Consider a force overrightarrow{F}=-x hat{i}+ | vedclass

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(B) The work done by a variable force is given by the line integral $W = \int_{A}^{B} \overrightarrow{F} \cdot d\overrightarrow{r}$.The equation of th

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(B) The work done by a variable force is given by the line integral $W = \int_{A}^{B} \overrightarrow{F} \cdot d\overrightarrow{r}$.The equation of the line segment connecting $A(1, 0)$ and $B(0, 1)$ is $y = -x + 1$,or $x + y = 1$.Thus,$d\overrightarrow{r} = dx \hat{i} + dy \hat{j}$.The work done is $W = \int_{A}^{B} (-x \hat{i} + y \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = \int_{A}^{B} (-x dx + y dy)$.Substituting the limits from $x=1$ to $x=0$ and $y=0$ to $y=1$:$W = \int_{1}^{0} -x dx + \int_{0}^{1} y dy$$W = \left[ -\frac{x^2}{2} ight]_{1}^{0} + \left[ \frac{y^2}{2} ight]_{0}^{1}$$W = -\left( \frac{0^2}{2} - \frac{1^2}{2} ight) + \left( \frac{1^2}{2} - \frac{0^2}{2} ight)$$W = -\left( 0 - 0.5 ight) + \left( 0.5 - 0 ight) = 0.5 + 0.5 = 1 	ext{ J}$.

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).

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