Work done and Energy Transfers

Derivation of the equations for gravitational Potential energy and also Kinetic energy:

Gravitational Potential Energy:

We begin with the equation for work done;

$Work\ Done = Fd$

Since gravitational potential energy only changes if the height of the object is altered, the distance, d, must be solely related to the height, h. So

$Work\ Done = Fh\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Now, any object that has a mass will have a weight, which is a force and acts downwards towards Earth. This can be written in the equation, $F = mg$

So, substituting this into equation (1) gives us:

$Work\ Done = mgh$

Since work done is just energy, we can say that this type of energy is the gravitational potential energy:

$G.P.E = mgh$

—————————————————————————————————————————–

Kinetic Energy

We begin with the equation for work done;

$Work\ Done = Fd$

In this case we can substitute in Newton’s 2nd law of motion; $F = ma$

$Work\ Done = mas\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

We then use SUVAT and appropriate values for each quantity;
$s = s$
$u = 0$
$v = v$
$a = a$ (we assume negligible air resistance)
$t =$ X

Therefore we use the equation $v^2 = u^2 + 2as$
Subbing the value from above in gives:
$v^2 = 2as$
$\frac{1}{2}v^2 = as$

Subbing this into equation (1) gives:

$Work\ Done = mas = m\frac{1}{2}v^2$

Since the object has a velocity, we can say that the work done the transfer into kinetic energy:

$Kinetic\ Energy = \frac{1}{2}mv^2$

## Comment List

• Pretty great post. I just stumbled upon your weblog and wished to say that I have truly enjoyed surfing around your weblog posts.
After all I’ll be subscribing for your feed and
I am hoping you write once more very soon!