Mechanics

Chapter 10: Energy of a System

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 nameSymbolEveryday picture
KineticKK"How fast it's moving"
Gravitational potentialUGU_G"How high it is above a reference level"
Spring (elastic) potentialUspU_{sp}"How stretched or squished a spring is"
ThermalEthE_{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

Wext=ΔK+ΔUG+ΔUsp+ΔEthW_{ext} = \Delta K + \Delta U_G + \Delta U_{sp} + \Delta E_{th}
  • WextW_{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 Wext=0W_{ext} = 0 and the equation collapses to

Ki+UGi+Uspi=Kf+UGf+Uspf+ΔEthK_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 ΔEth=0\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

Emech=K+U (all potential types)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 EmechE_{mech} without losses. That means

Ki+Ui=Kf+UfK_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 EthE_{th} or other non-mechanical forms. Once converted, it does not flow back.

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

  1. Draw two pictures - "before" and "after."
  2. List the energy banks that are non-zero in each picture.
  3. Write the energy equation (include WextW_{ext} or ΔEth\Delta E_{th} only if they actually exist).
  4. 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 MM on a rough table is attached to a spring (k=200 N/mk = 200 \text{ N/m}) and to a hanging 5 kg mass mm. 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 MM.

Try it yourself first, then click Reveal Answer.

  1. Choose the system: both blocks, spring, Earth. (Friction is inside the boundary because it acts on MM.)

  2. Initial state (i)

    • Ki=0K_i = 0 (released from rest)
    • UGi=mg(0.15 m)U_{G i} = m g (0.15 \text{ m})
    • Spring un-stretched → Uspi=0U_{sp i} = 0

  3. Final state (f)

    • Blocks instantaneously at rest → Kf=0K_f = 0
    • Hanging block is 0.10 m lower than before → UGf=mg(0.15+0.10)U_{G f} = m g (0.15 + 0.10)
    • Spring stretched by x=0.10 mx = 0.10 \text{ m} → Uspf=12kx2U_{sp f} = \tfrac{1}{2} k x^2
    • Friction has done work that shows up as ΔEth\Delta E_{th}.

  4. Energy equation (isolated system, so Wext=0W_{ext} = 0)

    UGi=UGf+Uspf+ΔEthU_{G i} = U_{G f} + U_{sp f} + \Delta E_{th}

    Thermal change equals the friction force times the slide distance d=0.10 md = 0.10 \text{ m}: ΔEth=fkd\Delta E_{th} = f_k d.

  5. Solve

    mg(0.15)=mg(0.25)+12(200)(0.10)2+fk(0.10)m g (0.15) = m g (0.25) + \tfrac{1}{2} (200)(0.10)^2 + f_k(0.10) fk=39 Nf_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 N/mk = 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 EmechE_{mech} is constant.

Ki+UGi=UspfK_i + U_{G i} = U_{sp f}
  • Start: Ki=0K_i = 0, UGi=mghU_{G i} = m g h (with hh the vertical drop)
  • End: Uspf=12kx2U_{sp f} = \tfrac{1}{2} k x^2

Geometry on the ramp: h=dsin⁡θh = d \sin\theta. We can eliminate θ\theta by noting the problem only asks for the ramp distance dd:

mg(dsin⁡θ)=12kx2⇒d=kx22mgsin⁡θm 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⁡θ\sin\theta comes from the drawn slope (given in the original slide), the numeric crunch leads to

d=0.821 md = 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.

Ki+UGi=Kf+UGfK_i + U_{G i} = K_f + U_{G f}
  • Initial: at rest   ⟹  \implies Ki=0K_i = 0; set UG=0U_G = 0 at the floor.
  • Final: both blocks move with speed vv; heavier block at zero potential.

After algebra, slides show v=4.43 m/sv = 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!
Back to Mechanics