A block of mass 5.0 kg slides down from the top of an inclined plane of length 3 m. The first 1 m of the plane is smooth and the next 2 m is rough. The block is released from rest and again comes to rest at the bottom of the plane. If the plane is inclined at 30^o with the horizontal, find the coefficient of friction on the rough portion.

 

Force method: 

Form P to Q: ${\mathrm{v}}^2=0^2+2{\mathrm{a}}_1{\mathrm{s}}_1$

where ${\mathrm{a}}_1=\mathrm{gsin}{30}^{\circ },{\mathrm{s}}_1=1\mathrm{m}$

and from Q to R: $0^2={\mathrm{v}}^2+2{\mathrm{a}}_2{\mathrm{s}}_2$

where ${\mathrm{a}}_2=\mathrm{gsin}{30}^{\circ }-\mathrm{\mu gcos}{30}^{\circ },{\mathrm{s}}_2=2\mathrm{m}$

solve to get $\mathrm{\mu }=\sqrt{3}/2$

Work-energy method:
 

In terms of energy considerations you can summarize the whole process as loss in gravitational potential energy of the block = work done against friction, or 

$\begin{array}{l}\left(\mathrm{mg}\right)\times \left(3\sin {30}^{\circ }\right)=(\mathrm{\mu mgcos}\mathrm{\theta })\times 2\\ 3\times 1/2\mathrm{\mu }\times \frac{\sqrt{3}}{2}\times 2\Rightarrow \mathrm{\mu }=\frac{\sqrt{3}}{2}\end{array}$