A force $\vec{f}=x^{2} y \hat{i}+y^{2} \hat{j}$ acts on a particle in a plane $x+y=10$. The work don

Question

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

Professor PC Thomas Classes · Answer

(152) : Given, Force $(\vec{f})=x^{2} y \hat{i}+y^{2} \hat{j}$ Also, $x+y=10 ; y=10-x$\ So, force $(\vec{f})=x^{2}(10-x) \hat{i}+y^{2} \hat{j}$