Physics

4.14 - Transfer Between KE & GPE

3 min

note 1 of 3

Falling and rising change energy between two forms

When an object falls, its gravitational potential energy changes into kinetic energy - it loses height and speeds up.
When an object rises - thrown upwards or rolled up a slope - the change runs the other way, from kinetic energy to gravitational potential energy - it slows down as it gains height.

A pink ball rolls down a green slope on the left, labelled gravitational potential energy with an arrow to kinetic energy; on the right it rolls up, with the two labels swapped.

With no air resistance

With no air resistance, none of the energy changes into other forms, so the amount of one form lost equals the amount gained by the other:

Gravitational potential energy lost = kinetic energy gained

Air resistance changes some energy to other forms

As an object moves through the air, air resistance changes some energy into thermal energy (heat) in the object and the surrounding air.
The kinetic energy gained is therefore less than the gravitational potential energy lost.

note 2 of 3

Speed of a falling object from conservation of energy

When an object falls freely with no air resistance, all the gravitational potential energy lost changes into kinetic energy:

$E_{k} = m g h$

Rearranging the kinetic energy equation $E_{k} = \frac{1}{2} m v^{2}$ gives the speed:

$v = \sqrt{\frac{2 E_{k}}{m}}$

Gravitational field strength on Earth is taken as 10 N/kg. Always use the value given in the question.

note 3 of 3

Worked example - Speed of a falling object

A stationary stone of mass 3.0 kg is dropped from a height of 5.0 m. Taking $g = 10$ N/kg and ignoring air resistance, find the kinetic energy it gains and its speed just before it lands.

Step 1 - Find the gravitational potential energy lost

$E_{p} = m g h = 3.0 \times 10 \times 5.0 = 150 J$

Step 2 - All of this energy changes into kinetic energy

With no air resistance, the kinetic energy gained is the full 150 J:

$E_{k} = 150 J$

Step 3 - Rearrange the kinetic energy equation to find the speed

$\begin{matrix} v & = \sqrt{\frac{2 E_{k}}{m}} \\ = \sqrt{\frac{2 \times 150}{3.0}} = \sqrt{100} \\ = 10 m/s \end{matrix}$

So the stone gains 150 J of kinetic energy and reaches a speed of 10 m/s just before it lands.

end of notes

Key points to remember

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4.15 - Conduction, Convection, RadiationNext up

Falling and rising change energy between two forms

When an object falls, its gravitational potential energy changes into kinetic energy - it loses height and speeds up.

When an object rises - thrown upwards or rolled up a slope - the change runs the other way, from kinetic energy to gravitational potential energy - it slows down as it gains height.

Speed of a falling object from conservation of energy

When an object falls freely with no air resistance, all the gravitational potential energy lost changes into kinetic energy:

E_{k} = m g h

Rearranging the kinetic energy equation

E_{k} = \frac{1}{2} m v^{2}

gives the speed:

v = \sqrt{\frac{2 E_{k}}{m}}

Gravitational field strength on Earth is taken as 10 N/kg. Always use the value given in the question.

Worked example - Speed of a falling object

A stationary stone of mass 3.0 kg is dropped from a height of 5.0 m. Taking

g = 10

N/kg and ignoring air resistance, find the kinetic energy it gains and its speed just before it lands.

Step 1 - Find the gravitational potential energy lost

$E_{p} = m g h = 3.0 \times 10 \times 5.0 = 150 J$

Step 2 - All of this energy changes into kinetic energy

With no air resistance, the kinetic energy gained is the full 150 J:

$E_{k} = 150 J$

Step 3 - Rearrange the kinetic energy equation to find the speed

$\begin{matrix} v & = \sqrt{\frac{2 E_{k}}{m}} \\ = \sqrt{\frac{2 \times 150}{3.0}} = \sqrt{100} \\ = 10 m/s \end{matrix}$

So the stone gains 150 J of kinetic energy and reaches a speed of 10 m/s just before it lands.