Engineering Mechanics: Statics & Dynamics (14th Edition)

$v_2=16.4~m/s$
We can determine the required speed as follows: According to the conservation of energy equation $T_1+V_1=T_2+V_2$ $\implies 0+\frac{1}{2}kx_1^2=\frac{1}{2}mv_2^2+\frac{1}{2}kx_2^2$ We plug in the known values to obtain: $\frac{1}{2}(50)(3.7)^2=\frac{1}{2}(2)v_2^2+\frac{1}{2}(50)(1.7)^2$ This simplifies to: $v_2=16.4~m/s$