Question

A particle moves from a point $(2\hat{i}+\hat{k})\text{m}$ to $(4\hat{i}+3\hat{j}-\hat{k})\text{m}$ when a force of $(3\hat{i}+\hat{j})\text{N}$ is applied. The work done by the force is

• A. $8\text{J}$
• B. $11\text{J}$
• C. $5\text{J}$
• D. $9\text{J}$

Answer 1

The correct option is D $9\text{J}$

Since the force is constant here, work done by the force will be given by
$W=\overrightarrow{F}.\overrightarrow{d}...\left(i\right)$
where displacement is given as $\overrightarrow{d}=\overrightarrow{r_2}-\overrightarrow{r_1}$
$\&{\overrightarrow{r}}_i$ represents the position vector of particle.
Given, $\overrightarrow{F}=3\hat{i}+\hat{j}\text{N}$
$\overrightarrow{r_2}=4\hat{i}+3\hat{j}-\hat{k}\text{m}\&\overrightarrow{r_1}=2\hat{i}+\hat{k}\text{m}$
$\overrightarrow{d}=\overrightarrow{r_2}-\overrightarrow{r_1}$
$=2\hat{i}+3\hat{j}-2\hat{k}\text{m}$
From Eq.$\left(i\right)$ we get,
$W=(3\hat{i}+\hat{j}).(2\hat{i}+3\hat{j}-2\hat{k})$
$\Rightarrow W=(3\times 2)+(1\times 3)+(0\times (-2\left)\right)$
$\therefore W=9\text{J}$
Work done by the force is $9\text{J}$.