Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
0 votes

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 $
Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App