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A force F=-K(yhati+xhatj) (where K is a ...

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

A

`-2 K a^2`

B

`2 Ka^2`

C

`-Ka^2`

D

`Ka^2`

Text Solution

Verified by Experts

The correct Answer is:
C
`w = int f_x dx + int f_y dy = -k int ydx - k int xdy`
From 0 to A, `dy = 0, y = 0`
So, `W_(QA) = 0`
From A to B : `dx = 0, x = a, W_(AB) = -ka int_(0)^a dy = -ka^2`
Hence, total work done `W_T = W_(OA) + W_(AB) = 0 - ka^2 = -ka^2`
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