Question

Forces of magnitude $5$ and $3$ units acting in the directions $6\hat{i}+2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+6\hat{k}$ respectively act on a particle which is displaced from the point $(2,2,-1)$ to $(4,3,1)$. The work done by the forces is

• A. $148$ units
• B. $\frac{148}{7}$
• C. $\frac{78}{7}$
• D. None of these

Answer 1

Answer: (B)

$\frac{148}{7}$

Let $F$ be the resultant force and $d$ be the displacement vector.
Then, $F=5\frac{(6\hat{i}+2\hat{j}+3\hat{k})}{\sqrt{36+4+9}}+3\frac{(3\hat{i}-2\hat{j}+6\hat{k})}{\sqrt{9+4+36}}$
$=\frac{1}{7}(39\hat{i}+4\hat{j}+33\hat{k})$
and $d=(4\hat{i}+3\hat{j}+\hat{k})-(2\hat{i}+2\hat{j}-\hat{k})$
$=(2\hat{i}+\hat{j}+2\hat{k})$
$\therefore$ Total work done $=F\cdot d$
$=\frac{1}{7}[(39\hat{i}+4\hat{j}+33\hat{k})\cdot (2\hat{i}+\hat{j}+2\hat{k})]$
$=\frac{1}{7}[78+4+66]=\frac{148}{7}$ units.