Revise Gravitational potential energy and kinetic energy for Cambridge 9702 (5.2)

Any object that is moving possesses kinetic energy. The faster an object travels and the more mass it has, the greater its kinetic energy. This energy is a scalar quantity, meaning it only has magnitude, not direction. Think of a bullet fired from a gun, a car on a motorway, or even electrons moving in a circuit - they all have kinetic energy!

$KE = \frac{1}{2}mv^2$

Where:

• $KE$ is kinetic energy (Joules, J)
• $m$ is the mass of the object (kilograms, kg)
• $v$ is the speed of the object (metres per second, m s\textsuperscript{-1})

The kinetic energy of an object is equal to the work done on the object to accelerate it from rest to its final speed. We can derive this from the definitions of work and acceleration. Consider an object of mass $m$ at rest ($u=0$) that is accelerated by a constant force $F$ over a distance $s$. The work done is $W = Fs$. From Newton's second law, $F=ma$. Substituting this gives $W = (ma)s$. Using the kinematic equation $v^2 = u^2 + 2as$, and since $u=0$, we have $v^2 = 2as$, which rearranges to $as = \frac{v^2}{2}$. Substituting this into our work equation gives $W = m(\frac{v^2}{2})$. Since the work done is equal to the kinetic energy gained, we arrive at the formula $KE = \frac{1}{2}mv^2$.

Gravitational potential energy (GPE) is the energy an object has stored due to its position within a gravitational field, specifically its vertical height. When you lift an object, you do work against gravity, and this work is stored as GPE in the object. This stored energy is ready to be converted into other forms, like kinetic energy, if the object is allowed to fall.

$GPE = mgh$

Where:

• $GPE$ is gravitational potential energy (Joules, J)
• $m$ is the mass of the object (kilograms, kg)
• $g$ is the acceleration due to gravity (metres per second squared, m s\textsuperscript{-2})
• $h$ is the vertical height above a chosen reference point (metres, m)

The change in gravitational potential energy is defined as the work done to move an object vertically against a uniform gravitational field. To lift an object of mass $m$ to a vertical height $h$, a force must be applied that is at least equal to its weight, $F = mg$. The work done ($W$) is the product of this force and the vertical distance moved, $h$. Therefore, $W = F \times h = (mg) \times h$. This work done against the gravitational field is stored as gravitational potential energy. Thus, the change in GPE is given by $\Delta E_p = mgh$.

One of the most fundamental laws in physics is the Principle of Conservation of Energy. It states that in a closed, isolated system, the total amount of energy remains constant. Energy cannot be created or destroyed; it can only be transformed from one form to another. In the context of mechanics, if there are no non-conservative forces like friction or air resistance, the total mechanical energy (the sum of GPE and KE) is conserved. This means that as an object moves, its GPE can become KE, or vice versa, while the total $GPE + KE$ stays the same.

In real-world scenarios, perfectly conserved mechanical energy (GPE + KE) is rare. Resistive forces, such as air resistance or friction, are non-conservative forces that do work on moving objects. This work done against resistance converts some of the mechanical energy into internal energy (heat and sound), which is dissipated into the surroundings. Therefore, the total energy of the universe is still conserved, but the mechanical energy of the specific system decreases. The energy lost is equal to the work done by the resistive forces.

Work done by resistive forces = Change in mechanical energy

$W_{resistive} = \Delta E_{mechanical} = E_{initial} - E_{final}$

Where $E = KE + GPE$