One of the forces acting on a certain particle depends on the particle's position in the `xy`-plane. This force `vec F` expressed in newtons, is given by the expression `vec F = (xy hat i+ xy hat j)` where `x` and `y` are in metres. The particle is moved from `O` to `C` through three different paths :-

The work done by this force on path `OBC` is

Options

A. `(1)/(2) J`
B. `-(1)/(2) J`
C. `(2)/(3) J`
D. `-(2)/(3) J`

Answer

A

Explanation

Along the straight line `OC, y = x` , so `dy = dx` and `vec F.vec(ds) = xydx +xydy =2x^(2)dx`
Hence, `W_(OC) int F.ds =int_(0)^(1) 2x^(2)dx = |(2x^(3))/(3)|_(0)^(1) = (2)/(3) J`
Although `W_(OAC) = w_(OBC) = (1)/(2) J`. The work done `W_(OC)` along `OC` is not equal to `W_(OAC)` or `W_(OBC)`. The force `vec F` is non-conservative, because work done between two points depends upon the particular path.