A particle is released one by one from the top of two inclined rough surfaces of height h each. The angles of inclination of the two planes are 30^o and 60^o, respectively. All other factors (e.g., coefficient of friction, mass of block, etc.) are same in both the cases. Let K₁ and K₂ be the kinetic energies of the particle at the bottom of the plane in the two cases. Then : 

\(K_1 < K_2\)

Work done by friction :

\(W = (\mu m g \cos {\theta})S = (\mu m g \cos {\theta}) \frac{h}{\sin {\theta}}\)

\(W = (\mu m g h) \cot {\theta}\)

\(\text{Now : } \)

\(\cot {\theta_1} = \cot {30^o} = \sqrt{3}\)

\(\cot {\theta_2} = \cot {60^o} = \frac{1}{\sqrt{3}}\)

\(\text{i.e. kinetic energy : } K_1 < K_2\)