Conservation of Energy

Mechanical energy conservation and energy transformations

🔋 Conservation of Energy

The Law of Conservation of Energy

Energy cannot be created or destroyed, only transformed from one form to another.

This is one of the most fundamental principles in all of physics!

Mechanical Energy

Mechanical energy is the sum of kinetic and potential energies:

$E_{mech} = KE + PE$

For many problems:

$E_{mech} = \frac{1}{2}mv^2 + mgh + \frac{1}{2}kx^2$

(kinetic + gravitational PE + elastic PE)

Conservation of Mechanical Energy

When only conservative forces do work:

$E_i = E_f$

$KE_i + PE_i = KE_f + PE_f$

$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$

💡 Key: Total mechanical energy stays constant - energy just transforms between KE and PE!

Conservative vs. Non-Conservative Forces

Conservative Forces

Forces for which mechanical energy is conserved:

Gravity - can define gravitational PE
Spring force - can define elastic PE
Electrostatic force - can define electric PE

Properties:

Work is path-independent (depends only on start/end points)
Potential energy can be defined
Work around closed path = 0

Non-Conservative Forces

Forces that dissipate mechanical energy:

Friction - converts mechanical energy to thermal energy
Air resistance - converts KE to thermal energy
Tension (usually) - can do positive or negative work

Important: When non-conservative forces do work, mechanical energy is NOT conserved!

Energy with Non-Conservative Forces

When friction or other non-conservative forces are present:

$E_i + W_{nc} = E_f$

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

where $W_{nc}$ = work done by non-conservative forces

Special case: Friction

$W_{friction} = -f_k \cdot d = -\mu_k N \cdot d$

(Negative because friction opposes motion)

This energy is converted to thermal energy (heat) and "lost" from mechanical energy.

Problem-Solving with Energy Conservation

Step-by-Step Strategy

Identify the system and choose reference points
Choose two points in the motion (initial and final)
List energies at each point:
- Kinetic: $\frac{1}{2}mv^2$
- Gravitational PE: $mgh$
- Elastic PE: $\frac{1}{2}kx^2$
Check for non-conservative forces
Write conservation equation:
- If only conservative forces: $E_i = E_f$
- If non-conservative forces present: $E_i + W_{nc} = E_f$
Solve for unknown

Common Energy Transformations

Pendulum

At highest points: Maximum PE, minimum KE (zero if released from rest)
At lowest point: Minimum PE, maximum KE
Continuous conversion: $PE \leftrightarrow KE$

Roller Coaster

Top of hill: High PE, low KE
Bottom of dip: Low PE, high KE
First hill must be highest (no external energy added)

Vertical Spring

Maximum compression: Maximum elastic PE
Equilibrium: Maximum KE
Includes both elastic and gravitational PE!

Projectile Motion

At launch: Some KE, some PE (if not ground level)
At peak: Horizontal KE only, maximum PE
At landing: Maximum KE (if returns to launch height)

Energy Bar Diagrams

A useful visualization tool showing energy distribution. Energy bar diagrams show the distribution of kinetic and potential energy:

Initial state: Mostly potential energy (at height)
Final state: Mostly kinetic energy (at bottom)
Total bar height stays same if energy conserved

⚠️ Common Mistakes

Mistake 1: Forgetting Energy Types

Don't forget to include ALL forms of energy! If there's a spring, include elastic PE. If there's height, include gravitational PE.

Mistake 2: Wrong Reference Point

Choose a consistent reference point for PE. Usually pick the lowest point as $h = 0$ .

Mistake 3: Ignoring Non-Conservative Work

If friction is present, you CANNOT use $E_i = E_f$ . Must account for work by friction!

Mistake 4: Sign Errors

PE increases as height increases (positive work against gravity)
Friction work is negative (opposes motion)
Spring PE is always positive

When to Use Energy vs. Force Methods

Use Energy Methods When:

Asked for speed/velocity (not acceleration)
Motion involves height changes
Springs are involved
Path is complicated but you only care about start/end

Use Force Methods (Newton's Laws) When:

Asked for acceleration or force
Need to find motion at specific point (not just start/end)
Need to analyze contact forces

Many problems can be solved either way!

Power: Rate of Energy Transfer

Power is how fast energy is transferred:

$P = \frac{W}{t} = \frac{\Delta E}{t}$

Also: $P = Fv$ (for constant force)

Units: Watt (W) = J/s

1 horsepower (hp) = 746 W

Applications

Hydroelectric Power

Gravitational PE of water → KE → Electrical energy

Regenerative Braking

KE of car → Electrical energy (charges battery)

Bungee Jumping

Gravitational PE → KE → Elastic PE (of cord) → KE → Gravitational PE

Earthquakes

Elastic PE stored in rock → KE of seismic waves → Damage

Key Formulas Summary

| Concept | Formula | When to Use | |---------|---------|-------------| | Mechanical Energy | $E = KE + PE$ | Always define | | Conservation (no friction) | $E_i = E_f$ | Only conservative forces | | With non-conservative forces | $E_i + W_{nc} = E_f$ | Friction present | | Power | $P = \frac{W}{t}$ or $P = Fv$ | Rate of work/energy |

📚 Practice Problems

1Problem 1medium

❓ Question:

A 0.50 kg ball is dropped from a height of 10 m. Ignore air resistance. (a) What is its initial potential energy? (b) What is its kinetic energy just before hitting the ground? (c) What is its speed just before impact?

