Q. The total work done by two forces ${\overrightarrow{\mathbf{F}}}_{1} = 2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}}$ and ${\overrightarrow{\mathbf{F}}}_{2} = 3\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$ acting on a particle when it is displaced from the point $3\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$ to $5\widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$ is
Explanation
(d)
$\therefore$ Total force , $\overrightarrow{\mathbf{F}} = {\overrightarrow{\mathbf{F}}}_{1} + {\overrightarrow{\mathbf{F}}}_{2}$
$= 5\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$
and displacement,$\overrightarrow{\mathbf{d}} = (5 - 3)\widehat{\dot{\mathbf{i}}} + (5 - 2)\widehat{\dot{\mathbf{j}}} + (3 - 1)\widehat{\mathbf{k}}$
$= 2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$
$\therefore W = \overrightarrow{\mathbf{F}} \bullet \overrightarrow{\mathbf{d}}$
$= \left( 5\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}} \right) \bullet (2\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}})$
$= 11$ units
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.