We first determine the total force ni he horizontal axis, that is:
[tex]\sum F_x=120cos(30)N-10N\Rightarrow\sum F_x=(60\sqrt{3}-10)N[/tex]
Now, we determine the acceleration:
[tex]\begin{gathered} (60\sqrt{3}-10)N=(32kg)a\Rightarrow a=\frac{(60\sqrt{3}-10)N}{32kg} \\ \\ \Rightarrow a=\frac{30\sqrt{3}-5}{16}m/s^2\Rightarrow a\approx2.935m/s^2 \end{gathered}[/tex]
So, the acceleration is approximately 2.935 m/s^2.
