Q.

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 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 on the particle is [AIIMS 2017]

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a

-2ka2

b

2ka2

c

-ka2

d

ka2

answer is C.

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Detailed Solution

Given,   F=-k(yi^+xj^)As, work done,   W=∫F·drSo,     W'=-k∫(yi^+xj^)·(dxi^+dyj^)⇒ W=-k∫(ydx+xdy)⇒W=-k∫(0,0)(a,a)d(xy)  ∵ydx+xdy=∫d(xy)Hence,    W=-k(xy)(0,0)(a,a)=-ka2
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Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
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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 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 on the particle is [AIIMS 2017]