A particle moves from a point ($-2\overrightarrow{i}+5\overrightarrow{j}$) m to $(4\overrightarrow{i}+3\overrightarrow{k})$ m when a force of $(4\overrightarrow{i}+3\overrightarrow{j}) \mathrm{N}$ is applied. How much work has been done by the force?

Displacement of the particle

$\overrightarrow{\mathrm{s}} = \overrightarrow{{\mathrm{r}}_2}-\overrightarrow{{\mathrm{r}}_1} = (4\overrightarrow{j}+3\overrightarrow{k})-(-2\overrightarrow{i}+5\overrightarrow{j}) =6\overrightarrow{i}-5\overrightarrow{j}+3\overrightarrow{k}$

Applied force $\overrightarrow{\mathrm{F}} = 4\overrightarrow{i}+3\overrightarrow{j}$

Hence, work done, W = $\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{s}} =24-15= 9\mathrm{J}$