Conservation of Mechanical Energy

Master the law of conservation of energy and distinguish between conservative and non-conservative forces.

5 min read

Prerequisites

Before studying this topic, review:

Conservation of Mechanical Energy

Roller Coasters in Oppenheimer!
Energy conservation is like the physics of roller coasters in every theme park scene in movies. As the cart goes UP, kinetic energy converts to potential. Coming DOWN, potential converts back to kinetic. In Oppenheimer (2023), the scientists understood this principle at the atomic level — energy is never destroyed, only transformed. That’s the terrifying power of E = mc²!

Statement

When only conservative forces do work, the total mechanical energy of a system remains constant:

$$\boxed{KE_i + PE_i = KE_f + PE_f}$$

or equivalently:

$$\boxed{E_{total} = KE + PE = \text{constant}}$$

Interactive Demo: Visualize Energy Conservation

See how energy transforms between kinetic and potential forms.

Conservative vs Non-Conservative Forces

Conservative Forces

A force is conservative if:

  1. Work done is path-independent (depends only on endpoints)
  2. Work done in a closed loop is zero
  3. Can be derived from a potential energy function
$$W_{A \to B} = -\Delta U = U_A - U_B$$

Examples:

  • Gravitational force
  • Spring force
  • Electrostatic force

Non-Conservative Forces

A force is non-conservative if work depends on the path taken.

Examples:

  • Friction
  • Air resistance
  • Viscous forces
  • Tension (when length changes)
graph LR
    A[Forces] --> B[Conservative]
    A --> C[Non-Conservative]
    B --> B1[Gravity]
    B --> B2[Spring force]
    B --> B3[Electrostatic]
    C --> C1[Friction]
    C --> C2[Air resistance]
    C --> C3[Viscous drag]
Key Difference
Conservative: Work depends only on initial and final positions Non-conservative: Work depends on the entire path taken

Modified Energy Conservation

When non-conservative forces are present:

$$\boxed{W_{nc} = \Delta E_{mech} = (KE_f + PE_f) - (KE_i + PE_i)}$$

or

$$KE_i + PE_i + W_{nc} = KE_f + PE_f$$

For friction specifically:

$$KE_i + PE_i - f \cdot s = KE_f + PE_f$$

where $f \cdot s$ is energy lost to heat.

Application Strategy

When to Use Energy Conservation

  • Problems involving heights and speeds
  • When acceleration is not needed
  • Projectile motion problems (max height, landing speed)
  • Roller coaster type problems

When NOT to Use

x When you need time or acceleration (use kinematic equations) x When friction is present (use modified equation) x When forces are impulsive (use impulse-momentum)

Problem-Solving Steps

  1. Identify system and choose reference level for PE
  2. List forces and classify as conservative or non-conservative
  3. Write energy equation (choose appropriate form)
  4. Substitute values and solve

Common Scenarios

1. Free Fall

Object falling from height $h$:

$$mgh + 0 = 0 + \frac{1}{2}mv^2$$ $$\boxed{v = \sqrt{2gh}}$$
Mission Impossible Dead Reckoning
When Tom Cruise drives off a cliff in Mission: Impossible – Dead Reckoning (2023), his speed when he falls depends ONLY on height, not on how he fell! That’s $v = \sqrt{2gh}$ — whether you fall straight down or slide down a slope, the final speed at the same height is identical (ignoring friction).

2. Projectile (Max Height)

Object launched vertically with speed $v$:

$$\frac{1}{2}mv^2 + 0 = 0 + mgh_{max}$$ $$\boxed{h_{max} = \frac{v^2}{2g}}$$

3. Spring Launch

Spring (compression $x$) launches mass $m$:

$$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$$ $$\boxed{v = \sqrt{\frac{k}{m}}x}$$

4. Inclined Plane (Smooth)

Block sliding down from height $h$:

$$mgh = \frac{1}{2}mv^2$$ $$v = \sqrt{2gh}$$

(Same as free fall — path doesn’t matter!)

5. Inclined Plane (Rough)

With friction coefficient $\mu$, incline angle $\theta$, length $s$:

$$mgh - \mu mg\cos\theta \cdot s = \frac{1}{2}mv^2$$

Worked Examples

Example 1: Simple Pendulum

Problem
A pendulum of length 2 m is released from horizontal position. Find the speed at the lowest point.

Solution: Taking lowest point as reference:

  • Initial: $KE_i = 0$, $PE_i = mgh = mg(2)$
  • Final: $KE_f = \frac{1}{2}mv^2$, $PE_f = 0$

Conservation:

$$mg(2) = \frac{1}{2}mv^2$$ $$v = \sqrt{4g} = \sqrt{4 \times 10} = \boxed{2\sqrt{10} \approx 6.32 \text{ m/s}}$$

Example 2: Block on Spring

Problem
A 0.5 kg block compresses a spring (k = 800 N/m) by 10 cm. What height does it reach after release?

Solution: Initial (compressed spring): $E_i = \frac{1}{2}kx^2 = \frac{1}{2}(800)(0.1)^2 = 4$ J

Final (maximum height): $E_f = mgh$

Conservation:

$$4 = (0.5)(10)h$$ $$h = \frac{4}{5} = \boxed{0.8 \text{ m}}$$

Example 3: Rough Incline

Problem
A 2 kg block slides 5 m down a 30° rough incline ($\mu_k = 0.2$) from rest. Find the final speed.

Solution: Height dropped: $h = 5\sin30° = 2.5$ m

Work by friction:

$$W_f = -\mu_k mg\cos30° \cdot s = -0.2(2)(10)(0.866)(5) = -17.32 \text{ J}$$

Energy equation:

$$mgh + W_f = \frac{1}{2}mv^2$$ $$2(10)(2.5) - 17.32 = \frac{1}{2}(2)v^2$$ $$50 - 17.32 = v^2$$ $$v = \sqrt{32.68} = \boxed{5.72 \text{ m/s}}$$

Energy Conservation in Connected Systems

For systems like pulleys (see Newton’s Laws for force analysis):

$$\Delta KE_{system} + \Delta PE_{system} + W_{friction} = 0$$
For Atwood's Machine

If masses $m_1$ and $m_2$ start from rest and $m_1$ falls height $h$:

$$m_1gh - m_2gh = \frac{1}{2}(m_1 + m_2)v^2$$

Practice Problems

  1. A ball is thrown upward with 20 m/s. Find maximum height using energy conservation.

  2. A 3 kg block compresses a spring (k = 600 N/m) by 20 cm, then is released. If the surface is smooth, find the block’s speed when spring is at natural length.

  3. A 5 kg block slides down a curved frictionless track from height 4 m, then travels 8 m on a rough horizontal surface ($\mu_k = 0.25$) before stopping. Verify energy conservation.