Answer:
The position is
[tex]d_{o}=1.706m[/tex]
Explanation:
The kinetic energy of the motion is
[tex]E_{K}=\frac{1}{2}*m*v^2[/tex]
[tex]E_{K}=\frac{1}{2}*0.8kg*8(\frac{m}{s})^2[/tex]
[tex]E_{K}=25.6 J[/tex]
So apply the conservation of energy the force of the spring is the same of the kinetic energy
[tex]F_{s}=\frac{1}{2}*k*d^2[/tex]
[tex]F_{s}=E_{K}[/tex]
[tex]25.6=\frac{1}{2}*30\frac{N}{m}*d_{o}^2[/tex]
Solve to do
[tex]d_{o}=\sqrt{\frac{2*25.6J}{30\frac{N}{m}}}[/tex]
[tex]d_{o}=1.706m[/tex]