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Grade 12Mechanics

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.

Profile image of Aiswarya Ram Gupta
15 Years agoGrade 12
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5 Answers

Profile image of Vinay Arya
15 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.

Profile image of vikas askiitian expert
ApprovedApproved Tutor Answer15 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

Profile image of vikas askiitian expert
15 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 ....

Profile image of Aiswarya Ram Gupta
15 Years ago

thnku Mr.Vikas  Smile

Profile image of Yash Chourasiya
6 Years 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.