Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 146: 55

$N=6.18KN$

The required normal force can be determined as $a_n=\frac{v^2}{\rho}$ $\implies v=\sqrt{\rho a_n}=\sqrt{(800)(78.5)}=251m/s$ Now, $\Sigma F_n=ma_n$ $\implies N-W=ma_n$ $\implies N=m(g+a_n)=m(g+8g)=9mg$ We plug in the known values to obtain: $N=9(70)(9.81)$ $\implies N=6.18KN$