⚡Conservation of Energy
The Most Important Law in Physics
The Law of Conservation of Energy: Energy cannot be created or destroyed — only converted from one form to another.
For a system with no friction (conservative forces only):
KE₁ + PE₁ = KE₂ + PE₂
½mv₁² + mgh₁ = ½mv₂² + mgh₂
This is incredibly powerful. You can analyze complex motions (rollercoasters, swinging pendulums, sliding objects) by just comparing energy at two points — no need to know the path!
When Energy is Lost to Friction
When friction acts, mechanical energy is not conserved — some is converted to thermal energy (heat).
The complete energy equation with friction:
KE₁ + PE₁ = KE₂ + PE₂ + W_friction
or equivalently:
E_initial = E_final + |W_friction|
W_friction = f_k × d (energy lost to heat)
Note: "energy lost" is a bit misleading — the energy doesn't disappear, it just becomes thermal energy that isn't useful for the motion.
Energy Bar Charts
Watch as KE and PE exchange as the pendulum swings. Total energy remains constant.
Energy Skate Park
Explore conservation of energy on a skate track. Watch KE and PE trade off — with and without friction.
Think About It
A rollercoaster starts from rest at the top of a 40 m hill. What is its speed at the bottom? What assumption are you making?
✏️ Worked Example
Problem: A 2 kg ball is dropped from 20 m height. Find its speed just before it hits the ground. (g = 10 m/s²)
📐 Key Equations
Conservation of Energy
KE_1 + PE_1 = KE_2 + PE_2 (no friction)(1)/(2)mv_1² + mgh_1 = (1)/(2)mv_2² + mgh_2E_1 = E_2 + W_frictionv = √(2gh) (free fall from height h)⚠️ Common Mistakes
Misconception: Applying conservation of energy when friction is present without accounting for the energy lost.
✓ Correct thinking: When friction acts, use E_initial = E_final + W_friction, where W_friction = f_k × d (energy converted to heat).
Why: Mechanical energy (KE + PE) is not conserved when non-conservative forces like friction act. Total energy is still conserved — it just includes thermal energy now.
Misconception: Forgetting that the mass cancels in many energy conservation problems, then struggling with missing mass values.
✓ Correct thinking: When all energy terms involve the same mass m (KE = ½mv² and PE = mgh), m cancels from both sides of the equation.
Why: This is why all objects in free fall hit the ground at the same speed regardless of mass — the result v = √(2gh) contains no mass term.
Misconception: Thinking "conservation of energy" means the speed stays constant throughout the motion.
✓ Correct thinking: Energy is conserved (total stays constant) but KE and PE trade off constantly. Speed changes as height changes — it only stays constant if KE stays constant (e.g., constant-height circular motion).
Why: Conservation means the sum KE + PE is fixed, not that individual terms are fixed. A falling ball speeds up because PE is converting into KE.
📝 Practice Problems
Try these problems. Check your answer when ready.
A 5 kg ball is dropped from 8 m height. Find its speed just before it hits the ground. (g = 10 m/s²)
mgh = (1)/(2)mv²A rollercoaster starts from rest at 30 m height. What is its speed at the bottom? (No friction, g = 10 m/s²)
(1)/(2)mv² = mghA 3 kg pendulum bob is released from a height of 0.8 m. What is its speed at the lowest point? (g = 10 m/s²)
mgh = (1)/(2)mv²A 2 kg block slides down a 5 m ramp (height 3 m). Friction does −12 J of work. Find the speed at the bottom. (g = 10 m/s²)
KE_1 + PE_1 = KE_2 + PE_2 + |W_friction|A rollercoaster is at the top of a 40 m hill (v = 0) and later reaches a 25 m hill. Find the speed at the top of the 25 m hill. (No friction, g = 10 m/s²)
mgh_1 = (1)/(2)mv² + mgh_2A 0.5 kg spring-launched ball (k = 600 N/m, compressed 0.2 m) is launched vertically. How high does it rise? (g = 10 m/s²)
(1)/(2)kx² = mghA 4 kg box slides from rest down a 6 m long incline (height 2 m). The coefficient of kinetic friction is 0.2 and the normal force is 33.9 N. Find the speed at the bottom. (g = 10 m/s²)
KE_f = PE_i - W_frictionA 1 kg ball on a 2 m string is released from horizontal (θ = 90° from vertical). What is the tension in the string at the bottom of the swing? (g = 10 m/s²)
T - mg = (mv²)/(r)Finished reading through this lesson?