Detailed Study Notes on Energy Conservation Concepts and Calculations

LESSON 5.1 - Potential Energy (PE), Kinetic Energy (KE), and Total Mechanical Energy (TME)

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• Key equations:

  • Potential Energy (PE):
    $U_1 = mgy$

  • Where:

    • $U_1$: potential energy (Joules or J)

    • $m$: mass (kilograms or kg)

    • $g$: acceleration due to gravity ($9.8 ext{ m/s}^2$)

    • $y$: height (meters or m)

  • Kinetic Energy (KE):
    $K = rac{1}{2}mv^2$

  • Where:

    • $K$: kinetic energy (J)

    • $m$: mass (kg)

    • $v$: velocity (meters per second or m/s)

• Energy transformations:

  • From point 1 to 2, energy converts from potential to kinetic.

  • From point 4 to 5, energy also transforms, confirming the conversion.

  • Position 3 has more potential energy than position 5 because it is at a greater height.

  • Position 4 has the greatest kinetic energy, typically the lowest point in a height differential.

Schedule for Current Lesson (1/22/2026)

• Due Today: Popper Lab

• Objectives:

  1. Notes on Mechanical Energy.

  2. Discuss Energy Conservation.

  3. Work through examples of KE to PE transformations.

• Assignments/Announcements:

  1. Complete the lab part started previously.

  2. If it is an e-Day (Monday), expect assignments from Pearson Practice 5.1.

• Goal:

  • Investigate conservation of energy examples related to kinetic and potential energy transforms.

  • Utilize Pear Deck for interactive learning.

Law of Conservation of Energy

• Statement: The law of conservation of energy asserts that:

  • Energy cannot be created or destroyed; it can only be transformed from one form to another.

• Types of energy involved:

  • Kinetic Energy (KE)

  • Potential Energy (PE)

  • Total Mechanical Energy (TME)

• Example Inputs:

  • Height: $0.40 ext{ m}$

  • Speed: $0.00 ext{ m/s}$

Conservation of Energy

• Energy Measurement:

  • Energy is measured in Joules (J).

• Types of Mechanical Energy:

  1. Potential Energy (Stored Energy):

  • Gravitational potential energy described by the formula: $U_1 = mgy$

    • Where:

      • $m$: mass (kg)

      • $g$: acceleration due to gravity ($9.8 ext{ m/s}^2$)

      • $y$: height (m)

  1. Kinetic Energy (Energy of Motion):

  • Translational kinetic energy defined by: $K = rac{1}{2} mv^2$

    • Where:

      • $m$: mass (kg)

      • $v$: velocity (cm/s)

  • Rotational kinetic energy defined as: $K_w = rac{1}{2} I heta^2$

    • Where:

      • $K_w$: rotational kinetic energy (J)

      • $I$: moment of inertia

      • $ heta$: angular velocity

Conservation of Energy Principle

• The Total Mechanical Energy (TME) is conserved in an isolated system:
  $TME<em>1 = TME</em>2$$</p></li><li><p>The total mechanical energy includes both potential and kinetic energy and is given by:<br>$TME = PE + KE$</p></li><li><p>This total is constant, indicating energy stored as potential can convert to kinetic energy in motion, and vice versa without any loss during the transformation.</p></li></ul><h4 id="2cdb216c-cfa6-41be-afb1-6956ada5988d" data-toc-id="2cdb216c-cfa6-41be-afb1-6956ada5988d" collapsed="false" seolevelmigrated="true">Conservation of Energy Math Concepts</h4><ul><li><p>Common scenarios in energy problems include:</p><ol><li><p>Starting with Potential Energy (PE, stationary mass at height) where it falls to convert to Kinetic Energy:</p></li></ol><ul><li><p>$PE1 = KE2$</p></li><li><p>$mgh_1 = rac{1}{2} mv^2$</p></li></ul><ol><li><p>Starting with Kinetic Energy (moving object) that ascends, converting to Potential Energy:</p></li></ol><ul><li><p>$KE1 = PE2$</p></li><li><p>$ rac{1}{2} m v_1^2 = mgh$</p></li></ul></li></ul><h4 id="24d1d1cc-99d4-4851-a57b-75f0b15658d8" data-toc-id="24d1d1cc-99d4-4851-a57b-75f0b15658d8" collapsed="false" seolevelmigrated="true">Group Practice (Case Study: Skier)</h4><ul><li><p>Scenario: A 65 kg skier starts from rest at the top of an 80 m tall hill and skis down.</p><ol><li><p>Calculate the speed at the bottom of the hill.</p></li></ol><ul><li><p>Use the potential energy at the top: <br>$PE = mgh = (65 ext{ kg})(9.8 ext{ m/s}^2)(80 ext{ m}) = 50,960 ext{ J}$</p></li><li><p>Equate to kinetic energy at the bottom:$KE = rac{1}{2} mv^2 ightarrow 50,960 = rac{1}{2}(65)v^2$</p><ul><li><p>Solve for$v$:</p></li><li><p>$v^2 = rac{50,960 imes 2}{65}
  ightarrow v^2 = 1,568
  ightarrow v = 40 ext{ m/s}$</p></li></ul></li></ul></li></ul><h4 id="8b5ad760-5b19-49da-85ba-fc5af0516276" data-toc-id="8b5ad760-5b19-49da-85ba-fc5af0516276" collapsed="false" seolevelmigrated="true">Additional Group Practice</h4><ul><li><p>A pendulum ball with a mass of 0.3 kg swings from rest at point A to point E without any work done. Calculating each state:</p><ul><li><p></p></li></ul><ul><li><p>Point A:</p></li><li><p>PE =$mgh = (0.3)(9.8)(0.5) = 1.47 ext{ J}$,</p></li><li><p>KE =$0 ext{ J}$,</p></li><li><p>ME at point A =$1.47 ext{ J}$.</p></li></ul><ul><li><p></p></li></ul><ul><li><p>Point E (lower position):</p></li><li><p>PE =$0 ext{ J}$,</p></li><li><p>KE = Calculate using$KE = ME - PE
  ightarrow KE = 1.47 ext{ J}$.</p></li><li><p>The velocity ($v$$) can also be calculated using the energy relations.