Question

The work done by the force $\overrightarrow{F}=A(y^2\hat{i}+2x^2\hat{j})$, where $A$ is a constant and $x$ & $y$ are in meters around the path shown is :

• A. $zero$
• B. $Ad$
• C. $A{d}^2$
• D. $A{d}^3$

Answer 1

The correct option is D $A{d}^3$

Variable force is apllied across the path $OA$ to $AB$ to $BC$ to $CO$

$W=\int \overline{F}.d\overline{r}$

$=\int A(y^2\overline{i}+2x^2\overline{j}).(dx\overline{i}+dy.\overline{j})$

$=\int A(y^{2}dx+2x^{2}dy)$

Now following the path of displacement along $OA$ the $y=0$

$W_{OA}={\int }_{x=0}^{x=d}A(0+2x^2.0)=0+0=0$

Similerly, for $W_{AB},x=d$

$W_{AB}=A[0+2d^{2}d]=2A{d}^3$

Similarly, for $W_{BC},x=d,y=d$

$W_{BC}={\int }_{x=d}^{x=0}A(d^{2}dx+0)$

$W_{BC}=A\left[d^2\right(-d)+0]=-A{d}^3$

$W_{CO}=A[0+0]$

$W=0+2A{d}^3-A{d}^3+0=A{d}^3$