Answer:
[tex]PE=0.625\ J[/tex]
Explanation:
Elastic Potential Energy
We find objects like springs that hold potential energy when stretched and they have the capacity to release it when left return to its equilibrium position. The elastic potential energy of a spring of constant k when is stretched a distance x is
[tex]\displaystyle PE=\frac{kx^2}{2}[/tex]
The spring constant can be obtained if we know the force N needed to stretch the spring a distance x, by using the Hook's Law
[tex]F=kx[/tex]
The question provides the information that a force of F = 25 N stretches a spring x= 5 cm = 0.05 m. Using the formula F=kx we can compute the value of k.
[tex]\displaystyle k=\frac{F}{x}[/tex]
[tex]k=\displaystyle \frac{25}{0.05}[/tex]
[tex]k=500\ N/m[/tex]
The potential energy of the spring in the compressed position (assumed 5 cm as well) if
[tex]\displaystyle PE=\frac{(500)(0.05)^2}{2}[/tex]
[tex]PE=0.625\ J[/tex]
