Atwood’s Machine Data Partner Name
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y = ½ a t2; e.g. 100 cm = ½a(3.08s)2 ; a = 21.1 cm/s2 y = _____ cm (this can change in Part A,B,C, but must be recorded each measurement if changing.) |
FNet in Dynes à “g” = 980 cm/s2 FNet in Newtons à “g” = 9.8 m/s2 |
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Part A: Total Mass Held Constant m1 + m2 = ____ grams Plotting FNet vs acceleration |
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mass 1 |
mass 2 |
m1–m2 |
FNet |
time |
accel |
Use LineFit: Plot FNet vs accel |
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98+50 = 148 g |
102+50 = 152 g |
0.004 kg |
9.8*0.004 = 0.0392 N |
s |
m/s2 |
y-int ( ) |
x-int (m/s2) |
slope (kg) |
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mpulley |
Ffriction |
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Part B: Net Force Held Constant (m1 – m2 = _____ grams) Plotting 1/a vs m1 + m2 |
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mass 1 (grams) |
mass 2 (grams) |
time (s) |
accel ( ) |
1/a (s2/m) |
m1 + m2 ( ) |
Hand Graph: Never force line through origin |
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y-int |
x-int |
slope (from data) |
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FNet à (m1 – m2) g |
FNet (from graph/data includes friction) |
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mpulley |
Ffriction |
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LineFit (or Excel) for comparison |
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y-int |
x-int |
slope |
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Part C: Predicting the mass distribution that will yield a given acceleration |
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eq. 4 (m1 – m2)g – Ff = mT a |
mtotal (grams) |
mass1 (grams) |
mass2 (grams) |
Slope (cm/s2) |
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Predicted (calculated) |
200 |
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60.0 |
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Experimental (Actual mass) |
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Since you have a limited number of masses in your mass holder set, actual m1 & m2 usually varies slightly from calculated m1 & m2 |
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For predicted Part C, please use mpulley from Part A and Ff from Part B; Reasoning: unknowns obtained from slopes typically have less uncertainty than from intercepts. |
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