• question_answer A force $\overset{\to }{\mathop{F}}\,=-k(y\hat{i}+x\hat{j})$ 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 j'-axis to the point (a, a). The total work done by the force is A) $-2k{{a}^{2}}$ B) $2\,\,k{{a}^{2}}$ C) $-k{{a}^{2}}$ D) $k{{a}^{2}}$

 [c] ${{W}_{1}}\int_{0}^{a}{\overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{dx}}\,}=\int_{0}^{a}{-k(y\hat{i}-x\hat{j}).\hat{i}dx}$ $=\int_{0}^{a}{-k(0\hat{i}}+x\hat{j}).\hat{i}dx=zero$ $W\int_{0}^{a}{\overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{dy}}\,}=\int_{0}^{a}{-k(y\hat{i}}+x\hat{j}).\hat{j}dy$ $=\int_{0}^{a}{-k(a\hat{i}}+a\hat{j}).\hat{j}dy$ $=-ka\int_{0}^{a}{dy=-k{{a}^{2}}}$ Total work done, $W={{W}_{1}}+{{W}_{2}}=0-k{{a}^{2}}=-k{{a}^{2}}$