Potential Energy & Equilibrium: Practice Problem & Solution
When a spring is subjected to \(4\text{ N}\) force its length is \(a\text{ metre}\). And if \(5\text{ N}\) is applied length is \(b\text{ metre}\). If \(9\text{ N}\) is applied length is: (1999)
Solution Explained:
To solve this problem, we apply the core principles of Potential Energy & Equilibrium. Understanding the underlying formula is key to arriving at the correct answer below:
Concept: Hooke's Law. Formula: \(F = k(L - L_0)\), where \(L_0\) is original length. We have: (1) \(4 = k(a - L_0)\), (2) \(5 = k(b - L_0)\). From (1) and (2), we find \(L_0 = 5a - 4b\) and \(k = \frac{1}{b - a}\). For \(F=9\text{ N}\), \(9 = k(L_3 - L_0)\). Substitute \(k\) and \(L_0\): \(9 = \frac{1}{b - a}(L_3 - (5a - 4b))\). Solving for \(L_3\), we get \(L_3 = 9(b - a) + 5a - 4b = 9b - 9a + 5a - 4b = 5b - 4a\).
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