Question

Three constant forces $\overrightarrow{F_1}=(2\hat{i}-3\hat{j}+2\hat{k})\text{N},\overrightarrow{F_2}=(\hat{i}+\hat{j}-\hat{k})\text{N}$ and $\overrightarrow{F_3}=(3\hat{i}+\hat{j}-2\hat{k})\text{N}$ displace a particle from $(1,-1,2)\text{m}$ to $(-1,-1,3)\text{m}$ and then to $(2,3,4)\text{m}$, displacement being measured in meters. Find the total work done by the forces.

• A. $\sqrt{40}\text{J}$
• B. $0\text{J}$
• C. $\sqrt{50}\text{J}$
• D. $24\text{J}$

Answer 1

The correct option is B $0\text{J}$

Net force acting on particle,
$\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}$
$=(2\hat{i}-3\hat{j}+2\hat{k})+(\hat{i}+\hat{j}-\hat{k})+(3\hat{i}+j-2\hat{k})$
$=(6\hat{i}-\hat{j}-\hat{k})\text{N}$
Initial position of particle $\overrightarrow{x_i}=\hat{i}-\hat{j}+2\hat{k}$
Final position of particle $\overrightarrow{x_f}=2\hat{i}+3\hat{j}+4\hat{k}$
Net displacement of particle:
$\overrightarrow{S}=\overrightarrow{x_f}-\overrightarrow{x_i}=(2-1)\hat{i}+(3-(-1\left)\right)\hat{j}+(4-2)\hat{k}$
$=(\hat{i}+4\hat{j}+2\hat{k})\text{m}$
Hence, Net work done by all acting forces:
$W=\overrightarrow{F}.\overrightarrow{S}=(6\hat{i}-\hat{j}-\hat{k}).(\hat{i}+4\hat{j}+2\hat{k})\text{J}$
$\therefore W=\left(6\right)\left(1\right)+(-1)\left(4\right)+(-1)\left(2\right)=0\text{J}$