## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 36 - Relativity - Exercises and Problems - Page 1060: 37

#### Answer

The rest energy is $9.0\times 10^{13}~J$ The kinetic energy is $6.0\times 10^{13}~J$ The total energy is $1.50\times 10^{14}~J$

#### Work Step by Step

We can find the rest energy: $E = mc^2$ $E = (1.0\times 10^{-3}~kg)(3.0\times 10^8~m/s)^2$ $E = 9.0\times 10^{13}~J$ The rest energy is $9.0\times 10^{13}~J$ We can find $\gamma$: $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-\frac{(0.80~c)^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-0.64}}$ $\gamma = \frac{1}{\sqrt{0.36}}$ $\gamma = 1.67$ We can find the kinetic energy: $K = (\gamma -1)~mc^2$ $K = (1.67 -1) (9.0\times 10^{13}~J)$ $K = 6.0\times 10^{13}~J$ The kinetic energy is $6.0\times 10^{13}~J$ We can find the total energy: $E_{total} = 9.0\times 10^{13}~J+6.0\times 10^{13}~J$ $E_{total} = 1.50\times 10^{14}~J$ The total energy is $1.50\times 10^{14}~J$

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