Question

What are the rest energy, the kinetic energy, and the total energy of a $1.0g$ particle with a speed of $0.80c$?

Short answer

$$\begin{gathered} KE=2.88\times {10}^{13}J \\ {E}_o=9.0\times {10}^{13}J \\ {E}_{total}=1.188\times {10}^{14}J \end{gathered}$$

Step-by-step solution

Given Information

We have given that

$$\begin{gathered} m=1.0g\to 0.001kg \\ v=0.80c \end{gathered}$$

Simplify

Kinetic Energy formula is

$KE=\frac{1}{2}m{v}^2$

Therefore substituting values in equation we get

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ KE=\frac{1}{2}(0.001kg){(0.8)}^2{(3\times {10}^8)}^{2}KE=2.88\times {10}^{13}J \end{gathered}$$

Now, to find rest energy we use the below formula

$E_o=m{c}^2$

Substituting values in rest energy formula we get

$$\begin{gathered} E_o=m{c}^2 \\ {E}_o=(0.001kg){(3\times {10}^8)}^2{E}_o=9.0\times {10}^{13}J \end{gathered}$$

Lastly, to find the total energy we will use the below formula

$E_{total}=KE+E_o$

Substitute the given values in the total energy formula

$$\begin{gathered} E_{total}=KE+E_o \\ {E}_{total}=2.88\times {10}^{13}J+9.0\times {10}^{13}J \\ {E}_{total}=1.188\times {10}^{14}J \end{gathered}$$