Forces of magnitudes 3 and 4 units acting along $6\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$ and $3\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}$ respectively act on a particle and displace it from $(2,2 - 1)$ to $(4,3,1)$. The work done is

Question

Dr. Abubacker · Accepted Answer

(a)
$Total\ force,\$
$\overrightarrow{\mathbf{F}} = 3\left( \frac{6\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}}{7} ight) + 4\left( \frac{3\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 6\widehat{\mathbf{k}}}{7} ight)$
$= \frac{(30\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 33\widehat{\mathbf{k}})}{7}$
$	herefore\ \overrightarrow{\mathbf{d}} = 4\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} - \left( 2\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}} ight)$
$= 2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$
$	herefore$ Work done $W = \overrightarrow{\mathbf{F}} \bullet \overrightarrow{\mathbf{d}}$
$= \left( \frac{30\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 33\widehat{\mathbf{k}}}{7} ight) \bullet (2\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}})$
$= \frac{60 - 2 + 66}{7} = \frac{124}{7}$

Explanation

Reviewed by Dr. Abubacker

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