Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 147: 59

$v=1.48m/s$

We know that $\Sigma F_n=ma_n$ $\implies N_s(\frac{3}{5})+0.2N_s(\frac{4}{5})=2(\frac{v^2}{\rho})$ $\implies 0.38N_s=\frac{v^2}{0.2}$ $\implies v^2=0.076N_s~~~$eq(1) and $\Sigma F_b=ma_b$ $\implies N_s(\frac{4}{5})-0.2N_s(\frac{3}{5})=mg$ $\implies 0.68N_s=2\times 9.81$ $\implies N_s=28.85N$ We plug in this value into eq(1) to obtain: $0.076\times 28.85=v^2$ $\implies v=1.48m/s$