Question

derive;

W_nc=(½mv_f²+mgh_f)-(½mv₀²+mgh₀)

Answer 1

For kinetic energy 

$m\frac{dv_x}{dt}=F_x\to\frac{d}{dt}(\frac{mv_x^2}{2})=F_xv_x \to d(\frac{mv_x^2}{2})=F_xdx\to$

$\frac{mv_{xf}^2}{2}-\frac{mv_{x0}^2}{2}=\int F_xdx$ . In general $\frac{mv_{f}^2}{2}-\frac{mv_{0}^2}{2}=\int \vec Fd\vec r$

For potential energy

$\vec Fd\vec r=-\frac{\partial E_p}{\partial x}dx-\frac{\partial E_p}{\partial y}dy-\frac{\partial E_p}{\partial z}dz \to \vec Fd\vec r=-dE_p \to$

$-(E_{p2}-E_{p1})=\int \vec Fd\vec r$

If the body has both potential and kinetic energy then

$W=\int \vec Fd\vec r=\frac{mv_{f}^2}{2}-\frac{mv_{0}^2}{2}-(-(E_{p2}-E_{p1}))=$

$=(\frac{mv_{f}^2}{2}+E_{p2})-(\frac{mv_{0}^2}{2}+E_{p1})$ .