The correct option is C −ka2
While moving from (0,0) to (a,0)
Along positive x - axis,y=0⇒→F=−kx^j
i.e. force is in negative y - direction while displacement is in positive x - direction.
∴W1=0
Because force is perpendicular to displacment.
Then particle moves from (a,0) to (a,a) along a line parallel to y - axis (x=+a) during this →F=−k(y^i+a^j)
The first component of force, −ky^i 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. −ka^j will perform negative work
∴W2=(−ka^j)(a^j)=(−ka)(a)=−ka2
So, net work done on the particle
W=W1+W2=0+(−ka2)=−ka2
Alternate solution
Given F=−k(y^i+x^j)
We know that
W=∫→F.→ds=∫(a,a)(0,0)−k(y^i+x^j).(dx^i+dy^j)=∫(a,a)(0,0)−k(ydx+xdy)W=−∫(a,a)(0,0)k d(xy)=−ka2