Question

A force $\overrightarrow{F}=6x\hat{i}+2y\hat{j}$ displaces a body from $\overrightarrow{r_1}=3\hat{i}+8\hat{j}$ to $\overrightarrow{r_2}=5\hat{i}-4\hat{j}$. Find the work done by the force.

• A. $0\text{J}$
• B. $75\text{J}$
• C. $64\text{J}$
• D. $16\text{J}$

Answer 1

The correct option is A $0\text{J}$

Given, force $\overrightarrow{F}=6x\hat{i}+2y\hat{j}$
Initial position, $\overrightarrow{r_1}=3\hat{i}+8\hat{j}$ and final position $\overrightarrow{r_2}=5\hat{i}-4\hat{j}$
We know that, work done
$W=\int \overrightarrow{F}.ds=\int \overrightarrow{F}.(dx\hat{i}+dy\hat{j})$
$={\int }_{S_1}^{S_2}(6x\hat{i}+2y\hat{j}).(dx\hat{i}+dy\hat{j})$
$={\int }_3^{5}6xdx+{\int }_8^{-4}2ydy$
$=[3x^2{]}_3^5+[y^2{]}_8^{-4}$
$=(75-27)+(16-64)=0$