Classical Mechanics · Chapter 4

Energy & Work

Energy is neither created nor destroyed — it only changes form. This conservation law is one of the most powerful tools in physics.

Prerequisites

Newton's laws \cdot Calculus (derivatives, integrals) \cdot Vectors

Learning Goals

Compute work using the dot product and interpret positive, negative, and zero work.
Use the work-energy theorem to relate net work to changes in kinetic energy.
Identify conservative forces and write the corresponding potential energy functions.
Apply mechanical energy conservation when nonconservative work is absent or accounted for.
Explain pendulum motion as an exchange between kinetic and gravitational potential energy.

4.1 Work and the Work-Energy Theorem

Work is done on an object when a force acts through a displacement. The formal definition uses the dot product, because only the component of force parallel to displacement does work:

W = \mathbf{F}\cdot\mathbf{d} = Fd\cos\theta

(4.1)

For a variable force, work is the integral of force over displacement. This leads directly to one of the most important results in mechanics:

Theorem 4.1 — Work-Energy Theorem

The net work done by all forces acting on an object equals the change in its kinetic energy:

W_{net} = \DeltaKE = \frac{1}{2}mv_f^{2} - \frac{1}{2}mv_i^{2}

This is derived directly from Newton's second law by integrating F = ma with respect to di

splacement. It holds regardless of the path taken.

4.2 Potential Energy and Conservation

Some forces — called conservative forces — have the property that the work they do is path-independent and can be stored as potential energy. Gravity and the spring force are conservative. Friction is not.

Definition 4.1 — Conservative Force and Potential Energy

A force F is conservative if the work it does around any closed path is zero. For such forces, we define potential energy U such that:

W_{conservative} = -\DeltaU \qquad or equivalently \qquad F = -dU/dx

Gravitational PE near Earth's surface: U_{g} = mgh. Spring PE: U_{s} = \frac{1}{2}kx^{2}

When only conservative forces act, the total mechanical energy is constant:

E_\mathrm{total} = K + U = \frac{1}{2}mv^2 + mgh = \mathrm{constant}

(4.2)

4.3 The Pendulum — Energy in Oscillatory Motion

The simple pendulum beautifully illustrates energy conservation. As it swings, kinetic and potential energy constantly exchange. At the lowest point, all energy is kinetic; at the turning points, all energy is potential.

The equation of motion follows from Newton's second law in the tangential direction. The restoring force is F = −mg sin(θ), so:

\theta'' = -\frac{g}{L}\sin\theta

(4.3)

For small angles (θ < ~15°), sin(θ) ≈ θ and we get simple harmonic motion with exact period:

T = 2\pi\sqrt{\frac{L}{g}} \qquad \text{(small-angle approximation)}

(4.4)

For large angles, equation (4.3) must be solved numerically. The simulation below uses fourth-order Runge-Kutta (RK4) integration, which is accurate to order Δt⁴. With damping b representing air resistance, the full equation is:

\theta'' = -\frac{g}{L}\sin\theta - b\theta'

(4.5)

String length1.5 m

Initial angle45°

Damping (air resistance)0.05

Properties

Period T2.457 s

Frequency f0.407 Hz

Length1.5 m

Initial angle45°

The phase space plot (top right of canvas) traces θ vs θ̇. An undamped pendulum draws a closed ellipse; damping spirals it inward.

Figure 4.1.

Pendulum simulation with RK4 integration. The phase space plot (\theta vs \thetȧ) in the upper righ

t traces the energy state — undamped motion is a closed ellipse; damping spirals inward.

Example 4.1 — Pendulum Period on Mars

$A pendulum clock keeps perfect time on Earth (g = 9.81 m/s^{2}, L = 0.993 m \to T = 2.00 s). Wh$ $at is its period on Mars (g = 3.72 m/s^{2}$ ?

Period formula:

T = 2\pi \sqrt(L/g

On Earth:

T_{E} = 2\pi \sqrt(0.993/9.81) = 2.00 s ✓

On Mars:

T_{M} = 2\pi \sqrt(0.993/3.72) = 2\pi \times 0.516 =

3.25 s

Ratio:

T_{M}/T_{E} = \sqrt(9.81/3.72) = 1.62 — the clock runs slow by factor 1.62 on Mars

Example 4.2 — Energy Conservation — Maximum Speed

$A 0.5 kg pendulum bob is released from rest at 60° from vertical. String length L = 1.2 m$ . What is its maximum speed?

Height gained:

h = L(1 - cos 60°) = 1.2 \times 0.5 = 0.6 m

Energy conservation:

mgh = \frac{1}{2}mv_max^{2} \to v_{max} = \sqrt(2gh) = \sqrt(2 \times 9.81 \times 0.6) =

3.43 m/s

Note:Mass cancelled — the max speed is independent of mass, just like the period.

Definition 4.2 — Common Traps

Work is not force: work requires displacement and only the parallel force component contributes.
Energy conservation needs a boundary: decide which objects belong to the system.
Friction does negative work: mechanical energy is transformed into thermal energy.
Potential energy zero is arbitrary: only changes in potential energy affect motion.
The pendulum period formula is approximate: $large amplitudes have longer periods than 2\pi\sqrt(L/g$

Exercises — 4.1–4.3 Energy

A 2 kg object accelerates from rest to 15 m/s. How much kinetic energy does it gain?

Straightforward

A 300 g ball is dropped from 5 m height. Using energy conservation, find its speed just before impact.

m/s

Straightforward

3.By what factor must you change a pendulum's length to double its period?

Intermediate

4.A 1500 kg car brakes from 20 m/s to rest over 80 m. Find the friction force using the work-energy theorem.

Intermediate

Describe qualitatively what the phase space plot (\theta vs \thetȧ) looks like for (a) undamped sma

ll oscillations, (b) undamped large oscillations, and (c) damped oscillations. Explain each shape physically.

Challenging

Key Takeaways

$W_{net} = \DeltaKE connects force (Newton) to energy — both describe the same physics$ .
Conservative forces permit potential energy; total mechanical energy is conserved when only conservative forces act.
$Pendulum period T = 2\pi\sqrt(L/g) depends only on length and gravity, not mass or amplitude (sm$ all-angle approx).
For large-angle pendulums, the period is longer than the formula predicts — the motion is nonlinear.
$The phase space portrait captures the entire energy state: ellipse = undamped, inward spir$ $al = damped$ .