Question

A constant force $\overrightarrow{F}=(\hat{i}-2\hat{j}+\hat{k})\text{N}$ is acting on a body to displace it from $(0,1,1)\text{m}$ to $(0,-2,3)\text{m}$, the amount of work done by force is

• A. $8\text{J}$
• B. $-8\text{J}$
• C. $-7\text{J}$
• D. $7\text{J}$

Answer 1

The correct option is A $8\text{J}$

Given constant force $\overrightarrow{F}=(\hat{i}-2\hat{j}+\hat{k})\text{N}$
Initial position $=(0,1,1)=(0\hat{i}+\hat{j}+\hat{k})\text{m}$
Final position $=(0,-2,3)=(0\hat{i}-2\hat{j}+3\hat{k})\text{m}$

Work done by constant force $\left(W\right)=\overrightarrow{F}.\overrightarrow{s}$

To find displacement $\overrightarrow{s}$:
Displacement $\overrightarrow{s}=$ change in position $=\overrightarrow{r_2}-\overrightarrow{r_1}$
$=(0\hat{i}-2\hat{j}+3\hat{k})-(0\hat{i}+\hat{j}+\hat{k})$
$=0\hat{i}-3\hat{j}+2\hat{k}$

We know that, if $\overrightarrow{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k};\overrightarrow{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$,
then $\overrightarrow{A}.\overrightarrow{B}=A_x{B}_x+A_y{B}_y+A_z{B}_z$

Work done $\left(W\right)=\overrightarrow{F}.\overrightarrow{s}$
$=(\hat{i}-2\hat{j}+\hat{k}).(0\hat{i}-3\hat{j}+2\hat{k})$
$=0+6+2=8\text{J}$

Hence option A is the correct answer