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The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. 

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Given : m = 2 kg, E = 40 J

The speed of the body while crossing the centre of the path (mean position) is vmax and the total energy is entirely kinetic energy.

  \(\frac{1}{2}\)mv2max = E

∴  vmax = \(\sqrt{\frac{2E}{m}}= \sqrt{\frac{2×40}{2}}\) = 6.324 m/s

 

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