The mass of the box is given as,
[tex]m=10\text{ kg}[/tex]The force acting at the 30 degree is given as,
[tex]F=40\text{ N}[/tex]The frictional force acting on the box is,
[tex]\begin{gathered} F^{\prime}=\mu mg \\ F^{\prime}=0.30\times10\times9.8 \\ F^{\prime}=29.4\text{ N} \end{gathered}[/tex]The applied force in the horizontal direction is,
[tex]\begin{gathered} F\cos (30^{\circ})=40\times\cos (30^{\circ}) \\ =34.64\text{ N} \end{gathered}[/tex]The net force acting on the box is,
[tex]\begin{gathered} F_{net}=F\cos (30^{\circ})-F^{\prime} \\ F_{net}=34.64-29.4 \\ F_{net}=5.24\text{ N} \end{gathered}[/tex]According to the newton second law,
[tex]\begin{gathered} F_{net}=ma \\ 5.24=10\times a \\ a=\frac{5.24}{10} \\ a=0.524ms^{-2} \end{gathered}[/tex]Thus, the acceleration of the box is 0.524.
