Question

A force F=-k(yi+xj) Vector where k is a positive constant acts on a particle moving in the X-Y plane. Starting from the origin the particle is taken along the positive X-axis to the point (a,0) then parallel to the y-axis to the point (a,a). Then total work done by the force?

Answer 1

Answer:

Explanation:

(please assume i to be i cap and j to be j cap)

Given,

Force, F= -K(yi+xj)

As the particle moves in the x-y plane,

the displacement would be, dr= dx i + dy j

now work done is, W=integration( F.dr)

=>W=-K(yi+xj).(dx i + dy j)

=>W=-K(ydx+xdy)

=>W=-Kd(xy). {as ydx+xdy=dxy)

therefore, on integration,

W= -K(xy)

W=-K(a*a).

{limit from (0,0) to (a,a); will give 0 for (0,0) and a^2 for (a,a)}

W=-Ka^2. (answer)

Answer 2

Answer:

The total work done by the force is $-ka^{2}$

Explanation:

Given, Force acting on a particle, $\vec F= -k(y\hat i+x\hat j)$

Displacement of particle in the x-y plane is $d\vec r= dx \hat i + dy \hat j$

As the force varying in magnitude and direction,

Work done is defined as force times displacement and is given by,

                         $W=\int\limits^{a,a}_{0,0} \vec Fd\vec r$

                              $=\int\limits^{a,a}_{0,0} -k(y\hat i+x\hat j)( dx \hat i + dy \hat j)$

                              $=-k\int\limits^{a,a}_{0,0} (ydx+xdy)$

                               $=-k\int\limits^{a,a}_{0,0} d(xy)$

                               $=-k\big|xy\big|\limits^{a,a}_{0,0}$

                               $=-k(a.a)$

                              $=-ka^{2}$

Therefore, the total work done is $-ka^{2}$