A block is attached to a spring of spring constant 50 N/m. The block is displaced by 0.2 m and released. What is the force acting on the block at x = 0.1 m?
Q1. A block is attached to a spring of spring constant 50 N/m. The block is displaced by 0.2 m and released. What is the force acting on the block at x = 0.1 m?
Answer: -5 N
Explanation: Force = -kx = -50 * 0.1 = -5 N, using Hooke's Law.