A block of mass 2.60 kg is placed against a horizontal spring of constant k = 725 N/m and pushed so the spring compresses by 0.0500 m.

What is the elastic potential energy of the block-spring system (in J)?
_______J

If the block is now released and the surface is frictionless, calculate the block's speed (in m/s) after leaving the spring.
_____s/s

Respuesta :

The elastic potential energy of the block-spring system is 0.906 J.

Velocity of the block , v = 0.83 m/s.

What is elastic potential energy?

Elastic potential energy is the energy stored in a stretched or compressed elastic material.

  • Elastic potential energy = Ke²/2

The elastic potential energy of the block-spring system = 725 * 0.05²/2 The elastic potential energy of the block-spring system = 0.906 J

The elastic potential energy of the spring is converted to kinetic energy of the block.

1/2 mv² = 0.906 J

where v is velocity

v = √(0.906 * 2)/2.6

v = 0.83 m/s.

In conclusion, elastic potential energy is present in compressed or stretched elastic materials.

Learn more about elastic potential energy at: https://brainly.com/question/20797227

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