A 2 kg block is attached to a spring of spring constant 100 N/m. If it is displaced by 0.2 m, what is the potential energy stored?
Q1. A 2 kg block is attached to a spring of spring constant 100 N/m. If it is displaced by 0.2 m, what is the potential energy stored?
Answer: 2 J
Explanation: Potential energy = 1/2 kx² = 1/2 * 100 * 0.2² = 2 J, using the formula for spring potential energy.