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Work and Energy - Law of Conservation of Energy

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Law of Conservation of Energy states that energy can neither be created nor destroyed; it can only be transformed from one form to another.

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The total energy of an isolated system remains constant before and after the transformation.

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Mechanical Energy is the sum of Kinetic Energy (EkE_k) and Potential Energy (EpE_p). In the absence of air resistance, the total mechanical energy of an object remains conserved: Ek+Ep=constantE_k + E_p = \text{constant}.

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In a free-falling body, as the object falls, its potential energy (mghmgh) decreases while its kinetic energy (12mv2\frac{1}{2}mv^2) increases. At any point during the fall, the sum of these energies remains the same.

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Energy transformation example: In an oscillating pendulum, energy constantly changes between potential energy (at extreme positions) and kinetic energy (at the mean position).

📐Formulae

Etotal=Ek+EpE_{total} = E_k + E_p

mgh+12mv2=Constantmgh + \frac{1}{2}mv^2 = \text{Constant}

ΔEp+ΔEk=0\Delta E_p + \Delta E_k = 0

Einitial=EfinalE_{initial} = E_{final}

💡Examples

Problem 1:

An object of mass 20 kg20\text{ kg} is dropped from a height of 4 m4\text{ m}. Calculate its potential energy and kinetic energy when it is half-way down. (Take g=10 m s−2g = 10\text{ m s}^{-2})

Solution:

  1. Initial Potential Energy at h=4 mh = 4\text{ m}: Ep=mgh=20×10×4=800 JE_p = mgh = 20 \times 10 \times 4 = 800\text{ J}.
  2. Total Energy at the start is 800 J800\text{ J} (since v=0v=0, Ek=0E_k=0).
  3. At half-way down, height h′=2 mh' = 2\text{ m}.
  4. New Potential Energy Ep′=mgh′=20×10×2=400 JE_p' = mgh' = 20 \times 10 \times 2 = 400\text{ J}.
  5. According to the Law of Conservation of Energy: Etotal=Ek′+Ep′E_{total} = E_k' + E_p'.
  6. 800=Ek′+400⇒Ek′=400 J800 = E_k' + 400 \Rightarrow E_k' = 400\text{ J}.

Explanation:

At the half-way point, exactly half of the initial gravitational potential energy has been converted into kinetic energy, keeping the total mechanical energy constant at 800 J800\text{ J}.

Problem 2:

Calculate the velocity of a 2 kg2\text{ kg} stone just before it hits the ground if it is dropped from a height of 5 m5\text{ m}.

Solution:

  1. Initial energy (at h=5 mh=5\text{ m}): Ep=mgh=2×10×5=100 JE_p = mgh = 2 \times 10 \times 5 = 100\text{ J} and Ek=0E_k = 0.
  2. Final energy (just before hitting ground): Ep=0E_p = 0 and Ek=12mv2E_k = \frac{1}{2}mv^2.
  3. By Law of Conservation of Energy: 100=12×2×v2100 = \frac{1}{2} \times 2 \times v^2.
  4. 100=v2⇒v=100=10 m s−1100 = v^2 \Rightarrow v = \sqrt{100} = 10\text{ m s}^{-1}.

Explanation:

All the initial potential energy is converted into kinetic energy at the point of impact, allowing us to solve for velocity vv.

Law of Conservation of Energy - Revision Notes & Key Formulas | CBSE Class 9 Science