A mass .60 $\mathit{kg}$ is attached to a spring with a spring constant of 130 $\mathit{\frac{N}{m}}$. There are no non-conservative forces acting, and the system is stretched .13 $\mathit{m}$. Find the force when it is instaneous released.
$$$$ From here is made the decision of calculating it like this:
\begin{align} \mathbf{Force} &= -k \triangle x \\ &= -130 \ \cdot \ .13 = -16.9 \mathit{N} \end{align}
Using that same expression for Force I calculated the acceleration in the x-axis:
\begin{align} F_{\mathit{Hooke's\ Law}}&= F_{\mathit{net}} \\ \frac{-k \triangle x}{m} &= a_{net} \\ \frac{-16.9}{.60} &= 30 \mathit{\frac{m}{s^2}} \end{align}
Correction:
$a_{net} = 28 \mathit{\frac{m}{s^s}}$
Is my attempt correct?
That looks right, but you may want to write $$\frac{-16.9}{0.60}\approx 28 \frac{\text{m}}{\text{s}^2}$$ instead of $30 \text{m}/\text{s}^2$, since the actual value is $28.166...$ which is closer to $28$.