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 positive x-axis, `y = 0 therefore vec(F) = -k x hat(j)`
i.e., force is in negative y-direction while displacement is in positive x-direction.
`therefore 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 `vec(F) = -k(y hat(i) + a hat(J))`
The first component of force, `-k y hat(i)` will not contribute any work because this component is along negative x-direction `(-hat(i))` while displacement is in positive y-direction (a, 0) to (a, a). The second component of force i.e., `-k a hat(j)` will perform negative work
`therefore W_(2) = (-ka hat(j))(a hat(j)) = (-k a)(a) = - k a^(2)`
So net work done on the particle `W = W_(1) + W_(2)`
`= 0 + (-k a^(2)) = -k a^(2)`