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Course: PHY 122, Spring 2008
School: ASU
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Word Count: 584

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Report Lab 1 Introduction The objective of this lab is to determine the static and kinetic friction of an object with a known mass being dragged across a table. Materials Force sensor Woodblock Weight String Data Studio Program Procedure 1. Calibrate force sensor with Data Studio Program using 1 kg weight 2. Attach string to woodblock and force sensor 3. Place weights on the woodblock 4. Press the Tare button to...

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Report Lab 1 Introduction The objective of this lab is to determine the static and kinetic friction of an object with a known mass being dragged across a table. Materials Force sensor Woodblock Weight String Data Studio Program Procedure 1. Calibrate force sensor with Data Studio Program using 1 kg weight 2. Attach string to woodblock and force sensor 3. Place weights on the woodblock 4. Press the Tare button to zero the force senor 5. Start recording on the Data Studio Program 6. Pull the force sensor, dragging the woodblock and weights. Try to make sure you are pulling at a consistent rate to get more accurate data. Pull for a few inches then stop, and then repeat this procedure 50 times. 7. You will have to stop the Data Studio Program and record the data by hand once the program stops recording. 8. Use the program to calculate the kinetic and static friction. Static friction is the peak of each slope, and the static friction is the average of each point on the plateau- like area following the peak. 9. Calculate the mean, median, and one standard deviation for static and kinetic friction based on your results using the Data Studio Program. 10. Calculate the uncertainty. 11. Create a histogram of your data using the program. 12. Print out the histogram and label the mean and + and one and two standard deviations. Data Analysis Our data supports the fact that static friction is notably greater than kinetic friction. Our average value for static friction is 4.76 N while the average for kinetic was only 3.16 N. Not only did our mean value support this, but all 50 trials we completed for the experiment did as well. Our median values for static and kinetic friction, 4.69 and 3.12, respectively, also stand up to this idea and are very close to the mean we values obtained. The coefficient for static friction, .430, is also greater than that of kinetic friction, .285. Although this experiment is anticipated to generate a high number of errors, a majority of our data fell within one standard deviation of the mean, showing that the experiment was run fairly consistently. This can be seen on the histogram created for kinetic friction. The reason that a high uncertainty in the data is expected from this experiment is that it is difficult to run each trial consistently. A human has to pull the force sensor, and it is very likely that the human conducting this part of the experiment is not perfectly consistent in the amount of force they pull with. Also, it is likely that friction changes on different points on the table. The surface may be more or less rough, or have some substance on it that could later the friction. This experiment was fairly consistent; however the histogram shows that some trials varied considerably. Fortunately, because so many trials were run, these out of the ordinary results did not have a very significant impact on the results. Conclusion It can confidently be concluded that static friction is indeed greater than kinetic friction. Not only did our calculations lead to this statement, but so did every individual trial. The mean, median, and coefficient all suggest that static friction is greater. The histogram verifies this conclusion because it shows that the experiment was fairly accurate and consistent. This is a reasonable statement to conclude because Newton's first law of motion states that an object in equilibrium will stay that way unless an external force affects it. Once the wooden block and weights are in motion, less force is required to maintain this velocity.
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. .&quot;.= 3
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To get z, andI-1RI4RII2 = 01-0- z, = -=(a)V 1 + 6 (1 (4+2) = 4 R I 1, I, = - I , ,1 . 2v, = 21,=I,To get z2, and z, , consider the circuit in Fig. (b).I1=04Rz, = -= 2 (1 (4 + 6) = 1.667 R v 212 I,'
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[f3* 3 1 3I bp3 - 3 3 34-cAv a'3.Wa\Jc
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A2\Jf&quot; 6O Y,\AI Iza +A4w ,=3 Rc A
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, J( l O + kJT==33.334