Potential energy of a stretched spring – Rankers Physics

Potential Energy & Equilibrium: Practice Problem & Solution

The potential energy of a long spring when stretched by \(2\text{ cm}\) is \(U\). If the spring is stretched by \(8\text{ cm}\) the potential energy stored in it is: (2006)
\(4U\)
\(U/8\)
\(16U\)
\(U/4\)

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: Potential energy stored in a spring. Formula: \(PE = \frac{1}{2}kx^2\). Potential energy is proportional to the square of the extension (\(PE \propto x^2\)). Given \(PE_1 = U\) for \(x_1 = 2\text{ cm}\). We need \(PE_2\) for \(x_2 = 8\text{ cm}\). \(\frac{PE_2}{PE_1} = \left(\frac{x_2}{x_1}\right)^2 = \left(\frac{8\text{ cm}}{2\text{ cm}}\right)^2 = (4)^2 = 16\). Therefore, \(PE_2 = 16U\).

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