# Work, energy, power?

A body of mass $\displaystyle m$ slides down a plane inclined at an angle $\displaystyle \alpha$. The coefficient of friction is $\displaystyle \mu$. Find rate at which kinetic plus gravitational potential energy is dissipated at any time $\displaystyle t$.
Motion exists along the surface of the plane only (call this direction horizontal). Analyse the weight force $\displaystyle B=mg$ in two components: the vertical and horizontal forces, $\displaystyle N=mg\cos(a)$ and $\displaystyle F=mg\sin(a)$ respectively. Along the horizontal direction, two forces are applied: $\displaystyle F$ and the friction $\displaystyle T=\mu m g \cos(a)$. Call $\displaystyle v(t)$ the velocity at time $\displaystyle t$. By Newton's law, $\displaystyle F-T=mv'(t)$, and solve to get $\displaystyle v(t)=g(\sin(a)-\mu \cos(a))t$ (assume the body was resting at $\displaystyle t=0$).
If $\displaystyle h(t)$ is the height and $\displaystyle S(t)=\int_0^t v(s)ds$ the length traveled at time $\displaystyle t$, then a simple argument shows $\displaystyle h(t)=S(t)\sin(a)$. Now, the total energy is $\displaystyle E(t)=mgh(t)+1/2mv^2(t)=mgS(t)\sin(a)+1/2mv^2(t)$. Differentiate to get $\displaystyle E'(t)=mgv(t)\sin(a)+mv(t)v'(t)$...