Question

A force $\overrightarrow{F}=\hat{i}+3\hat{j}+2\hat{k}$ acts on a particle to displace it from the point $(\widehat{i+2}\hat{j}-3\hat{k})$ to the point $B(3\hat{i}-\hat{j}+5\widehat{k).}$ The work done by the force will be

• A. $5$ units
• B. $7$ units
• C. $9$ units
• D. $10$ units

Answer 1

Answer: (C)

$9$ units

Given

$\overrightarrow{F}=\hat{i}+3\hat{j}+2\hat{k}$

$\overrightarrow{B}=3\hat{i}-\hat{j}+5\hat{k}$

Initial point P

$\overrightarrow{P}=\hat{i}+2\hat{j}-3\hat{k}$

$Displacement=\overrightarrow{B}-\overrightarrow{P}$

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

$\overrightarrow{S}=3\hat{i}-\hat{j}+5\hat{k}-\hat{i}-2\hat{j}+3\hat{k}$

$\overrightarrow{S}=2\hat{i}-3\hat{j}+8\hat{k}$

Work Done

$\overrightarrow{W}=Force\cdot Displacement$

$\overrightarrow{W}=\overrightarrow{F}\cdot \overrightarrow{S}$

$\overrightarrow{W}=(\hat{i}+3\hat{j}+2\hat{k})\cdot (2\hat{i}-3\hat{j}+8\hat{k})$

$\overrightarrow{W}=2-9+16$

$\overrightarrow{W}=9$

Hence, the answer is $9$.