Force $\left(F\right)$ acting on a particle varies will displacement $\left(x\right)$ as shown in figure. The work done by this force on the particle from $x=0$ to $x=6m$ is:

The correct option is B $40J$
Work done by the given force = area under the graph

= area of rectangle + area of 1st triangle + area of 2nd triangle.

$\begin{array}{c}=\left[\left(10-0\right)\times \left(4-0\right)\right]+\left[\frac{1}{2}\times \left(5-4\right)\times \left(10-0\right)\right]+\left[\frac{1}{2}\times \left(6-5\right)\times \left(-10-0\right)\right]\\ =\left[10\times 4\right]+\left[\frac{1}{2}\times 1\times 10\right]+\left[\frac{1}{2}\times 1\times -10\right]\\ =40+5+\left(-5\right)\\ =40J\end{array}$