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Question

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 the y - axis to the point (a,a). The total work done by the force F on the particle is

A
2ka2
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B
2ka2
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C
ka2
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D
ka2
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Solution

The correct option is C ka2
While moving from (0,0) to (a,0)
Along positive x - axis,y=0F=kx^j
i.e. force is in negative y - direction while displacement is in positive x - direction.
W1=0
Because force is perpendicular to displacment.
Then particle moves from (a,0) to (a,a) along a line parallel to y - axis (x=+a) during this F=k(y^i+a^j)
The first component of force, ky^i 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. ka^j will perform negative work
W2=(ka^j)(a^j)=(ka)(a)=ka2
So, net work done on the particle
W=W1+W2=0+(ka2)=ka2

Alternate solution

Given F=k(y^i+x^j)
We know that
W=F.ds=(a,a)(0,0)k(y^i+x^j).(dx^i+dy^j)=(a,a)(0,0)k(ydx+xdy)W=(a,a)(0,0)k d(xy)=ka2

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