Question

Consider a particle on which constant forces $\overrightarrow{F_1}=\hat{i}+2\hat{j}+3\hat{k}N$ and $\overrightarrow{F_2}=4\hat{i}-5\hat{j}-2\hat{k}N$ act together resulting in a displacement from position $\overrightarrow{r_1}=20\hat{i}+15\hat{j}cm$ to $\overrightarrow{r_2}=7\hat{k}$. The total work done on the particle is

• A. $-0.48J$
• B. $+0.48J$
• C. $-4.8J$
• D. $+4.8J$

Answer 1

The correct option is A $-0.48J$

Net force acting $\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}=5\hat{i}-3\hat{j}+\hat{k}$
Position vector $\overrightarrow{S}=\overrightarrow{r_2}-\overrightarrow{r_1}=-20\hat{i}-15\hat{j}+7\hat{k}$ $\times {10}^{-2}m$
$W=\overrightarrow{F}\overrightarrow{S}=(5\hat{i}-3\hat{j}+\hat{k}).(-20\hat{i}-15\hat{j}+7\hat{k})\times {10}^{-2}$
$W=(-100+45+7)\times {10}^{-2}=-0.48J$