Derive the work energy theorem for a variable force exerted on a body in one dimension .

Explanation

If a body of mass m and speed v moving in X - direction in one dimensions then its kinetic energy .
`K = 1/2 mv^(2)`
Intergating on both the side ,
`(dK)/(dt) =d/(dt) (1/2 mv^(2))`
` =1/2 mxx2v . (dv)/(dt)`
` :. (dK)/(dt) =m . (dv)/(dt) xxv`
` :. (dK)/(dt) = mav [ :. (dv)/(dt) =a] `
` :. (dK)/(dt) = F (dx)/(dt) [ :. ma = F and V = (dx)/(dt)] `
` :. dK = F dx `
Intergating on both side from intial position `x_(i)` to final position `x_(f)`
`int_(K_(i))^(K_(f)) dK = int_(x_(i))^(x_(f)) F dx`
`K_(f) - K_(i) = F(x_(f)-x_(i))`
` = FDeltac ` where `Deltax = x_(f) -x_(i)`
which is a work energy theorem for a variable force .
Work energy theorem does not incorporate the complete dynamical information of Newton.s second law .
Work energy theorem involves an integral over an interval of time , the time information contained in the Newton.s second law is integrated over and is not available explicity .
Newton.s second law for two or three dimensions is in vector form whereas the work energy theorem is in scalar form .