Question

A position dependent force $\overrightarrow{F}=(y^{2}x\hat{i}+x^{2}y\hat{j})N$ acts on a particle and displaces it from $x=1m,y=2m$ to $x=2m,y=1m$, The work done by the force is

• A. $\frac{10}{3}J$
• B. $10J$
• C. $20J$
• D. Zero

Answer 1

The correct option is D Zero

By definition,
$W={\int }_{x_i}^{x_f}{F}_{x}dx+{\int }_{x_i}^{y_f}{F}_{y}dy$
$=\int y^{2}xdx+\int x^{2}ydy$
$=\int [y^{2}xdx+x^{2}ydy]$
$={\int }_4^4\frac{d\left(x^2{y}^2\right)}{2}(\because d(x^2{y}^2)=2y^{2}xdx+2x^{2}ydy)$

since
$(x^2{y}^2{)}_i=(1^2\times 2^2)=4$
and
$(x^2{y}^2{)}_f=(2^2\times 1^2)=4$

$W=-\frac{x^2{y}^2}{2}{|}_4^4=0.$