A 2 kg block is attached to a spring with k = 50 N/m. If it is displaced by 0.1 m, what is the elastic potential energy?
Q1. A 2 kg block is attached to a spring with k = 50 N/m. If it is displaced by 0.1 m, what is the elastic potential energy?
Answer: 0.25 J
Explanation: Elastic potential energy = 0.5 * k * x^2 = 0.5 * 50 N/m * (0.1 m)^2 = 0.25 J. Option B is incorrect due to miscalculation.