Question

A particle is displaced from point $\text{A}(1,2,3)$ to $\text{B}(1,3,4)$ under action of two constant forces ${\overrightarrow{F}}_1=$$\hat{i}+2\hat{j}+\hat{k}$ and ${\overrightarrow{F}}_2=$$\hat{i}+\hat{j}$. Find the total work done on the particle to move it from point $\text{A}$ to $\text{B}$.

• A. $2\text{J}$
• B. $3\text{J}$
• C. $4\text{J}$
• D. $5\text{J}$

Answer 1

The correct option is C $4\text{J}$

As particle is shifted from point $(1,2,3)$ to $(1,3,4)$.

So, we can write given points as position vectors,

$\overrightarrow{r_1}=\hat{i}+2\hat{j}+3\hat{k}\overrightarrow{r_2}=\hat{i}+3\hat{j}+4\hat{k}$

Displacement vector$\left(\Delta \overrightarrow{r}\right)$ is,

$\Delta \overrightarrow{r}=\overrightarrow{r_2}-\overrightarrow{r_1}$
$\Rightarrow \Delta \overrightarrow{r}=(\hat{i}+3\hat{j}+4\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})=(\hat{j}+\hat{k})$

$\because$ Force is given in terms of components so, we can add them,
$\overrightarrow{F_{net}}=\overrightarrow{F_1}+\overrightarrow{F_2}=(\hat{i}+2\hat{j}+\hat{k})+(\hat{i}+\hat{j})=(\hat{2}i+3\hat{j}+\hat{k})$

Work done:
$W=\overrightarrow{F_{net}}\cdot \Delta \overrightarrow{r}=(2\hat{i}+3\hat{j}+\hat{k})\cdot (\hat{j}+\hat{k})=4\text{J}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(c\right)$ is correct.

Why this question ? In case of multiple forces, the total work done is, $W=\overrightarrow{F_{net}}\cdot \Delta \overrightarrow{S}$ $W=(F_x\hat{i}+F_y\hat{i}+F_z\hat{k}).(S_x\hat{i}+S_y\hat{j}+S_z\hat{k})$ $W=F_x{S}_x+F_y{S}_y+F_z{S}_z$