A force $$f=x^{2}y\widehat{i}+y^{2}\widehat{j}$$ acts on a particle in a plane x+y=10. The work done by this force during a displacement from (0,0) to (4m,2m) is Joule
(round off to the nearest integer)

The work done by $$\vec{F} = x^2 y\,\hat{i} + y^2\,\hat{j}$$ along a displacement from $$(0, 0)$$ to $$(4, 2)$$ is $$W = \int_C x^2 y\,dx + \int_C y^2\,dy$$.

Since the force acts in the plane $$x + y = 10$$, we substitute $$y = 10 - x$$ in the $$x$$-component of the force. The first integral becomes $$\int_0^4 x^2(10 - x)\,dx = \int_0^4 (10x^2 - x^3)\,dx = \left[\dfrac{10x^3}{3} - \dfrac{x^4}{4}\right]_0^4 = \dfrac{640}{3} - 64 = \dfrac{448}{3}$$.

The second integral evaluates directly as $$\int_0^2 y^2\,dy = \left[\dfrac{y^3}{3}\right]_0^2 = \dfrac{8}{3}$$.

Adding both contributions: $$W = \dfrac{448}{3} + \dfrac{8}{3} = \dfrac{456}{3} = \boxed{152}$$ Joule.