Question

A horizontal spring-mass system oscillates on a frictionless table. If the ratio of the mass to the spring constant is 0.042 kg·m/N, and the maximum speed of the mass was measured to be 5.9 m/s, the maximum extension of the spring will be 25 cm

Answer 1

$$0.5mv^2=0.5kx^2\\x=v\sqrt{\frac{m}{k}}=5.9\sqrt{0.042}=1.23\ m$$