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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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If a particle is compelled to move on a given smooth plane curve under the action of given forces in the plane F⃗=xi^+yj^\mathbf{\vec F} = x\hat i + y\hat j , then
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Correct option: A
F⃗.dr→=x  dx+y  dy\mathbf{\vec F}.\overrightarrow {\mathbf{dr}} = x\,\,dx + y\,\,dy
Option A (F⃗.dr→=x  dx+y  dy\mathbf{\vec F}.{\overrightarrow {\mathbf{dr}}} = x\,\,dx + y\,\,dy) is correct because it represents the dot product of the force vector and the displacement vector, which equals the work done.

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