The block is in equilibrium, so the net forces parallel and perpendicular to the surface are
∑ F[para] = F[normal] - mg = 0
∑ F[perp] = F[pull] - F[friction] = 0
where mg is the weight of the block. It follows that
F[normal] = mg = (5 kg) (9.8 m/s²) = 49 N
and so the frictional force has magnitude
F[friction] = 0.25 F[normal] = 12.25 N ≈ 12 N
