A horizontal spring with stiffness 0.5 N/m has a relaxed length of 19 cm (0.19 m). A mass of 22 grams (0.022 kg) is attached and you stretch the spring to a total length of 26 cm (0.26 m). The mass is then released from rest. What is the speed of the mass at the moment when the spring returns to its relaxed length of 19 cm (0.19 m)?

 v = 0.0147 m / s

Explanation:

For this exercise let's use energy conservation

Starting point. Fully stretched spring

            Em₀ = Ke = ½ k (x-x₀)²

Final point. Unstretched position

          Emf = K = ½ m v²

          Emo = Emf

         ½ k (x- x₀)² = ½ m v²

           v = √m/k    (x-x₀)

Let's calculate

            v = √(0.022 / 0.5)      (0.26-0.19)

            v = 0.0147 m / s

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