Question

Three forces $\overrightarrow{F_1}=2\hat{i}-4\hat{j}+5\hat{k}$, $\overrightarrow{F_2}=\hat{i}+3\hat{j}-8\hat{k}$ and $\overrightarrow{F_3}=-6\hat{i}-10\hat{j}+7\hat{k}$ are acting on a particle present on the positive $x$ axis, $5\text{units}$ away from the origin. The three forces combined displace the body to a point $6\hat{i}-3\hat{j}+10\hat{k}$. What is the net work done by the three forces ?

• A. $76\text{J}$
• B. $70\text{J}$
• C. $-76\text{J}$
• D. $-70\text{J}$

Answer 1

The correct option is B $70\text{J}$

Initial position of the particle = $5\hat{i}$
Final position of the particle = $6\hat{i}-3\hat{j}+10\hat{k}$
Displacement = $(6\hat{i}-3\hat{j}+10\hat{k})-5\hat{i}=i-3\hat{j}+10\hat{k}$

Net force = ${\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3=(2+1-6)\hat{i}+(-4+3-10)\hat{j}+(5-8+7)\hat{k}=-3\hat{i}-11\hat{j}+4\hat{k}$

Work done is given by,
$W=\overrightarrow{F}.\overrightarrow{s}\Rightarrow W=(-3\hat{i}-11\hat{j}+4\hat{k}).(\hat{i}-3\hat{j}+10\hat{k})\Rightarrow W=-3+33+40=70J$