A particle of mass 2kg is moving along  $x$ axis. Its velocity at origin is  $\sqrt{13}{\text{\hspace{0.17em}ms}}^{-1}$. Find its velocity at  $x=10\text{\hspace{0.17em}m}$. Force (F) varies with displacement (d) as shown in figure ?

Total work done = ${\text{W}}_1+{\text{W}}_2+{\text{W}}_{\text{3}}$ 
                              = 8 + 16 – 12
                              =  $\boxed{\text{12\hspace{0.17em}\hspace{0.17em}J}}$
(b) Velocity at origin = $\sqrt{13}{\text{\hspace{0.17em}ms}}^{-1}$ 
At $x_1=0$                         $x_2=10\text{\hspace{0.17em}m}$    
${\text{V}}_1=\sqrt{13}{\text{\hspace{0.17em}ms}}^{-1}$          ${\text{V}}_2=?$   
Total work done =  $\Delta \text{\hspace{0.17em}KE}$
$12={\text{KE}}_{\text{2}}-{\text{KE}}_{\text{1}}$

$12=\frac{1}{2}{\text{MV}}_2^2-\frac{1}{2}{\text{MV}}_1^2$

$12=\frac{\text{M}}{\text{2}}\left[{\text{V}}_{\text{2}}^{\text{2}}-{\text{V}}_{\text{1}}^{\text{2}}\right]$

$12=\frac{2}{2}\left[{\text{V}}_{\text{2}}^{\text{2}}-13\right]$

$12={\text{V}}_{\text{2}}^{\text{2}}-13$

${\text{V}}_2^2=25$

${\text{V}}_{\text{2}}=5\text{m}/\text{sec}$