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

Aiswarya Ram Gupta Grade: 12

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

7 years ago

Answers : (4)

Vinay Arya

37 Points

The path of the particle cannot be like this due to this force.It is moving in straight lines.It should be some curve.

7 years ago

vikas askiitian expert

510 Points

										F = -k(xi+yj)
W = F.dr              (dot product of force & displacement)
displacement vector(dr) is of line joining initial & final position vectors...
W = -k(xi+yj).(dxi+dyj)
W = -2k[xy] lim (0,0) to (a,a)
W = -2ka²
this is the amount of work

7 years ago

Approved

vikas askiitian expert

510 Points

										 there is a mistake in previous sol so dont be confused n go for this result ...
W = -K(yi+xj) . (dxi+dyj)
dW = -k(ydx + xdy)
dW = -kd(xy)
integrating both sides ,
W = -k [xy] lim from (0,0) to (a,a)
    = -ka²
this is correct ....

7 years ago