Potential Energy & Equilibrium: Practice Problem & Solution
Two similar springs \(P\) and \(Q\) have spring constants \(K_P\) and \(K_Q\) such that \(K_P > K_Q\). They stretched first by the same amount (case a), then by the same force (case b). The work done by the springs \(W_P\) and \(W_Q\) are related as in case (a) and case (b), respectively: (2015)
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: Work done to stretch a spring. Formula: \(W = \frac{1}{2}Kx^2\) and \(W = \frac{F^2}{2K}\). Given \(K_P > K_Q\). Case (a): Same extension \(x\). \(W propto K\), so \(W_P > W_Q\). Case (b): Same force \(F\). \(W \propto 1/K\), so \(W_P W_P\). Combining these, the correct option is \(W_P > W_Q\); \(W_Q > W_P\).
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