Force And Extension Experiment Coursework

Force And Extension Experiment Coursework-16
Parallel 0.045 * 47 = 2.12 we have closest force to what we originally added.Evaluation Many errors could have occurred during this process.Here, you can then take out the common factor, F and divide both sides by k F which gives you a final formula: 1 / k = 1 / k1 1 / k2 .

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Prediction Single Spring: Hooke's law, where F = kx.

I predict that I if I plot Force on the Y axis and extension, x, on the X axis, it will be a straight line and the gradient will be the spring constant.

When we do the maths for double the mass 0.025 * 364 In series 0.011 * 202 = 2.22 The force is just over what would be expected.

This is by about the same amount as a single spring.

Hypothesis The change in length of spring is directly proportional to the applied so that it will cause greater change in length of the spring for greater force applied.

Force And Extension Experiment Coursework Bear Writing Paper

It is supported by the formula of force, F = kx, where F is the applied force, k is the spring constant of the spring, and x is the change in length or extension of the spring.Hooke's Law I have designed the experiment to measure spring constants when the springs are in series and in parallel.The theory is based on Hooke's law which is: F = kx where F = Force, k = Constant and x = Extension [Ref. Unfortunately with the springs I have, I can only measure extension, not compression for which Hooke's is also valid.Still fairly clear is that two identical springs in series will extend twice as much for the same force (k/2)...So draw your graphs, find your gradients, give the values for each plus an uncertainty, tell us how well your results fit the predictions above.Parallel Springs: In the diagram above, T1 is the tension in spring 1, and T2 is the tension is spring 2. I assume the two springs are originally the same length and the extensions are the same in both springs. So by knowing this, you get the formula: F = k1x k2x = x (k1 k2).So overall, the spring constant for the two springs is k1 k2.Since the spring used is the same, the spring constant will always be the same for any value of force applied and extension of the spring.Theory The relationship between a load force and a light spring (F = kx) was the first determined by Robert Hooke in the 17th Table of final length of spring for 200.00 ± 0.01g of load Trial | Final Length of String (m), Δl = 0.0005m | 1 | 0.539 | 2 | 0.539 | 3 | 0.540 | 4 | 0.540 | 5 | 0.538 | 6 | 0.542 | 7 | 0.540 | 8 | 0.540 | 9 | 0.540 | 10 | 0.541 | Data Processing Table 6.Would some one be so kind as to give a little help.Single spring Mass (kg) Weight (N) Length (mm) Extension (mm) 0 0 406 0 0.05 0.5 418 12 = 12 0.1 1.0 438 20 = 32 0.15 1.5 458 20 = 52 0.2 2.0 477 19 = 71 0.25 2.5 497 20 = 91 0.3 3.0 537 40 = 131 0.35 3.5 553 16 = 147 0.4 4.0 569 16=163 0.45 4.5 590 21= 184 0.5 5.0 613 23 = 207 0.55 5.5 624 11 = 218 0.6 6.0 645 21 = 239 In parellel Mass (kg) Weight (N) Length(mm) Extension (mm) 0 0 403 0 0.05 0.5 421 18 0.1 1.0 430 9 = 27 0.15 1.5 441 11=38 0.2 2.0 450 9=47 0.25 2.5 462 12= 59 0.3 3.0 471 9= 68 0.35 3.5 482 11=79 0.4 4.0 492 10=89 0.45 4.5 502 10=99 0.5 5.0 512 10=109 0.55 5.5 523 11= 120 0.6 6.0 534 11= 131 In series Mass (kg) Weight (N) Length (mm) Extension (mm) 0 0 451 0 0.05 0.5 552 72 =72 0.1 1.0 562 40 =112 0.15 1.5 603 41 = 153 0.2 2.0 652 49 = 202 0.25 2.5 691 39 = 241 0.3 3.0 730 39 = 280 0.35 3.5 775 45 =325 0.4 4.0 814 39 = 364 0.45 4.5 848 34 = 398 Conclusion Single spring.

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