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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.

9 years ago

Answers : (5)

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.

9 years ago
vikas askiitian expert
509 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

9 years ago
vikas askiitian expert
509 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 ....

9 years ago
Aiswarya Ram Gupta
35 Points
							

thnku Mr.Vikas  Smile

9 years ago
Yash Chourasiya
askIITians Faculty
246 Points
							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.
4 months ago
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