#### Answer

The distance that the box slides across the rough surface is 0.54 meters.

#### Work Step by Step

The box's kinetic energy when it leaves the spring will be equal to the initial energy stored in the spring. The magnitude of the work done by friction to bring the box to rest is equal to the box's kinetic energy. Therefore, the magnitude of the work done by friction is equal to the initial energy stored in the spring.
$W_f = U_s$
$F_f~d = U_s$
$mg~\mu_k~d = \frac{1}{2}kx^2$
$d = \frac{kx^2}{2mg~\mu_k}$
$d = \frac{(100~N/m)(0.20~m)^2}{(2)(2.5~kg)(9.80~m/s^2)(0.15)}$
$d = 0.54~m$
The distance that the box slides across the rough surface is 0.54 meters.