Answer
The effective spring constant of the web is 11 N/m.
Work Step by Step
$v = (60 \frac{km}{h})(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s})$
$v = 16.67 ~m/s$
The potential energy in the web will equal the original kinetic energy of the train. Therefore,
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
$k = \frac{mv^2}{x^2}$
$k = \frac{(10^4 ~kg)(16.67 ~m/s)^2}{(500 ~m)^2}$
$k = 11.11\approx11 ~N/m$
Therefore, the effective spring constant of the web is 11 N/m.
