Conservation of energy
Work, Energy & Power - OCR A-Level Physics
Principle of Conservation of Energy
Energy cannot be created or destroyed. It can only be transferred between stores and objects. The total energy of a closed systemA group of objects that has no external forces acting on it from outside the system, so no energy enters or leaves. stays constant.
Case 1 : Falling object, no resistive forces
When an object falls through height $h$ with negligible air resistance, the loss of GPE equals the gain in KE.
$$mgh = \tfrac{1}{2}mv^{2}$$
Case 2 : Sliding down a slope with friction
An object slides a distance $x$ down a slope, falling a vertical height $h$. Friction force $F$ acts up the slope. Energy conservation gives loss of GPE = gain of KE + work done against friction.
$$mgh = \tfrac{1}{2}mv^{2} + Fx$$
- For a frictionless drop, mass cancels: $v = \sqrt{2gh}$. The speed at the bottom depends only on the height and $g$, not the mass.
- A pendulum swings between maximum GPE at the top of its arc (where $v = 0$) and maximum KE at the bottom (where $h = 0$, taking the lowest point as the reference).
- Work done against resistive forces (friction, air resistance, drag) is transferred to the thermal store of the object and surroundings. Energy is conserved overall, but useful energy is lost.
- Real systems are never 100% efficient because some energy is always dissipated to thermal stores during a transfer.
Common Mistake
MEDIUM
Wrong: Equating GPE loss to KE gain in a problem that includes friction or drag, then being surprised when the speed comes out larger than the measured value.
Right: Always check whether the question says "smooth", "frictionless" or "negligible air resistance". If it doesn't, include a $W_{\text{friction}} = Fx$ term so that $mgh = \tfrac{1}{2}mv^{2} + Fx$.
Right: Always check whether the question says "smooth", "frictionless" or "negligible air resistance". If it doesn't, include a $W_{\text{friction}} = Fx$ term so that $mgh = \tfrac{1}{2}mv^{2} + Fx$.
Worked Example [4 marks]
A $0.50 \text{ kg}$ ball is released from rest and rolls down a frictionless slope, dropping through a vertical height of $2.0 \text{ m}$. Calculate its speed at the bottom. Use $g = 9.81 \text{ m s}^{-2}$.
Show Solution
1
Apply conservation of energy (no friction)
Loss in GPE = gain in KE, so $mgh = \tfrac{1}{2}mv^{2}$.
[1]2
Mass cancels: $gh = \tfrac{1}{2}v^{2}$, so $v^{2} = 2gh$.
[1]3
$v^{2} = 2 \times 9.81 \times 2.0 = 39.24 \text{ m}^{2} \text{ s}^{-2}$.
[1]4
$v = \sqrt{39.24} = 6.26 \text{ m s}^{-1} \approx 6.3 \text{ m s}^{-1}$.
[1]Answer
$6.3 \text{ m s}^{-1}$
Examiner Tips and Tricks
- In "show that" questions, work symbolically first. Cancel the mass if it appears on both sides before substituting numbers.
- If you're unsure whether acceleration is constant, do not use the suvat equations. Use conservation of energy instead.