Question

Work done when a force $\overrightarrow{F}=(\hat{i}+2\hat{j}+3\hat{k})\text{N}$ acting on a particle displaces it from the point $r_1=(\hat{i}+\hat{j}+\hat{k})$ to the point $r_2=(\hat{i}-\hat{j}+2\hat{k})$ is

• A. $-3\text{J}$
• B. $7\text{J}$
• C. $-1\text{J}$
• D. $1\text{J}$

Answer 1

The correct option is C $-1\text{J}$

Given constant force $\overrightarrow{F}=\hat{i}+2\hat{j}+3\hat{k}\text{N}$
Initial position $\overrightarrow{r_1}=(\hat{i}+\hat{j}+\hat{k})\text{m}$
Final position $\overrightarrow{r_2}=(\hat{i}-\hat{j}+2\hat{k})\text{m}$
Work done $\left(W\right)=\overrightarrow{F}.\overrightarrow{s}$


To find the displacement $\overrightarrow{s}$:
$\overrightarrow{s}=\overrightarrow{r_2}-\overrightarrow{r_1}=$ change in position
$\overrightarrow{s}=(\hat{i}-\hat{j}+2\hat{k})-(\hat{i}+\hat{j}+\hat{k})$
$\overrightarrow{s}=(-2\hat{j}+\hat{k})\text{m}$

We know that, if $\overrightarrow{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k};\overrightarrow{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$,
then $\overrightarrow{A}.\overrightarrow{B}=A_x{B}_x+A_y{B}_y+A_z{B}_z$

$\therefore W=\overrightarrow{F}.\overrightarrow{s}=(\hat{i}+2\hat{j}+3\hat{k}).(-2\hat{j}+\hat{k})$
$W=1\left(0\right)+2(-2)+3\left(1\right)$
$W=-4+3=-1$
$\therefore$ Workdone $\left(W\right)=\overrightarrow{F}.\overrightarrow{s}=-1\text{J}$

Hence option C is the correct answer.