Answer
a) The magnitude of the spring's displacement is $3.3cm$
b) The magnitude of the spring's displacement is $7.6cm$
Work Step by Step
a) The acceleration of the box is zero when the box's weight balances the restoring force from the spring, so $$mg=ky$$ $$y=\frac{mg}{k}=\frac{1.5\times9.8}{450}=0.033m=3.3cm$$
b) When the box just makes contact with the spring, we have $v_0=0.49m/s$, $h_0=y_0=0$. When the spring is fully compressed, we have $v_f=0$ and $h_f=y_f$. Therefore, because energy is conserved, $$\frac{1}{2}mv_0^2=mgy_f+\frac{1}{2}ky_f^2$$ $$\frac{1}{2}(450)y_f^2+(1.5\times9.8)y_f-\frac{1}{2}(1.5\times0.49^2)=0$$ $$225y_f^2+14.7y_f-0.18=0$$
Because the spring is compressed, $y_f\lt0$. Solving the equation, we get $y_f=-0.076m=-7.6cm$, so the magnitude of the spring's displacement is $7.6cm$
