Question

Answer the Question
If you don't know then you can spam -_-​

Answer 1

While moving from (0,0) to (a,0)

ĐỸ - k Along positive x-axis, y = 0 =>

i.e. force is in negative y-direction while displacement is in positive x-direction.

W₁ = 0

Because force is perpendicular to displacement.

Then particle moves from (a,0) to (a, a) along a line parallel to y-axis (x=+a) during this F-k(vi+aj)

The first component of force, -kyi will not contribute any work because this component is along negative x-direction (-i) while displacement is in positive y-direction (a,0) to (a,a).

The second component of force i.e. -kaj will perform

negative work

W₂ = (-kaj) (aj) = (-ka)(a) = -ka²

So net work done on the particle W=W₁ + W₂=0+ (-ka²) = -ka²

Answer 2

Explanation:

= (please assume i to be i cap and j to be j cap)

Given,

= Force, F= -K(yi+xj)

= As the particle moves in the x-y plane,

= the displacement would be, dr=dx i + dy j now work done is, W-integration( F.dr)

=>W=K(yi+xj).(dxi + dy j)

=>W=K(ydx+xdy)

=>W=Kd(xy). [as ydx+xdy=dxy)

therefore, on integration,

= W = -K(xy)

= W = -K(a×a).

= (limit from (0,0) to (a,a); will give 0 for (0,0) and a² for (a,a)]

= W = -Ka². (answer)

hope it was helpful to you ❣️☺️

mark me brainlist ☺️☺️