A particle gets displaced by $\Delta \bar{r}=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \mathrm{m}$ under the action of…
A particle gets displaced by $\Delta \bar{r}=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \mathrm{m}$ under the action of a force $\vec{F}=(7 \hat{i}+4 \hat{j}+3 \hat{k})$. The change in its kinetic energy is
$38 \mathrm{~J}$
$70 \mathrm{~J}$
$52.5 \mathrm{~J}$
$126 \mathrm{~J}$
Solution
According to work-energy theorem, Change in kinetic energy $=$ work done $=\vec{F} \cdot \Delta \vec{r}=(7 \hat{i}+4 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})$ $=14+12+12=38 \mathrm{~J}$