Question

A force $F=-K(y\hat{i}-x\hat{j})$, (where $K$ is a positive constant) acts on a particle moving in the $XY-$plane. Starting from the origin, the particle is taken along the positive $X-$axis to the plane $(a,0)$ and then parallel to the $Y-$axis to the point $(a,a)$. The total work done by the force $F$ on the particle is

• A. $-2K{a}^2$
• B. $2K{a}^2$
• C. $-K{a}^2$
• D. $K{a}^2$

Answer 1

The correct option is C $-K{a}^2$

The expression of work done by the variable force $F$ on the particle is given by

In going from $(0,0)$ to $(a,0)$, the coordinate of $x$ varies from $0$ to $a$, while that of $y$ remains zero.Hence, the work done along this path is

$W=\int \overrightarrow{F}.\overrightarrow{dl}$

$W_1={\int }_0^a(-kx\hat{j}).dx\hat{i}=0$ since $\hat{j}.\hat{i}=0$

In going from $(a,0)$ to $(a,a)$, the coordiante of $x$ remains constant$(=a)$ while that of $y$ changes from

$0$ to $a$.Hence, the work done along the path is

$W_2={\int }_0^a\left[(-k(y\hat{i}-x\hat{j}))dy\hat{j}\right]$ by putting $x=a$

$=Ka{\int }_0^{a}dy=-K{a}^2$

Hence,$W=W_1+W_2=-K{a}^2$