Question

A force $\overrightarrow{F}=2x{y}^2\hat{i}+2x^{2}y\hat{j}$ acts on a particle. The particle starts from point $(1,2)$ and moves to the point $(3,4)$. The total work done by the force $\overrightarrow{F}$ on the particle is

• A. $150\text{J}$
• B. $148\text{J}$
• C. $140\text{J}$
• D. $144\text{J}$

Answer 1

The correct option is C $140\text{J}$

Given variable force $\overrightarrow{F}=2x{y}^2\hat{i}+2x^{2}y\hat{j}$
So, let us check whether the form is of perfect differential or not
We have
$F_x=\int 2x{y}^{2}dx=x^2{y}^2$
Similarly, $F_y=\int 2x^{2}ydy=x^2{y}^2$
Since, both are equal, the form of force given is perfect differential and we can say that the force is conservative
Thus, work done $W=\int d\left(x^2{y}^2\right)=\int d\left(K\right)$
Also we have, $K_{initial}=1^2\times 2^2=4$ and $K_{final}=3^2\times 4^2=144$
Hence, we have
$W=\int d\left(K\right)=[K{]}_4^{144}=(144-4)=140\text{J}$