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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

Edexcel A level Physics-Study Notes- All Topics

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 

  • 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:

\( \tfrac{1}{2}mv^2 = 185.2 \Rightarrow v^2 = \dfrac{185.2 \times 2}{4} = 92.6 \)

\( v = 9.62\, \mathrm{m\,s^{-1}} \)

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