A3-3. Conservation of Energy

The Principle of Conservation of Energy

The fundamental premise governing all physical processes is that energy is never truly lost.

The Principle of Conservation of Energy states:

Energy cannot be created or destroyed, it can only be transferred from one form to another.

This means the total amount of energy in a closed system remains perfectly constant, even though the energy may shift between different stores (e.g., from kinetic to gravitational potential to thermal).

Sankey Diagrams

A Sankey diagram is a visual tool used to represent energy transfers. It elegantly displays how energy is distributed into useful and wasted forms.

  • The width of the arrows is strictly proportional to the amount of energy transferred.
  • The straight horizontal arrow usually represents useful energy.
  • The arrow curving downwards represents wasted energy (typically thermal energy escaping to the surroundings).

Example 1: Interpreting a Sankey Diagram

Problem: An electric motor is used to lift a weight. The Sankey diagram below represents the energy transfers in the motor. By reading the diagram, how much energy is wasted?

Input: 500 J Useful: 270 J (GPE) + 110 J (KE) Wasted Energy

Solution:

  • Analyze the given values:
    Total Input = $500\text{ J}$
    Total Useful Output = $270\text{ J (GPE)} + 110\text{ J (KE)} = 380\text{ J}$
  • Calculate the wasted energy:
    Due to the conservation of energy: Input = Useful + Wasted
    Wasted Energy = Input - Useful = $500\text{ J} - 380\text{ J} = \mathbf{120\text{ J}}$

Conservation of Mechanical Energy

When an object is in free fall or sliding without friction, mechanical energy (the sum of kinetic and potential energy) is perfectly conserved. Energy simply shifts seamlessly back and forth between $E_k$ and $E_p$.

Mechanical Energy Equation:

$$E = E_k + E_p = \text{Constant}$$
$$\dfrac{1}{2}mv_i^2 + mgh_i = \dfrac{1}{2}mv_f^2 + mgh_f$$

Note: Because mass ($m$) exists in every term of this equation for objects in freefall, it perfectly cancels out! Therefore, the final speed of a falling object is entirely independent of its mass.

Example 2: Diving Board Velocity

Problem: A diver of mass 70 kg dives from a 10 m high diving board. Calculate the speed of the diver just before hitting the water. (Assume no air resistance and $g = 9.81\text{ m/s}^2$).

Solution:

  • Step 1: Calculate the initial GPE.
    At the top of the board, all energy is stored as GPE ($E_k = 0$).
    $$E_p = mgh = 70 \times 9.81 \times 10 = 6867\text{ J}$$
  • Step 2: Relate to Kinetic Energy at the bottom.
    Right before hitting the water, all $E_p$ has transferred entirely into $E_k$.
    $$E_k = 6867\text{ J}$$
  • Step 3: Solve for velocity ($v$).
    $$\dfrac{1}{2} m v^2 = 6867$$
    $$\dfrac{1}{2} \times 70 \times v^2 = 6867 \implies 35 v^2 = 6867$$
    $$v^2 = \dfrac{6867}{35} \approx 196.2$$
    $$v = \sqrt{196.2} \approx \mathbf{14\text{ m/s}}$$
  • Alternative method: $mgh = \dfrac{1}{2}mv^2 \implies v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 10} = \mathbf{14\text{ m/s}}$. Notice that the 70 kg mass wasn't strictly necessary to solve the problem!