## College Physics (4th Edition)

$K = 2.7\times 10^{15}~J$
We can find the kinetic energy of the object: $K = (\gamma-1)mc^2$ $K = (\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)mc^2$ $K = (\frac{1}{\sqrt{1-\frac{(1.80\times 10^8~m/s)^2}{(3.0\times 10^8~m/s)^2}}}-1)(0.12~kg)(3.0\times 10^8~m/s)^2$ $K = (1.25-1)(0.12~kg)(3.0\times 10^8~m/s)^2$ $K = 2.7\times 10^{15}~J$