Answer
$v_B=42.2ft/s$, $N=50.6lb$, $a_t=26.2ft/s^2$
Work Step by Step
According to the principle of work and energy
$\frac{1}{2}mv_A^2+\Sigma U_{A\rightarrow B}=\frac{1}{2}mv_B^2$
$\implies \frac{1}{2}mv_A^2+W(y_A-y_B)=\frac{1}{2}mv_B^2$
We plug in the known values to obtain:
$\frac{1}{2}(\frac{150}{32.2})(5)^2+150(50-22.7)=\frac{1}{2}(\frac{150}{32.2})v_B^2$
This simplifies to:
$v_B=42.2ft/s$
We know that
$\Sigma F_n=ma_n=m\frac{v^2}{\rho}$
$\implies -N+150cos(54.44^{\circ})=(\frac{150}{32.2})(\frac{(42.2)^2}{227.024})$
This simplifies to:
$N=50.6lb$
We can find the acceleration as
$\Sigma F_t=ma_t$
$\implies 150sin(54.44)=(\frac{150}{32.2})a_t$
This simplifies to:
$a_t=26.2ft/s^2$
