Question

A particle of mass 0.5 Kg is displaced from position ${\overrightarrow{r}}_1(2,3,1)to{\overrightarrow{r}}_2(4,3,2)$ by applying of force magnitude 30 N which acting $(\hat{i}+\hat{j}+\hat{k})$. The work done by the force is-

• A. $10\sqrt{3}J$
• B. $30\sqrt{3}J$
• C. 30 J
• D. None of these

Answer 1

The correct option is B $30\sqrt{3}J$

Given that,

$m=0.5$kg

$r_1=(2,3,1)$

$r_2=(4,3,2)$

$F=30$ N

$\overrightarrow{F}=\hat{i}+\hat{j}+\hat{k}$

The work done is the dot product of the force and displacement.

The vector

$r_2{r}_1=r_2-r_1$

$\left[\begin{array}{ccc}4& 3& 2\end{array}\right]-\left[\begin{array}{ccc}2& 3& 1\end{array}\right]=\left[\begin{array}{ccc}2& 0& 1\end{array}\right]$

The magnitude of the force

$|\overrightarrow{F}|=30N$

The unit vector

$\hat{f}=\frac{1}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}}\times \left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\end{array}\right)$

Now, the force

$\overrightarrow{F}=F\times \hat{f}$

$\overrightarrow{F}=\left|\overrightarrow{F}\right|\cdot \hat{f}=\left(\begin{array}{c}\frac{30}{\sqrt{3}}\\ \frac{30}{\sqrt{3}}\\ \frac{30}{\sqrt{3}}\end{array}\right)$

Now, the work done is

$W=\overrightarrow{F}\cdot \left(r_2{r}_1\right)$

$W=\left(\begin{array}{c}\frac{30}{\sqrt{3}}\\ \frac{30}{\sqrt{3}}\\ \frac{30}{\sqrt{3}}\end{array}\right)\left(\begin{array}{ccc}2& 0& 1\end{array}\right)$

$W=(\frac{60}{\sqrt{3}}+0+\frac{30}{\sqrt{3}})$

$W=30\sqrt{3}J$

Hence, the work done is $30\sqrt{3}J$