The work done by the force F→=6i^+2j^ N in displacing an object from r_1→=3i^+8j^ m to r_2→=5i^−4j^ m, is

Correct answer: (D) -12 J. F→=6i^+2j^ r→=r→_2-r→_1=5i^-4j^-3i^-8j^ =2i^-12j^ W=F→·S→=(6i^+2j^)·(2i^-12j^) =-12J

Q40 / 92 · NTA Abhyas (JEE Main) 2020 · Physics · Work Power Energy

The work done by the force $\vec{F} = 6 \hat{i} + 2 \hat{j} N$ in displacing an object from $\vec{r_{1}} = 3 \hat{i} + 8 \hat{j} m$ to $\vec{r_{2}} = 5 \hat{i} - 4 \hat{j} m$ , is

$\vec{F} = 6 \hat{i} + 2 \hat{j}$

$\vec{r} = {\vec{r}}_{2} - {\vec{r}}_{1} = 5 \hat{i} - 4 \hat{j} - 3 \hat{i} - 8 \hat{j}$

$= 2 \hat{i} - 12 \hat{j}$
$W = \vec{F} \cdot \vec{S} = (6 \hat{i} + 2 \hat{j}) \cdot (2 \hat{i} - 12 \hat{j})$
$= - 12 J$

Hence, the correct answer is option: 'D'.