A particle is moving in a force field given by `F = y^2 hati - x^2hatj`. Starting from A, the particle has to reach C either along ABC or ADC. Let the workdone along the two paths be `W_1 and W_2`, respectively. Then, `(W_1,W_2)` are

Options

A. (-1,1)
B. (-1,0)
C. (1,1)
D. (-1,1)

Answer

A

Explanation

`W_(ABC)=W_(AB)+w_(BC)`
`= int_(x=0)^(x=1)(y^2hati-x^2 hatj).dxhati+int_(y=0)^(y=1)(y^2hati-x^2hatj).dyhatj`
For AB , y = 0 and BC , x = 1
`W_(ABC) = int_(x=0)^(x=1)(0hati-x^2hatj).dxhati+int_(y=0)^(y=1)(y^2hati-hatj).dx hatj`
`=0+(-1)int_0^1 dy =-1`
Now , `W_(ADC) = W_(AD) +W_(DC)`
`=int_(y=0)^(y=1)(y^2hati-x^2hatj).dxhati+int_(y=0)^(x=1)(y^2hati-x^2hatj).dxhati`
For AD , `x = 0 and ` for DC , y = 1
`implies W_(ADC) = int_0^1(y^2 hati-0)dyhatj+int_0^1(1hati-x^2hatj)dxhati`
`=0+int_0^1 dx = 1`
Hence `(W_1,W_2) = (-1,1)`