Answer
$$
\begin{aligned} v_{B} &=3.5 \mathrm{m} / \mathrm{s} \end{aligned}
$$
Work Step by Step
From Eq. $8-17,$ we find the speed it has at point $C$ to be
$$
v_{C}=\sqrt{v_{A}^{2}-2 g h}=\sqrt{(8.0)^{2}-2(9.8)(2.0)}=4.980 \approx 5.0 \mathrm{m} / \mathrm{s} \text { . }
$$
Thus, we see that its kinetic energy right at the beginning of its "rough slide" (heading
uphill towards $B$ ) is
$$
K_{C}=\frac{1}{2} m(4.980 \mathrm{m} / \mathrm{s})^{2}=12.4 m
$$
$$K_{\mathrm{C}}=m g y+f_{k} d \quad \Rightarrow \quad 12.4 m=m g d \sin \theta+\mu_{k} m g d \cos \theta$$
$$
\frac{1}{2} m v^{2}=K_{C}-\left(m g L \sin \theta+\mu_{k} m g L \cos \theta\right)
$$
This determines the speed with which it arrives at point $B :$
$$
\begin{aligned} v_{B} &=\sqrt{v_{C}^{2}-2 g L\left(\sin \theta+\mu_{k} \cos \theta\right)} \\ &=\sqrt{(4.98 \mathrm{m} / \mathrm{s})^{2}-2\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)(0.75 \mathrm{m})\left(\sin 30^{\circ}+0.4 \cos 30^{\circ}\right)}=3.5 \mathrm{m} / \mathrm{s} \end{aligned}
$$
