9781118230725.pdf

go the other way; that is, we know the potential energy function U(x) and want

to find the force.

For one-dimensional motion, the work Wdone by a force that acts on a parti-

cle as the particle moves through a distance xisF(x)x. We can then write

Eq. 8-1 as

U(x)WF(x)x. (8-21)

Solving for F(x) and passing to the differential limit yield

(one-dimensional motion), (8-22)

which is the relation we sought.

We can check this result by putting , which is the elastic poten-

tial energy function for a spring force. Equation 8-22 then yields, as expected,

F(x)kx, which is Hooke’s law. Similarly, we can substitute U(x)mgx,

which is the gravitational potential energy function for a particle – Earth system,

with a particle of mass mat height xabove Earth’s surface. Equation 8-22 then

yieldsFmg, which is the gravitational force on the particle.

The Potential Energy Curve

Figure 8-9ais a plot of a potential energy function U(x) for a system in which a

particle is in one-dimensional motion while a conservative force F(x) does work

on it. We can easily find F(x) by (graphically) taking the slope of the U(x) curve at

various points. (Equation 8-22 tells us that F(x) is the negative of the slope of the

U(x) curve.) Figure 8-9bis a plot of F(x) found in this way.

Turning Points

In the absence of a nonconservative force, the mechanical energy Eof a system

has a constant value given by

U(x)K(x)Emec. (8-23)

HereK(x) is the kinetic energy functionof a particle in the system (this K(x)

gives the kinetic energy as a function of the particle’s location x). We may

rewrite Eq. 8-23 as

K(x)EmecU(x). (8-24)

Suppose that Emec(which has a constant value, remember) happens to be 5.0 J. It

would be represented in Fig. 8-9cby a horizontal line that runs through the value

5.0 J on the energy axis. (It is, in fact, shown there.)

Equation 8-24 and Fig. 8-9dtell us how to determine the kinetic energy Kfor

any location xof the particle: On the U(x) curve, find Ufor that location xand

then subtract UfromEmec. In Fig. 8-9efor example, if the particle is at any point

to the right of x 5 , then K1.0 J. The value of Kis greatest (5.0 J) when the parti-

cle is at x 2 and least (0 J) when the particle is at x 1.

SinceKcan never be negative (because v^2 is always positive), the particle can

never move to the left of x 1 , where EmecUis negative. Instead, as the particle

moves toward x 1 fromx 2 ,Kdecreases (the particle slows) until K0 at x 1 (the

particle stops there).

Note that when the particle reaches x 1 , the force on the particle, given by

Eq. 8-22, is positive (because the slope dU/dxis negative). This means that the

particle does not remain at x 1 but instead begins to move to the right, opposite its

earlier motion. Hence x 1 is a turning point,a place where K0 (because UE)

and the particle changes direction. There is no turning point (where K0) on

the right side of the graph. When the particle heads to the right, it will continue

indefinitely.

U(x)^12 kx^2

F(x)

dU(x)

dx

188 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY