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

Answer 1

The correct option is D $K_1<K_2$

Draw a labelled diagram.


Find the kinetic energy.
Work done in friction is, From the FBD,
$N=mg\cos \theta$

Friction force, $f=\mu N=\mu mg\cos \theta$

Displacement, $s=\frac{h}{sin\theta }$

$W=\left(\mu mg\cos \theta \right)s$

$W=\left(\mu mg\cos \theta \right)\left(\frac{h}{sin\theta }\right)=\mu mgh\cot \theta$

Now,
$\cot {\theta }_1=\cot {30}^{\circ }=\sqrt{3}$

$\cot {\theta }_2=\cot {60}^{\circ }=\frac{1}{\sqrt{3}}$

$\therefore W_1>W_2$

As we know, From W-E theorem $K=mgh-W$

i.e., kinetic energy in first case will be less.
$K_1<K_2$

Final Answer:(c)