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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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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

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$x=0$ 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

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