Question

A particle acted upon by constant forces $4\hat{i}+\hat{j}-3\hat{k}$ and $3\hat{i}+\hat{j}-\hat{k}$ is displaced from the point $\hat{i}+2\hat{j}+3\hat{k}$ to point $5\hat{i}+4\hat{j}+\hat{k}$. The total work done by the forces in SI unit is:

• A. $20$
• B. $40$
• C. $50$
• D. $60$

Answer 1

Answer: (B)

$40$

Here, $\overrightarrow{F_1}=4\hat{i}+\hat{j}-3\hat{k},\overrightarrow{F_2}=3\hat{i}+\hat{j}-\hat{k}$
$\overrightarrow{r_1}=\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{r_2}=5\hat{i}+4\hat{j}+\hat{k}$
Displacement, $\overrightarrow{r}=\overrightarrow{r_2}-\overrightarrow{r_1}$
$=(5\hat{i}+4\hat{j}+\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})=4\hat{i}+2\hat{j}-2\hat{k}$
Work done by the forces,
$W=[\overrightarrow{F_1}+\overrightarrow{F_2}]\cdot \overrightarrow{r}$
$=\left[\right(4\hat{i}+\hat{j}-3\hat{k})+(3\hat{i}+\hat{j}-\hat{k}\left)\right]\cdot (4\hat{i}+2\hat{j}-2\hat{k})$
$=(7\hat{i}+2\hat{j}-4\hat{k})\cdot (4\hat{i}+2\hat{j}-2\hat{k})=28+4+8=40J$.