Energy stored in springs with different constants and same applied force – Rankers Physics

Potential Energy & Equilibrium: Practice Problem & Solution

Two springs \(A\) and \(B\) having spring constant \(K_A\) and \(K_B\) (\(K_A = 2K_B\)) are stretched by applying force of equal magnitude. If energy stored in spring \(A\) is \(E\) then energy stored in \(B\) will be: (2001)
\(2E\)
\(E/4\)
\(E/2\)
\(4E\)

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: Energy stored in a spring under constant force. Formula: \(PE = \frac{F^2}{2K}\). When the same force \(F\) is applied, potential energy is inversely proportional to the spring constant (\(PE \propto 1/K\)). Given \(K_A = 2K_B\). The ratio \(\frac{PE_B}{PE_A} = \frac{K_A}{K_B}\). Substituting \(K_A = 2K_B\), we get \(\frac{PE_B}{E} = \frac{2K_B}{K_B} = 2\). Therefore, \(PE_B = 2E\).

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