Question

A force $\overrightarrow{F}=(2x\hat{i}+3y^2\hat{j})\text{N}$ acts on a body and displaces it from the origin $(0,0)$ to $(2\text{m},4\text{m})$.The work done (in $\text{J}$) by the force is

Answer 1

Given, $\overrightarrow{F}=(2x\hat{i}+3y^2\hat{j})\text{N}$
Position vector of initial position ${\overrightarrow{r}}_1=(0\hat{i}+0\hat{j})$
Position vector of final position of the body is, ${\overrightarrow{r}}_2=(2\hat{i}+4\hat{j})\text{m}$
Work done by a variable force is given by,
$W=\underset{{\overrightarrow{r}}_1}{\overset{{\overrightarrow{r}}_2}{\int }}\overrightarrow{F}.\overrightarrow{dr}$
$W=\underset{{\overrightarrow{r}}_1}{\overset{{\overrightarrow{r}}_2}{\int }}(2x\hat{i}+3y^2\hat{j}).(dx\hat{i}+dy\hat{j})$
[$\because \overrightarrow{dr}=dx\hat{i}+dy\hat{j}$]
Putting the limits of $(x,y)$ as $(0,0)\text{to}(2,4)$
$W=\underset{(0,0)}{\overset{(2,4)}{\int }}2xdx+3y^{2}dy...\left(1\right)$
$W={[x^2+y^3]}_{(0,0)}^{(2,4)}$
$W=\left[\right(4+64)-(0+0\left)\right]$
$\therefore W=68\text{J}$
Work done by the force is $68\text{J}$.