A uniform force of $(3 \hat{i}+\hat{j})$ newton acts on a particle of mass 2 kg . Hence the particle

Question

A uniform force of $(3 \hat{i}+\hat{j})$ newton acts on a particle of mass 2 kg . Hence the particle is displaced from position $(2 \hat{i}+\hat{k})$ metre to position $(4 \hat{i}+3 \hat{j}-\hat{k})$ metre. The work done by the force on the particle is\

Professor PC Thomas Classes · Answer

(c) : Here, $\vec{F}=(3 \hat{i}+\hat{j}) \mathrm{N}$ Initial position, $\vec{r}_{1}=(2 \hat{i}+\hat{k}) \mathrm{m}$\ Final position, $\vec{r}_{2}=(4 \hat{i}+3 \hat{j}-\hat{k}) \mathrm{m}$\ Displacement, $\vec{r}=\vec{r}_{2}-\vec{r}_{1}$\ $\vec{r}=(4 \hat{i}+3 \hat{j}-\hat{k}) \mathrm{m}-(2 \hat{i}+\hat{k}) \mathrm{m}=2 \hat{i}+3 \hat{j}-2 \hat{k} \mathrm{~m}$\ Work done, $W=\vec{F} \cdot \vec{r}=(3 \hat{i}+\hat{j}) \cdot(2 \hat{i}+3 \hat{j}-2 \hat{k})$ $$ =6+3=9 \mathrm{~J} $$