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

Explanation

Parameterize the Path:

The straight line from A(1,0) to B(0,1) can be parameterized as:

r(t) = (1 - t)i + tj, where t ranges from 0 to 1.

1. Compute the Differential Element:

   The differential displacement vector dr is obtained by differentiating r(t) with respect to t:

   dr = d(r(t))/dt dt = (-i + j) dt.

2. Evaluate the Force Along the Path:

   Substitute r(t) into the force F:

   F(r(t)) = -xi + yj = -(1 - t)i + tj.

3. Compute the Dot Product:

   The dot product F · dr is:

   F · dr = [-(1 - t)i + tj] · [-i + j] = (1 - t) + t = 1.

4. Integrate Along the Path:

   Integrate the dot product over the parameter t from 0 to 1 to find the work done:

   W = ∫ from 0 to 1 of 1 dt = [t] from 0 to 1 = 1.

Final Answer

The work done by the force in moving the particle from point A to point B along the line segment is 1 joule.