Question

Two forces $P=3\hat{i}-2\hat{j}+\hat{k}$ and $Q=\hat{i}+3\hat{j}-5\hat{k}$ acting on a particle $A$ move it to $B$. If the position vectors of $A$ and $B$ are respectively $-2\hat{i}+5\hat{k}$ and $3\hat{i}-7\hat{j}+2\hat{k}$, then the total work done is

• A. $20units$
• B. $30units$
• C. $40units$
• D. $25units$

Answer 1

Answer: (D)

$25units$

Given two forces are

$\overrightarrow{P}=3\hat{ı}-2\hat{ȷ}+\hat{k}\text{and}\overrightarrow{Q}=\hat{ı}+3\hat{ȷ}-5\hat{k}$

Net force on the particle is, $\overrightarrow{F}=\overrightarrow{P}+\overrightarrow{Q}$

$\overrightarrow{F}=(3\hat{ı}-2\hat{j}+\hat{k})+(\hat{ı}-3\hat{ȷ}+5\hat{k})$

$\overrightarrow{F}=4\hat{ı}+\hat{ȷ}-4\hat{k}$

Position vectors of $A$ and $B$ are given as

$\overrightarrow{A}=-2\hat{ı}+5\hat{k}$ and $\overrightarrow{B}=3\hat{ı}-7\hat{ȷ}+2\hat{k}$

Net displacement of the particle, $\overrightarrow{S}=\overrightarrow{B}-\overrightarrow{A}$

$\overrightarrow{S}=(3\hat{ı}-7\hat{j}+2\hat{k})-(-2\hat{ı}+5\hat{k})$

$\overrightarrow{S}=5\hat{i}-7\hat{j}-3\hat{K}$

Total work done, $W=\overrightarrow{F}\cdot \overrightarrow{S}$

$\Rightarrow W=(4\hat{ı}+\hat{j}-4\hat{k})\cdot (5\hat{ı}-7\hat{ȷ}-3\hat{k})$

$=20-7+12$

$=25$ units

$\text{(D) is the correct option.}$