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# A particle is released from top of two inclined rough surfaces of height 'h' each. The angle of inclination of two planes are $30^{\circ}$ and $60^{\circ}$ respectively. All other factor (eg. coefficient of friction, mass of block etc) are same in both cases. Let $k_1$ and $k_2$ be kinetic energies of the particle at the bottom of the plane in two cases. Then

$(a)\;k_1=k_2 \quad (b)\;k_1 < k_2 \quad (c)\; k_1 > k_2 \quad (d)\;data\;insufficient$

Work done against friction
$W=\mu mg \cos \theta\;s$
$W=\mu mg \cos \theta\;\large\frac{h}{\sin \theta}$
$W=\mu mgh\; \cot \theta$
$\cot \theta_1=\cot 30^{\circ}=\sqrt 3$
$\cot \theta_2=\cot 60^{\circ}=\large\frac{1}{\sqrt 3}$
$W_1 > W_2$
Kinetic energy $k=mgh-W$
$\therefore$ kinetic energy in 1st case $(\theta=30^{\circ})$ will be less than kinetic energy when $(\theta=60^{\circ})$
$k_1 < k_2$
Hence b is the correct answer.

edited Feb 10, 2014 by meena.p