If the constant force $2\widehat{\dot{\mathbf{i}}} - 5\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}$ and $- \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$ act on a particle due to which it is displaced from a point $A(4, - 3, - 2)$ to a point $B(6,1, - 3)$ then the work done by the force is

Question

Dr. Abubacker · Accepted Answer

(d)
Resultant force $\overrightarrow{\mathbf{F}} = \left( 2\widehat{\dot{\mathbf{i}}} - 5\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}} ight) + \left( - \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}} ight)$
$= \widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} + 5\widehat{\mathbf{k}}$
and displacement, $\overrightarrow{\mathbf{d}} = \left( 6\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - 3\widehat{\mathbf{k}} ight) - \left( 4\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}} ight)$
$= 2\widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$
$	herefore\$work done $W = \overrightarrow{\mathbf{F}} \bullet \overrightarrow{\mathbf{d}}$
$= \left( \widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} + 5\widehat{\mathbf{k}} ight) \bullet \left( 2\widehat{\dot{\mathbf{i}}} + 4\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}} ight)$
$= - 15$
$= 15$ units [neglecting $–$ ve sign]

Explanation

Reviewed by Dr. Abubacker

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