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Why can energy change form but never be created or destroyed?

The principle of conservation of energy, interconversion of kinetic and gravitational potential energy, energy dissipated by resistive forces, and applying conservation of energy to falling and oscillating systems.

A focused answer to AQA A-Level Physics 3.4.1.8, covering the principle of conservation of energy, the interconversion of kinetic and gravitational potential energy, energy dissipated by resistive forces, and applying conservation of energy to falling and oscillating systems.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The principle of conservation of energy
  3. Interchange of kinetic and potential energy
  4. Energy dissipated by resistive forces
  5. Try this

What this dot point is asking

AQA specification point 3.4.1.8 wants you to state the principle of conservation of energy, follow energy changing between kinetic and gravitational potential forms, account for energy dissipated by resistive forces, and apply conservation of energy to problems such as falling objects and pendulums.

The principle of conservation of energy

This is one of the most powerful principles in physics: it lets you relate the start and end of a process without tracking the messy detail in between. To use it, identify all the energy stores at the start and end and set the total equal, accounting for any energy transferred out as heat, light or sound.

Interchange of kinetic and potential energy

When an object falls a height hh, its gravitational potential energy is converted to kinetic energy. If no energy is dissipated,

The mass cancels, so all objects gain the same speed after falling the same distance in the absence of air resistance, consistent with Galileo's observation. The same energy interchange runs in reverse when an object is thrown upwards: kinetic energy converts back to potential energy until the object is momentarily at rest at its highest point.

Energy dissipated by resistive forces

In real systems friction and air resistance do negative work, transferring mechanical energy to thermal energy (and some sound). The object reaches a lower final speed than the ideal calculation predicts. The total energy is still conserved, but it is no longer all in useful mechanical form. The energy dissipated equals the loss in potential energy minus the gain in kinetic energy, which is the standard exam method for friction problems.

Try this

Q1. State the principle of conservation of energy. [1 mark]

  • Cue. Energy cannot be created or destroyed, only transferred or changed in form.

Q2. A 1.0 kg1.0 \text{ kg} object falls 5.0 m5.0 \text{ m} from rest with no air resistance. Calculate its speed on landing. Take g=9.8 m s−2g = 9.8 \text{ m s}^{-2}. [2 marks]

  • Cue. v=2gh=2×9.8×5.0=9.9 m s−1v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5.0} = 9.9 \text{ m s}^{-1}.

Q3. State the two forms of energy that interchange in a swinging pendulum. [1 mark]

  • Cue. Kinetic energy and gravitational potential energy.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20183 marksA ball of mass 0.20 kg0.20 \text{ kg} is dropped from a height of 1.8 m1.8 \text{ m} and rebounds to a height of 1.2 m1.2 \text{ m}. Calculate the energy dissipated during the bounce. Take g=9.8 m s−2g = 9.8 \text{ m s}^{-2}.
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The gravitational potential energy lost between the drop height and the rebound height equals the energy dissipated, since the ball is momentarily at rest at both heights.

Energy at the start: mgh1=(0.20)(9.8)(1.8)=3.53 Jmgh_1 = (0.20)(9.8)(1.8) = 3.53 \text{ J}.

Energy after the bounce: mgh2=(0.20)(9.8)(1.2)=2.35 Jmgh_2 = (0.20)(9.8)(1.2) = 2.35 \text{ J}.

Energy dissipated =3.53−2.35=1.2 J= 3.53 - 2.35 = 1.2 \text{ J}, transferred mostly to thermal energy and sound during the impact.

Markers reward using the difference in gravitational potential energy and identifying where the dissipated energy goes.

AQA 20214 marksA skier of mass 70 kg70 \text{ kg} starts from rest and descends a slope of vertical height 25 m25 \text{ m}. She reaches a speed of 18 m s−118 \text{ m s}^{-1} at the bottom. Calculate the energy dissipated by friction and air resistance. Take g=9.8 m s−2g = 9.8 \text{ m s}^{-2}.
Show worked answer →

Gravitational potential energy lost: mgh=(70)(9.8)(25)=1.72×104 Jmgh = (70)(9.8)(25) = 1.72 \times 10^4 \text{ J}.

Kinetic energy gained: 12mv2=12(70)(18)2=12(70)(324)=1.13×104 J\tfrac{1}{2}mv^2 = \tfrac{1}{2}(70)(18)^2 = \tfrac{1}{2}(70)(324) = 1.13 \times 10^4 \text{ J}.

By conservation of energy, the energy dissipated is the difference: 1.72×104−1.13×104=5.9×103 J1.72 \times 10^4 - 1.13 \times 10^4 = 5.9 \times 10^3 \text{ J}.

Markers reward calculating the GPE lost, the KE gained, and taking the difference as the dissipated energy.

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