Answer
The object will move an additional 0.240 meters before stopping momentarily and moving back to the left.
Work Step by Step
We can find the value of $\frac{k}{m}$
$kx = ma$
$\frac{k}{m} = \frac{a}{x}$
$\frac{k}{m} = \frac{8.40~m/s^2}{0.600~m}$
$\frac{k}{m} = 14.0~s^{-2}$
We can find $\omega$.
$\omega = \sqrt{\frac{k}{m}}$
$\omega = \sqrt{14.0~s^{-2}}$
$\omega = 3.74~rad/s$
We can find the amplitude.
$\omega~\sqrt{A^2-x^2} = v$
$A = \sqrt{x^2+\frac{v^2}{\omega^2}}$
$A = \sqrt{(0.600~m)^2+\frac{(2.20~m/s)^2}{(3.74~rad/s)^2}}$
$A = 0.840~m$
From $x = 0.600~m$, the distance to $A = 0.840~m$ is $0.240~m$. Therefore, the object will move an additional 0.240 meters before stopping momentarily and moving back to the left.
