A force `F=-K (yhati + xhatj)`, 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 is 

Options

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

Answer

C

Explanation

In taking the particle along the positive x-axis from (0,0) to the point (a, 0)
`W_1 = int_0^a vecF .d vecx = int_0^a - K(y hati + x hatj).hati dx`
`= int_0^a -K(0 hati + x hatj).hati dx = ` Zero.
And then taking the particle parallel to y-axis to the point (a,a)
`W_2 = int_0^a vecF . dvecy = int_0^a - K(y hati + x hatj) .hatj dy`
`= int_0^a - K(y hati + a hatj).hatj dy = -Ka int_0^a dy = -Ka^2`
Total work done,
`W = W_1 + W_2 = 0 - Ka^2 = -Ka^2`