Question

The two forces together ${\overline{F}}_1=2\hat{i}+\hat{j}+3\hat{k}$ and ${\overline{F}}_2=\hat{i}+2\hat{j}+\hat{k}$ worked to move a particle from $\overline{a}=4\hat{i}+6\hat{j}+\hat{k}$ to $\overline{b}=2\hat{i}+4\hat{j}+\hat{k}$. Find the work done

• A. $14$
• B. $16$
• C. $12$
• D. None of these

Answer 1

Answer: (C)

$12$

Net force acting on the particle

$=(2\hat{i}+\hat{j}+3\hat{k})+(\hat{i}+2\hat{j}+\hat{i})=3\hat{i}+3\hat{j}+4\hat{k}$

The displacement vector$($Final position $-$ Initial position$)$
$=(2\hat{i}+4\hat{j}+\hat{k})-(4\hat{i}+6\hat{j}+\hat{k})=-2\hat{i}-2\hat{j}$

Now work done

$=\overline{F}.\overline{d}=(3\hat{i}+3\hat{j}+4\hat{k}).(-2\hat{i}-2\hat{j})$
$=-6-6$

$=12$ units.