Distinguish allowed positions from reachable motion

The logarithmic threshold-time argument below is optional mathematical enrichment. Core energy reasoning uses potential, kinetic energy, turning points and allowed regions; the improper-integral proof is not presented as a required AP exam formula.

Convert an energy statement into a position constraint

For a conservative one-dimensional mass, K=E−U(x) must be nonnegative. Thus U(x)≤E identifies allowed positions. It does not specify the sign of velocity or elapsed time. For mass 2 kg in the supplied units, speed magnitude is sqrt(E−U), but its sign must come from an initial condition and subsequent turning behavior.

Solve both turning branches

The assigned double-well potential is U=(x²−1)² J with numerical x in metres and dimensional coefficients understood. Setting U=E gives x²=1±sqrt(E). The outer roots ±sqrt(1+sqrt(E)) always exist. Inner roots ±sqrt(1−sqrt(E)) exist only for E≤1. When E is below 1, the allowed intervals around the two minima are disconnected.

Use the starting component

The workbench chooses the initial well minimum x=−1 or +1. For E below the barrier, the object remains in that well in the conservative model. It cannot jump to the other interval merely because both satisfy U≤E. Which turning point it reaches first still depends on the initial velocity direction, which the controls deliberately do not provide.

Examine the threshold without inventing a crossing

At E=1 the allowed set touches at the unstable center. Near x=0, K=2x²−x⁴ and speed approaches sqrt(2)|x|. The time integral of dx/|x| diverges, so an approach from a well reaches the center only asymptotically in this ideal model. Equality with barrier energy is not a finite-time passage. For E>1, speed at the center is positive and passage is permitted.

Challenge the model boundary

Turning points have zero speed but can have nonzero acceleration; the sign of the potential slope determines the subsequent direction. At the stationary barrier, both speed and force can be zero, requiring the special analysis above. Dissipation, a time-dependent drive or a perturbation changes the energy argument and must be stated rather than hidden inside an energy plot.

Worked original case

At E=.25 J, inner and outer turning magnitudes are sqrt(.5) and sqrt(1.5) m. Starting in the left well selects only the negative interval. At E=2 J the center speed is 1 m/s.

Investigation

Compare E=.5,1 and 1.5 at the same starting well. Sketch allowed components, state whether finite-time crossing is possible, and identify the missing initial-direction information. Retain the threshold case rather than rounding it into either side.

Public numerical transfer practice

Try a new practice case

For mass 2 kg and E=2 J in this potential, what is the speed at x=0 in m/s?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original teaching, values and calculations, checked 30 October 2026 against the official calculus-based Mechanics framework and OpenStax work and kinetic energy. The numerical packets are supplied or generated, not measured apparatus records. No physical launch, machinery, loading or field intervention is requested. Public feedback and virtual captures do not establish independent mastery or supervised laboratory participation.