The work done by the force `vec(F)=A(y^(2) hati+2x^(2)hatj)`, where A is a constant and x & y are in meters around the path shown is :

Options

A. zero
B. A d
C. `A d^(2)`
D. `A d^(3)`

Answer

D

Explanation

`int barF. Bar(dx)=int A (y^(2)hat i+2x^(2) hat j). (dx hat i=dy hat j)`
`=A int (y^(2) dx+2x^(2) dy)`
` W_(OA)=+0, W_(AB)=A [0+2d^(2)d]`
.
`W_(BC)=A [d^(2) (-d) + 0], W_(CD) = A [0+0]`
`W=0 +2Ad^(3) -Ad^(3) +0=Ad^(3)`.