A block of mass 3.20 kg is placed against a horizontal spring of constant k = 865 N/m and pushed so the spring compresses by 0.0650 m. HINT (a) What is the elastic potential energy of the block-spring system (in J)? (b) If the block is now released and the surface is frictionless, calculate the block's speed (in m/s) after leaving the spring. n/s

a) The initial elastic potential energy of the block-spring system is 28.113 joules.

b) The final speed of the block is approximately 4.192 meters per second.

Explanation:

a) By applying Hooke's law and definition of work, we define the elastic potential energy (), measured in joules, by the following formula:

(1)

Where:

- Spring constant, measured in newtons per meter.

- Deformation of the spring, measured in meters.

If we know that and , then the elastic potential energy is:

The initial elastic potential energy of the block-spring system is 28.113 joules.

b) According to the Principle of Energy Conservation, the initial elastic potential energy of the block-spring system becomes into translational kinetic energy, that is:

(2)

Where:

- Mass, measured in kilograms.

- Final speed, measured in meters per second.

Then, the final speed is cleared:

If we know that and , then the final speed of the block is:

The final speed of the block is approximately 4.192 meters per second.

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