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 the, 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. 30
• E. 35

Answer 1

Answer: (B)

Here, $\vec{F_{1}} = 4\hat{i}+\hat{j}-3\hat{k}, \vec{F_{2}} = 3\hat{i}+\hat{j}-\hat{k}$
$\vec{r_{1}} = \hat{i}+2\hat{j}+3\hat{k}$, $\vec{r_{2}} = 5\hat{i}+4\hat{j}+\hat{k}$
Displacement, $\vec{r} = \vec{r_{2}}-\vec{r_{1}}$
$= \left(5\hat{i}+5\hat{j}+\hat{k}\right)-\left(\hat{i}+2\hat{j}+3\hat{k}\right)$
$= 4\hat{i}+2\hat{j}-2\hat{k}$
Work done by the forces,
$W = \left[\vec{F_{1}}+\vec{F_{2}}\right]\cdot\vec{r}$
$= \left[\left(4\hat{i}+\hat{j}-3\hat{k}\right)\left(3\hat{i}+\hat{j}-\hat{k}\right)\right]\cdot\left(4\hat{i}+2\hat{j}-2\hat{k}\right)$
$= \left(7\hat{i}+2\hat{j}-4\hat{k}\right)\cdot\left(4\hat{i}+2\hat{j}-2\hat{k}\right)$
$= 28+4+8 = 40\,J$