Question

A particle is released from the top of two inclined rough surfaces of height ' $h$ ' each. The angle of inclination of the two planes are $30^{\circ }$ and $60^{\circ }$ respectively. All other factors (e.g. coefficient of friction, mass of block etc.) are same in both the cases. Let $K_1$ and $K_2$ be the kinetic energies of the particle at the bottom of the plane in two cases. Then

• A. $K_1=K_2$
• B. $K_1>K_2$
• C. $K_1<K_2$
• D. data insufficient

Answer 1

Answer: (C)

Work done in friction is
$W=(\mu mg\cos \theta )s$
$=(\mu mg\cos \theta )\left(\frac{h}{\sin \theta }\right)=\mu mgh\cot \theta$
Now $\cot {\theta }_1=\cot 30^{\circ }=\sqrt{3}$
and $\cot {\theta }_2=\cot 60^{\circ }=\frac{1}{\sqrt{3}}$
$\therefore W_1>W_2$
i.e., kinetic energy in first case will be less.
$(K=mgh-W)$
Or $K_1<K_2$