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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

In the arrangement as shown a mass m is released from rest with spring unstreched.What is the speed of block at equillibrium position

$a)\; v=\sqrt {\large\frac{gm}{k}}\\ b)\;v=g \sqrt {\large\frac{m}{R}} \\ c)\;v=\sqrt {\large\frac{m}{Rg}}\\ d)\;v=\sqrt {\large\frac{mk}{g}}$

1 Answer

At equillibrium position when net forcea acting on the block is zero.
$kx_0=mg$
$x_0=\large\frac{mg}{k}$
Let v be speed of block in this position.
Applying conservation of mechanical energy
$mgx_0=\large\frac{1}{2}$$mv_0^2+\large\frac{1}{2}$$kx_0^2$
$v=\sqrt {2gx_0-\large\frac{k}{m} \normalsize x_0^2}$
Substituting $x_0=\large\frac{mg}{k}$
We get $v=g \sqrt {\large\frac{m}{k}}$
Hence b is the correct answer. 
answered Aug 7, 2013 by meena.p
edited Jun 11, 2014 by lmohan717
 

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