Answer
$v_{A}=\sqrt {\dfrac {2U_{0}}{m}}=\sqrt {\dfrac {2\times 5.76}{2}}\approx 2.4\dfrac {m}{s}$
Work Step by Step
İnitial total potential energy relative to point A (shown in picture) is
$U_{0}=mg\Delta h+\dfrac {k\Delta x^{2}}{2}=mg\Delta x\sin \theta +\dfrac {k\Delta x^{2}}{2}=2\times 9.8\times 0.2\times \sin37 -\dfrac {170\times 0.2}{2}\approx 5.76J$
When it reaches point A, this potential energy will convert to kinetic energy:
$\dfrac {mv^{2}_{A}}{2}=U_{0}\Rightarrow v_{A}=\sqrt {\dfrac {2U_{0}}{m}}=\sqrt {\dfrac {2\times 5.76}{2}}\approx 2.4\dfrac {m}{s}$
