logo

Ask Questions, Get Answers

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

A block is placed on a rough horizontal plane attached with an elastic spring as shown.

Finally the spring is unstretched, If the plane is mow gradually lifted from $\theta=0^{\circ}$ to $\theta=90^{\circ}$ then the graph showing extension in the spring(x) versus angle $(\theta)$ is 

Can you answer this question?
 
 

1 Answer

0 votes
$x=0$ till $mg \sin \theta < \mu mg \cos\theta$
ie no extension takes place is less than the frictional force.
gradually x will increase at angle $\theta > \tan ^{-1} \mu$
$kx+\mu mg \cos \theta =mg \sin \theta$
$x=\large\frac{mg \sin \theta-\mu mg \cos \theta}{k}$
where k is spring constant
Hence a is the correct answer.

 

answered Jul 11, 2013 by meena.p
edited Jan 26, 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
...