A 2 kg block is attached to a spring of spring constant 200 N/m. If it is displaced by 0.1 m, what is the potential energy stored?
Q1. A 2 kg block is attached to a spring of spring constant 200 N/m. If it is displaced by 0.1 m, what is the potential energy stored?
Answer: 1 J
Explanation: Potential energy = 1/2 kx² = 1/2 * 200 * 0.1² = 1 J, using the formula for spring potential energy.