JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Sample Paper Topic Test - Work Energy Power

  • 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) \[k{{a}^{2}}\]

    Correct Answer: C

    Solution :

    [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,
      \[W={{W}_{1}}+{{W}_{2}}=0-k{{a}^{2}}=-k{{a}^{2}}\]


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