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Questions 4 and 5 QUESTIONS: 1.Plot a graph of the force acting on the spring versus...

Questions 4 and 5
QUESTIONS: 1.Plot a graph of the force acting on the spring versus spring extension. 2. (a) Use the graph to calculate the sp

Force Acting on a Spring (N) vs Spring Extension (cm) LINEAR x-range: 3.92-637 N y=mx+b Spring winston lom) me 7,638 be -25.8

QUESTIONS: 1.Plot a graph of the force acting on the spring versus spring extension. 2. (a) Use the graph to calculate the spring constant including units. () Does the spring obey Hooke's Law over the range of applied forces used? Justify your response. (e) Write the equation for the spring which describes the relationship between extension and applied force to the spring. 3. (a) Use the spring equation to calculate the extension of the spring when a force of 20 kg was attached to the spring, (b) What is the name of the process if the answer to part (a) was obtained using the graph? (c) Does the answer obtained in part (a) represent a reasonable answer to the question in part(a)? Justify your response. 4. If the area beneath the graph line represents the work done on the spring and the elastic potential stored in the spring, develop an equation for the stored potential energy in a stretched or compressed spring, where E, -f(x,k). 5. Why was the length of the unstretched spring measured when it was in the horizontal position rather than the vertical position?
Force Acting on a Spring (N) vs Spring Extension (cm) LINEAR x-range: 3.92-637 N y=mx+b Spring winston lom) me 7,638 be -25.8 0.9993 RMSE 0 2928 40 42 50 52 54 56 58 60 62 Force Acting on Spring (NO
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Answer #1

[4] the work done + potential energy due to extension of spring= Area beneath the graph plotted between Force vs spring extension

\int_{0}^{x}F.dl + V = \frac{1}{2}\times (dF)\times (dx) = \frac{1}{2}\times (2.5)\times (17.5) =21.875

As we know that the force = kx for spring hence,

\int_{0}^{x}kx.dx + V = 21.875

0.5 kx2 + V = 21.875

hence V = 21.875- 0.5kx2

[5] In vertical position extension in the spring will also be due to the gravity as mg will act on the spring downwards. Hence it was measured in horizontal position where gravity does not affect the spring's extension.

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