Force And Extension Experiment Coursework

Force And Extension Experiment Coursework-16
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).

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).

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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.

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.

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.

Also read your specification as to what they expect from conclusions, it varies arbitrarily between exam boards To add to the very sound advice above (#2), I would also suggest, if it's possible, that you go back and measure the extension of the 2nd spring separately.

From the results it does look as though the springs were not identical.

You could then confirm this, and at the same time show that your results with the 2 springs, both series and parallel, are consistent with this.

Even if it's not possible to take these measurements, you should still be able to conclude that the springs were not of equal stiffness.

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