Question

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

Answer 1

Given that

a uniform force ,$\overrightarrow{F}=3\hat{i}+\hat{j}$

Mass ,$m=2kg$

Initial position of particle , $\overrightarrow{r_1}=2\hat{i}+\hat{k}$ m

Final position of particle, $\overrightarrow{r_2}=4\hat{i}+3\hat{j}-\hat{k}$ m

Net Displacement of particle , $\overrightarrow{\Delta r}=\overrightarrow{r_2}-\overrightarrow{r_1}=2\hat{i}+3\hat{j}-2\hat{k}$ m

Work done by the force on the particle , $W=\overrightarrow{F}\cdot \overrightarrow{\Delta r}=6+3=9J$