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 $B(0,1)$ along the line segment is?
(all quantities are in SI units)

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

Answer 1

Answer: (C)

$1$

Given,

$\overrightarrow{F}=-x\hat{i}+y\hat{i}$

Along $AB$

$ds=dx\hat{i}+dy\hat{j}$

work done $=\int \overrightarrow{F}.\overrightarrow{ds}$

$=\int (-x\hat{i}+y\hat{j})(dx\hat{i}+dy\hat{j})$

W.D $={\int }_1^0-x.dx+{\int }_0^{1}y.dy$

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

$=\frac{1}{2}+\frac{1}{2}=1$

$W.D=1J$

Option (C)