Energy calculations
In a nutshell
Gravitational potential energy is stored when work is done against the force of gravity. Kinetic energy is stored when there is movement. Through mechanical work, the two energy stores can transfer.
Equation
You will be learning about the following equations in this lesson.
Word Equation  Symbol Equation 
$gravitational \space potential \space energy = mass \times gravitational \space field \space strength \times height$  $E_p = m \times g \times h$ 
$kinetic \space energy = {1\over 2} \times mass \times velocity^2$  $E_k = {1\over 2} \times m \times v^2$ 
Variable definitions
These are the symbols and units for the equations above.
QUANTITY NAME  SYMBOL  UNIT NAME  UNIT 
$gravitational \space potential \space energy$  $E_p$  $joule$  $J$ 
$mass$  $m$  $kilogram$  $kg$ 
$gravitational \space field \space strength$  $g$  $newton \space per \space kilogram$  $N/kg$ 
$height$  $h$  $metre$  $m$ 
$kinetic \space energy$  $E_k$  $joule$  $J$ 
$velocity$  $v$  $metre \space per \space second$  $m/s$ 
Gravitational potential energy ($E_p$)
Gravitational potential energy ($E_p$) is stored when when work is done against the force of gravity. Work done is the energy transferred when a force is applied to an object.
Gravitational potential energy increases with increasing height.
Example
Fox number two has more gravitational potential energy than fox number one as it is higher on the stairs.
 1.
 Fox at the bottom of the stairs  2.  Fox at the top of the stars 

To calculate gravitational potential energy, you will need to use the following equation
$\begin{aligned}gravitational \space potential \space energy &= mass \times gravitational \space field \space strength \times height \\ \ \\ E_p &= m \times g \times h\end{aligned}$
Gravitational field strength ($g$)
Gravitational field strength is used to represent how much force acts upon one kg of mass. For example, Earth's gravitational field strength is $10 \, N/kg$. For every one $kg$ of mass on Earth, a force of $10 \, N$ from the Earth pulls down on the mass.
Gravitational field strength varies depending on the mass of the object. Jupiter's gravitational field strength is about $2.5$ times bigger than Earths, because Jupiter has a much larger mass.
Example
A ball is thrown into the air, and reaches a height of $4 \, m$. The ball has a mass of $500 \, g$. What is the gravitational potential energy of the ball?
Firstly, write down the quantities and make sure they are in the correct units:
$\begin{aligned}m &= 500 \space g = 0.5 \space kg \\ \ \\ g &= 10 \space N/kg \\ \ \\ h &= 4 \space m\end{aligned}$
Next, write down the equation you need to use:
$E_p = m \times g \times h$
Then, substitute the values into the equation:
$E_p = 0.5 \times 10 \times 4$
Don't forget to include your units:
$20 \, J$
At a height of $4 \, m$, the ball has $\underline {20 \, J}$ of gravitational potential energy.
Kinetic energy ($E_k$)
Kinetic energy ($E_k$) is stored when an object is moving.
The faster something is going, the more kinetic energy there is.
Example
Fox number two is a running fox and is moving faster than fox number one. Therefore, fox number two has more kinetic energy.
 1.
 Stationary fox  2.  Running fox 

To calculate kinetic energy, you will need to use the following equation.
$\begin{aligned}kinetic \space energy &= {1\over 2} \times mass \times velocity^2 \\ \ \\ E_k &= {1 \over 2} \times m \times v^2\end{aligned}$
Hint: When gravitational potential energy has completely transferred into kinetic energy, they are the same value!
Example
Assume all gravitational potential energy from the previous example is converted into kinetic energy. What is the velocity of the ball when it hits the ground? Give your answer to two significant figures.
Firstly, write down the quantities and make sure they are in the correct units:
$\begin{aligned}E_k &= 20 \space J \newline m &= 0.5 \space kg\end{aligned}$
Secondly, write down the equation you need to use:
$E_k = {1\over2} \times m \times v^2$
Next, start rearranging the formula. Divide E_{k} by half and mass:
$\frac{E_k}{\frac{1}{2}\times m} = v^2$
Now, you need to square root both sides in order to get $v$, rather than $v^2$:
$\sqrt{\frac{E_k}{\frac{1}{2}\times m}} = v$
Then, you should substitute the values into the equation:
$\sqrt{\frac{20}{\frac{1}{2}\times 0.5}} = v$
Don't forget to include your units and to round your answer to two significant figures:
$8.9 \space m/s$
The ball hits the ground at a velocity of $\underline {8.9 \, m/s}$.
Energy transfer between $E_p$ and $E_k$
Kinetic and gravitational potential stores are often transferred to each other. When there is a gravitational potential energy store, work has been done against gravity which will be transferred from a kinetic energy store. This is because movement must have happened.