A force F = -k(yi + xj ) (where, k is a positive constant) acts on a particle moving in the XY – plane. Starting from the origin, the particle is taken ….

Q: A force $ \vec{F} = -k(y \hat{i} + x \hat{j} ) $(where, k is a positive constant) acts on a particle moving in the XY – 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) -2ka2

(b) 2ka2

(c) -ka2

(d) ka2

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Ans: (c)
Sol: $\large dW = \vec{F}.\vec{ds} $

$\large dW = -k(y \hat{i} + x \hat{j}).(dx\hat{i}+ dy \hat{j} + dz \hat{k})$

$\large dW = -k(ydx + x dy) = – k d(x y)$

$\large W = -k\int_{0,0}^{a,a} d(x y)$

$\large = -k[x y]_{0,0}^{a,a}$

W = –ka2