2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Vectors

JEE Main & Advanced Physics Vectors Question Bank Critical Thinking

  • question_answer
    A force \[\overrightarrow{F}=-K(y\hat{i}+x\hat{j})\] (where K is a positive constant) 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 y-axis to the point (a, a). The total work done by the forces \[\overrightarrow{F}\] on the particle is                    [IIT-JEE 1998]

    A)             \[-2\,K{{a}^{2}}\]

    B)                                         \[2\,K{{a}^{2}}\]

    C)             \[-K{{a}^{2}}\]                      

    D)             \[K{{a}^{2}}\]

    Correct Answer: C

    Solution :

                           For motion of the particle from (0, 0) to (a, 0)                    \[\overrightarrow{F}=-K(0\,\hat{i}+a\,\hat{j})\]\[\Rightarrow \,\,\,\,\overrightarrow{F}=-Ka\hat{j}\]                    Displacement  \[\overrightarrow{r\,}=(a\,\hat{i}+0\,\hat{j})-(0\,\hat{i}+0\,\hat{j})=a\hat{i}\]                    So work done from (0, 0) to (a, 0) is given by                    \[W=\overrightarrow{F}\,.\,\overrightarrow{r\,}\]\[=-Ka\hat{j}\,.\,a\hat{i}=0\]                    For motion (a, 0) to (a, a)                    \[\overrightarrow{F}=-K(a\hat{i}+a\hat{j})\] and displacement                    \[\overrightarrow{r\,}=(a\hat{i}+a\hat{j})-(a\hat{i}+0\hat{j})=a\hat{j}\]                    So work done from (a, 0) to (a, a) \[W=\overrightarrow{F}\,.\,\overrightarrow{r\,}\]                    \[=-K(a\hat{i}+a\hat{j})\,.\,a\hat{j}=-K{{a}^{2}}\]             So total work done\[=-K{{a}^{2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner