Checking Units
According to the equation for kinetic energy, the unit should be kg⋅m⋅s<sup>−2</sup>. We can prove that this unit is equal to the joule, the unit for energy. When checking units, analyse the question to determine what information is provided. Also, analyse the ...
Checking Units
According to the equation for kinetic energy, the unit should be \(\text{kg·m·s$^{-2}$}\). We can prove that this unit is equal to the joule, the unit for energy.
\begin{align*} \left(\text{kg}\right){\left(\text{m·s$^{-1}$}\right)}^{2} & = \left(\text{kg·m·s$^{-2}$}\right)·\text{m} \\ & = \text{N·m} \left(\text{ because Force }\left(\text{N}\right) = \text{ mass }\left(\text{kg}\right) \times \text{ acceleration }\left(\text{m·s$^{-2}$}\right)\right) \\ & = \text{J} \left(\text{ Work }\left(\text{J}\right) = \text{ Force }\left(\text{N}\right) \times \text{ distance }\left(\text{m}\right)\right) \end{align*}
We can do the same to prove that the unit for potential energy is equal to the joule:
\begin{align*} \left(\text{kg}\right)\left(\text{m·s$^{-2}$}\right)\left(\text{m}\right) & = \text{N·m} \\ & = \text{J} \end{align*}
Example: Mixing Units and Energy Calculations
Question
A bullet, having a mass of \(\text{150}\) \(\text{g}\), is shot with a muzzle velocity of \(\text{960}\) \(\text{m·s$^{-1}$}\). Calculate its kinetic energy.
Step 1: Analyse the question to determine what information is provided
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We are given the mass of the bullet m = \(\text{150}\) \(\text{g}\). This is not the unit we want mass to be in. We need to convert to \(\text{kg}\).
\begin{align*} \text{Mass in grams } ÷ 1000 & = \text{ Mass in kg} \\ \text{150}\text{ g} ÷ 1000 & = \text{0.150}\text{ kg} \end{align*} -
We are given the initial velocity with which the bullet leaves the barrel, called the muzzle velocity, and it is \(v = \text{960}\text{ m·s$^{-1}$}\).
Step 2: Analyse the question to determine what is being asked
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We are asked to find the kinetic energy.
Step 3: Substitute and calculate
We just substitute the mass and velocity (which are known) into the equation for kinetic energy:
\begin{align*} {E}_{K} & = \frac{1}{2}m{v}^{2} \\ & = \frac{1}{2}\left(\text{0.150}\text{ kg}\right){\left(\text{960}\text{ m·s$^{-1}$}\right)}^{2} \\ & = \text{69 120}\text{ J} \end{align*}This lesson is part of:
Work, Energy and Power