Potential Energy & Equilibrium: Practice Problem & Solution

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:

The energy stored in a spring is \(U = \frac{1}{2}kx^2\). If stretched by 1 cm, \(U_1 = 50\text{ J}\). If further stretched by 1 cm, total stretch becomes 2 cm, so the stored energy is \(U_2 = \frac{1}{2}k(2x)^2 = 4U_1 = 200\text{ J}\).