Chapter 39, Problem 1P

To determine

The approximate total energy of the particle.

Explanation of Solution

Introduction:

The total relativistic energy of a particle is the sum of its kinetic energy and rest energy. The mass of particle is $m$ and it is moving with velocity $u$ with is very less than speed of light.

Write expression for total energy of the particle.

  $E=K+m{c}^2$

Here, $K$ is kinetic energy or particle and $m{c}^2$ is rest energy of particle.

Substitute $\frac{1}{2}m{v}^2$ for $K$ in above expression.

  $E=\frac{1}{2}m{v}^2+m{c}^2$

Conclusion:

The total energy of particle is $m{c}^2+\frac{1}{2}m{v}^2$ . Thus, option (a) is correct.

The total energy is the sum of kinetic energy and rest energy. Thus, option (b) is incorrect.

The $cmu$ does not indicates any type of energy. Thus, option (c) is incorrect.

The total energy is sum of kinetic energy and rest energy. Thus, option (d) is incorrect.

The $\frac{1}{2}cmu$ does not indicates any type of energy. Thus, option (e) is incorrect.