1.5.5Dynamics
Conservation of Energy Calculations
5 min read 6 parts Lesson 20 of 22
Section 1 of 6The Method
Energy conservation lets you find speeds without knowing the forces or times involved.
Idea
- If no energy is wasted, the potential energy lost equals the kinetic energy gained.
Equation
- This gives m g h = ½ m v², so the mass cancels and v = √(2 g h).
Mass cancels
- The speed of a freely falling object is independent of its mass, if there is no air resistance.
With friction
- If energy is wasted, then Ep lost = Ek gained + energy transferred as heat.
Section 2 of 6Typical Situations
Typical Situations
Falling object
- Potential energy transfers to kinetic energy as it falls;
- the speed at the bottom depends only on the height dropped.
Pendulum
- At the highest point the bob has maximum Ep and zero Ek; at the lowest point the reverse is true.
Roller coaster
- Energy is exchanged between Ep and Ek as the car goes over hills and dips, with some heat lost to friction.
Slope
- An object sliding down a slope without friction gains Ek equal to m g multiplied by the vertical height.
| Point | Ep | Ek |
|---|---|---|
| Top of fall | Maximum | Zero |
| Halfway | Half | Half |
| Just before ground | Zero | Maximum |
Section 3 of 6Solution Steps
1
Identify the starting and finishing points and note which energies exist at each.
2
Write the energy equation, for example Ep lost = Ek gained.
3
Substitute m g h on one side and ½ m v² on the other.
4
Cancel the mass if it appears on both sides, then solve for the unknown.
5
If the question mentions friction or air resistance, include the heat energy in the equation.
Section 4 of 6Common Problem Types
Common Problem Types
Dropped object
- Find the final speed from the height using v = √(2 g h).
Thrown upwards
- The initial kinetic energy gives the maximum height from m g h = ½ m v².
Frictional slope
- Ek at the bottom equals Ep lost minus the heat transferred by friction.
Pendulum
- At the bottom of the swing, Ek equals the Ep lost between the highest point and the lowest point.
Section 5 of 6Worked Example
Question
- A 0.50 kg ball is dropped from a height of 20 m.
- Find its potential energy at the start and its speed just before landing, ignoring air resistance (g = 9.8 N/kg).
- Ep = m g h = 0.50 × 9.8 × 20 = 98 J.What the symbols mean: Ep = gravitational potential energy gained (J) · m = mass (kg) · g = 9.81 N kg−1 · h = change in height (m)
- All of this becomes kinetic energy½ m v² = 98, so v² = (2 × 98) ÷ 0.50 = 392.
- v = √392 = 19.8 m/s (about 20 m/s).
Answer 98 J; 19.8 m/s.
Section 6 of 6Exam Checklist & Tips
1.5.5 – Key Points for the Exam
- Conservation of energy Total energy stays the same during a transfer.
- Potential energy Energy stored due to position.
- Kinetic energy Energy due to motion.
- Dissipated energy Energy spread out as heat in the surroundings.
Common Mistakes
- Using the mass twice or forgetting to cancel it correctly.
- Forgetting that Ek = ½ m v² has a one half.
- Ignoring the energy wasted as heat when friction is mentioned.
- Using the slope length instead of vertical height.