Hooke's Law Experiment Report

3886 Words Mar 2nd, 2013 16 Pages
Hooke’s Law Experiment Report
Done by Yovaphine Wijaya – 11 Science 1

Aim
To investigate Hooke’s law for simple strings or rubber.

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. 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. 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
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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. Calculation table of the spring constant of spring for 30.00g ± 0.01g of load Trial | Force (N) | Final Length of Spring (m), Δl = 0.0005m | Extension (m), Δx = 0.0010m | Spring Constant of Spring (N/m), k | k - k ̅ (N/m) | (k - k ̅)2 (N/m) | 1 | 0.29 | 0.200 | 0.060 | 4.9 | -1 x 10-2 | 1 x 10-4 | 2 | | 0.200 | 0.060 | 4.9 | -1 x 10-2 | 1 x 10-4 | 3 | | 0.198 | 0.058 | 5.1 | 2 x 10-1 | 3 x 10-2 | 4 | | 0.201 | 0.061 | 4.8 | -9 x 10-2 | 8 x 10-3 | 5 | | 0.200 | 0.060 | 4.9 | -1 x 10-2 | 1 x 10-4 | 6 | | 0.198 | 0.058 | 5.1 | 2 x 10-1 | 3 x 10-2 | 7 | | 0.200 | 0.060 | 4.9 | -1 x 10-2 | 1 x 10-4 | 8 | | 0.202 | 0.062 | 4.7 | -2 x 10-1 | 3 x 10-2 | 9 | | 0.200 | 0.060 | 4.9 | -1 x 10-2 | 1 x 10-4 | 10 | | 0.200 | 0.060 | 4.9 | -1 x 10-2 | 1 x 10-4 | | TOTAL | | | 49.1 | | 9 x 10-2 |

Example of the calculation of the spring constant of spring for 30.00g ± 0.01g of load, 1st trial

Extension x = x1 – x0 = (0.200m ± 0.0005m) – (0.140m ± 0.0005m) = 0.060m ± 0.0010m

Force
F = mg = (0.030kg ± 0.00001kg) x 9.8m/s2 = 0.294N = 0.29N (2 SF)

Spring constant of spring k = Fx = 0.294N0.060m =

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