Conservation of energy; block on incline

Principle of conservation of energy

The total mechanical energy (kinetic + potential) of a system remains constant when only conservative forces (gravity, springs) do work:

$$T_1+V_1=T_2+V_2$$

Example: a body of mass $m$ dropped from height $h$. At the top $T_1=0$ and $V_1=mgh$; just before it hits the ground $V_2=0$ and $T_2=\tfrac12mv^{2}$. So $mgh=\tfrac12mv^{2}$ and $v=\sqrt{2gh}$: the potential energy changes completely into kinetic energy. When friction acts, mechanical energy is not conserved, and the work done by friction must be subtracted.

Block on the incline

Data: incline 3 m high and 4 m horizontal, so length $AB=5$ m and $\cos\theta=4/5=0.8$. $\mu_k=0.3$ (the block slides because $\tan\theta=0.75>\mu_s=0.35$).

  A
  |\
  | \  5 m
3 m|  \
  |   \  theta
  +----B==========C
          friction

A to B (work-energy):

$$mg(3)-\mu_k(mg\cos\theta)(5)=\tfrac12mv_B^{2}$$ $$v_B^{2}=2g\left[3-0.3(0.8)(5)\right]=2(9.81)(1.8)=35.32\ \text{m}^2/\text{s}^2$$ $$v_B=5.943\ \text{m/s}$$

B to C (comes to rest):

$$\tfrac12mv_B^{2}=\mu_k mg\,d\ \Rightarrow\ d=\frac{v_B^{2}}{2\mu_kg}=\frac{35.32}{2(0.3)(9.81)}=6.0\ \text{m}$$

Answer: the block travels $6.0$ m along the horizontal plane before stopping.