Question

A particle is moved from $(0,0)$ to $(a,a)$ under a force $\overrightarrow{F}=(3\hat{i}+4\hat{j})$ from two paths. Path $1$ is $OP$ and path $2$ is $OQP$. Let $W_1$ and $W_2$ be the works done by this force along two paths. Then

• A. $W_1=W_2$
• B. $W_1=2W_2$
• C. $W_2=2W_1$
• D. $W_2=4W_1$

Answer 1

The correct option is A $W_1=W_2$

Given, force $\overrightarrow{F}=3\hat{i}+4\hat{j}\text{N}$
Initial position $\left(\overrightarrow{r_1}\right)=0\hat{i}+0\hat{j}$
Final position $\left(\overrightarrow{r_2}\right)=a\hat{i}+a\hat{j}$

In path $1\left(OP\right)$:

$W_1=\overrightarrow{F}.\overrightarrow{s}$
$\overrightarrow{s}=\overrightarrow{r_2}-\overrightarrow{r_1}=a\hat{i}+a\hat{j}$
$W_1=(3\hat{i}+4\hat{j})(a\hat{i}+a\hat{j})=3a+4a=7a\text{J}$

In path $2\left(OQP\right)$:

$W_2=W_{OQ}+W_{QP}$
Along $OQ:$ Initial position $\overrightarrow{r_1}=0\hat{i}+0\hat{j}$
Final position $\overrightarrow{r_2}=a\hat{i}+0\hat{j}$
${\overrightarrow{s}}_{OQ}=\overrightarrow{r_2}-\overrightarrow{r_1}=a\hat{i}$
$\Rightarrow W_{OQ}=\overrightarrow{F}.{\overrightarrow{s}}_{OQ}=(3\hat{i}+4\hat{j})\left(a\hat{i}\right)=3a$

Along $QP:$ Initial position $\left(\overrightarrow{r_1}\right)=a\hat{i}+0\hat{j}$
Final position $\left(\overrightarrow{r_2}\right)=a\hat{i}+a\hat{j}$
$\therefore {\overrightarrow{s}}_{QP}=\overrightarrow{r_2}-\overrightarrow{r_1}=a\hat{j}$
$\Rightarrow W_{QP}=\overrightarrow{F}.{\overrightarrow{s}}_{QP}=(3\hat{i}+4\hat{j})\left(a\hat{j}\right)=4a$
$\therefore W_2=3a+4a=7a$

$\Rightarrow W_1=W_2$

Hence option A is the correct answer

(We can also conclude that work done by conservative forces does not depend on the path, it depends on the initial and final position)