The fractional change in length produced in an elastic material of spring constant 680Nm\(^{-1}\) when a force of 306N is applied to stretch it is 1.5. Calc...
Question 1 Report
The fractional change in length produced in an elastic material of spring constant 680Nm\(^{-1}\) when a force of 306N is applied to stretch it is 1.5. Calculate the original length of the material.
Answer Details
From Hooke's law, F = kx where x = extension, k = force constant
K = 680Nm\(^{-1}\), F = 306N
x = \(\frac{\text{F}}{\text{x}}\) = \(\frac{306}{680}\) ≈ 0.450m
Fractional Change in Length: Given the fractional change is 1.5:
1.5 = \(\frac{\text{x}}{\text{L}}\)
L = \(\frac{0.450}{1.5}\) = 0.30m.
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