Answer
(b) $vmc^{2}$
Work Step by Step
The total energy E of a free particle of mass m moving with speed v is the sum of its relativistic kinetic energy (K) and rest energy (Eo).
The relativistic kinetic energy
$E=(v-1)mc^{2}$, where $v=\frac{1}{\sqrt (1-\frac{v^{2}}{c^{2}})}$, c is the speed of light.
The rest energy is $Eo=mc^{2}$
So, total energy = $E+Eo=(v-1)mc^{2}+mc^{2}=vmc^{2}$
Hence, (b) is the answer.
