Question

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

• A. $24J$
• B. $33J$
• C. $83J$
• D. $45J$

Answer 1

The correct option is C

$83J$

The explanation for correct option

Step 1: Given

1. The object is displaced from position $\overrightarrow{r_1}=\left(2\hat{i}+3\hat{j}\right)m$ to $\overrightarrow{r_2}=\left(4\hat{i}+6\hat{j}\right)m$
2. Applied force $\overrightarrow{F}=\left(3x^2\hat{i}+2y\hat{j}\right)N$

Step 2: Formula used

$W=\int \overrightarrow{F·}dr=\int \overrightarrow{F·}\left(dx\hat{i}+dy\hat{j}\right)$

Where $W$ is the work done and $\overrightarrow{F}$ is the applied force

Step 3: Find the work done

The object is displaced from position $\overrightarrow{r_1}=\left(2\hat{i}+3\hat{j}\right)m$ to $\overrightarrow{r_2}=\left(4\hat{i}+6\hat{j}\right)m$

Therefore $x_1=2,y_1=3,x_1=4,y_1=6$

$$\begin{gathered} W={\int }_{R_1}^{R_2}\overrightarrow{F·}dr \\ ={\int }_{R_1}^{R_2}\left(3x^2\hat{i}+2y\hat{j}\right)·\left(dx\hat{i}+dy\hat{j}\right) \\ ={\int }_2^{4}3x^{2}dx+{\int }_3^{6}2ydy \\ ={\left[x^3\right]}_2^4+{\left[y^2\right]}_3^6 \\ =64-8+36-9 \\ =83J \end{gathered}$$

Hence option C is correct