#### Answer

The box was moving with a speed of 0.49 m/s

#### Work Step by Step

The spring brings the box to rest when the spring compresses by 6.5 cm. The work done by the spring on the box is $W = -\frac{1}{2}kx^2$. We can use the work-energy theorem to find the initial speed of the box.
$KE_1+W = KE_2$
$\frac{1}{2}mv^2-\frac{1}{2}kx^2 = 0$
$\frac{1}{2}mv^2=\frac{1}{2}kx^2$
$v^2=\frac{kx^2}{m}$
$v=\sqrt{\frac{kx^2}{m}}$
$v=\sqrt{\frac{(85~N/m)(0.065~m)^2}{1.5~kg}}$
$v = 0.49~m/s$
The box was moving with a speed of 0.49 m/s