Question

The total work done by three forces on a block is $60\text{J}$. If block is displaced by a force from position $A(1,2,3)$ to $B(7,6,5)$. Find out the third force if the other two forces are $\overrightarrow{F_1}=(7\hat{i}+3\hat{j}+2\hat{k})\text{N}$, and $\overrightarrow{F_2}=(-2\hat{i}+2\hat{j}+\hat{k})\text{N}$.
Assume that $\overrightarrow{F_3}$ is perpendicular to $XY$ plane.

• A. $\overrightarrow{F_3}=\hat{k}$
• B. $\overrightarrow{F_3}=2\hat{k}$
• C. $\overrightarrow{F_3}=3\hat{k}$
• D. $\overrightarrow{F_3}=4\hat{k}$

Answer 1

The correct option is B $\overrightarrow{F_3}=2\hat{k}$

Displacement of the block
$\overrightarrow{ds}=\overrightarrow{B}-\overrightarrow{A}$
$\overrightarrow{ds}=(7\hat{i}+6\hat{j}+5\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})$
$\overrightarrow{ds}=(6\hat{i}+4\hat{j}+2\hat{k})$

Net force
$F_{net}=\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}$
$=(7\hat{i}+3\hat{j}+2\hat{k})+(-2\hat{i}+2\hat{j}+\hat{k})+m\hat{k}$
$=5\hat{i}+5\hat{j}+(3+m)\hat{k}$
$W={\overrightarrow{F}}_{net}.\overrightarrow{ds}$
$\Rightarrow 60=(6\hat{i}+4\hat{j}+2\hat{k}).[5\hat{i}+5\hat{j}+(3+m\left)\hat{k}\right]$
$\Rightarrow 60=6\times 5+4\times 5+2(3+m)$
$\Rightarrow 10=2(3+m)$
$\Rightarrow m=2$
Therefore, $\overrightarrow{F_3}=2\hat{k}\text{N}$