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 work done by this force in these 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 that,

Position vector of point $P=a\hat{i}+a\hat{j}=\overrightarrow{s}$

Work done $W_1$ is

$W_1=\overrightarrow{F}\cdot \overrightarrow{s}$

$W_1=(3\hat{i}+4\hat{j})\cdot (a\hat{i}+a\hat{j})$

$W_1=3a+4a$

$W_1=7a$

Now, work done in two different paths

$OQ=a\hat{i}$

$QP=a\hat{j}$

Now, work done $W_2$ is

$W_2=\overrightarrow{F}\cdot \overrightarrow{OQ}+\overrightarrow{F}\cdot \overrightarrow{QP}$

$W_2=(3\hat{i}+4\hat{j})\cdot \left(a\hat{i}\right)+(3\hat{i}+4\hat{j})\cdot \left(a\hat{j}\right)$

$W_2=3a+4a$

$W_2=7a$

So, $W_1=W_2$

Hence, the work done is equal for both paths