1.

A body is displaced from prigin to (1m,1m) by force `F=(2yhati + 3x^(2)hatj)` along two paths (a) `x=y` (b) `y=x^(2)` Find the work done along both paths.

Answer» `F=(2yhati + 3x^(2)hatj)`
`dr=(dxhati + dyhatj)`
`F.dr=(2ydx + 3x^(2)dy)`
We cannot integrate F.dr or `(2ydx + 3x^(2)dy)` as such to find the work done. But along the given paths we can change this expression.
(a) Along the path `x = y`
`(2y dx + 3x^(2)dy) = (2x dx + 3y^(2)dy)`
`W_(1) = int_(0,0)^(1m,1m,)F.dr =int_(0,0)^(1m,1m)(2xdx + 3y^(2)dy)`
`=[x^(2) + y^(3)] _(0,0)^(1m,1m)`
`=(1)^(2) + (1)^(3)=2J`
(b) Along the path `y=x^(2)`
`(2ydx + 3x^(2)dy) =(2x^(2)dx + 3ydy)`
`:. W=(2)=int_(0,0)^(1m,1m)F.df =int(2x^(2)dx + 3y dy)`
`=[2/3x^3 + 3/2y^(2)]_(0,0)^(1m,1m)`
`=(13)/6 J`


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