Answer
$4.19\dfrac {m}{s}$
Work Step by Step
İnitial potential Energy relative to point B is:
$U_{0}=\dfrac {k\Delta x^{2}}{2}+mg\left( D\sin \theta +\Delta h\right)$
$ =\dfrac {170\times 0.2 ^{2}}{2}+2\times 9.8\times \left( 1\times \sin 37+0.2\times \sin 37\right)$
$ \approx 17.56J$
All this potential energy will be converted to kinetic energy. So:
$\dfrac {mv^{2}_{B}}{2}=v_{0}\Rightarrow v_{B}=\sqrt {\dfrac {2U_{0}}{m}}=\sqrt {\dfrac {2\times 17.56}{2}}\approx 4.19\dfrac {m}{s}$
