A spring is stretched by 0.2 m. If its spring constant is 50 N/m, what is the energy stored?
Q1. A spring is stretched by 0.2 m. If its spring constant is 50 N/m, what is the energy stored?
Answer: 1 J
Explanation: Energy stored = 1/2 kx² = 1/2 * 50 * 0.2² = 1 J, using the formula for energy stored in a spring.