A particle moves in x-y plane under action of a path dependent force `F=yhati+xhatj.` The work done by the force as it moves in x-y plane can be evaluated by solving the integral `intvecF.vecdr,`where `vec(dr)=dxhati+dyhatj.` The position coordinates x and y will vary according to some constraint determined by teh path followed by the particle. For example, if particle moves along a straight line from one position to other, then an possible rarticle. For example, if particle moves along a straight line from one position to other, then a possible relation is `y=mx+c.` NOw. try to solve the following questions.
1. The particel moves along a straight line from origin to (a ,a) . The work done by the force on the particle is
`(1) a^(2)" "(2)2s" "(3)(a^(2))/(2)" "(4)`Zero
2. The particle moves from `(0,0)` to (a,0) and then from, `(a,0)` to `(a,a),` in straight line paths. The work done by the force on the particle is
`(1)a^(2)" "(2)2a" "(3)(a^(2))/(2)" "(4)` Zero
3. The differential work by the given force over an elementary displacement is given as `dW=vecF.vec(dr)=ydx+xdy`
This can be expressed as `dW=d(xy).` From this, it can be inferred that the work done by the force
(1) Depends only on initial and final values of x and y
(2) Depends on initial and final values and on the path followed
(3) Is zero for any values of x and y
(4) Is zero when object returns to original position after following any path

Explanation

1. Answer (1) As the particle moves straight from (0,0) to (a,a) it moves along the line
`y=ximplies(dy)/(dx)=1ordy=dx`
Now, work done is calculated using
`W=intvecF.vec(dr)=int(yhati+xhatj)(dxhati+dyhatj)`
`int(ydx+xdy)`
Taking `y=xand dy=dx` and limits from (0,0) to (a,a),
`W=underset(0)overset(a)int2xdx=[x^(2)]_(0)^(a)=a^(2)`
2. Answer (1)
The path followed is shown in the figure. For the path between (0,0) to (a,0), y remains constant
So, `dy=0.` Hence

`W_(1)=int(ydx+xdy)=intydx`
`=yintdx`
`=0underset(0) overset(a) intdx=0`
For the between (a,0) to (a,a) x remains constant and `dx=0`
`impliesW_(2)=int(ydx+xdy)=x intdy=a underset(0)overset(a)intdy=a^(2)impliesW=0+a^(2).`
3. Answer (1,4)
`dW =vecF.vec (dr)=ydx+xdy=d(xy)`
`impliesW=intd(xy)=[xy]_((x_(1)y_(1)))^(""(x_(2),y_(2)))`
`=x_(2)y_(2)-x_(1)y_(1)`
In this case, work depends only on `(x_(1),y_(1))` the initial position, and `(x_(2),y_(2))` the final position. If `(x_(2),y_(2))=(x_(1),y_(1)),W=0.`