9781118230725.pdf

8-3 READING A POTENTIAL ENERGY CURVE 189

6 5 4 3 2 1

U(J)

x x
(a) x 1 2 x 3 x 4 x 5

U(x)

+

• 
  

x

(b)

x 1 x 2 x 3 x 4 x 5

F (N)

Mild force, –x direction

Strong force, +xdirection

This is a plot of the potential

energyU versus position x.

Force is equal to the negative of

the slope of the U(x) plot.

6 5 4 3 2 1

U (J),Emec (J)

x x

(c) x 1 2 x 3 x 4 x 5

Emec = 5.0J

U(x)

The flat line shows a given value of

the total mechanical energy Emec.

The difference between the total energy

and the potential energy is the

kinetic energy K.

6 5 4 3 2 1

U (J),Emec (J)

x x

(d) x 1 2 x 3 x 4 x 5

Emec = 5.0J

U(x)

K 6 5 4 3 2 1 x

(f) x 1 x 2 x 3 x 4 x 5

U (J),Emec (J)

At this position, K is greatest and

the particle is moving the fastest.

At this position, K is zero (a turning point).

The particle cannot go farther to the left.

For either of these three choices for Emec,

the particle is trapped (cannot escape

left or right).

6 5 4 3 2 1

U (J),Emec (J)

x x

(e) x 1 2 x 3 x 4 x 5

Emec = 5.0J

K = 1.0 J at x > x 5

K = 5.0 J at x 2

A

Figure 8-9(a) A plot of U(x), the potential energy function of a system containing a particle confined to move along an xaxis. There is no
friction, so mechanical energy is conserved. (b) A plot of the force F(x) acting on the particle, derived from the potential energy plot by
taking its slope at various points. (c)–(e) How to determine the kinetic energy. (f) The U(x) plot of (a) with three possible values of Emec
shown.InWileyPLUS, this figure is available as an animation with voiceover.