# Energy

As we mentioned earlier, energy is the capacity to do work. When positive work is done on an object, the system doing the work loses energy. In fact, the energy lost by a system is exactly equal to the work done by the system.
Like work (W) the unit of energy (E) is the joule (J). This follows as work is just the transfer of energy.
A very important property of our universe which was discovered around 1890 is that energy is conserved.
   * Energy is never created nor destroyed, but merely
   * transformed from one form to another.
Energy conservation and the conservation of matter are the principles on which classical mechanics is built.
   * IN THE ABSENCE OF FRICTION
   * When work is done on an object by a system:
   * -the object gains energy equal to the work done by the system
   * Work Done = Energy Transferred
Thermal energy (heat) is the disorganized movement of microscopic particles. Once energy is converted to this form, it has limited usefulness for doing further work in the system. Friction is the general name for forces that converts energy to heat.
   * IN THE PRESENCE OF FRICTION
   * When work is done by a system:
   * -only some of the energy lost by the system is transferred into useful energy
   * -the rest of the energy transferred is lost to heat by friction
   * Total Work Done = Useful Work Done + Work Done Against Friction

## Types of Energy

So what different types of energy exist? Kinetic, mechanical, thermal, chemical, electrical, radiant, and atomic energy are just some of the types that exist. By the principle of conservation of energy, whenwork is done energy is merely transferred from one object to another and from one type of energy to another.

### Kinetic Energy

Kinetic energy is the energy of motion that an object has. Objects moving in straight lines possess translational kinetic energy, which we often abbreviate as E$_k$.
The translational kinetic energy of an object is given by
   $E_{k}={\frac {1}{2}}mv^{2}$ :
   | E$_k$ | : kinetic energy (J) |
   | m | : mass of object (kg) |
   | v | : speed of the object (m.s$^{-1}$) |
Note the dependence of the kinetic energy on the speed of the object- kinetic energy is related to motion. The faster an object is moving the greater its kinetic energy.

#### Worked Example 41 Calculation of Kinetic Energy

Question: If a rock has a mass of 1kg and is thrown at 5m/s, what is its kinetic energy?
Answer:
Step 1 : Analyse the question to determine what information is provided
* The mass of the rock m = 1kg
* The speed of the rock v = 5m/s
These are both in the correct units so we do not have to worry about unit conversions.
Step 2 : Analyse the question to determine what is being asked
* We are asked to find the kinetic energy. From the definition we know that to work out E$_k$, we need to know the mass and the velocity of the object and we are given both of these values.
Step 3 : Substitute and calculate the kinetic energy
${\begin{matrix}E_{k}&=&{\frac {1}{2}}mv^{2}\\&=&{\frac {1}{2}}(1kg)(5{\frac {m}{s}})^{2}\\&=&12.5{\frac {kg\cdot m^{2}}{s^{2}}}\\&=&12.5\ J\end{matrix}}$
To check that the units in the above example are in fact correct:
${\begin{matrix}{\frac {kg\cdot m^{2}}{s^{2}}}&=&\left({\frac {kg\cdot m}{s^{2}}}\right)\cdot m=N\cdot m\\&=&J\end{matrix}}$
The units are indeed correct!
   * Study hint: Checking units is an important cross-check
   * and
   * you should get into a habit of doing this. If you, for example,
   * finish an exam early then checking the units in your calculations is a very good idea.

