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A block rests on an inclined plane. A spring is attached is a pulley and is pulled downward gradually with increasing force. $\mu $ is the coefficient of friction between block and inclined plane . Find the potential energy of the spring at the momentum when block begins to move for a force constant k of spring $[\theta= 30^{\circ}]$

$a)\; k\;mg \bigg[\frac{1}{2}+\frac{\sqrt 3 \mu}{2} \bigg]^2 \\ b)\;\frac{m^2g^2\bigg[\frac{\sqrt 3}{2}+\frac{\mu}{2}\bigg]^2}{2k} \\ c)\; \frac{k^2 m^2}{g^2}\bigg[\frac{1}{2}+\frac{\sqrt 3}{2} \mu\bigg]^2 \\ d)\; \frac{m^2g^2\bigg[\frac{1}{2}+\frac{\sqrt 3}{2} \mu\bigg]^2}{2 k}$

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1 Answer

Let x be the extension in the spring
When the block just begins to move
$kx=mg \sin \theta+\mu mg \cos \theta$
$x= \large\frac{mg (\sin \theta+ \mu \cos \theta)}{k}$
The potential energy $=\large\frac{1}{2}$$kx^2$
$\qquad= \large\frac{1}{2}$$k \bigg[\large\frac{mg(\sin \theta+\mu \cos \theta)}{k}\bigg]^2$
$\qquad=\large\frac{m^2g^2(\sin \theta+\mu \cos \theta)}{2k}$
$\qquad=\large\frac{m^2g^2\bigg[\Large\frac{1}{2}+\mu \frac{\sqrt 3}{2}\bigg]^2}{2k}$
Hence b is the correct answer. 


answered Aug 7, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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