Force acting on a particle have magnitude 5,3 and 1 unit act in the direction of the vectors $6\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$, $3\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}$ and $2\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} - 6\widehat{\mathbf{k}}$ respectively. They remain constant while the particle is displaced from the point $A(2, - 1, - 3)$ to $B(5 - 1,1)$. The work done is

Question

Dr. Abubacker · Accepted Answer

(b)
Total force,
$\overrightarrow{\mathbf{F}} = \frac{5\left( 6\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}} ight)}{7} + \frac{3\left( 3\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}} ight)}{7} + \frac{1\left( 2\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} - 6\widehat{\mathbf{k}} ight)}{7}$
$= \frac{1}{7}\left( 41\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 27\widehat{\mathbf{k}} ight)$
$and\ \ \overrightarrow{\mathbf{AB}} = 5\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} - 2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$
$= 3\widehat{\dot{\mathbf{i}}} + 4\widehat{\mathbf{k}}$
$	herefore\$Work done $= \overrightarrow{\mathbf{F}} \bullet \overrightarrow{\mathbf{AB}}$
$= \frac{1}{7}\left\lbrack 41\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 27\widehat{\mathbf{k}} ightbrack \bullet \left\lbrack 3\widehat{\dot{\mathbf{i}}} + 4\widehat{\mathbf{k}} ightbrack$
$= \frac{1}{7}\lbrack 123 + 108brack = 33\ units$

Explanation

Reviewed by Dr. Abubacker

Want more practice?