A block of mass 1 kg is kept at rest on a rough horizontal surface. The coefficient of friction between the block and surface is 0.5. A horizontal force of 10 N is applied on the block for 6 seconds after which the direction of force is reversed keeping the magnitude same. Find the speed (in m/s) with which the block returns to its starting point. $({\text{Take\hspace{0.17em}g = 10m/s}}^2,\sqrt{3}=1.732)$

 From   $A(t=0)toB(t=6)$
   $a=\frac{F-f}{m}={10}^{-5}=5{\text{\hspace{0.17em}m/s}}^{\text{2}}$
 $s_1=\frac{1}{2}a{t}^2=\frac{1}{2}5\left(36\right)=90\text{\hspace{0.17em}m}$  
  Velocity at  $B=at=5\left(6\right)=30\text{\hspace{0.17em}m/s}$ 
After $B,$  the direction of $F$ is reversed.
As velocity is in forward direction, friction acts backwards till block comes to instantaneous rest at $C.$
From $B$ to $C$
$a=\frac{(F+f)}{m}=-(10+5)=-15{\text{\hspace{0.17em}m/s}}^{\text{2}}$   
$u=30\text{\hspace{0.17em}m/s}$   
$0={\left(30\right)}^2+2(-15)s_2\Rightarrow s_2=30\text{m}$   
While returning from $C$ to $A,F$ acts towards left but friction acts toward right (as velocity is towards left)
$s=-(s_1+s_2)=-120m$  

$a=-\frac{(F-f)}{m}=-5{\text{m/s}}^{\text{2}}$
$v^2=u^2+2as0+2(-120)(-5)$  
$v^2=1200\Rightarrow v=20\sqrt{3}=20\left(1.732\right)=34.64\text{\hspace{0.17em}m/s}$