Q160 / 160 · TS EAMCET 2017 · Physics
Consider a particle on which constant forces $\mathbf{F}_{\mathbf{1}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} N$ and $\mathbf{F}_{\mathbf{2}}=4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} N$ act together resulting in a displacement from position $\mathbf{r}_1=20 \hat{\hat{i}}+15 \hat{\mathrm{j}} \mathrm{cm}$ to $\mathbf{r}_2=7 \hat{\mathrm{k}} \mathrm{cm}$. The total work done on the particle is
- Correct answerA$-0.48 \mathrm{~J}$
- B$+0.48 \mathrm{~J}$
- C$-4.8 \mathrm{~J}$
- D$+4.8 \mathrm{~J}$
Solution
Given,
$$
\begin{aligned}
& \mathbf{F}_1=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \mathrm{N} \\
& \mathbf{F}_2=(4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \mathrm{N} \\
& \mathbf{r}_1=20 \hat{\mathbf{i}}+15 \hat{\mathbf{j}} \mathrm{cm} \\
& \mathbf{r}_2=7 \hat{\mathbf{k}} \mathrm{cm}
\end{aligned}
$$
Total force on particle, $\mathbf{F}=\mathbf{F}_1+\mathbf{F}_2$
$$
\begin{aligned}
& =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\
& =(5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \mathrm{N}
\end{aligned}
$$
Displacement of particle, $\mathbf{s}=\mathbf{r}_2-\mathbf{r}_1$
$$
\begin{aligned}
& =7 \hat{\mathbf{k}}-20 \hat{\mathbf{i}}-15 \hat{\mathbf{j}} \\
& =(-20 \hat{\mathbf{i}}-15 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}) \mathrm{cm}
\end{aligned}
$$
We know that,
$$
\begin{aligned}
\operatorname{Work}(\infty) & =\mathbf{F} \cdot \mathbf{s} \\
& =(5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})(-20 \hat{\mathbf{i}}-15 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}) \times 10^{-2} \\
& =(-100+45+7) \times 10^{-2} \\
& =-0.48 \mathrm{~J}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.