Question

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

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

Answer 1

Answer: (C)

$\text{F}=-\text{K}\left(\text{y} \hat{\text{i}} + \text{x} \hat{\text{j}}\right)$
On moving from origin to (a, 0).
$\text{Work done}=\displaystyle \int _{0}^{\text{a}}-\text{K}\left(\text{y} \hat{\text{i}} + \text{x} \hat{\text{j}}\right)\text{.}\text{dx}$
= 0
as y = 0
Work done from (a, 0) to (a, 0)
Work done $\text{w}=\displaystyle \int _{0}^{\text{a}}-\text{K}\left(\text{y} \hat{\text{i}} + \text{x} \hat{\text{j}}\right)\text{dy}$ but x = a
$= \displaystyle \int _{0}^{\text{a}} - \text{K} \left(\text{a}\right) \text{dy}$
$= - \text{Ka}^{2}$