§7.5 — Energy Conservation

One of the most powerful principles in physics is the conservation of mechanical energy: when only conservative forces act, the total mechanical energy (kinetic plus potential) remains constant. This section derives this principle and explores what happens when friction and other dissipative forces are present.

The principle of mechanical energy conservation

Define total mechanical energy as:

\[ E_{\text{mech}} = KE + U = \frac{1}{2} m v^2 + U(x) \]

Principle: If only conservative forces act on a system, mechanical energy is conserved:

\[ E_{\text{mech}} = \text{constant} \]

Equivalently, at any two times:

\[ KE_i + U_i = KE_f + U_f \]

or

\[ \frac{1}{2} m v_i^2 + U_i = \frac{1}{2} m v_f^2 + U_f \]

Why this is true: Recall from §7.3 (work–energy theorem) and §7.4 (potential energy):

\[ W_{\text{net}} = \Delta KE = KE_f - KE_i \]

When only conservative forces act:

\[ W_{\text{conservative}} = -\Delta U = U_i - U_f \]

Therefore:

\[ KE_f - KE_i = U_i - U_f \]

\[ KE_i + U_i = KE_f + U_f \]

The total energy is constant.

Converting between kinetic and potential energy

Energy conservation tells us that kinetic and potential energy can convert into each other, but the total stays the same. This is the source of many physics phenomena.

Example 1: Pendulum (simple oscillation)

A pendulum bob of mass 1 kg swings from rest at height 1.0 m above the lowest point.

At the highest point (release position):

• Height: 1.0 m
• Velocity: 0 m/s
• \( U_i = mgh = 1 \times 9.8 \times 1.0 = 9.8 \) J
• \( KE_i = 0 \) J
• \( E_{\text{mech}} = 9.8 \) J

At the lowest point (bottom of swing):

• Height: 0 m (reference point)
• Velocity: \( v_f \) (unknown)
• \( U_f = 0 \) J
• \( KE_f = \frac{1}{2} m v_f^2 \) (unknown)
• \( E_{\text{mech}} = KE_f + 0 = KE_f \)

By energy conservation:

\[ 9.8 = \frac{1}{2} \times 1 \times v_f^2 \]

\[ v_f^2 = 19.6 \]

\[ v_f = \sqrt{19.6} \approx 4.43 \text{ m/s} \]

The bob reaches maximum speed (4.43 m/s) at the bottom, where all potential energy has converted to kinetic energy.

At height 0.5 m (midway):

• \( U = 1 \times 9.8 \times 0.5 = 4.9 \) J
• \( E_{\text{mech}} = 9.8 = KE + 4.9 \), so \( KE = 4.9 \) J
• \( \frac{1}{2} \times 1 \times v^2 = 4.9 \) → \( v = \sqrt{9.8} \approx 3.13 \) m/s

At the midpoint, potential and kinetic energies are equal.

Example 2: Projectile (parabolic motion)

A ball is thrown upward from height h₀ = 0 with initial speed v₀ = 20 m/s.

At release (t = 0):

• Height: 0 m
• Velocity: 20 m/s
• \( U_i = 0 \) J
• \( KE_i = \frac{1}{2} \times m \times (20)^2 = 200m \) J
• \( E_{\text{mech}} = 200m \) J

At maximum height (v = 0):

• Velocity: 0 m/s
• \( KE_f = 0 \) J
• \( E_{\text{mech}} = U_f \), so \( mgh_{\max} = 200m \) J
• \( h_{\max} = \frac{200}{9.8} \approx 20.4 \) m

At the peak, all kinetic energy converts to potential energy.

At some intermediate height h = 10 m:

• \( U = mg \times 10 \)
• \( E_{\text{mech}} = 200m = KE + 10mg \)
• \( KE = 200m - 10mg = m(200 - 10 \times 9.8) = m(200 - 98) = 102m \) J
• \( v = \sqrt{\frac{2 \times 102m}{m}} = \sqrt{204} \approx 14.3 \) m/s

The ball has speed 14.3 m/s both on the way up and on the way down at height 10 m (by symmetry, the speeds are the same).

Effect of friction (non-conservative forces)

In the real world, friction is always present. Friction is non-conservative: it dissipates mechanical energy into heat and cannot restore it.

When friction acts, mechanical energy is not conserved. Instead:

\[ E_{\text{mech,final}} = E_{\text{mech,initial}} + W_{\text{friction}} \]

Since friction opposes motion, \( W_{\text{friction}} < 0 \) (negative work), so:

\[ E_{\text{mech,final}} < E_{\text{mech,initial}} \]

The energy "lost" is converted to heat.

