#### Answer

$0.81\dfrac {m}{s}$

#### Work Step by Step

Lets assume that $H_{0}$ is the initial height of the box relative to the ground.
The total energy of the spring and box system initially will be :
$E_{tot}=mgH_{0}..............(1)$
since there is no speed of the box initially and the spring is relaxed.
When the box starts to move, its potential energy will be converted to the box’s kinetic energy and the spring’s potential energy. Lets assume that the box will descend distance $L$ on the incline (so spring will stretch by distance $L$). Therefore, lets write total energy equation again:$E_{tot}=E_{spring}+E_{box}=\dfrac {kL^{2}}{2}+\dfrac {mv^{2}_{1}}{2}+mgH_{1}...........(2)$
$H_{1}$ is the height of the box relative to the ground after descending distance $L$.
From the incline, we get:
$\Delta H=H_{0}-H_{1}=L\sin \theta \left................( 3\right) $
From (1) and (2), we get:
$mg\left( H_{0}-H_{1}\right) -\dfrac {kL^{2}}{2}=\dfrac {mv^{2}_{1}}{2}\Rightarrow v_{1}=\sqrt {2g\left( Ho-H_{1}\right) -\dfrac {k}{m}L^{2}}\left...( 4\right)
$ From (3) and (4), we get:
$v_{1}=\sqrt {L\left( 2g\sin \theta -\dfrac {kL}{m}\right) }$
Given the numbers $m=2kg;k=120\dfrac {N}{m};\theta =40^{0};L=10cm=0,1m;g=9.8\dfrac {m}{s^{2}}$,
we get $v_{1}\approx0.81\dfrac {m}{s}$