Question

a block of mass 2kg is on a horizontal surface .the coefficient of static and kinetic frictions are 0.6 n0.2 the minimum horizontal force required to start the motion is aplied an dif it is continued the velocity acquired by the body at the end of 2nd second if g=10 is

Answer 1

mass = 2kg
coefficient of static friction (s)= 0.6
coefficient of kinetic friction (k)= 0.2

Minimum horizontal force required to move the block is the same as static friction. Because unless you overcome static friction, you can't move the block. 

So force(F) = smg = 0.6×2×10 = 12N

12N force continues to act. Once the block starts moving, Instead of static friction, kinetic friction will start acting.
Kinetic friction(K) = kmg = 0.2×2×10 = 4N

Net force = F-K = 12-4 = 8N
Acceleration(a) = Net force/mass = 8/2 = 4m/s²

Initial velocity (u)= 0
After 2 seconds, velocity = u + at = 0+4×2 = 8m/s

Answer 2

given μ for
$f_{s} = 0.6$
$f_{k} = 0.2$

Force to move the block (F) ≥ Limiting static friction($f_{s}$)
$f_{s}=2*10*0.6$
$f_{s}=12N$
so F ≥ 12 N
Minimum force = 12 N
so Kinetic friction = $f_{k}=2*10*0.2$
                            =  $f_{k}=4N$

Net force acting on the body is F -  $f_{k}$ = 8 N

We know 

F = ma 
⇒8 = 2a
⇒a = 4 ms⁻²

so velocity after 2 second is
 v = u + at
⇒v = 4 × 2 = 8 m/s