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7-6 Potential energy is energy related to reversible changes in a system’s configuration 319

drops the barbell, y decreases, and so the gravitational potential energy of the system
decreases.
Potential Energy for a Curved Path

Here’s how the work-energy theorem describes what happens to a dropped isolated
barbell (just the object, not the Earth–barbell system) in the absence of air resistance: AP ® Exam Tip
As the barbell falls, the gravitational force does work on it, and this work goes into When solving problems, it
changing the barbell’s kinetic energy. is important to distinguish
Now let’s generalize to the case in which the barbell follows a curved path clearly between the potential
(as would be the case if the barbell was lifted at an angle upward, rather than energy of a system when it is
dropped). The barbell in Figure 7-18 moves along a curve from an initial y coordi- in a particular configuration
nate i y to a final coordinate .y As we did for the swinging spider in Example 7-6 (for a certain value of y, the
f
(Section 7-5), we divide the path into a large number of short segments, each of separation of Earth and an
which is small enough that we can treat it as a line. The displacement along such a object, for instance) and the

segment is d, with horizontal component ∆x and vertical component ∆.y The work change in potential energy

done along each segment by the force of gravity g F is exactly the same as it was in that comes from a change
Example 7-6 (because the barbell is moving upward and the gravitational force is in that configuration.
downward):
∆
W grav =−mgy
The total work W grav done by the force of gravity along the entire curved path is
the sum of the ∆W grav terms for each segment. The sum of all the ∆y terms is the total
y
, i
change in the y coordinate, f − y so
∆ )
∆ ) (
W grav =− ( mgy 1 + −mg y 2 +−mgy 3 +
∆ ) (
∆ (
=−mg y 1 + ∆ 2 y + ∆ 3 y + )
=−mg (y f − y ) i = −mgy f + mgy i
Barbell
We can reinterpret this statement in terms of the gravitational potential energy as ( nal)
given by Equation 7-14, U grav = mgy:
d
Earthexternal: W grav =− mgy + mgy i = − (mgy − mgy ) i q
f
f
Earthinsystem: − (U grav,f − U grav,i ) = ∆U grav (7-15) F g = –mg
In other words, the work done by gravity on an object if we choose our system to y f
be the isolated object is equal to the negative of the change in gravitational potential Barbell
energy of the Earth–object system if we choose our system to be Earth and the object. (initial)
If an object descends, we can think of it as the downward gravitational force doing
positive work on the object or as the gravitational potential energy of the Earth–object y i
system decreasing (its change is negative). If an object rises, we can describe it as the
downward gravitational force doing negative work on it or as the gravitational poten-
tial energy of the Earth–object system increasing (its change is positive). If an object Figure 7-18 Raising a barbell
begins and ends its motion at the same height, we can describe it as the gravitational along a curved path The work done
force doing zero net work on it or as there being zero net change in the Earth–object on the barbell by the gravitational

system’s gravitational potential energy. Remember, we have to define our system, so we force F g along a short segment d of
W
must choose only one of these interpretations: either Earth is in (use potential energy) this curved path is grav = g Fd cos θ.
Adding up the work done along all
or out (use work) of the system. However we define our system, we will get the same such segments gives the net work
results, as we must. We usually choose our system to make the problem we are inter- done by the gravitational force
ested in easier to solve. between height i y and height f y .
WATCH OUT !

The choice of ==y 0 for gravitational potential energy doesn’t affect what happens. AP ® Exam Tip
y
The value of U grav = mgy depends on what height you choose to be = 0. But Equa- Be sure to understand how the
tion 7-15 shows that what matters is the change in gravitational potential energy, W done by a force external to the
and that does not depend on your choice of = 0.y That’s because the change in gravita- system is related to the change
tional potential energy depends only on the difference between the initial and final in potential energy when the
heights, not the heights themselves: ∆U grav = U f − U i = mgy f − mgy i = mgy − ). interaction instead is defined
y
( f
i
Example 7-8 illustrates this important point. as internal to the system.

Uncorrected proofs have been used in this sample. Copyright © Bedford, Freeman & Worth Publishers.
Distributed by Bedford, Freeman & Worth Publishers. For review purposes only. Not for redistribution.

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