Example: Sliding block with friction

A 5 kg block slides across a table with coefficient of kinetic friction \( \mu_k = 0.3 \). Initial velocity: 10 m/s. The block travels 20 m before stopping.

Check using energy:

Initial kinetic energy: \[ KE_i = \frac{1}{2} \times 5 \times (10)^2 = 250 \text{ J} \]

Friction force: \[ f = \mu_k \times N = 0.3 \times (mg) = 0.3 \times 5 \times 9.8 = 14.7 \text{ N} \]

Work by friction over 20 m: \[ W_{\text{friction}} = -14.7 \times 20 = -294 \text{ J} \]

If there were no friction: the block would reach final kinetic energy of 250 J (traveling indefinitely).

With friction: the final kinetic energy is: \[ KE_f = KE_i + W_{\text{friction}} = 250 - 294 = -44 \text{ J} \]

But kinetic energy cannot be negative! This means the block stops before reaching 20 m.

Stopping distance: Setting \( KE_f = 0 \): \[ 0 = 250 - 14.7 \times d \]

\[ d = \frac{250}{14.7} \approx 17 \text{ m} \]

The block stops after 17 m (not 20 m as I assumed). The 250 J of initial kinetic energy is entirely dissipated as heat via friction.

Bound and unbound motion

Energy conservation predicts whether an object can escape from a potential well.

Example: Escape from a gravity well

A rocket has mass 1000 kg and sits on Earth's surface. To escape to infinity (where gravitational potential energy is zero), what minimum speed is needed?

At Earth's surface:

• Height: \( h = R_{\text{Earth}} = 6.37 \times 10^6 \) m
• Velocity: \( v_{\text{escape}} \) (unknown)
• \( U = -\frac{GMm}{R_{\text{Earth}}} \) (negative, since we set \( U = 0 \) at infinity)
• \( KE = \frac{1}{2} m v_{\text{escape}}^2 \)

At infinity (if barely escaping):

• Velocity: 0 m/s (just reaches infinity with zero speed)
• \( U = 0 \) J
• \( KE = 0 \) J

By energy conservation:

\[ \frac{1}{2} m v_{\text{escape}}^2 - \frac{GMm}{R_{\text{Earth}}} = 0 \]

\[ v_{\text{escape}} = \sqrt{\frac{2GM}{R_{\text{Earth}}}} \approx 11.2 \text{ km/s} \]

Physical meaning: Any object launched from Earth at less than 11.2 km/s is bound to Earth — it will return. At or above 11.2 km/s, it escapes to infinity. This critical speed is the escape velocity.

Comparison: energy vs. force approaches

Two ways to solve mechanics problems:

Using forces (Chapter 5, Newton's laws):

• Write Newton's second law: \( F = ma \)
• Integrate to find \( v(t) \) and \( x(t) \)
• Often requires solving differential equations

Using energy (this chapter):

• Apply conservation of energy: \( E_{\text{mech}} = \text{constant} \)
• Solve algebraically for unknown speeds or positions
• No need to find \( v(t) \) or \( x(t) \) — only initial and final states matter

Advantage of energy: Simpler algebra, avoids time-dependence.

Disadvantage: Cannot find details like the trajectory shape or the time to reach a point — only speeds and positions.

Connection to §7.3 and §7.4

This section synthesizes everything from §7.1–§7.4:

• §7.1–§7.2: Work by forces (constant and variable)
• §7.3: Work–energy theorem (net work = change in kinetic energy)
• §7.4: Potential energy (conservative forces)
• §7.5: Mechanical energy conservation (combination of KE and U)

Energy conservation is the deepest principle here. It shows that energy can hide in different forms (kinetic, gravitational potential, elastic potential, heat, etc.), but the total always balances.

What you should be able to do

After this section, you should be able to:

• Apply conservation of mechanical energy to find unknown speeds or positions
• Determine whether mechanical energy is conserved in a given scenario (only conservative forces?)
• Calculate the energy dissipated by friction
• Distinguish between bound and unbound motion based on total energy
• Solve problems using energy conservation where force-based approaches would require differential equations

Preview of §7.6: Power

Energy conservation tells us how much energy is transferred, but not how fast. Power measures the rate of energy transfer:

\[ P = \frac{dE}{dt} \]

For a force doing work:

\[ P = \mathbf{F} \cdot \mathbf{v} \]

This is the topic of §7.6.

📚 See also: Halliday Vol 1, Ch 7, §7.6 — Conservation of Mechanical Energy. 📖 Open reference: OpenStax University Physics Vol 1 — Chapter 7.4: Elastic Potential Energy. 🎓 Video: MIT 8.01 Lecture 11 (Walter Lewin) — "Conservation of Energy in Macroscopic Mechanical Systems".

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