As work done by force $=$ change in K.E. of the particle

From the graph,

When distance $= 2\text{\ }m$, Force $= 2\text{\ }N$

When distance $= 3\text{\ }m$, Force $= 3\text{\ }N$

Work done by the force on the particle will be area under the curve F vs x

$$W = 2 \times 2 + 1/2(2 + 3) \times (3 - 2) = 6.5\text{\ }J$$

$$W = \Delta K = K_{f} - K_{i}$$

(Here $\left. \ K_{i} = 0 \right)$

$$\therefore K_{f} = 6.5\text{\ }J$$