A block of mass $2 kg$ is free to move along the $x-axis$ . It is at rest and from $t = 0$ onwards it is subjected to a time- dependent force $F\left(t\right)$ in the $x direction$. The force F(t) varies with $t$ as shown in the figure. The kinetic energy of the block after $4.5 seconds$ is 

 

Area under $F – t$ graph gives the impulse or the change in the linear momentum of the body. As the initial velocity (and therefore the initial linear momentum) of the body is zero, the area under $F – t$  graph gives the final linear momentum of the body.

$$\begin{gathered} Area of ∆AOB \\ =\frac{1}{2}\times 3\times 4=6 N-s \end{gathered}$$

$Also \frac{OA}{OB}=\frac{CD}{CB}$

$\Rightarrow \frac{4}{3}=\frac{CD}{1.5}f$

$\Rightarrow CD=2$

$\therefore Area of ∆BCD =-\left[\frac{1}{2}\times 1.5\times 2\right] =-1.5 N-s$

$\therefore$ The final linear momentum $= 6 – 1.5 = 4.5 N-s$

$\therefore$  Kinetic energy of the block $=\frac{p^2}{2m}=\frac{{\left(4.5\right)}^2}{2\times 2}=5.06 J$