A force of 100 N is applied on a block of mass 3 kg as shown in the figure. The coefficient of friction between the surface and the block is $\mathrm{\mu } = \frac{1}{\sqrt{3}}$. The acceleration of block is

Normal reaction on the block  N = 100 cos $$\begin{gathered} {30}^0 = 50 \sqrt{3} N \\ \Rightarrow f_{max}=\mu N = 50 N \\ Vertical component of applied force = 100 \sin {30}^0 N =50 N \\ Since weight of the block is equal to f_{max }, net force on the block is zero . \end{gathered}$$