Answer
The amplitude of the resulting motion is $7.2cm$
Work Step by Step
Initially, $v_0=8m/s$, and since the spring is unstrained, $x_0=0$.
The amplitude $A$ of the resulting motion is equal to the spring's maximum displacement $x_f=A$, where the block's speed $v_f=0$
Since total energy is conserved, we can find $x_f$: $$\frac{1}{2}mv_0^2=\frac{1}{2}kx_f^2$$ $$x_f=v_0\sqrt{\frac{m}{k}}$$
The block's mass $m=0.01kg$ and spring constant $k=124N/m$. The amplitude of the motion is $$A=x_f=0.072m=7.2cm$$
