Question

Consider a force $\overrightarrow{F}=-x\hat{i}+y\hat{j}$. The work done by this force in moving a particle from point $A(1,0)$ to $B(1,1)$,along the line segment is ( all quantities are in SI units )

• A. $2$
• B. $\frac{3}{2}$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $1$

Formula used: ${\int }_{{\overrightarrow{r}}_i}^{\overrightarrow{r}f}\overrightarrow{F}.d\overrightarrow{r}$

Given, Force $\overrightarrow{F}=-x\hat{i}+y\hat{j},$

Position at $A{\overrightarrow{r}}_i=(1,0)$

Position at $B{\overrightarrow{r}}_f=(0,1)$

Work done ${\int }_{{\overrightarrow{r}}_i}^{\overrightarrow{r}f}\overrightarrow{F}.d\overrightarrow{r}$

$W={\int }_{{\overrightarrow{r}}_i}^{\overrightarrow{r}f}(-x\check{i}+yj).(dx\check{i}+dy\check{j})$

$W={\int }_1^0-xdx+{\int }_0^{1}ydy$

$=(\frac{-x^2}{2}{|}_1^0+\frac{y^2}{2}{|}_0^1)$

$=-(\frac{0^2}{2}-\frac{1^2}{2})+(\frac{1^2}{2}-\frac{0^2}{2})$

$W=1J$

Final Answer:(b)