💡 Show Solution

Solution:

Given: m = 0.50 kg, h = 10 m, g = 10 m/s²

(a) Initial potential energy: PE_i = mgh = 0.50 × 10 × 10 = 50 J (KE_i = 0 since dropped from rest)

(b) Kinetic energy at ground: By conservation of energy: E_total = constant PE_i + KE_i = PE_f + KE_f 50 + 0 = 0 + KE_f KE_f = 50 J

Check using kinematics: v² = v₀² + 2gh = 0 + 2(10)(10) = 200 → v = 14.1 m/s ✓

2Problem 2easy

❓ Question:

A 2 kg ball is dropped from a height of 5 m. What is its speed just before hitting the ground? (Use energy conservation)

💡 Show Solution

Given Information:

Mass: $m = 2$ kg
Initial height: $h_i = 5$ m
Final height: $h_f = 0$ m
Initial velocity: $v_i = 0$ (dropped from rest)

Find: Final speed $v_f$

Step 1: Choose reference point

Let ground be $h = 0$ .

Step 2: Calculate initial energy

At height 5 m:

$KE_i = \frac{1}{2}mv_i^2 = 0$ (at rest)
$PE_i = mgh_i = (2)(9.8)(5) = 98$ J

Total: $E_i = 98$ J

Step 3: Calculate final energy

At ground:

$KE_f = \frac{1}{2}mv_f^2$ (unknown)
$PE_f = mgh_f = 0$ J

Total: $E_f = \frac{1}{2}mv_f^2$

Step 4: Apply conservation of energy

$E_i = E_f$

$98 = \frac{1}{2}(2)v_f^2$

$98 = v_f^2$

$v_f = \sqrt{98} = 7\sqrt{2} \approx 9.90 \text{ m/s}$

Check with kinematics:

$v_f^2 = v_i^2 + 2g\Delta h = 0 + 2(9.8)(5) = 98$

$v_f = \sqrt{98}$ ✓

Answer: The speed just before hitting the ground is approximately 9.90 m/s or $7\sqrt{2}$ m/s.

Note: The mass cancels out - all objects fall at the same rate (ignoring air resistance)!

3Problem 3medium

❓ Question:

💡 Show Solution

Solution:

Given: m = 0.50 kg, h = 10 m, g = 10 m/s²

(a) Initial potential energy: PE_i = mgh = 0.50 × 10 × 10 = 50 J (KE_i = 0 since dropped from rest)

(b) Kinetic energy at ground: By conservation of energy: E_total = constant PE_i + KE_i = PE_f + KE_f 50 + 0 = 0 + KE_f KE_f = 50 J

Check using kinematics: v² = v₀² + 2gh = 0 + 2(10)(10) = 200 → v = 14.1 m/s ✓

4Problem 4hard

❓ Question:

A 2.0 kg block slides down a frictionless curved ramp from a height of 5.0 m. At the bottom, it encounters a horizontal surface with μₖ = 0.40. (a) What is the block's speed at the bottom of the ramp? (b) How far does it slide on the horizontal surface before stopping?

💡 Show Solution

Solution:

Given: m = 2.0 kg, h = 5.0 m, μₖ = 0.40, g = 10 m/s²

(a) Speed at bottom (frictionless ramp): Conservation of energy: PE_top = KE_bottom mgh = ½mv² gh = ½v² v² = 2gh = 2(10)(5.0) = 100 v = 10 m/s

(b) Distance on horizontal surface: On horizontal surface, friction does negative work: KE_bottom = W_friction ½mv² = f_k × d

Friction force: f_k = μₖN = μₖmg = 0.40(2.0)(10) = 8.0 N

½(2.0)(10)² = 8.0 × d 100 = 8.0d d = 12.5 m

Alternative using kinematics: Deceleration: a = -f_k/m = -8.0/2.0 = -4.0 m/s² v² = v₀² + 2ad 0 = (10)² + 2(-4.0)d 8d = 100 d = 12.5 m ✓

5Problem 5hard

❓ Question:

💡 Show Solution

Solution:

Given: m = 2.0 kg, h = 5.0 m, μₖ = 0.40, g = 10 m/s²

(a) Speed at bottom (frictionless ramp): Conservation of energy: PE_top = KE_bottom mgh = ½mv² gh = ½v² v² = 2gh = 2(10)(5.0) = 100 v = 10 m/s

