Question

An object is displaced from position ${\overrightarrow{r}}_1=\hat{i}+2\hat{j}\text{m}$ to $\overrightarrow{r_2}=2\hat{i}+4\hat{j}\text{m}$ under the action of force $\overrightarrow{F}=(6x^2\hat{i}+4y\hat{j})\text{N}$. Find the work done by this force.

• A. $76\text{J}$
• B. $46\text{J}$
• C. $19\text{J}$
• D. $38\text{J}$

Answer 1

The correct option is D $38\text{J}$

Since the force is variable in nature, we have to solve by integration technique using the fundamental formula for work done and substituting proper limits.
$\left(W\right)={\int }_{\overrightarrow{r_1}}^{\overrightarrow{r_2}}\overrightarrow{F}.\overrightarrow{dr}......\left(i\right)$

$\overrightarrow{F}=(6x^2\hat{i}+4y\hat{j})\text{N}$
The displacement vector $\overrightarrow{dr}$ can be expressed as :

$\overrightarrow{dr}=dx\hat{i}+dy\hat{j}+dz\hat{k}$, substituting in the Eq. $\left(i\right)$

$\Rightarrow W={\int }_{\overrightarrow{r_1}}^{\overrightarrow{r_2}}(6x^2\hat{i}+4y\hat{j}).(dx\hat{i}+dy\hat{j}+dz\hat{k})$

$W={\int }_{\overrightarrow{r_1}}^{\overrightarrow{r_2}}(6x^{2}dx+4ydy)$

Since particle is displaced from $\overrightarrow{r_1}=\hat{i}+2\hat{j}\text{to}\overrightarrow{r_1}=2\hat{i}+4\hat{j}$, hence limits for $x$-direction displacement is $x=1\text{m}$ to $x=2\text{m}$ and $y$-direction displacement is $y=2\text{m}$ to $y=4\text{m}$

$W=\frac{6}{3}{\left[x^3\right]}_1^2+\frac{4}{2}{\left[y^2\right]}_2^4$

$\Rightarrow W=2[8-1]+2[16-4]$

$\therefore W=38\text{J}$