Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.4 - Equations of Motion: Rectangular Coordinates - Problems - Page 134: 35

$t=5.66s$

We can determine the required time as follows: $a_{max}=\frac{\mu_s N}{m}=\mu_s g$ We plug in the known values to obtain: $a_{max}=0.3(9.81)=2.943m/s^2$ Now $v=at$ $\implies t=\frac{v}{a}$ $\implies t=\frac{50/3\space m/s}{2.943m/s^2}$ This simplifies to: $t=5.66s$