Question

A force of magnitude of $30N$ acting along $\left(\hat{i}+\hat{j}+\hat{k}\right)$, displaces a particle from a point$(2,4,1)to(3,5,2)$. The work done during this displacement is:

• A. $90J$
• B. $30J$
• C. $30\sqrt{3}J$
• D. $20J$

Answer 1

The correct option is C

$30\sqrt{3}J$

Step 1. Given data:

Force, $\overrightarrow{F}=30N$ acting along $\left(\hat{i}+\hat{j}+\hat{k}\right)$

Displacement of body, $\overrightarrow{dS}$ from point $A(2,4,1)toB(3,5,2)$

$$\begin{gathered} \Rightarrow \overrightarrow{dS}=\overrightarrow{BA} \\ \Rightarrow \overrightarrow{dS}=(3\hat{i}+5\hat{j}+2\hat{k})-(2\hat{i}+4\hat{j}+\hat{k}) \\ \therefore \overrightarrow{dS}=\left(\hat{i}+\hat{j}+\hat{k}\right) \end{gathered}$$

Step 2. Calculating $\overrightarrow{F}$ and the work done:

$\overrightarrow{F}=30\frac{\left(\hat{i}+\hat{j}+\hat{k}\right)}{\sqrt{3}}$ $\left[\hat{A}=\frac{\overrightarrow{A}}{\left|A\right|}\right]and\left[\left|A\right|=\sqrt{1+1+1}=\sqrt{3}\right]$

Now, Work done, $W=\overrightarrow{F}·\overrightarrow{dS}$

$\begin{array}{rcl}& \Rightarrow & W=30\frac{\left(\hat{i}+\hat{j}+\hat{k}\right)}{\sqrt{3}}·\left(\hat{i}+\hat{j}+\hat{k}\right)\\ & \Rightarrow & W=\frac{30\left(1+1+1\right)}{\sqrt{3}}\\ & \Rightarrow & W=\frac{30\times 3}{\sqrt{3}}\\ \therefore W& =& 30\sqrt{3}J\end{array}$

Hence. correct option is (C).