A3-3. Conservation of Energy and Collisions

1. 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).

2. 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 = $500\text{ J} - 380\text{ J} = \mathbf{120\text{ J}}$

3. Conservation of Mechanical Energy

Mechanical energy ($E$) is the sum of a system's kinetic and potential energy ($E = K + U$). It is perfectly conserved only when conservative forces (like gravity or spring forces) do work.

Conditions for Mechanical Energy Conservation:

  • Only Conservative Forces do work: Gravity and elastic forces simply transfer energy between kinetic and potential stores.
  • Non-Conservative Forces do ZERO work: Forces like Normal Force or String Tension often exist in these problems, but because they act perpendicularly to the motion ($\theta = 90^\circ$), they do no work.
    Example: A skier sliding down a smooth slope (Normal force does 0 work) or a swinging pendulum (Tension does 0 work).
$$E = K_i + U_i = K_f + U_f = \text{Constant}$$
$$\Delta E = \Delta K + \Delta U = 0$$

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: Set up Conservation of Mechanical Energy.
    At the top, all energy is GPE ($U_i = mgh, K_i = 0$).
    At the bottom, all energy is Kinetic ($U_f = 0, K_f = \dfrac{1}{2}mv_f^2$).
  • Step 2: Equate and solve for velocity ($v_f$).
    $$mgh = \dfrac{1}{2} m v_f^2$$
    (Notice the mass $m$ cancels out entirely! The speed is independent of mass.)
    $$v_f = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 10} = \sqrt{196.2} \approx \mathbf{14\text{ m/s}}$$

Work Done by Non-Conservative Forces

If friction or air resistance is present, they drain mechanical energy from the system (turning it into heat). The mechanical energy is no longer conserved. Instead, the work done by non-conservative forces ($W_{nc}$) equals the change in mechanical energy.

$$W_{nc} = \Delta E = (K_f + U_f) - (K_i + U_i)$$

Example 3: Block Sliding Up a Rough Incline

Problem: A wooden block is launched up a fixed, rough incline ($\theta = 30^\circ$) with an initial velocity of $v = 8.0\text{ m/s}$. The coefficient of kinetic friction between the block and the incline is $\mu_k = 0.60$. Calculate the distance $d$ the block slides up the incline before coming to a complete stop. ($g = 10\text{ m/s}^2$)

30° v = 8.0 m/s v = 0 Distance = d h = d sin(30°)

Solution:

  • 1. Work done by non-conservative force (friction):
    The normal force is $N = mg \cos 30^\circ$.
    Friction $f_k = \mu_k N = \mu_k mg \cos 30^\circ$.
    Because friction opposes motion, it does negative work: $W_{nc} = -f_k \cdot d = -\mu_k mg d \cos 30^\circ$.
  • 2. Change in Mechanical Energy:
    Initial Energy ($E_i$): $K_i = \dfrac{1}{2}mv^2$ and $U_i = 0$ (at the bottom).
    Final Energy ($E_f$): $K_f = 0$ (it stops) and $U_f = mgh = mg(d \sin 30^\circ)$.
    $\Delta E = E_f - E_i = mgd \sin 30^\circ - \dfrac{1}{2}mv^2$
  • 3. Equate and solve for $d$:
    $$W_{nc} = \Delta E$$ $$-\mu_k mg d \cos 30^\circ = mg d \sin 30^\circ - \dfrac{1}{2}mv^2$$ Divide everything by mass $m$ (it cancels out):
    $$-\mu_k g d \cos 30^\circ = g d \sin 30^\circ - \dfrac{1}{2}v^2$$ Rearrange to solve for $d$:
    $$\dfrac{1}{2}v^2 = d(g \sin 30^\circ + \mu_k g \cos 30^\circ)$$ $$d = \dfrac{0.5 \times (8.0)^2}{10 \times 0.5 + 0.60 \times 10 \times \frac{\sqrt{3}}{2}} = \dfrac{32}{5 + 5.196} = \dfrac{32}{10.196} \approx \mathbf{3.1\text{ m}}$$