#### Worked Example 42 Mixing Units and Kinetic Energy Calculations 1

Question: If a car has a mass of 900kg and is driving at 60km/hr, what is its kinetic energy?
Answer:
Step 1 : Analyse the question to determine what information is provided
* The mass of the car m = 900kg
Step 2 : Analyse the question to determine what is being asked
* We are asked to find the kinetic energy.
Step 3 : Substitute and calculate
We know we need the mass and the speed to work out E$_k$ and we are given both of these quantities. We thus simply substitute them into the equation for E$_k$:
${\begin{matrix}E_{k}&=&{\frac {1}{2}}mv^{2}\\&=&{\frac {1}{2}}(900kg)(16.67{\frac {m}{s}})^{2}\\&=&125\ 000{\frac {kgm^{2}}{s^{2}}}\\&=&125\ 000\ J\end{matrix}}$

#### Worked Example 43 Mixing Units and Kinetic Energy Calculations 2

Question: If a bullet has a mass of 150kg and is shot at a muzzle velocity of 960m/s, what is its kinetic energy?
Answer:
Step 1 : Analyse the question to determine what information is provided
* We are given the muzzle velocity which is just how fast the bullet leaves the barrel and it is v = 960m/s.
Step 2 : Analyse the question to determine what is being asked
* 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 E$_k$:
${\begin{matrix}E_{k}&=&{\frac {1}{2}}mv^{2}\\&=&{\frac {1}{2}}(150kg)(960{\frac {m}{s}})^{2}\\&=&69\ 120{\frac {kgm^{2}}{s^{2}}}\\&=&69\ 120\ J\end{matrix}}$

### Potential Energy

If you lift an object you have to do work on it. This means that energy is transferred to the object. But where is this energy? This energy is stored in the object and is called potential energy. The reason it is called potential energy is because if we let go of the object it would move.
   * Definition:
   * Potential energy is the energy an object has due to its position or state.
As an object raised above the ground falls, its potential energy is released and transformed into kinetic energy. The further it falls the faster it moves as more of the stored potential energy is transferred into kinetic energy. Remember, energy is never created nor destroyed, but merely transformed from one type to another. In this case potential energy is lost but an equal amount of kinetic energy is gained.
In the example of a falling mass the potential energy is known as gravitational potential energy as it is the gravitational force exerted by the earth which causes the mass to accelerate towards the ground. The gravitational field of the earth is what does the work in this case.
Another example is a rubber-band. In order to stretch a rubber-band we have to do work on it. This means we transfer energy to the rubber-band and it gains potential energy. This potential energy is called elastic potential energy. Once released, the rubber-band begins to move and elastic potential energy is transferred into kinetic energy.

### Gravitational Potential Energy

As we have mentioned, when lifting an object it gains gravitational potential energy. One is free to define any level as corresponding to zero gravitational potential energy. Objects above this level then possess positive potential energy, while those below it have negative potential energy. To avoid negative numbers in a problem, always choose the lowest level as the zero potential mark. The change in gravitational potential energy of an object is given by:
   $\Delta E_{P}=mg\Delta h$ :
   | $\Delta E_{P}$ | : Change in gravitational potential energy (J) |
   | m | : mass of object (kg) |
   | g | : acceleration due to gravity (m.s$^{-2}$) |
   | $\Delta h$ | : change in height (m) |
When an object is lifted it gains gravitational potential energy, while it loses gravitational potential energy as it falls.

#### Worked Example 44 Gravitational potential energy

Question: How much potential energy does a brick with a mass of 1kg gain if it is lifted 4m.
Answer:
Step 1 : Analyse the question to determine what information is provided
* The mass of the brick is m = 1kg
* The height lifted is $\Delta h=4m$
These are in the correct units so we do not have to worry about unit conversions.
Step 2 : Analyse the question to determine what is being asked
* We are asked to find the gain in potential energy of the object.
Step 3 : Identify the type of potential energy involved
Since the block is being lifted we are dealing with gravitational potential energy. To work out $\Delta E_{P}$, we need to know the mass of the object and the height lifted. As both of these are given, we just substitute them into the equation for $\Delta E_{P}$.
Step 4 : Substitute and calculate
${\begin{matrix}\Delta E_{P}&=&mg\Delta h\\&=&(1kg)\left(10{\frac {m}{s^{2}}}\right)(4m)\\&=&40{\frac {kg\cdot m^{2}}{s^{2}}}\\&=&40\ J\end{matrix}}$