A force `F=-K(yhati+xhatj)` (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 point `(a, a)`. The total work done by the force F on the particle is

Options

A. `-2Ka^2`
B. `2Ka^2`
C. `-Ka^2`
D. `Ka^2`

Answer

C

Explanation

While moving from (0,0) to (a,0)
Along positivve x-axis `y=0`,`vecF=-kxhatj` i.e., force is in negative y-direction while displacement is in positive x-direction.
`W_1=0`
Because 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 `vecF=-k(yhati+ahatj)` The first component of force, `-kyhati` will not contribute any work because this component is along negative x-direction `(-hati)` while displacement is in positive y-direction (a,0) to (a,a). The second component of force i.e. `-kahatj` will perform negative work
`W_2=(-kahatj)(ahatj)=(-ka)(a)=-ka^2`
so net work done on the particle `W=W_1+W_2`
`=0+(-ka^2)=-ka^2`