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Physics

1.18 - Transfer Between KE & GPE

5 min

note 1 of 6

Falling and rising transfer energy between two stores

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

Two panels side by side: a ball rolling down a green slope, labelled gravitational potential energy with an arrow to kinetic energy, and a ball rolling up a slope with the arrow reversed.

note 2 of 6

Energy conservation between the stores

With no air resistance

With no air resistance, none of the energy escapes to other stores, so the amount lost from one store equals the amount gained by the other:

Energy lost from gravitational potential energy store = energy gained by kinetic energy store

Air resistance transfers energy away

As an object moves through the air, air resistance transfers some energy to the thermal energy stores of the object and the surrounding air.
The kinetic energy store therefore gains less energy than the gravitational potential energy store loses.

note 3 of 6

Speed of a falling object from conservation of energy

When an object falls freely with no air resistance, all the energy lost from the gravitational potential energy store is gained by the kinetic energy store:

$E_{k} = m g h$

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 about 9.8 N/kg (sometimes rounded to 10 N/kg or given as 9.81 N/kg). Always use the value given in the question.

note 4 of 6

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 $g = 9.8$ N/kg and ignoring air resistance, find the energy transferred to its kinetic energy store and its speed just before it lands.

Step 1 - Find the energy lost from the gravitational potential energy store

$E_{p} = m g h = 3.0 \times 9.8 \times 5.0 = 147 J$

Step 2 - This energy is all gained by the kinetic energy store

With no air resistance, the kinetic energy store gains the full 147 J:

$E_{k} = 147 J$

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

$\begin{matrix} v & = \sqrt{\frac{2 E_{k}}{m}} \\ = \sqrt{\frac{2 \times 147}{3.0}} = \sqrt{98} \\ = 9.9 m/s \end{matrix}$

So the stone reaches 147 J of kinetic energy and a speed of about 9.9 m/s just before it lands.

note 5 of 6

Resistive force from the energy lost to air resistance

When air resistance is present, the kinetic energy gained is less than the gravitational potential energy lost.
The difference between them is the work done against the resistive force as the object falls.

Steps to find the average resistive force

Find the gravitational potential energy lost: $E_{p} = m g h$ .
Find the kinetic energy actually gained from the measured landing speed: $E_{k} = \frac{1}{2} m v^{2}$ .
Find the energy transferred to air resistance: $Energy transferred = E_{p} - E_{k}$ .
Divide this energy by the distance fallen to get the average resistive force: $F = \frac{energy transferred}{distance}$ .

note 6 of 6

Worked example - Resistive force on a falling object

A stone of mass 3.0 kg is dropped from a height of 5.0 m. This time air resistance is not ignored, and the stone is measured landing at 9.0 m/s. Taking $g = 9.8$ N/kg, find the average resistive force on it.

Step 1 - Energy lost from the gravitational potential energy store

$E_{p} = m g h = 3.0 \times 9.8 \times 5.0 = 147 J$

Step 2 - Kinetic energy actually gained

$\begin{matrix} E_{k} & = \frac{1}{2} m v^{2} \\ = 0.5 \times 3.0 \times {9.0}^{2} \\ = 121.5 J \end{matrix}$

Step 3 - Energy transferred to air resistance

147 J - 121.5 J = 25.5 J

Step 4 - Average resistive force over the 5.0 m fall

$\begin{matrix} F & = \frac{energy transferred}{distance} \\ = \frac{25.5}{5.0} \\ = 5.1 N \end{matrix}$

So the average resistive force on the stone as it falls is about 5.1 N.

end of notes

Key points to remember

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1.19 - Projectile MotionNext up

contents

Physics

1.18 - Transfer Between KE & GPE

5 min

note 1 of 6

Falling and rising transfer energy between two stores

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

note 2 of 6

Energy conservation between the stores

With no air resistance

With no air resistance, none of the energy escapes to other stores, so the amount lost from one store equals the amount gained by the other:

Energy lost from gravitational potential energy store = energy gained by kinetic energy store

Air resistance transfers energy away

As an object moves through the air, air resistance transfers some energy to the thermal energy stores of the object and the surrounding air.
The kinetic energy store therefore gains less energy than the gravitational potential energy store loses.

note 3 of 6

Speed of a falling object from conservation of energy

When an object falls freely with no air resistance, all the energy lost from the gravitational potential energy store is gained by the kinetic energy store:

$E_{k} = m g h$

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 about 9.8 N/kg (sometimes rounded to 10 N/kg or given as 9.81 N/kg). Always use the value given in the question.

note 4 of 6

Worked example - Speed of a falling object

Step 1 - Find the energy lost from the gravitational potential energy store

$E_{p} = m g h = 3.0 \times 9.8 \times 5.0 = 147 J$

Step 2 - This energy is all gained by the kinetic energy store

With no air resistance, the kinetic energy store gains the full 147 J:

$E_{k} = 147 J$

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

$\begin{matrix} v & = \sqrt{\frac{2 E_{k}}{m}} \\ = \sqrt{\frac{2 \times 147}{3.0}} = \sqrt{98} \\ = 9.9 m/s \end{matrix}$

So the stone reaches 147 J of kinetic energy and a speed of about 9.9 m/s just before it lands.

note 5 of 6

Resistive force from the energy lost to air resistance

When air resistance is present, the kinetic energy gained is less than the gravitational potential energy lost.
The difference between them is the work done against the resistive force as the object falls.

Steps to find the average resistive force

Find the gravitational potential energy lost: $E_{p} = m g h$ .
Find the kinetic energy actually gained from the measured landing speed: $E_{k} = \frac{1}{2} m v^{2}$ .
Find the energy transferred to air resistance: $Energy transferred = E_{p} - E_{k}$ .
Divide this energy by the distance fallen to get the average resistive force: $F = \frac{energy transferred}{distance}$ .

note 6 of 6

Worked example - Resistive force on a falling object

Step 1 - Energy lost from the gravitational potential energy store

$E_{p} = m g h = 3.0 \times 9.8 \times 5.0 = 147 J$

Step 2 - Kinetic energy actually gained

$\begin{matrix} E_{k} & = \frac{1}{2} m v^{2} \\ = 0.5 \times 3.0 \times {9.0}^{2} \\ = 121.5 J \end{matrix}$

Step 3 - Energy transferred to air resistance

147 J - 121.5 J = 25.5 J

Step 4 - Average resistive force over the 5.0 m fall

$\begin{matrix} F & = \frac{energy transferred}{distance} \\ = \frac{25.5}{5.0} \\ = 5.1 N \end{matrix}$

So the average resistive force on the stone as it falls is about 5.1 N.

end of notes

Key points to remember

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1.19 - Projectile MotionNext up