Energy Exercise 3

3. What is the kinetic energy of an electron, in joules, that has a mass of 9.109 x 10-28 g and moves at a speed of 2.2 x 103 km/s?

\(\displaystyle\mathbf{E_K\;=\;\frac{1}{2}\;m\textit{v}^2}\)

Convert the mass of the electron from g to kg.

9.109 x 10-28 g = 9.109 x 10-31 kg

The speed of the electron is converted from km/s to m/s

2.2 x 103 km/s = 2.2 x 106 m/s

Now we can calculate the kinetic energy.

\(\displaystyle \frac{1}{2}\times\;(9.109\times\;10^{-31}\;kg)\times\;(2.2\times\;10^6\;m/s)^2\;=\;2.2\times\;10^{-18}\frac{kg⋅m^2}{s^2}\)
 
\(\displaystyle one\;\frac{kg⋅m^2}{s^2}\;=\;1\;J\)

The amount of kinetic energy for this electron is 2.2 x 10-18 J.

Back to Energy: A Derived Unit

One thought on “Energy Exercise 3

  1. Dr. O'Connor Post author

    Thank you. The answer has been fixed. The problem was set up correctly but the v was not squared when entered into the calculator.

    Reply

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