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

A Block of mass 'm' are kept on a smooth horizontal plane, and attached to two unstretched spring with spring constants $k_1$ and $k_2$ as shown. If the block be displaced by a distance x on either side and released then the velocity of block as it passes through the mean position is

$ a)\; x \sqrt {\large\frac{m}{k_1}+\frac{m}{k_2}} \\ b)\; x \sqrt {\large\frac{k_1k_2}{m(k_1+k_2)}} \\ c)\; x \sqrt {\large\frac{k_1+k_2}{m}} \\ d) zero $

1 Answer

Spring $k_1$ is compressed, and $k_2$ expanded by distance x
Using conservation of energy
the total potential energy of the spring= Kinetic energy of the mass
$v=x \sqrt {\large\frac{k_1+k_2}{m}}$
Hence  cis the correct answer.


answered Jul 19, 2013 by meena.p
edited Feb 10, 2014 by meena.p

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