## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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