Q. A particle is acted on by a force of 6 units in the direction $9\widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$ and is displaced from the point $3\widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} - 15\widehat{\mathbf{k}}$ to the point $7\widehat{\dot{\mathbf{i}}} - 6\widehat{\dot{\mathbf{j}}} + 8\widehat{\mathbf{k}}$. The work done is

A) 18
B) 15
C) 12
D) 9
Explanation

(c)
$Here,\ force\ \overrightarrow{\mathbf{F}} = 6 \times \frac{(9\widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}})}{\sqrt{81 + 36 + 4}}$
$= \frac{6(9\widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}})}{11}$
Displacement vector $\overrightarrow{\mathbf{d}}$
$= (7 - 3)\widehat{\dot{\mathbf{i}}} + ( - 6 - 4)\widehat{\dot{\mathbf{j}}} + (8 + 15)\widehat{\mathbf{k}}$
$= 4\widehat{\dot{\mathbf{i}}} - 10\widehat{\dot{\mathbf{j}}} + 23\widehat{\mathbf{k}}$
$\therefore$Work done $= \overrightarrow{\mathbf{F}} \bullet \overrightarrow{\mathbf{d}}$
$= \frac{6}{16}\left( 9\widehat{\dot{\mathbf{i}}} + 6\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right) \bullet \left( 4\widehat{\dot{\mathbf{i}}} - 10\widehat{\dot{\mathbf{j}}} + 23\widehat{\mathbf{k}} \right)$
$= \frac{6}{11}(36 - 60 + 46) = 12$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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