Question

A particle of mass $m$ is displaced from a position $P_1$ to $P_2$ with position vectors ${\overrightarrow{r}}_1=a\hat{i}+b\hat{j}+c\hat{k};{\overrightarrow{r}}_2=c\hat{i}+a\hat{j}+b\hat{k}$ by a force $\overrightarrow{F}=b\hat{i}+c\hat{j}+a\hat{k}$. Find the work done by the force.

Answer 1

If the displacement of the particle is $\overrightarrow{S}$, the work done $W$ by the given force $\overrightarrow{F}$ is equal to $\overrightarrow{F}.\overrightarrow{S}$ where $\overrightarrow{F}$ is a constant force
$\Rightarrow \overrightarrow{W}=\overrightarrow{F}.\overrightarrow{S}....\left(1\right)$
where $\overrightarrow{S}=\Delta \overrightarrow{r}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1$
$\overrightarrow{S}=[(c\hat{i}+a\hat{j}+b\hat{k})-(a\hat{i}+b\hat{j}+c\hat{k})]$
$\overrightarrow{S}=[(c-a)\hat{i}+(a-b)\hat{j}+(b-c)\hat{k}]....\left(2\right)$
and $\overrightarrow{F}=b\hat{i}+c\hat{j}+a\hat{k}.....\left(3\right)$
Using (1), (2) and (3)
$W=(b\hat{i}+c\hat{j}+a\hat{k})[(c-a)\hat{i}+(a-b)\hat{j}+(b-c)\hat{k}]$
$\Rightarrow W=(c-a)b+(a-b)c+(b-c)a\Rightarrow W=bc-ab+ac-bc+ab-ac=0$
Net work done by the force $\overrightarrow{F}$ for the given displacement is zero.