Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

Starting from rest a body slides down a $45^{\circ}$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between body and plane is

\[(a)\;0.33 \quad (b)\;0.75 \quad (c)\;0.25 \quad (d)\;0.80 N \]
Can you answer this question?

1 Answer

0 votes
on a rough surface acceleration of a body down a plane
$a=(g \;sin \theta-\mu g \cos \theta)$
'2t' is time taken to go down a rough inclined plane of length 's'
't' is time taken to go down a smooth inclined plane of length 's'
$s=\large\frac{1}{2}$$ [g \sin \theta -\mu g \cos \theta](2t)^2$-----(1)
On a smooth surface $a=g \sin \theta$
$s= \large\frac{1}{2} $$(g \sin \theta)t^2$-----(2)
Substituting (2) in (1) and solving
$\mu =1- \large\frac{1}{2^2}$
$\mu =\large \frac{3}{4}$
Hence b is the correct answer. 
answered Jul 11, 2013 by meena.p
edited May 27, 2014 by lmohan717

Related questions

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