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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.

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.

Grade:12

5 Answers

Vinay Arya
37 Points
10 years ago

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

vikas askiitian expert
509 Points
10 years ago

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

vikas askiitian expert
509 Points
10 years ago

 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 ....

Aiswarya Ram Gupta
35 Points
10 years ago

thnku Mr.Vikas  Smile

Yash Chourasiya
askIITians Faculty 256 Points
one year ago
Hello Student

F = -k(yi + xj)

W = F.dr (dot product of force & displacement)

displacement vector(dr) is of line joining initial & final position vectors...

W = -k(yi + xj).(dxi + dyj)

dW = -k(ydx + xdy)

dW = -kd(xy) (exact)

integrating both sides ,

W = -k [xy] lim from (0,0) to (a,a)

W = -ka2

I hope this solution will help you.

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