 Exercise 2: Mass added m2 1/m2 A g/a m1/m2 + 1 (kg) (kg) (1/kg) (m/s^2) 0.02 0.0699 14.3 1.646 5.95 5.72 0.04 0.0899 11.1 1.979 4.95 4.67 0.06 0.1099 9.10 2.304 4.25 4.00 0.08 0.1299 7.69 2.678 3.66 3.54 0.10 0.1499 6.67 2.926 3.35 3.2 0.12 0.1699 5.89 3.173 3.09 2.94 0.14 0.1899 5.27 3.401 2.76 2.74 0.16 0.2099 4.76 3.540 2.74 2.57 Experimental m1: 0.347 Percent Difference: 8.87% Predicted y-intercept: 1.00 Actual y-intercept: 1.03 Percent difference: 2.96% Experimental Value of m1 g/a vs. 1/m2 (acceleration due to gravity/acceleration) vs. inverse of mass 2 7 g/a = 0.3472(1/m2) + 1.0309 6 R2 = 0.9973 5 4 g/a 3 2 1 0 0 2 4 6 8 10 12 14 16 1/m2 Roberts 4 Data Analysis: This experiment uses a computer program, air track, glider, and weights to determine accelerations and an initial mass. This experiment had two parts. Each part involved a specific derivation of Newton’s Second Law. Experiment 1 measures the acceleration due to gravity. The formula a=gsin(x) is used to determine g. Sin(x) is determined by recording the values for the height and distance of the legs in each portion of the experiment. This is shown in table 1. A computer program is used to calculate acceleration by plotting a graph of velocity versus time. This data is also in table 1. Graph 1 is acceleration vs. sin(x). This graph is used to get the arbitrary value of g, which is the slope of the line. This value is 9.981m/s^2. Since the actual value of acceleration due to gravity is 9.8m/s^2, the percent error in this portion of the experiment is 1.83%. Experiment 2 finds the mass of the glider that is sliding on the air track. First, the mass is found using a balance for comparison sake. Then, the same computer program used in Experiment 1 is used to plot a velocity vs. time graph for different masses of force. This data is used to determine the mass of a glider. This data is in table 2, and the graph used to determine the mass of the glider is in graph 2. The graph is also used to experimentally calculate a y-intercept. The y-intercept tells the initial velocity of the object. Since the object is being “pulled” by mass 2 and equation 5.10 represents this fact, the y-intercept was predicted to be one. The actual numbers calculated were quite similar to the predicted values. The calculated mass is .347kg whereas the mass determined by the balance was .33015 kg. The percent difference between the two was 8.87%. The y intercept found experimentally was 1.03 and the predicted y-intercept was 1. This percent difference was 2.96%. Roberts 5 Error Analysis: In Exercise 1, the percent error of the acceleration due to gravity calculated to the true is found. This is done by . For example (9.8-9.981)/(9.8)*100= 1.83%. Since this percent is much less than 15%, the experiment is valid. In Exercise 2, the percent difference between the experimental and calculated value of m1 and the percent difference between the predicted and graphed y- intercept value are found. The percent difference is calculated by The percent difference of the m1 values was 8.87% and the percent difference of the y- intercept values was 2.96%. Sources of Error: Random Error: Arbitrary error in the computer calculations or one specific data point being off in either of the two experiments. Experimental error: The y-intercept of the first graph is less than 0 and the y- intercept of the second graph is greater than 1, therefore, some degree of experimental error occurred. Questions: 1. Newton’s Second Law states that the sum of F=ma. If the mass of the glider in Exercise 1 were to double, how would the glider’s acceleration be affected? What if the mass were halved? a. If the mass were doubled, the acceleration would be halved, and if it were halved, the acceleration would double. Roberts 6 2. In Exercise 1, the computer calculated the slope and y-intercept of the velocity versus time plot. The slope is the average acceleration that the glider experienced. What does the y-intercept tell us? a. It tells us the initial velocity. 3. In Exercise 1, you found the acceleration due to gravity by determining the slope of your graph. What was the y-intercept of your graph? Explain what a non-zero y-intercept would indicate about the experimental setup. a. The y-intercept is -.0061 and this means that there was an initial velocity due to experimental error. 4. Derive equations 5.8 and 5.9 for the acceleration and tension in Exercise 2. a. Since ∑Fx = T = m1ax and ∑Fy = m2g-T = m2ay and ax=ay=a, then m2g- m1a = m2a and then a=m2g/(m1 + m2) and T = (m1m2g)/(m1 + m2) 5. Examine equation 5.8. What can be inferred about the mass of the glider, m1, if the acceleration of the glider equals that of gravity a=g? a. It can be inferred that m1 is equal to m2 in this instance because then all that is left in the equation is a=g. 6. Examine equation 5.8. What can be inferred about the mass of the glider, m1, if the acceleration of the glider is zero? a. It can be inferred that m1 is much greater than m2 and m2’s mass is no longer significant in applying a “pull” on m1 great enough to accelerate the object. 7. Show that equation 5.8 can be rewritten as equation 5.10. a. Since a=m2g/(m1+m2) then a(m1 + m2) = m2g and (m1 + m2) / m2 = m2g. Then, (m1/m2) + 1 = (g/a) which is equation 5.10.
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