we need to use positive and negative signs carefully for the direction
of the force.
In general, suppose that forces are acting on a particle — we
can think of them as coming from other objects that are “off stage”
— and that the interaction between the particle and the off-stage
objects can be characterized by an interaction energy, U, which
depends only on the particle’s position,x. Using the kinetic energy
theorem, we have dK=Fdx. (It’s not necessary to writeKcm, since
a particle can’t have any other kind of kinetic energy.) Conservation
of energy tells us dK+ dU = 0, so the relationship between force
and interaction energy is dU=−Fdx, or
F=−
dU
dx[relationship between force and interaction energy].Force exerted by a spring example 40
.A mass is attached to the end of a spring, and the energy of the
spring isU= (1/2)k x^2 , wherexis the position of the mass, and
x = 0 is defined to be the equilibrium position. What is the force
the spring exerts on the mass? Interpret the sign of the result.
.Differentiating, we findF=−dU
dx
=−k x.Ifxis positive, then the force is negative, i.e., it acts so as to bring
the mass back to equilibrium, and similarly forx <0 we have
F>0.
Most books do theF=−k xform before theU= (1/2)k x^2 form,
and call it Hooke’s law. Neither form is really more fundamental
than the other — we can always get from one to the other by
integrating or differentiating.
Newton’s law of gravity example 41
.Given the equationU=−Gm 1 m 2 /rfor the energy of gravita-
tional interactions, find the corresponding equation for the grav-
itational force on massm 2. Interpret the positive and negative
signs.
.We have to be a little careful here, because we’ve been taking
rto be positive by definition, whereas the position,x, of massm 2
could be positive or negative, depending on which side ofm 1 it’s
on.
For positivex, we haver=x, and differentiation givesF=−dU
dx
=−Gm 1 m 2 /x^2.Section 3.2 Force in one dimension 173