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A horizontal spring with stiffness 0.3 N/m has a relaxed length of 17 cm (0.17 m). A mass of 22 grams (0.022 kg) is attached and you stretch the spring to a total length of 24 cm (0.24 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 17 cm (0.17 m)?

Respuesta :

Answer:

The speed of the coin is 0.2585 m/s when the spring returns to its relaxed position.

Explanation:

The length the spring is extended by = 0.24 - 0.17 = 0.07 m

Stiffness of spring = 0.3 N/m

In order to solve this, we simply need to take convert all the potential energy of the spring into kinetic energy for the coin. This is:

Potential energy = 0.5 * k * x^2

Potential energy = 0.5 * 0.3 * 0.07^2

Potential energy = 0.000735 Joules

Since this is equal to kinetic energy we have:

Kinetic energy = 0.000735

0.5*m*V^2 = 0.000735

0.5 * 0.022 * V^2 = 0.000735

V^2 = 0.06682

V = 0.2585 m/s Or 28.85 cm/s

The speed of the mass at the moment when the spring returns to its relaxed length of 17 cm is 28.85 cm/s.

"Energy"

Total length is X= 0.24 - 0.17

Total length is X= 0.07 m

Horizontal spring is K= 0.3 N/m

Formula:

Potential energy= 1/2 * k * x^2

Potential energy = 0.5 * 0.3 * 0.07^2

Potential energy = 0.000735 Joules

Therefore ,

Potential energy = Kinetic energy

Kinetic energy = 0.000735

0.5*m*V^2 = 0.000735

0.5 * 0.022 * V^2 = 0.000735

V^2 = 0.06682

V = 28.85 cm/s

The speed of the mass at the moment when the spring returns to its relaxed length of 17 cm is 28.85 cm/s.

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