Gravitational potential and kinetic energy

As we have seen in this unit already, the work done in moving an object is equal to the energy transferred to the object when it moves.

The equation used for work done is:

$\text{work done} = \text{force} \times \text{distance}$

The amount of work done in moving the object is proportional to the force needed to move the object and the distance by which the object is moved.

The gravitational potential energy of an object is calculated using the equation:

$\text{GPE} = \text{mass} \times \text{gravitational field strength} \times \text{height}$

These two equations can be compared to show that the amount of work done in moving an object is equal to the amount of energy in the gravitational potential energy store of an object.

The weight of an object is a measure of the force of gravity with which it is pulled towards the ground. Weight is calculated using the equation:

$\text{Weight} = \text{mass} \times \text{gravitational field strength}$

The weight of an object is the force needed to lift the object away from the ground. This means that the work done in lifting the object through a distance can also be given as:

$\text{Work done} = \text{mass} \times \text{gravitational field strength} \times \text{distance}$

This equation could be applied to the act of lifting an object through a height by simply using height lifted as the distance moved. The equation becomes:

$\text{Work done} = \text{mass} \times \text{gravitational field strength} \times \text{height}$

This equation is now exactly the same as the one used to calculate GPE, thus showing that GPE and work done are equivalent.

When an object is released and falls, the energy is transferred from the gravitational potential energy store to the kinetic energy store of the object. According to the principle of conservation of energy, the amount of energy transferred to the kinetic energy store is equivalent to the amount of energy transferred from the gravitational potential energy store.

This means that the previous equation could also be used to calculate the amount of energy transferred to the kinetic store of an object.

The diagram illustrates a ball being thrown vertically into the air at point a, to a height of d metres where it comes to a stop. At point b the ball remains still at that height and momentarily has zero kinetic energy and maximum gravitational potential energy. At point b all of the energy has been transferred from the kinetic energy store into the gravitational potential energy store. At point c the ball begins to fall back to Earth. As the ball falls, an increasing amount of energy is transferred from the gravitational energy store back to the kinetic energy store.

At point d, the ball hits the ground and stops moving. Just before the ball stops moving, all of the energy has been transferred from the gravitational potential energy store to the kinetic energy store and the ball reaches its maximum speed.

The relationship between GPE and KE allows us to calculate factors such as the speed at which an object is travelling or the distance travelled by an object.

Example

A ball was thrown up into the air with an initial speed of 10 m/s. The gravitational field strength on Earth is 10 N/kg.

Calculate the maximum height reached by the ball before it started to fall back down to the ground.

Gravitational potential energy is calculated using the equation:

$\text{GPE} = m \times g \times h$

Kinetic energy is calculated using the equation:

$\text{KE} = \frac{1}{2} \times m \times v^2$

At the highest point reached by the ball, the kinetic energy is equal to gravitational potential energy. At this height we can say therefore that:

$\text{GPE} = \text{KE}$

We can also show this using the two equations as:

$m \times g \times h = \frac{1}{2} \times m \times v^2$

where is the mass of the object, is the speed, is gravitational field strength and is the maximum height reached.

The mass of the ball does not alter so the m on either side of the equation can be cancelled out. We can then substitute the values given from the question into the equation:

$\frac{10N}{kg} \times h = \frac{1}{2} \times \frac{10m}{s^2}$