A force $\mathrm{F}=-\mathrm{K}(\mathrm{yi}+\mathrm{xj})$  (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 point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is

detailed solution

Correct option is C

While moving from (0,0) to (a,0) Along positive x-axis, y = 0  ∴ F→=−kxj^i.e.  force is in negative y-direction while displacement is in positive x-direction.∴W1=0Because force is perpendicular to displacement Then particle moves from (a, 0) to (a, a) along a line parallel to y-axis (x=+a) during this  F→=−k(yi^+aJ^)The first component of force, −kyi^ will not contribute any work because this component is along negative    x-direction (−i^) while displacement is in positive           y-direction (a,0) to (a,a). The second component of force i.e. −kaj^ will perform negative work∴W2=(−kaj^) (aj^)=(−ka) (a)​ =−ka2So net work done on the particle  W=W1+W2=0+(−ka2)=−ka2