Given : m = 2 kg, E = 40 J

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

\(\frac{1}{2}\)mv^{2}_{max} = E

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