#### Answer

The spring constant of the spring is $1.96\times 10^5~N/m$

#### Work Step by Step

The energy stored in the spring when it is compressed will be equal to the initial potential energy of the safe. Note that the initial height of the safe is 2.5 meters above its final position. Therefore;
$U_s = PE$
$\frac{1}{2}kx^2 = mgh$
$k = \frac{2mgh}{x^2}$
$k = \frac{(2)(1000~kg)(9.80~m/s^2)(2.5~m)}{(0.50~m)^2}$
$k = 1.96\times 10^5~N/m$
The spring constant of the spring is $1.96\times 10^5~N/m$