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Answer:
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 ([tex]U_{g}[/tex]), measured in joules, by the following formula:
[tex]U_{g} = \frac{1}{2}\cdot k\cdot x^{2}[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]x[/tex] - Deformation of the spring, measured in meters.
If we know that [tex]k = 865\,\frac{N}{m}[/tex] and [tex]x = 0.065\,m[/tex], then the elastic potential energy is:
[tex]U_{g} = \frac{1}{2}\cdot \left(865\,\frac{N}{m} \right) \cdot (0.065\,m)[/tex]
[tex]U_{g} = 28.113\,J[/tex]
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:
[tex]U_{g} = \frac{1}{2}\cdot m\cdot v^{2}[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Final speed, measured in meters per second.
Then, the final speed is cleared:
[tex]v = \sqrt{\frac{2\cdot U_{g}}{m} }[/tex]
If we know that [tex]U_{g} = 28.113\,J[/tex] and [tex]m = 3.20\,kg[/tex], then the final speed of the block is:
[tex]v = \sqrt{\frac{2\cdot (28.113\,J)}{3.20\,kg} }[/tex]
[tex]v \approx 4.192\,\frac{m}{s}[/tex]
The final speed of the block is approximately 4.192 meters per second.