Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 146: 52

$v=3.13m/s$

The required speed can be determined as follows: $F_n=m\frac{mv^2}{r}$ $\implies \mu N=m\frac{v^2}{r}$ $\implies \mu mg=m\frac{v^2}{r}$ This can be rearranged as: $v=\sqrt{\mu g r}$ We plug in the known values to obtain: $v=\sqrt{(0.2)(9.81)(5)}$ This simplifies to: $v=3.13m/s$