Question

An object is displaced from position vector $\overrightarrow{r_1}=(2\hat{i}+3\hat{j})$ $\text{m}$ to $\overrightarrow{r_2}=(4\hat{i}+6\hat{j})$$\text{m}$ under a force $\overrightarrow{F}=(3x^2\hat{i}+2y\hat{j})$ $\text{N}$. Find the work done.

• A. $83\text{J}$
• B. $38\text{J}$
• C. $-83\text{J}$
• D. $-38\text{J}$

Answer 1

The correct option is A $83\text{J}$

Given, $\overrightarrow{F}=3x^2\hat{i}+2y\hat{j}\text{N}$
Initial position, $\overrightarrow{\left(r_1\right)}=(2\hat{i}+3\hat{j})\text{m}\Rightarrow (2,3)\to (x_1,y_1)$
Final position, $\overrightarrow{\left(r_2\right)}=(4\hat{i}+6\hat{j})\text{m}\Rightarrow (4,6)\to (x_2,y_2)$
Since, the given force is variable in x and y, displacement of object, $\overrightarrow{ds}=dx\hat{i}+dy\hat{j}$

Then, work done $W=\int \overrightarrow{F}.\overrightarrow{ds}={\int }_{x_1}^{x_2}{F}_{x}dx+{\int }_{y_1}^{y_2}{F}_{y}dy={\int }_2^{4}3x^{2}dx+{\int }_3^{6}2ydy=x^3{|}_2^4+y^2{|}_3^6=(4^3-2^3)+(6^2-3^2)=(64-8)+(36-9)=83\text{J}$