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Nonconservative Forces & Energy Accounting

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The test is path independence

A force is conservative if the work it does between two points is the same along every path — equivalently, if the work around any closed loop is zero. Gravity and spring forces pass this test, which is exactly why a potential energy function can be defined for them: U depends only on position. Friction and drag fail it: dragging a block around a loop back to its start does negative work the whole way, so no potential energy function exists for friction, and its effect must be tracked separately.

The generalized work–energy statement
K₁ + U₁ + W_applied = K₂ + U₂ + |ΔE_thermal| with ΔE_thermal = f_k d for friction
The distance d in f_k d is the actual path length traveled, not the displacement. That distinction matters whenever an object slides back and forth.

Energy is not lost, it is relocated

Saying friction "destroys" energy is loose and leads to errors. The energy is converted to thermal energy in the surfaces and eventually radiated away — the total is conserved, but it has left the mechanical account and cannot be recovered as organized motion. This is why the useful formulation is a balance sheet rather than a conservation statement: write what the system had, add what was put in, subtract what was dissipated, and the remainder is what it has.

Path length, not displacement

Because friction always opposes motion, the energy it dissipates is f_k times the total distance traveled along the surface. A block that slides 3 m forward and 3 m back dissipates f_k(6 m) even though its displacement is zero. This is the single most common error in multi-stage energy problems, and it is easy to catch: any time the object reverses direction, add the path segments rather than subtracting them.

Worked example

A 2.0 kg block is launched at 6.0 m/s along a horizontal surface with μ_k = 0.25, and slides to rest. Find the distance it travels and confirm the energy balance.

  1. 1.Initial kinetic energy: K = ½(2.0)(6.0)² = ½(2.0)(36) = 36 J.
  2. 2.Friction force: f_k = μ_k mg = 0.25(2.0)(9.8) = 4.9 N.
  3. 3.All kinetic energy is dissipated: 36 J = f_k d = 4.9d, so d = 36/4.9 ≈ 7.3 m.
  4. 4.Check with kinematics: a = −μ_k g = −2.45 m/s², and v² = v₀² + 2ad gives 0 = 36 − 4.9d, the same equation.
Answer: The block travels about 7.3 m; the 36 J of kinetic energy becomes 36 J of thermal energy in the block and surface
Checkpoint

A block slides down a rough incline, then back up to its starting height is impossible. The best explanation is that —

On the exam

Free-response energy questions are graded on the accounting, not the arithmetic. Write an explicit initial-equals-final statement naming every term before substituting numbers. A correct number arrived at without a stated energy equation routinely loses the setup points; a stated equation with a small numerical slip usually keeps most of them.

Checkpoint

The energy dissipated by kinetic friction is calculated using —

Answer the 2 checkpoints as you read.

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