A force of $(4 i+2 j+k) \mathrm{N}$ is acting on a particle of mass 2 kg displaces the particle from a…

A force of $(4 i+2 j+k) \mathrm{N}$ is acting on a particle of mass 2 kg displaces the particle from a position of $(2 \bar{i}+2 \bar{j}+\bar{k}) \mathrm{m}$ to a position of $(4 \bar{i}+3 \bar{j}+2 \bar{k}) m$. The work done by the force on the particle in joules is
  1. 21 J
  2. 11 J
  3. 14 J
  4. 18 J

Solution

$\overrightarrow{\mathrm{F}}=(4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \mathrm{N}, \mathrm{m}=2 \mathrm{~kg}$ $\overrightarrow{r_1}=(2 \hat{i}+2 \hat{j}+\hat{k}) m, \overrightarrow{r_2}=(4 \hat{i}+3 \hat{j}+2 \hat{k}) m$ $\begin{aligned} & \therefore \text { Work done, } W=\vec{F} \cdot \vec{S}=\vec{F} \cdot\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right) \\ & =(4 \hat{i}+2 \hat{j}+\hat{k}) \cdot(2 \hat{i}+\hat{j}+\hat{k}) \\ & =4 \times 2+2 \times 1+1 \times 1=11 \mathrm{~J}\end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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