Unit 4 · Topic 08

Energy diagrams

An energy diagram compares U(x) with mechanical energy E. The difference E-U determines kinetic energy and reveals accessible regions, turning points, barriers, and speed changes.

Unit 4Energy diagramsOpen navigation

Concept 01

The condition U≤E

Essential The minimum you should retain

Because K=E−U K=E-U cannot be negative, classical motion is possible only where U(x)≤E.

UnderstandInterpret and connect

A region with U>E is inaccessible at that mechanical energy in the conservative model.

DeepenFormulation and conditions

The E line sets an energy limit and the U curve determines available configurations.

ExploreConnections for further study

Changing the zero shifts U and E together without changing E-U or allowed regions.

Mathematical relation

Force and motion from U(x)

Fx=−dUdx,K=E−U,v=2(E−U)m F_x=-\frac{dU}{dx},\quad K=E-U,\quad v=\sqrt{\frac{2(E-U)}m}
Represents

Conservative force as negative slope and allowed speed in a 1D energy diagram.

Physical interpretation

Force points toward lower U and only U≤E U\le E is allowed.

DeepenVariables, conditions, and checks

Variables

F_x
conservative force; usual unit: N
U
potential energy; usual unit: J
E
total mechanical energy; usual unit: J
v
speed; usual unit: m/s

Conditions of application

  • One-dimensional conservative system.
  • The speed expression requires E≥U.

Dimensional check

J/m=N; the square root gives m/s.

Errors it helps prevent

  • Using F=dU/dx.
  • Allowing U>E with real K.

K is the E-U separation

A potential parabola intersects an energy line at two turning points.Eturning pointstable minimumturning point
  • U(x)=2x²

Only U≤E U\le E is allowed; U=E marks turning points and the minimum of U is stable equilibrium.

Concept 02

Where speed vanishes

Essential The minimum you should retain

At a turning point U=E and K=0.

UnderstandInterpret and connect

If the particle reaches that point, its motion reverses under the potential force.

DeepenFormulation and conditions

Turning points bound allowed intervals for confined motion.

ExploreConnections for further study

A turning point need not have F=0; the slope of U may be nonzero.

Concept 03

Vertical separation represents K

Essential The minimum you should retain

K(x)=E-U(x).

UnderstandInterpret and connect

A larger vertical separation between E and U means greater kinetic energy and speed.

DeepenFormulation and conditions

v(x)=2[E−U(x)]m v(x)=\sqrt{2[E-U(x)]/m} is physically defined only when E≥U.

ExploreConnections for further study

The height of U alone is not speed; its difference from E matters.

Worked example

Turning points in a parabolic potential

A particle moves with U(x)=2x² J and mechanical energy E=18 J.

Given
  • U(x)=2x² J
  • E=18 J
Target

Find the turning points and where speed is greatest.

  1. Turning points

    K=0 requires U=E.

  2. Calculation

    2x²=18 gives x=±3 m.

  3. Speed

    K=E−U K=E-U is greatest where U is smallest.

  4. Interpretation

    U is smallest at x=0, where speed is greatest.

Conclusion

Turning points are x=-3 m and x=+3 m; maximum speed occurs at x=0.

Concept 04

Reading the global shape of U

Essential The minimum you should retain

Minima of U identify stable equilibria and maxima identify unstable equilibria.

UnderstandInterpret and connect

A barrier higher than E separates regions the particle cannot classically connect.

DeepenFormulation and conditions

Increasing E can open new allowed intervals or overcome a barrier.

ExploreConnections for further study

The same potential can produce trapped oscillation, turning, or transit depending on E.

Concept review

Common errors

Each warning includes a concrete way to review the reasoning, not only an incorrect-answer marker.

Reading the height U(x) as speed.

Speed depends on K=E−U K=E-U , the vertical separation between E and U.

Allowing a particle to cross U>E at the same E.

That would require K<0; the region is inaccessible in the classical conservative model.