Answer
The book slides a distance of 1.06 meters.
Work Step by Step
The potential energy stored in the spring will be equal in magnitude to the work done by friction.
$F_f~d= \frac{1}{2}kx^2$
$mg~\mu_k~d = \frac{1}{2}kx^2$
$d = \frac{kx^2}{2mg~\mu_k}$
$d = \frac{(250~N/m)(0.250~m)^2}{(2)(2.50~kg)(9.80~m/s^2)(0.30)}$
$d = 1.06~m$
The book slides a distance of 1.06 meters.
