## College Physics (4th Edition)

$E = 0.64~MeV$
We can find the total energy: $E = \gamma~m~c^2$ $E = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~m~c^2$ $E = \frac{1}{\sqrt{1-\frac{(0.60~c)^2}{c^2}}}~\times (9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$ $E = (1.25)(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$ $E = 1.024\times 10^{-13}~J$ $E = (1.024\times 10^{-13}~J)\times (\frac{1~MeV}{1.60\times 10^{-13}~J})$ $E = 0.64~MeV$