Energy Conservation in Practice

A guided lesson on applying conservation of mechanical energy to roller coasters, pendulums, and falling objects.

The Energy Bank Account

Think of mechanical energy as a bank account: money (energy) can move between checking (kinetic) and savings (potential), but the total stays the same (if no fees = no friction).

$E_{\text{total}} = \frac{1}{2}mv^2 + mgh = \text{constant}$

Strategy for Energy Problems

1
Identify the system and reference height
Choose a convenient reference point where $h = 0$ .
2
List energy at two key points
Write $E_k$ and $E_p$ at start and end positions.
3
Apply conservation
Set total energy equal at both points.
4
Solve for the unknown
Usually velocity or height.

Roller coaster: starts at rest at 30 m height. Find speed at 10 m height.

$mgh_1 = \frac{1}{2}mv^2 + mgh_2$

$g(30) = \frac{1}{2}v^2 + g(10)$

$v = \sqrt{2g \times 20} = \sqrt{2 \times 9.8 \times 20} \approx 19.8 \text{ m/s}$

Warning

Energy conservation only applies when no energy is lost to friction or air resistance. In real systems, use $W_{\text{friction}} = \Delta E_{\text{mechanical}}$ .

Real world: roller coaster design

A roller coaster car (mass 500 kg) starts from rest at 45 m height. At the bottom of the first drop ( $h = 0$ ): $v = \sqrt{2 \times 9.8 \times 45} \approx 29.7\text{ m/s} \approx 107\text{ km/h}$ . Engineers use energy conservation to determine maximum speeds and design safe braking zones, without a motor anywhere on the track.

Real world: hydroelectric power station

Water falls 100 m at a dam. Each kilogram of water converts potential energy $E_p = mgh = 1 \times 9.8 \times 100 = 980\text{ J}$ into kinetic energy at the turbine. A flow rate of 500 kg/s gives power $P = 500 \times 980 = 490\,000\text{ W} = 490\text{ kW}$ . Hydroelectric design is applied energy conservation.

Real world: electric vehicles — regenerative braking

An electric car converts kinetic energy back into electrical (stored potential) energy when braking. A 1500 kg car travelling at 20 m/s has $KE = \frac{1}{2} \times 1500 \times 400 = 300\,000\text{ J}$ . Regenerative braking recovers a significant fraction of this energy, extending battery range — energy conservation makes this recapture possible.

Tip Set your reference height (

h = 0

) at the lowest point of the problem so that

E_p = 0

there and you only deal with kinetic energy — this simplifies the algebra.

Practice Problems

A 2 kg ball is dropped from 10 m. Find its KE just before hitting the ground.

Show answer

KE = mgh = 2 × 9.8 × 10 = 196 J

A pendulum swings from height 0.5 m. Find speed at lowest point.

Show answer

v = √(2gh) = √(2 × 9.8 × 0.5) = √9.8 ≈ 3.13 m/s

A 1 kg object has KE = 50 J at the bottom. How high does it rise?

Show answer

h = KE/(mg) = 50/(1 × 9.8) ≈ 5.1 m