Chapter 10: Energy of a System

Evan LuoMay 12, 2025

5 min read

1. What's Inside a "System"?

When we solve energy problems, we first draw an imaginary boundary around the objects we care about-blocks, springs, Earth (for gravity), maybe some air if drag matters. Inside that boundary we track four common energy "banks":

Energy name	Symbol	Everyday picture
Kinetic	$K$	"How fast it's moving"
Gravitational potential	$U_G$	"How high it is above a reference level"
Spring (elastic) potential	$U_{sp}$	"How stretched or squished a spring is"
Thermal	$E_{th}$	"Microscopic jiggling-rises when things rub (friction)"

If we later decide to shrink or enlarge that boundary, we simply rewrite the bookkeeping to match.

2. The Master Energy-Balance Equation

For any system we can write

W_{ext} = \Delta K + \Delta U_G + \Delta U_{sp} + \Delta E_{th}

$W_{ext}$ is work done by forces outside the boundary (a push you supply, a motor, wind, ...).
The 4 (delta) means "change = final - initial."

Special case: isolated system

If our boundary is so generous that no outside forces do work on it, then $W_{ext} = 0$ and the equation collapses to

K_i + U_{G i} + U_{sp i} = K_f + U_{G f} + U_{sp f} + \Delta E_{th}

That is the law of conservation of energy: total energy stays the same, it only moves between the banks.

If there is no friction (so $\Delta E_{th} = 0$ ), the right-hand thermal term simply drops out.

3. Mechanical Energy and Conservative Forces

Mechanical energy is just the "nice" pair

E_{mech} = K + U \text{ (all potential types)}

A conservative force (gravity, ideal spring, electrostatic) acts like a rechargeable battery: it shifts energy back and forth inside $E_{mech}$ without losses. That means

K_i + U_i = K_f + U_f

whenever only conservative forces are present.

A non-conservative force (friction, air drag, a car engine) leaks energy out of the mechanical account into $E_{th}$ or other non-mechanical forms. Once converted, it does not flow back.

4. Problem-Solving Recipe (Before-and-After Method)

Draw two pictures - "before" and "after."
List the energy banks that are non-zero in each picture.
Write the energy equation (include $W_{ext}$ or $\Delta E_{th}$ only if they actually exist).
Plug numbers; solve for the unknown.

(Tip: pick the zero of gravitational potential where it makes algebra easiest-often the lower of the two heights.)

5. Worked Examples

Example 1 - Block-Spring-Friction Combo

A 10 kg block $M$ on a rough table is attached to a spring ( $k = 200 \text{ N/m}$ ) and to a hanging 5 kg mass $m$ . The hanging mass starts 15 cm below the tabletop and falls another 10 cm before everything momentarily stops. Find the magnitude of the friction force on $M$ .

Try it yourself first, then click Reveal Answer.

Choose the system: both blocks, spring, Earth. (Friction is inside the boundary because it acts on $M$ .)
Initial state (i)
- $K_i = 0$ (released from rest)
- $U_{G i} = m g (0.15 \text{ m})$
- Spring un-stretched → $U_{sp i} = 0$
Final state (f)
- Blocks instantaneously at rest → $K_f = 0$
- Hanging block is 0.10 m lower than before → $U_{G f} = m g (0.15 + 0.10)$
- Spring stretched by $x = 0.10 \text{ m}$ → $U_{sp f} = \tfrac{1}{2} k x^2$
- Friction has done work that shows up as $\Delta E_{th}$ .
Energy equation (isolated system, so $W_{ext} = 0$ )
$U_{G i} = U_{G f} + U_{sp f} + \Delta E_{th}$
Thermal change equals the friction force times the slide distance $d = 0.10 \text{ m}$ : $\Delta E_{th} = f_k d$ .
Solve
$m g (0.15) = m g (0.25) + \tfrac{1}{2} (200)(0.10)^2 + f_k(0.10)$ $f_k = 39 \text{ N}$

Click to reveal

Example 2 - Slope to Spring (No Friction)

A 2.0 kg block slides down a frictionless ramp and hits a spring ( $k = 400 \text{ N/m}$ ). The spring compresses 0.22 m before the block stops. How far did the block slide along the ramp?

Only conservative forces act (gravity + spring), so $E_{mech}$ is constant.

K_i + U_{G i} = U_{sp f}

Start: $K_i = 0$ , $U_{G i} = m g h$ (with $h$ the vertical drop)
End: $U_{sp f} = \tfrac{1}{2} k x^2$

Geometry on the ramp: $h = d \sin\theta$ . We can eliminate $\theta$ by noting the problem only asks for the ramp distance $d$ :

m g (d\sin \theta) = \tfrac{1}{2} k x^2 \quad\Rightarrow\quad d = \frac{k x^2}{2 m g \sin\theta}

But because $\sin\theta$ comes from the drawn slope (given in the original slide), the numeric crunch leads to

d = 0.821 \text{ m}

(See slide for the specific angle used.)

Click to reveal

Example 3 - Two-Block Pulley Drop

Two blocks are connected over a light pulley. The 12 kg block falls to the floor; find its speed just before impact.

Treat both blocks + Earth as the system. No friction mentioned $\implies$ mechanical energy conserved.

K_i + U_{G i} = K_f + U_{G f}

Initial: at rest $\implies$ $K_i = 0$ ; set $U_G = 0$ at the floor.
Final: both blocks move with speed $v$ ; heavier block at zero potential.

After algebra, slides show $v = 4.43 \text{ m/s}$ .

(Underlying idea: gravitational potential of the falling block converts into kinetic energy of both masses.)

Click to reveal

6. Key Take-Aways

Energy is book-keeping: choose a boundary, write "before-after," and balance the accounts.
Isolated system → total energy constant.
Conservative forces (gravity, springs) shuffle energy between kinetic and potential but keep it mechanical.
Non-conservative forces (friction, engines) drain mechanical energy into thermal or other forms.
Always match the math to the picture-that's the secret to stress-free energy problems!

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Last edited May 17, 2025