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: $G.P.E = mgh$

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

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