A force `F = -K(y hatI + x hatj)` (where `K` is a posive constant ) acts on a particle moving in the `xy` plane . Starting form the original , the partical is taken along in the positive `x` axis to the point `(a,0)` and then partical to they axis the point `(a,a)` . The total work done by the force `F` on the particls is

Options

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

Answer

C

Explanation

While moving from `(0.0)` to `(a.0)`
Along positive x-axis , `y = 0 :. vecF = - kx hatj`
i.e. , force is in negative y-direction while dicplacement is in positive direction.
`:. W_(1) = 0`
Because force is perpendicular to displacement .
Then partical moves from `(a.0)` to `(a.a)` during this
`vec F = - k(y hati + a hatj)`
The first componenet of force, `- ky hatj` will not contribitute 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. `- ka hatj` will perform negative work
`:. W_(2) = (- ka hatj) (a hatj) = (-ka) (a) = - ka^(2)`
So net work done on the partical `W = W_(1) + W_(2)`
`= 0 + (-ka^(2)) = - ka^(2)`