Physics Lab Spring

Physics lab 16 November 2015 Introduction Springs were investigated by Hooke about 300 years ago. He found that the extension was proportional to the stretching force so long as the spring was not permanently stretched. This means that doubling the force doubles the extension trebling the force treble the extension and so on. Hooke’s law is true only if the elastic limit of the spring is not exceeded, i.e. if the spring returns to its original length when the force is removed. This experiment/observation was to find out if the extension is proportional to the stretching force. Hypothesis I predict that when more weight is added, the spring’s extension would increase thus following the Hooke’s Law. Apparatus -Clamp -Boss -Stand -Load/weight/hanger - (Steel) spring -Ruler/scale Method 1. 2. 3. 4. 5.

Arrange a steel spring as in fig.1. Read the scale opposite the bottom of the hanger. Add 100g loads one at a time (thereby increasing the stretching force by steps of 1 N). Take the reading after each one. Enter the readings in the table for loads up to 600g.

Data and results Single spring Total hanging mass (g)

Total hanging mass (kg)

Equilibrium length Total force=mg Stretched length G=10N/kg (N) (m) 1

Extension (m)

0 100 200 300 400 500 600 700 800

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0N 1N 2N 3N 4N 5N 6N 7N 8N

2 cm 3.5 cm 5.5 cm 7.6 cm 9.6 cm 11.5 cm 13.5 cm No data

0 1.5 cm= 0.015m 3.5 cm=0.035m 5.6 cm= 0.056m 7.6 cm=0.076m 9.5 cm=0.095m 11.5 cm= 0.115m

Analysis (see graph attached at back) From the graph, we obtained a straight line. This shows that the extension of the spring is directly proportional to the total force (mg); Slope=rise/run = (6-1)/ (0.115-0.015) =50N/m Therefore, the extension(x) α total force (F) that is: F=k(x+0.005), where k=50N/m, because the line passes through the x-axis at -0.005m. We may use this result to measure the mass of an object by using this formula. For example: If we put a mass (M) on the spring and the extension of the spring is 10cm (0.1m), then the mass would be given by F=k(x+0.005), Mg=k(x+0.005) M=k(x+0.005)/g= (50 N/m) *0.105m/ (10 N/kg) =0.525kg

Conclusion From the results, a straight line is obtained. However, the place where the line intersects the xintercept is at -0.005m.Which means that the spring still followed the Hooke’s law except that 2

from the results, the equation will differ from mg=kx to mg=k(x+0.005). The outcome is most likely affected due to our initial measurements. Maybe, we have some error in measuring the length of the spring at 0 N and therefore the x-axis will move to the left a bit. When we doubled or tripled the load, the strength doubled and tripled proportionally following the Hooke’s Law. From the graph, we can work out the gradient: Gradient =rise/run = (6-1)/ (0.115-0.015) = 50 N/m When large loads are added to the spring, the extension would be stretched longer depending on the mass of the load till it exceeds it limit of proportionality and reached its elastic limit where the load is too large and therefore damaging the spring to a point where it could not return its original form and thus leading the straight line on the graph to curve if we continue this experiment by adding more larger loads to the hanger. If we replace the spring with a stiffer spring, the results would change. The line of best fit would go flatter closer to the x-axis as the extension would be less than its original spring. On the other hand, if you replaced the spring with a weaker spring, the result would differ in an opposite way. The line of best fit would be steeper as the extension would increase because there is a weaker force holding the spring together even at its original form.

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