Conservation of energy
Work, Energy & Power - OCR A-Level Physics
Key Definition
Conservation of energy
Energy cannot be created or destroyed, only transferred from one form to another. The total energy of a closed system remains constant.
Energy cannot be created or destroyed, only transferred from one form to another. The total energy of a closed system remains constant.
- In an ideal system (no friction): loss in GPE = gain in KE, so $mg\Delta$h = $\frac{1}{2}$$mv^{2}$
- The mass cancels: v = $\sqrt{2g\Delta h}.$ The speed at the bottom depends only on height and g, not mass
- With friction or drag: loss in $GPE = gain$ in KE + energyThe capacity to do work. Measured in joules (J). transferred to thermal stores
- EnergyThe capacity to do work. Measured in joules (J). is always conserved, but useful energyThe capacity to do work. Measured in joules (J). may be dissipated as heat, sound, etc.
- In a pendulum: energy transfers continuously between KE and GPE. Maximum KE at lowest point; maximum GPE at highest point
Worked Example [4 marks]
A 500 g ball is released from rest and rolls down a frictionless slope, dropping through a vertical height of 2.0 m. Calculate its speed at the bottom. Use g = 9.81 $m s^{-2}$.
Show Solution
1
By conservation of energyEnergy cannot be created or destroyed, only transferred from one form to another. The total energy of a closed system remains constant. (no friction)
loss in $GPE = gain$ in KE
[1]2
$mg\Delta h = \frac{1}{2}mv^{2}. The mass cancels: g\Delta h = \frac{1}{2}v^{2}$
[1]3
$v^{2} = 2g\Delta h = 2 \times 9.81 \times 2.0 = 39.24$
[1]4
$v = \sqrt{39.24} = 6.26 \text{ m s}^{-1} \approx 6.3 \text{ m s}^{-1}$
[1]Answer
$6.3 $m \(s^{-1}\)$$
Examiner Tips and Tricks
- In 'show that' questions using energy conservation, work symbolically first (cancel m if it appears on both sides), then substitute numbers.
- Examiners want to see the algebraic reasoning.