Question

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)?

Answer 1

Answer:

 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

Answer 2

The speed of the mass at the mean position is 0.333 m/s

Conservation of energy:

The potential energy stored in a fully stretched spring

PE = ½ kx²

where x is the stretch of the spring  = 26 -19 = 7 cm = 0.07 m

At the mean position, where x = 0, the PE stored in sprig is zero,

So according to the law of conservation of energy total energy must remain conserved so all the energy is converted into kinetic energy KE of the mass

KE = ½ mv²

where m is the mass and v is the velocity

½ kx² = ½ mv²

where k is the spring constant = 0.5 N/m

and m is the mass = 0.022 kg

$v=\sqrt{\frac{k}{m} } x$

$v=\sqrt{\frac{0.5}{0.022} } 0.07$

v = 0.333 m/s

Learn more about conservation of energy:

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