Question

A force $\overrightarrow{F}=4x\hat{i}+2y\hat{j}$ displaces a body from $r_1=3\hat{i}+8\hat{j}$ to $\overrightarrow{r_2}=9\hat{i}-6\hat{j}$. Find the work done by the force.

• A. $125\text{J}$
• B. $120\text{J}$
• C. $116\text{J}$
• D. Zero

Answer 1

The correct option is C $116\text{J}$

Gives force $=\overrightarrow{F}=4x\hat{i}+2y\hat{j}$
initial position $\overrightarrow{r_1}=3\hat{i}+8\hat{j}$
and portion $\overrightarrow{r_2}=9\hat{i}-6\hat{j}$
We know work done
$W={\int }_{s1}^{s_2}\overrightarrow{F}.d\overrightarrow{s}$
$={\int }_{s_1}^{s_2}\overrightarrow{F}.(dx\hat{i}+dy\hat{j})$
$W={\int }_{s1}^{s_2}(4x\hat{i}+2y\hat{j}).(dx\hat{i}+dy\hat{j})$
$={\int }_3^{9}4xdx+{\int }_8^{-6}2ydy$
$W=[2x^2{]}_3^9+[y^2{]}_8^{-6}$
$W=144+(-28)$
$W=116\text{J}$