Question

A force of magnitude $12\text{N}$ acts on a particle in the direction of a vector $\overrightarrow{A}=2\hat{i}-4\hat{j}+4\hat{k}$, which results in the change of position of the particle from $(3,3,5)\text{m}$ to $(2,-1,4)\text{m}$. The work done by the force is (in Joule )

• A. $25$
• B. $40$
• C. $20$
• D. $30$

Answer 1

The correct option is C $20$

Given, a force is acting in the direction of the vector $\overrightarrow{A}=2\hat{i}-4\hat{j}+4\hat{k}$,
So, we have
$\overrightarrow{F}=|\overrightarrow{F}|.\hat{F}=|\overrightarrow{F}|.\hat{A}=|\overrightarrow{F}|.\frac{\overrightarrow{A}}{|\overrightarrow{A}|}$
$\Rightarrow \overrightarrow{F}=\left(12\right)\left(\frac{2\hat{i}-4\hat{j}+4\hat{k}}{\sqrt{2^2+(-4{)}^2+4^2}}\right)=4\hat{i}-8\hat{j}+8\hat{k}$
Now, we have position vectors as
$\overrightarrow{r_1}=3\hat{i}+3\hat{j}+5\hat{k}$ and $\overrightarrow{r_2}=2\hat{i}-\hat{j}+4\hat{k}$
So, the displacement vector is given as
$\overrightarrow{S}=\overrightarrow{r_2}-\overrightarrow{r_1}=-\hat{i}-4\hat{j}-\hat{k}$
Thus, the work done by the force is given as
$W=\overrightarrow{F}.\overrightarrow{S}=(4\hat{i}-8\hat{j}+8\hat{k}).(-\hat{i}-4\hat{j}-\hat{k})$
$\Rightarrow W=-4+32-8=20\text{Joule}$