Edexcel A Level (IAL) Physics-1.20 The Principle of Conservation of Energy- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -1.20 The Principle of Conservation of Energy- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -1.20 The Principle of Conservation of Energy- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 1.20 know, and understand how to apply, the principle of conservation of energy including use of work done, gravitational potential energy and kinetic energy
Principle of Conservation of Energy
Principle of Conservation of Energy
Energy cannot be created or destroyed — it can only be transferred from one form to another. In a closed system, the total energy remains constant.

\(\text{Total Energy Initially} = \text{Total Energy Finally}\)
In mechanics, the main forms of energy we use are:
- Kinetic energy \( E_k = \tfrac{1}{2}mv^2 \)
- Gravitational potential energy \( E_{\text{grav}} = mg\Delta h \)
- Work done by forces \( W = F\Delta s \)
Energy Transfers Involving Work, GPE, and KE
- Work done on an object increases its energy.
- Work done by an object decreases its energy.
- Falling objects lose GPE and gain KE.
- Raising objects increases GPE and requires work to be done.
- If no energy is lost to friction,
- \(\Delta E_k = -\Delta E_{\text{grav}}\)
- \(\Delta E_{\text{grav}} + \Delta E_k + W_{\text{external}} = 0\)
- In real situations, some energy becomes thermal energy (not recoverable as mechanical energy).
Using Conservation of Energy in Calculations
For a system where only GPE and KE are involved (no friction):
\( mg\Delta h + \tfrac{1}{2}mu^2 = \tfrac{1}{2}mv^2 \)
With external work:
\( W_{\text{external}} + \text{Initial Energy} = \text{Final Energy} \)
With friction:
\( \text{Useful Energy Output} = \text{Energy Input} – \text{Work Against Friction} \)
Energy Diagrams
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- At top of a fall: maximum GPE, minimum KE.
- At bottom of a fall: maximum KE, minimum GPE.
- Total mechanical energy remains (approximately) constant.
Example (Easy)
A \( 2.0\, \mathrm{kg} \) object is raised by \( 3.0\, \mathrm{m} \). Find the increase in GPE.
▶️ Answer / Explanation
\( \Delta E_{\text{grav}} = mg\Delta h = 2.0 \times 9.8 \times 3.0 = 58.8\, \mathrm{J} \)
Energy input = gain in GPE.
Example (Medium)
A \( 5.0\, \mathrm{kg} \) object is dropped from a height of \( 10\, \mathrm{m} \). Find its speed just before it hits the ground (ignore air resistance).
▶️ Answer / Explanation
GPE lost = KE gained.
\( mg\Delta h = \tfrac{1}{2}mv^2 \)
Mass cancels:
\( 9.8 \times 10 = \tfrac{1}{2}v^2 \Rightarrow v^2 = 196 \)
\( v = 14\, \mathrm{m\,s^{-1}} \)
Example (Hard)
A block of mass \( 4.0\, \mathrm{kg} \) slides down a rough slope. It starts from rest at a height of \( 6.0\, \mathrm{m} \). The work done against friction is \( 50\, \mathrm{J} \). Find the speed of the block at the bottom.
▶️ Answer / Explanation
Initial GPE:
\( mg\Delta h = 4.0 \times 9.8 \times 6.0 = 235.2\, \mathrm{J} \)
Energy lost to friction = \( 50\, \mathrm{J} \).
KE at bottom:
\( E_k = 235.2 – 50 = 185.2\, \mathrm{J} \)
Use KE formula:
\( \tfrac{1}{2}mv^2 = 185.2 \Rightarrow v^2 = \dfrac{185.2 \times 2}{4} = 92.6 \)
\( v = 9.62\, \mathrm{m\,s^{-1}} \)
