The Atwood's Machine
Pe Procedure A (100+120+107) Measured system mass M 327_kg +0.001 kg Initial value of m, 120 kg 120 gram une value. Proportional error in M 3.058 x 10 Initial value of m,- 107 kg 107grar Proportional error in Am 7.7x10-30.00 0.013 kg +0.0001 kg Initial mass difference Am -AV a= At a- 1,(s) v (m/s) t(s) v (m/s) a (m/s) 0.227 0.77 0.358 0.820.38 156 10.49 1.4820.61 0.37 10.603 0.63 0.827/0.71 10.36 0.141 10.15 0.3520.82 0.33 mean value of a 0.36 mean m/s2 sample standard deviation in a m/s2
Procedure B exponential m, (kg) (m/s) Table for Additional Measurements m, (kg) Am (kg) 1 () , (m/s) 1. (S) v, (m/s) . 120 10.7·013 0.0271 0.77 0.258 0.82 0.38 . 130.112 . 018 1.008 10.93 . 118 99 1.58 1.135 • 124 .011 235 .62 uus .67 0.24 SS . 135 .02 1.212 .75 1.584 .१५ o.si 130 • 120 olo 2.068.46 2.53 .57 0.24 Note: Transcribe the data from Part A into the first row of the table above and include that point in your 10 E- graph. Note: Include appropriate units in your answers below. २६. slope of graph -Pisica. FFO FC calculated system mass (from slope) _DSCH OF.0 80.000 percent error in calculated system mass horizontal intercept of graph calculated friction force (from intercept) Does the value of the friction force resulting from this calculation lie within the range (calculated value plus or minus uncertainty) for the friction force that you found in Procedure A? & 0x80P 3-10
The Atwood's Machine
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