Question

A particle moves from position $\overrightarrow{r_1}=(3\hat{i}+2\hat{j}-6\hat{k})m$ to position $\overrightarrow{r_2}=(14\hat{i}+13\hat{j}+9\hat{k})m$ under the action of force $(4\hat{i}+\hat{j}+3\hat{k})N$. Find the work done.

Answer 1

$Displacements=\overrightarrow{r_2}-\overrightarrow{r_1}=14\hat{i}+13\hat{j}+9\hat{k}-(3\hat{i}+2\hat{j}-6\hat{k})=11\hat{i}+11\hat{j}+15\hat{k}$

As we know work done is a scalar quantity therefore W.D=F.S

$=(4\hat{i}+\hat{j}+3\widehat{k)}.(11\hat{i}+11\hat{j}+15\hat{k})=44+11+45=100$ Joule