Answer
$k=15MN/m^2$
Work Step by Step
We know that
$T_1+\Sigma U_{1-2}=T_2$
$\implies \frac{1}{2}mv^2+\int_0^{0.2} \Sigma U_{1-2}=(0)^2ds$
We plug in the known values to obtain:
$\frac{1}{2}(5000)(4)^2-k\frac{(0.2)^3}{3}=0$
This simplifies to:
$k=15MN/m^2$
