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Conservation of Energy

Source lecture(s): SC133 Lec 9

Intuition

Energy is the universe's strictest accountant. It changes form constantly — motion to height, height to spring compression, everything eventually to heat — but the total never budges. In mechanics this gives a superpower: compare any two moments of a process without solving for anything in between. Roller coaster, pendulum, planet — if you know the energy budget, you know the speeds.

The statement

For a system where only conservative forces do work:

\[\boxed{\,E = K + U = \text{constant}\,} \qquad \tfrac12 mv_i^2 + U_i = \tfrac12 mv_f^2 + U_f\]

With non-conservative forces (friction, drag) the mechanical energy leaks into thermal energy, but the total still balances:

\[K_i + U_i = K_f + U_f + E_\text{thermal}, \qquad E_\text{thermal} = f_k\, d\]

Energy conservation isn't broken by friction — mechanical energy just stops being the whole story.

Where it comes from

Mechanical energy conservation is the work–energy theorem plus the definition of potential energy: \(W_\text{cons} = -\Delta U\) and \(W_\text{net} = \Delta K\) combine into \(\Delta(K + U) = W_\text{non-cons}\). Deeper still (Noether's theorem, for later courses): energy conservation is the consequence of physics being the same today as tomorrow — time-translation symmetry.

Worked example: roller coaster loop

A cart starts from rest at height \(h\) and must maintain contact at the top of a loop of radius \(R\). Minimum \(h\)?

At the loop top, gravity alone must supply the centripetal force: \(mg = mv^2/R \Rightarrow v^2 = gR\). Energy from start to loop top (height \(2R\)):

\[mgh = \tfrac12 mv^2 + mg(2R) = \tfrac12 mgR + 2mgR \;\Rightarrow\; \boxed{h = \tfrac52 R}\]

No forces along the track ever entered the calculation — that's the power of the method.

Worked example: friction included

A 2 kg block slides from rest down a ramp of height 1.5 m, arriving at the bottom at 4 m/s. How much energy went to heat?

\(E_\text{thermal} = mgh - \tfrac12 mv^2 = 2(9.8)(1.5) - \tfrac12(2)(16) = 29.4 - 16 = 13.4\,\text{J}\) — the books always balance.

When to use energy vs Newton

Question asks about... Best tool
speed at a position energy (path-independent)
time, or force at an instant Newton / kinematics
direction of motion at a point Newton (energy is a scalar — it forgot direction)
systems with friction over known distance energy with \(f_k d\) term

Common mistakes

  • Using \(E\) conservation across friction without the heat term. Check for non-conservative forces before writing \(K_i + U_i = K_f + U_f\).
  • Expecting energy methods to give direction or time. Energy is a scalar; it yields speeds, not velocity vectors or durations.
  • Double counting: if you include a force via potential energy, don't also add its work.
  • Mixing zero-points mid-problem. Choose where \(U = 0\) once.

Knowledge graph position

Prerequisites: Work & kinetic energy, Potential energy. Leads to: Collisions, SHM, thermodynamics, Bernoulli.

Quiz

Q1 (computational). A pendulum is released from rest with the string horizontal (length \(L\)). Speed at the bottom?

Answer

Drop height \(= L\): \(v = \sqrt{2gL}\). The string tension does no work (⊥ motion), so pure energy conservation applies.

Q2 (conceptual). A ball bounces, each bounce reaching 80% of the previous height. Where does the energy go?

Answer

Into thermal energy and sound during each inelastic contact (deforming the ball and floor). Mechanical energy shrinks by 20% per bounce; total energy is conserved throughout.

Q3 (multiple choice). Two ramps, one steep and one gentle, connect the same two heights (frictionless). A block slides down each. At the bottom: (a) steep ramp gives higher speed (b) equal speeds, different times (c) equal speeds and equal times

Answer

(b). Same \(\Delta U\) ⇒ same speed. The steep ramp is quicker, though — time is Newton's department, not energy's.

Q4 (conceptual). Why does the loop-the-loop answer \(h = \frac52 R\) not depend on the cart's mass?

Answer

Every term in the budget (\(mgh\), \(\frac12 mv^2\), \(mgR\)) is proportional to \(m\) — gravity accelerates all masses alike, so \(m\) cancels. The same cancellation behind Galileo's tower experiment.