A block of mass 2 kg is attached to a spring of spring constant 100 N/m. The block is displaced by 0.2 m and released. What is its kinetic energy when it passes through the equilibrium position?
Q1. A block of mass 2 kg is attached to a spring of spring constant 100 N/m. The block is displaced by 0.2 m and released. What is its kinetic energy when it passes through the equilibrium position?
Answer: 2 J
Explanation: Kinetic energy = 1/2 kx² = 1/2 * 100 * 0.2² = 2 J, using conservation of energy.