Answer:
v = 2.57 m /sec
Explanation: See Annex Free Body Diagram
From free body diagram and Newton´s second law we have
There is not movements in the y axis direction
cos 30°  =  √3/2       sin 30°  =  1/2
We have  P  = mg   =  3 Kg  *  9.8 m/sec²
P  =   29.4  Kg*m/ sec²      P  =  29.4 [N]
Py  =  P * cos 30°    Py =  29.4 [N] * √3/2   ⇒    Py = 25.43 [N]
Px  =   P * sin 30°    Px =  29.4 [N] * 1/2      ⇒     Px = 14.7  [N]
∑ F   =  m* a         ⇒    ∑ Fy   =  0       ∑ Fx  =  m *a
 ∑ Fy   =  Fn  -  Py   =  0         Py   = P*cos30°       Py = 25.43 [N]
Fn  =  25.43 [N]
Fr  =  μk * Fn      ⇒   Fr  =  0.19 * 25.43   ⇒ Fr  =  4.83 [N]    
Now
∑ Fx  =  m *a       mg sin30° - Fr =  m*a    ⇒   Px  - Fr  = m*a
14.7 [N]   -  4.83 [N]   =  3 [kg] * a       ⇒   9.87 /3   = a [m /sec²]
a = 3.29 [m/sec²]
From uniformly accelerated movement
distance  =  x₀  + V₀*t ± at²/2     but  x₀   and  V₀    =  0
Then
d = ( 1/2 )*a*t²     ⇒  1 [m]  * 2  =  3.29 [m/sec²] * t²
t  =  0.78 sec
And finally
v =  a*t      ⇒   v  =  3.29 *(.78)     ⇒   v = 2.57 m /sec