$\begin{array}{l}=\mu N=\mu mg\\ =0.6\times 1\times 9.8=5.88\mathrm{N}\end{array}$

The truck is accelerating at $5\mathrm{m}{\mathrm{s}}^{-2}$.

The pseudo force acting on the mass as seen by the observer on the truck is  $m\times a=1\times 5=5 \mathrm{N}$.  The frictional force will try to oppose the movement of the mass by this force. Therefore the frictional force acting will be $5 \mathrm{N}$.