Answer
$v_B=18m/s$, $N_B=12.5KN$
Work Step by Step
According to the principle of work and energy
$\frac{1}{2}mv_A^2+WS=\frac{1}{2}mv_B^2$
We plug in the known values to obtain:
$\frac{1}{2}(250)(3)^2+(250)(9.81)(16)=\frac{1}{2}(250)v_B^2$
This simplifies to:
$v_B=18m/s$
We know that
$\Sigma F_n=m\frac{v^2}{\rho}$
$\implies N_B-W=m\frac{v^2}{\rho}$
$\implies N_B=W+m\frac{v^2}{\rho}$
We plug in the known values to obtain:
$N_B=250(9.81)+250(\frac{(18)^2}{8})$
This simplifies to:
$N_B=12.5KN$
