UY1: Potential Energy & Conservative Forces

Potential energy is not something you can define for every force. It exists cleanly only for conservative forces — the forces whose work depends only on the endpoints, not the path. This lesson shows how that idea leads to energy conservation and to “energy diagram” intuition.

At a glance

• Prerequisites: Work-Energy Theorem (UY1); Concept of Work (UY1); basic integration (Integration Techniques)
• You will learn: how to test if a force is conservative; how to define U from work; how to use energy conservation with and without non-conservative work; how to read equilibrium and turning points from U(x)
• Key result: W_c = -Δ U, Kᵢ + Uᵢ + Wₒₜₕₑᵣ = K_f + U_f
• Common trap: treating potential energy as absolute (only differences matter) or using K + U = constant even when friction/drag does work

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Setup (system + reference)

We consider a particle/system moving between two states “1” and “2”. We will:

• separate forces into conservative (can be captured by U) and other (friction, drag, driving forces),
• choose a convenient reference for U = 0 (only changes in U matter).

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1) Conservative forces (definition + tests)

A force is conservative if the work it does depends only on the endpoints: W_c(1 → 2) is path independent.

Equivalent tests:

• Closed-loop test: ∮ vector F · d vector r = 0.
• Potential-energy test: there exists a scalar U(vector r) such that vector F = -∇ U (in 1D: Fₓ = -dU/dx).

Examples:

• Conservative: gravity, ideal spring force.
• Non-conservative: kinetic friction, air resistance (drag).

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2) Potential energy from work

For a conservative force, define potential energy so that its work equals minus the change in potential energy: W_c(1 → 2) = U₁-U₂ = -Δ U.

One common definition is: U(vector r) = -∫_(vector r₀)^(vector r) vector F · d vector r + U(vector r₀).

In 1D: U(x) = -∫_x₀^xFₓ(x) dx + U(x₀).

Common potentials you will use a lot:

• Near-Earth gravity (choose y upward): U(y) = mgy + constant.
• Spring (choose U(0) = 0): U(x) = (1/2)kx².
• Newtonian gravity (radial): U(r) = -GMm/r.

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3) Energy conservation with non-conservative work

Start from the work–energy theorem: Δ K = Wₙₑₜ = W_c + Wₒₜₕₑᵣ.

For conservative forces, W_c = -Δ U, so: Δ K = -Δ U + Wₒₜₕₑᵣ. Rearrange into the common “energy accounting” form: Kᵢ + Uᵢ + Wₒₜₕₑᵣ = K_f + U_f.

Special cases:

• If Wₒₜₕₑᵣ = 0, then K + U is constant (mechanical energy conserved).
• For kinetic friction on a surface with constant fₖ, over distance d: Wₒₜₕₑᵣ = -fₖ d.

If the system is closed and Wₒₜₕₑᵣ is purely dissipative, the lost mechanical energy appears as internal energy: Δ Uᵢₙₜ = -Wₒₜₕₑᵣ.

Mini-example: energy with friction (quick stopping-distance model)

A block of mass m slides on a rough horizontal surface with coefficient μₖ. If its initial speed is vᵢ, how far does it travel before stopping?

• There is no change in potential energy ( Δ U = 0).
• Friction does negative work: Wₒₜₕₑᵣ = -μₖ mg d.

Energy accounting: Kᵢ + Wₒₜₕₑᵣ = K_f ⇒ (1/2)mvᵢ²-μₖ mg d = 0. So: d = vᵢ²/(2μₖ g).

Checks: larger μₖ gives shorter stopping distance; units are metres.

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4) Energy diagrams & equilibrium (1D intuition)

In 1D, the force is the negative slope of the potential-energy curve: Fₓ(x) = -dU/dx.

Equilibrium points satisfy Fₓ = 0, i.e. dU/dx = 0.

• Stable equilibrium: U(x) has a local minimum ( d²U/dx² > 0).
• Unstable equilibrium: U(x) has a local maximum ( d²U/dx² < 0).
• Neutral equilibrium: U is constant over a region.

Turning points occur where the kinetic energy is zero, so E = K + U satisfies E = U(x).

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Worked example

Example: turning points in a spring potential

Given: a mass m = 0.50 kg in 1D has potential energy U(x) = (1/2)kx² with k = 4.0 N m⁻¹. Total mechanical energy is E = 2.0 J.

Find: the turning points and the maximum speed.

Working

Turning points occur when K = 0, so U = E: (1/2)kx² = E ⇒ x = ± square root of (2E/k) = ± 1.0 m.

Maximum speed occurs where U is minimum (here at x = 0), so K = E: (1/2)mvₘₐₓ² = E ⇒ vₘₐₓ = square root of (2E/m) = square root of (4.0/0.50) ≈ 2.83 m s⁻¹.

Answer: turning points at x = ± 1.0 m; vₘₐₓ ≈ 2.83 m s⁻¹.

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Practice set (with hints + answers)

1. Is kinetic friction conservative? Use the closed-loop test to justify your answer (1–2 sentences).
2. A spring force is F = -kx with k = 10 N m⁻¹ and U(0) = 0. Find U(x) and evaluate U(0.30 m).
3. A 1D potential is U(x) = ax² with a = 3 J m⁻². Find F(x) and state whether x = 0 is stable or unstable.
4. A 2.0 kg block with speed 5.0 m s⁻¹ slides on a rough horizontal surface with μₖ = 0.20 for 3.0 m. Find its speed after 3.0 m and the increase in internal energy (assume it all goes to heat).
5. In 1D motion with total energy E, what condition on U(x) must hold for the particle to be able to reach position x?

Hints

1. Consider going out and back to the starting point.
2. Use U(x) = -∫₀^x F dx (or F = -dU/dx).
3. Use F = -dU/dx and look at the curvature of U(x) near 0.
4. Use Wₒₜₕₑᵣ = -μₖ mgd and Kᵢ + Wₒₜₕₑᵣ = K_f (since U is unchanged).
5. Motion is allowed only where kinetic energy is non-negative.

Answers

1. No; ∮ vector F · d vector r ≠ 0 for friction (work depends on path length).
2. U(x) = (1/2)kx²; U(0.30) = 0.45 J.
3. F(x) = -2ax = -6x (N); x = 0 is stable (minimum of U).
4. Wₒₜₕₑᵣ = -μ mgd = -11.76 J; v ≈ 3.64 m s⁻¹; Δ Uᵢₙₜ = 11.76 J.
5. E ≥ U(x).

Summary + next steps

• Conservative forces have path-independent work; equivalently ∮ vector F · d vector r = 0.
• For conservative forces, define potential energy so W_c = -Δ U.
• In 1D, F = -dU/dx (force is minus the slope of U).
• With non-conservative work: Kᵢ + Uᵢ + Wₒₜₕₑᵣ = K_f + U_f.
• Energy diagrams: minima are stable equilibria; turning points satisfy E = U.

Next: Linear Momentum, Impulse & Collisions Previous: Work-Energy Theorem Back To Mechanics (UY1)