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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.
The path of the particle cannot be like this due to this force.It is moving in straight lines.It should be some curve.
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
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
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 ....
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 ....
thnku Mr.Vikas
Hello StudentF = -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 = -ka2I hope this solution will help you.
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