The concept required to develop this problem is Hook's Law and potential elastic energy.
By definition the force by Hooke's law is defined as
[tex]F = kx[/tex]
Where,
k = Spring Constant
x = Displacement
On the other hand, the elastic potential energy is defined as
[tex]E = \frac{1}{2} k\Delta x^2[/tex]
With the given values we can find the value of the spring constant, that is,
[tex]F = kx[/tex]
[tex]k=\frac{F}{x}[/tex]
[tex]k= \frac{83}{0.02}[/tex]
[tex]k = 4120N/m[/tex]
Applying the concepts of energy conservation then we can find the position of the block, that is,
[tex]E = \frac{1}{2} k\Delta x^2[/tex]
[tex]E = \frac{1}{2} k\Delta x^2[/tex]
[tex]4.8 = \frac{1}{2} (4120)(x^2-(-0.02)^2)[/tex]
[tex]x = \sqrt{\frac{2*4.8}{4120}+(-0.02)^2}[/tex]
[tex]x = \pm 0.05225m[/tex]
Therefore the position of the block can be then,
[tex]x_1 = 5.225cm[/tex]
[tex]x_2 = -5.225cm[/tex]
