Conservation of Mechanical Energy
Senior 2 · Physics
Conservation of Mechanical Energy
Lesson 3 of 4
22 min
A ripe mango hangs on a branch 5.0 m above the ground in Rusizi. It has potential energy and no kinetic energy. Just before it hits the ground it has almost no potential energy and a lot of kinetic energy. The energy did not disappear and it did not appear from nowhere: it changed form.
If you can keep track of that exchange, you can predict how fast something will be moving at any height without knowing how long it took to get there. That is the power of the conservation law in this lesson.
The mango has a mass of 0.30 kg and falls from rest. Ignore air resistance and take g = 9.8 m/s².
| Height above ground | PE = mgh | KE | Total ME | Speed |
|---|---|---|---|---|
| 5.0 m (on the branch) | 14.7 J | 0 J | 14.7 J | 0 m/s |
| 2.5 m (half-way) | 7.35 J | 7.35 J | 14.7 J | 7.0 m/s |
| 0 m (just before landing) | 0 J | 14.7 J | 14.7 J | 9.9 m/s |
How the speeds were found: at half-way, KE = 7.35 J, so v = √(2 × 7.35 ÷ 0.30) = √49 = 7.0 m/s. At the ground, v = √(2 × 14.7 ÷ 0.30) = √98 ≈ 9.9 m/s.
Every joule of potential energy the mango loses becomes a joule of kinetic energy. The total stays 14.7 J at every height.
The law of conservation of energy says that energy cannot be created or destroyed; it can only change from one form to another. The total energy of an isolated system is constant.
The law of conservation of mechanical energy is a special case of it: in a system where only gravity (or an elastic force) does work, with no friction, no air resistance and no engine, the total mechanical energy stays constant.
KE₁ + PE₁ = KE₂ + PE₂
Here "1" and "2" are any two points on the motion.
An object falls from rest through a height h. All its PE becomes KE:
mgh = ½mv², so v = √(2gh).
The mass cancels. Without air resistance, a heavy stone and a light stone dropped from the same height land at the same speed (though the heavy one carries more kinetic energy, because KE depends on m as well).
Ishimwe throws a ball straight up at 14 m/s. How high does it rise? (g = 9.8 m/s², no air resistance)
At the top, the ball stops for an instant, so all its KE has become PE: ½mu² = mgh.
h = u² ÷ (2g) = 14² ÷ (2 × 9.8) = 196 ÷ 19.6 = 10 m.
A pendulum bob is pulled aside so that it is 0.20 m above its lowest point, then released. At the ends of the swing the bob stops for an instant: all its energy is PE. At the lowest point it moves fastest: all its energy is KE.
Speed at the lowest point: v = √(2gh) = √(2 × 9.8 × 0.20) ≈ 2.0 m/s.
Without friction the bob would rise to exactly 0.20 m on the other side, and keep swinging for ever.
Open Energy Skate Park: Basics. On the Intro screen, switch on the Bar Graph and Speed. Release the skater from different heights on the track and watch the kinetic-energy and potential-energy bars trade places while the total bar stays the same. Then go to the Friction screen, add some friction, and watch a new thermal energy bar grow while the mechanical energy shrinks. That is the next lesson.
Tip
The two-point method. Pick two points, usually "start" and "finish". Write KE₁ + PE₁ = KE₂ + PE₂. Then cross out what is zero: KE = 0 wherever the object is at rest (released, or at the top of its flight), and PE = 0 at your reference level. What remains is usually one line of algebra.
Energy works for any path shape: a slide, a curved track, a swing. The equation v² = u² + 2as only works for a straight line with constant acceleration.
Take g = 9.8 m/s² and ignore air resistance.
A 2.0 kg stone falls from rest from a height of 20 m. Find its speed and its kinetic energy just before it lands.
Solution: v = √(2gh) = √(2 × 9.8 × 20) = √392 ≈ 19.8 m/s. KE = PE lost = mgh = 2.0 × 9.8 × 20 = 392 J.
A ball is thrown straight up at 7.0 m/s. How high does it rise?
Solution: h = u² ÷ (2g) = 49 ÷ 19.6 = 2.5 m.
A child on a swing starts from rest 0.80 m above the lowest point. How fast is the child moving at the bottom?
Solution: v = √(2 × 9.8 × 0.80) = √15.68 ≈ 4.0 m/s. The child's mass is not needed.
A 0.40 kg ball rolls at 3.0 m/s across a smooth shelf 2.0 m high, then down a smooth curved ramp to the floor. Find its speed at the floor.
Solution: KE₁ + PE₁ = KE₂ + 0. Divide every term by m: ½u² + gh = ½v². ½ × 9 + 9.8 × 2.0 = 4.5 + 19.6 = 24.1, so v² = 48.2 and v ≈ 6.9 m/s.
The mango in this lesson falls from 5.0 m. At what height are its kinetic and potential energies equal?
Solution: When KE = PE, each is half of the total, 7.35 J. PE = mgh gives h = 7.35 ÷ (0.30 × 9.8) = 2.5 m, exactly half-way down.
Remember
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