Question

A force $\overrightarrow{F}=(2\hat{i}+3\hat{j})N$ acts ona particle and displaces it form initial position $\overrightarrow{x_1}=(3\hat{i}+4\hat{j})m$ to final position $\overrightarrow{x_2}=(2\hat{i}+3\hat{j})m$, then work done by force, is

• A. $-19J$
• B. $-5J$
• C. $12J$
• D. $19J$

Answer 1

The correct option is B $-5J$

A force $\overrightarrow{F}=(2\hat{i}+3\hat{j})N$

acts on a particle and displaces it form initial position $\overrightarrow{x_1}=(3\hat{i}+4\hat{j})m$ to final position

$\overrightarrow{x_2}=(2\hat{i}+3\hat{j})m$

The work done$=?$

Solution

Now,

Work done

$W=force\times displacement$

Given data

$force=(\hat{i}+)\hat{j}$

Now, displacement

$d=(\overrightarrow{x_2}-\overrightarrow{x_1})=(2\hat{i}+3\hat{j})-(3\hat{i}+4\widehat{j)}=-\hat{i}-\hat{j}$

$w=F\times d$

$W=(2\hat{i}+3\hat{j})\times (-\hat{i}-\hat{j})=-2\hat{i}-3\hat{j}=-5J$