logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
0 votes

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

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

Can you answer this question?
 
 

1 Answer

0 votes
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.

 

answered Aug 5, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...