Question

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

Answer 1

Dear Student,

Given,
Force acting on the particle, $\overrightarrow{F}=3\hat{i}+\hat{j}\mathrm{N}$
Mass of the particle, m = 2 kg

Initial position of the particle, ${\overrightarrow{r}}_i=2\hat{i}+\hat{k}\mathrm{m}$
Final position of the particle, ${\overrightarrow{r}}_f=4\hat{i}+3\hat{j}-\hat{k}\mathrm{m}$
Work done by the force,
$$\begin{gathered} W=\overrightarrow{F}.\left({\overrightarrow{r}}_f-{\overrightarrow{r}}_i\right)=\left(3\hat{i}+\hat{j}\right).\left[\left(4\hat{i}+3\hat{j}-\hat{k}\right)-\left(2\hat{i}+\hat{k}\right)\right] \\ \Rightarrow W=\left(3\hat{i}+\hat{j}\right).\left(2\hat{i}+3\hat{j}-2\hat{k}\right) \\ \Rightarrow W=6+3=9\mathrm{J} \end{gathered}$$

Regards,