Potential energy, conservation of mechanical energy, and conservative forces
Potential energy is stored by position: mgh under gravity, ½kx² in a spring. Under conservative forces, kinetic plus potential energy stays constant.
Potential energy
Bodies can store energy because of their position (or shape). This is potential energy. Lift a mass m from a height h₁ to h₂, with g taken as constant. The work done against gravity is mg(h₂ − h₁) = mgh. It is stored as gravitational potential energy. Let the mass go and it falls. As it falls, it can do work, such as lifting another mass tied to a string over a pulley. Only the change in height matters, so the zero level of potential energy is our choice; usually it is the ground. A spring stretched or compressed by x stores elastic potential energy ½kx². When released, this changes into kinetic energy of the attached body.
Proof: mechanical energy is conserved in free fall
- 1A body of mass m is held at rest at a height h. Here U = mgh and K = 0, so the total energy E = mgh.
- 2It falls freely through a distance h₁ to a point P, which is at a height h₂ = h − h₁.
- 3At P, the potential energy U = mgh₂.
- 4Its speed at P: v² = 0 + 2gh₁, so K = ½mv² = ½m × 2gh₁ = mgh₁.
- 5★Total at P = mgh₁ + mgh₂ = mg(h₁ + h₂) = mgh.
- 6Just before it reaches the ground, U = 0 and v² = 2gh, so K = mgh. So, the total stays mgh at every point: mechanical energy is conserved.
★ key step
How to draw the energy graph for a freely falling body
- Put the distance fallen (from 0 to h) on the horizontal axis and energy on the vertical axis.
- Draw a horizontal line at mgh. Label it total energy E.
- Draw a straight line falling from mgh (at 0) to zero (at h). Label it potential energy U.
- Draw a straight line rising from zero (at 0) to mgh (at h). Label it kinetic energy K.
- Mark where the two lines cross, at h/2, with K = U = mgh/2. Write K + U = constant.
Worked example: a block on a smooth slide
A 2 kg block slides from rest down a smooth curved slide whose top is 5 m above the bottom. Find its energy at the top and its speed at the bottom. Take g = 9.8 m s⁻².
- At the top: U = mgh = 2 kg × 9.8 m s⁻² × 5 m = 98 J, and K = 0. Total energy = 98 J.
- At the bottom: U = 0, so all 98 J is kinetic energy: ½mv² = 98 J.
- v² = 2 × 98 J ÷ 2 kg = 98 m² s⁻².
- v = √98 ≈ 9.9 m s⁻¹. The shape of the smooth slide does not matter, only its height.
98 J at the top; about 9.9 m s⁻¹ at the bottom
Conservative and non-conservative forces
The total energy is still conserved. The law of conservation of energy says that the total energy of an isolated system stays constant: energy only changes form. A thermal power station turns the chemical energy of coal into electrical energy. In nuclear reactions, mass itself changes into energy, so mass and energy are conserved together.
Worked example: energy lost to friction
A 2 kg block slides from rest down a rough slope 5 m long. It drops through a height of 3 m and reaches the bottom at 6 m s⁻¹. Find the energy lost to friction and the average friction force. Take g = 9.8 m s⁻².
- Loss of potential energy = mgh = 2 × 9.8 × 3 = 58.8 J.
- Gain of kinetic energy = ½mv² = ½ × 2 × 6² = 36 J.
- Energy lost to friction = 58.8 J − 36 J = 22.8 J (it becomes heat).
- Work against friction = f × distance, so f = 22.8 J ÷ 5 m = 4.56 N.
22.8 J lost to friction; the average friction force is about 4.6 N
Hydroelectric dams such as Bhakra Nangal hold water high up; as it falls, its potential energy becomes kinetic energy that turns turbines. A child on a park slide turns height into speed, and friction turns part of it into heat.
3 marks: “Show that the total mechanical energy of a freely falling body stays constant.” At height h at rest: E = mgh. After falling h₁ (height h₂): U = mgh₂, v² = 2gh₁, K = mgh₁, so E = mgh. At the ground: K = mgh, U = 0. Total = mgh everywhere.
2 marks: “Distinguish between conservative and non-conservative forces, with one example of each.” Conservative: work independent of path and zero round a closed path; mechanical energy conserved; e.g. gravity. Non-conservative: work depends on path; mechanical energy turns into heat; e.g. friction.
3 marks: “A 0.5 kg ball is dropped from a height of 20 m. Find its potential and kinetic energy 5 m above the ground, and its speed there. (g = 9.8 m s⁻²)” Total E = 0.5 × 9.8 × 20 = 98 J. At 5 m: U = 0.5 × 9.8 × 5 = 24.5 J, so K = 98 − 24.5 = 73.5 J. v² = 2 × 73.5 ÷ 0.5 = 294, so v ≈ 17.1 m s⁻¹.
Height into speed, speed into height: mgh + ½mv² stays the same unless friction steals some as heat.