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The net force on the block perpendicular to the floor is
∑ F[perp] = F[normal] - mg = 0
so that
F[normal] = (5 kg) g = 49 N
Then
F[friction] = 0.1 F[normal] = 4.9 N
so that the net force parallel to the floor is
∑ F[para] = -4.9 N = (5 kg) a
Solve for the acceleration a :
a = (-4.9 N) / (5 kg) = -0.98 m/s²
Starting with an initial velocity of 5 m/s, the box comes to a stop after time t such that
0 = 5 m/s - (0.98 m/s²) t
⇒ t ≈ 5.1 s