A force of magnitude $\sqrt{6}$ acting along the line joining the points $A(2, - 1,1)$ and $B(3,1,2)$ displaces a particle from $A$ to $B$. The work done by the force is

Question

Dr. Abubacker · Accepted Answer

(a)
$\because\$Displacement, $\overrightarrow{\mathbf{AB}} = (3 - 2)\widehat{\dot{\mathbf{i}}} + (1 + 1)\widehat{\dot{\mathbf{j}}} + (2 - 1)\widehat{\mathbf{k}}$
$= \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$
$and\ force,\ \overrightarrow{\mathbf{F}} = \frac{\sqrt{6}(\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}})}{\sqrt{6}}$
$= \left( \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} ight)$
$	herefore$ Work done $= \overrightarrow{\mathbf{F}} \bullet \overrightarrow{\mathbf{AB}} = \left( 1 + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} ight) \bullet \left( \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} ight) = 6$

Q. A force of magnitude $\sqrt{6}$ acting along the line joining the points $A(2, - 1,1)$ and $B(3,1,2)$ displaces a particle from $A$ to $B$. The work done by the force is

Explanation

Reviewed by Dr. Abubacker

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