(b) Distance on horizontal surface: On horizontal surface, friction does negative work: KE_bottom = W_friction ½mv² = f_k × d

Friction force: f_k = μₖN = μₖmg = 0.40(2.0)(10) = 8.0 N

½(2.0)(10)² = 8.0 × d 100 = 8.0d d = 12.5 m

Alternative using kinematics: Deceleration: a = -f_k/m = -8.0/2.0 = -4.0 m/s² v² = v₀² + 2ad 0 = (10)² + 2(-4.0)d 8d = 100 d = 12.5 m ✓

6Problem 6medium

❓ Question:

A roller coaster car (mass 500 kg) starts from rest at point A (height 30 m). It descends to point B (height 10 m). (a) What is its speed at point B? (b) What is its speed at ground level (point C)?

💡 Show Solution

Given Information:

Mass: $m = 500$ kg
Point A: $h_A = 30$ m, $v_A = 0$ (starts from rest)
Point B: $h_B = 10$ m
Point C: $h_C = 0$ m
Assume no friction

(a) Find speed at point B

Step 1: Energy at point A

$E_A = KE_A + PE_A = 0 + mgh_A$

$E_A = (500)(9.8)(30) = 147,000 \text{ J}$

Step 2: Energy at point B

$E_B = KE_B + PE_B = \frac{1}{2}mv_B^2 + mgh_B$

$E_B = \frac{1}{2}(500)v_B^2 + (500)(9.8)(10)$

$E_B = 250v_B^2 + 49,000$

Step 3: Apply conservation of energy

$E_A = E_B$

$147,000 = 250v_B^2 + 49,000$

$98,000 = 250v_B^2$

$v_B^2 = 392$

$v_B = \sqrt{392} = 14\sqrt{2} \approx 19.8 \text{ m/s}$

(b) Find speed at point C (ground)

Step 4: Energy at point C

$E_C = KE_C + PE_C = \frac{1}{2}mv_C^2 + 0$

Step 5: Apply conservation

$E_A = E_C$

$147,000 = \frac{1}{2}(500)v_C^2$

$147,000 = 250v_C^2$

$v_C^2 = 588$

$v_C = \sqrt{588} = 14\sqrt{3} \approx 24.2 \text{ m/s}$

Answers:

(a) Speed at point B (10 m high): 19.8 m/s
(b) Speed at point C (ground): 24.2 m/s

Check: Notice speed increases as height decreases - gravitational PE converts to KE!

7Problem 7hard

❓ Question:

A 3 kg block slides down a 37° incline from a height of 4 m. If the coefficient of kinetic friction is 0.25, what is the block's speed at the bottom?

💡 Show Solution

Given Information:

Mass: $m = 3$ kg
Initial height: $h_i = 4$ m
Final height: $h_f = 0$ m
Angle: $\theta = 37°$
Coefficient of kinetic friction: $\mu_k = 0.25$
Initial velocity: $v_i = 0$ (starts from rest)

Find: Final speed $v_f$

Note: Friction is present, so mechanical energy is NOT conserved. Must account for work by friction!

Step 1: Find distance along incline

$d = \frac{h}{\sin\theta} = \frac{4}{\sin(37°)} = \frac{4}{0.6} \approx 6.67 \text{ m}$

Step 2: Find normal force

On incline, $N = mg\cos\theta$ :

$N = (3)(9.8)\cos(37°) = 29.4(0.8) = 23.52 \text{ N}$

Step 3: Calculate work by friction

$W_f = -f_k \cdot d = -\mu_k N \cdot d$

$W_f = -(0.25)(23.52)(6.67)$

$W_f = -39.2 \text{ J}$

(Negative because friction opposes motion)

Step 4: Set up energy equation

$E_i + W_{friction} = E_f$

$KE_i + PE_i + W_f = KE_f + PE_f$

$0 + mgh_i + W_f = \frac{1}{2}mv_f^2 + 0$

Step 5: Substitute and solve

$(3)(9.8)(4) + (-39.2) = \frac{1}{2}(3)v_f^2$

$117.6 - 39.2 = 1.5v_f^2$

$78.4 = 1.5v_f^2$

$v_f^2 = 52.27$

$v_f \approx 7.23 \text{ m/s}$

Check without friction:

If no friction: $mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh} = \sqrt{2(9.8)(4)} = \sqrt{78.4} \approx 8.85$ m/s

With friction, speed is less (7.23 < 8.85) ✓

Answer: The block's speed at the bottom is approximately 7.23 m/s.

Energy accounting:

Initial PE: 117.6 J
Lost to friction: 39.2 J
Final KE: 78.4 J
Total: 117.6 - 39.2 = 78.4 ✓

🎴

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