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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.

6 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.

6 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

6 years ago
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 ....

6 years ago
Aiswarya Ram Gupta
35 Points
										

thnku Mr.Vikas  Smile

6 years ago
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