Conservation of Energy- Practice Problems and Solutions

Conservation of Energy: What It Actually Means

The law of conservation of energy states that energy cannot be created or destroyed—only converted from one form to another. That's it. No magic, no exceptions (within closed systems). The total energy before a process equals the total energy after.

This principle shows up everywhere in physics: falling objects, roller coasters, pendulums, springs. If you're solving mechanics problems and ignoring energy conservation, you're making your life harder.

The Core Equation

For mechanical systems without friction or other non-conservative forces:

KE₁ + PE₁ = KE₂ + PE₂

Where:

• KE = Kinetic energy = ½mv²
• PE = Potential energy (gravitational = mgh, elastic = ½kx²)
• m = mass
• v = velocity
• g = gravitational acceleration (9.8 m/s² on Earth)
• h = height
• k = spring constant
• x = displacement from equilibrium

Energy Types at a Glance

Practice Problems

Problem 1: The Falling Ball

A 2 kg ball drops from a height of 10 meters. What's its velocity just before hitting the ground?

Step 1: Identify your reference points. At the top: maximum height, zero velocity, maximum PE. At the bottom: zero height, maximum velocity, zero PE.

Step 2: Apply conservation of energy.

Initial: KE₁ + PE₁ = 0 + (2)(9.8)(10) = 196 J

Final: KE₂ + PE₂ = ½(2)v² + 0

Step 3: Solve.

196 = v²

v = √196 = 14 m/s

Problem 2: The Roller Coaster Car

A 500 kg coaster car starts at rest at height 40 m. Calculate its speed at height 10 m.

Step 1: Initial energy at 40 m.

E₁ = 0 + (500)(9.8)(40) = 196,000 J

Step 2: Energy at 10 m.

E₂ = ½(500)v² + (500)(9.8)(10)

196,000 = 250v² + 49,000

Step 3: Solve for v.

147,000 = 250v²

v² = 588

v = 24.2 m/s

Problem 3: The Spring Launcher

A spring with k = 200 N/m is compressed 0.3 m. It launches a 0.5 kg block. What's the block's launch speed?

Step 1: Spring's elastic PE converts to block's KE.

PE_spring = ½(200)(0.3)² = ½(200)(0.09) = 9 J

Step 2: Set equal to KE and solve.

9 = ½(0.5)v²

9 = 0.25v²

v² = 36

v = 6 m/s

Problem 4: The Pendulum

A pendulum bob (1 kg) swings from rest at 0.5 m height to its lowest point. What's its speed at the bottom?

This one's straightforward. All PE at the top becomes KE at the bottom.

PE = (1)(9.8)(0.5) = 4.9 J

4.9 = ½(1)v²

v² = 9.8

v = 3.13 m/s

How to Solve Any Conservation of Energy Problem

1. Identify two moments in the process—one where you know most values, one where you need to find something.
2. Write out total energy for both moments. Include every energy type that applies (KE, gravitational PE, elastic PE).
3. Set them equal. E₁ = E₂
4. Plug in known values. Leave one variable unknown.
5. Solve algebraically. Isolate the variable.
6. Check your work. Does the answer make physical sense?

Where Students Screw Up

• Forgetting to square the velocity. KE = ½mv², not ½mv. This trips people up constantly.
• Using wrong height reference. PE depends on your zero point. Pick one and stick with it.
• Ignoring units. Keep everything in kg, m, s, J. Mixing units gives garbage answers.
• Overcomplicating it. If a ball falls, you don't need to consider horizontal motion. Only vertical displacement matters for gravitational PE.
• Forgetting friction exists. In real problems, some energy converts to thermal energy. If the problem doesn't mention friction losses, assume 100% conversion.

When to Use This vs. Newton's Laws

Conservation of energy shines when:

• You're tracking an object through multiple points
• Forces are complicated but energy paths are simple
• You need velocity but don't need time

Newton's Second Law works better when:

• You're solving for acceleration
• You need to find time
• Forces are the focus

Both methods work. Energy conservation often gets you to the answer faster. 🔧