Here $\mathrm{m}=5 \mathrm{~kg}, F_1=\mathrm{OA}=8 \mathrm{~N}$ and $\mathrm{F}_2=\mathrm{OB}=6 \mathrm{~N}$ From the adjoining figure,
$$
\begin{aligned}
&\text { Resultant force, } \begin{aligned}
F &=\sqrt{\mathrm{F}_1^2+\mathrm{F}_2^2}=\sqrt{8^2+6^2} \\
&=10 \mathrm{~N} \\
\tan \mathrm{COA}=\tan \theta &=\mathrm{AC} / \mathrm{OA}=\mathrm{OB} / \mathrm{OA}=6 / 8=0.75 \\
\therefore \quad \theta=36^{\circ} 52^{\prime}
\end{aligned}
\end{aligned}
$$
which is the direction of the resultant force, and hence the direction of the acceleration.
$\therefore \quad a=F / m=10 / 5=2 \mathrm{~ms}^{-2}$