Answer
$$v_{B}=1.69 \mathrm{m} / \mathrm{s} \approx 1.7 \mathrm{m} / \mathrm{s}$$
Work Step by Step
From this result, we find $l_{0}=x_{0}=0.347 \mathrm{m}-0.055 \mathrm{m}=0.292 \mathrm{m},$ which means that the block has descended a vertical distance
$$
|\Delta y|=h_{A}-h_{B}=l_{0} \sin \theta=(0.292 \mathrm{m}) \sin 30^{\circ}=0.146 \mathrm{m}
$$
in sliding from point $A$ to point $B .$ Thus, using Eq. $8-18,$ we have
$$
0+m g h_{A}=\frac{1}{2} m v_{B}^{2}+m g h_{B} \Rightarrow \frac{1}{2} m v_{B}^{2}=m g|\Delta y|
$$
which yields $$v_{B}=\sqrt{2 g|\Delta y|}=\sqrt{2\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(0.146 \mathrm{m})}=1.69 \mathrm{m} / \mathrm{s} \approx 1.7 \mathrm{m} / \mathrm{s}$$
