For one-dimensional conservative motion, the total mechanical energy is

$$E=K+U(x).$$

Since kinetic energy cannot be negative,

$$K=E-U(x)\ge0.$$

Therefore the motion is possible only where

$$\boxed{U(x)\le E}.$$

A graph of potential energy $U(x)$ together with a horizontal line at total energy $E$ can therefore reveal much of the motion without solving the differential equation.

Allowed and forbidden regions

Where the energy line lies above the potential curve,

$$E>U(x),$$

the particle has positive kinetic energy and can move.

Where

$$E<U(x),$$

the required kinetic energy would be negative, so that region is classically forbidden for a particle with that total energy.

If the particle has mass $m$, its speed at an allowed position is

$$\boxed{|v|=\sqrt{\frac{2(E-U(x))}{m}}}.$$

The vertical separation between the energy line and the potential curve therefore measures kinetic energy. The particle moves faster where that separation is larger.

Turning points

A turning point occurs where

$$E=U(x).$$

There,

$$K=0$$

and hence

$$v=0.$$

If the force then points back into the allowed region, the particle reverses direction. Turning points are therefore boundaries of a classically allowed interval.

Bound and unbound motion

If an allowed region is trapped between two turning points, the particle cannot escape that interval while its energy remains fixed. Its motion is bound.

If the allowed region extends indefinitely in some direction, the energy is sufficient for unbound motion in that direction.

A potential barrier can separate allowed regions. If the total energy lies below the top of the barrier, the particle cannot cross it in classical mechanics. If the energy exceeds the barrier, crossing becomes energetically possible.

Worked example

Consider

$$U(x)=\frac12kx^2$$

with total energy

$$E=\frac12kA^2.$$

The allowed condition is

$$\frac12kx^2\le\frac12kA^2,$$

so

$$|x|\le A.$$

The turning points are

$$x=\pm A.$$

At $x=0$ the potential is smallest, so the kinetic energy is largest:

$$K(0)=E,$$

and the speed is maximal. At $x=\pm A$, the kinetic energy and speed vanish and the particle reverses direction.

Force information from the same graph

Because

$$F(x)=-\frac{dU}{dx},$$

the slope of the potential curve determines the force direction. A downward slope as $x$ increases means $F>0$; an upward slope means $F<0$. Minima correspond to stable equilibria under the usual nondegenerate conditions, while maxima correspond to unstable equilibria.

Potential-energy diagrams thus combine energy conservation and the force-potential relation into a qualitative map of the motion: they reveal where motion is possible, where it stops and reverses, how its speed changes, and whether it is trapped or able to escape.