# Forces of magnitudes $3$ and $4$ units acting in the directions $\overrightarrow{6i}+ \overrightarrow{2j} +\overrightarrow{3k}$ and $\overrightarrow{3i}-\overrightarrow{2j}+\overrightarrow{6k}$ respectively act on a particle which is displaced from the point $(2,2,-1)$to$( 4,3,1)$. find the work done by the forces.

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• If a number of forces $\overrightarrow F_1, \overrightarrow F_2, \overrightarrow F_3, ....$ act on a particle and displace it through $\overrightarrow d$, then the work done by the force is $( \overrightarrow F_1+\overrightarrow F_2+\overrightarrow F_3...).\overrightarrow d$
The unit vector in the direction $6\overrightarrow i+2\overrightarrow j+3\overrightarrow k$ is $\large\frac{6\overrightarrow i+2\overrightarrow j+3\overrightarrow k}{\sqrt{36+4+9}} = \large\frac{6\overrightarrow i+2\overrightarrow j+3\overrightarrow k}{7}$
$\therefore$ the first force $\overrightarrow F_1 = \large\frac{3}{7} (6\overrightarrow i+2\overrightarrow j+3\overrightarrow k)$
The unit vector in the direction $3\overrightarrow i-2\overrightarrow j+6\overrightarrow k \: is \: \large\frac{3\overrightarrow i-2\overrightarrow j+6\overrightarrow k}{\sqrt{9+4+36}} = \large\frac{3\overrightarrow i-2\overrightarrow j+6\overrightarrow k}{7}$
$\therefore$ the second force $\overrightarrow F_2 = \large\frac{4}{7} (3\overrightarrow i-2\overrightarrow j+6\overrightarrow k)$
The resultant force $\overrightarrow F = \overrightarrow F_1+\overrightarrow F_2 = \large\frac{1}{7} [ 30\overrightarrow i-2\overrightarrow j+33\overrightarrow k]$
The particle is displaced from $A(2,2,-1) (\overrightarrow {OA}=2\overrightarrow i+2\overrightarrow j-\overrightarrow k) \: to \: B(4,3,1) (\overrightarrow {OB}=4\overrightarrow i+3\overrightarrow j+\overrightarrow k)$
$\therefore \overrightarrow d = \overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA} = 2\overrightarrow i+\overrightarrow j+2\overrightarrow k$
work done = $\overrightarrow f.\overrightarrow d = \large\frac{1}{7} [ 60-2+66] = \large\frac{124}{7}$ units
answered Jun 4, 2013