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7-5 The work-energy theorem is also valid for curved paths and varying forces 311

the displacement of the center of mass and the point of contact where a force is being
exerted can be different, so types of energy other than kinetic can change.

Work is the amount of energy transferred to the
system by a force exerted on a point on the system Angle between the
directions of EQUATION IN WORDS
W = Fd contact cos(q) F and d contact (7-11) Work for a system

The magnitude of the The magnitude of the displacement
force exerted on a of the point of contact where the
point on the system force is exerted on the system in
the direction of the force.

For instance, in the series of chapter-opening photos, the wall is exerting a normal
force on the ball, but the point of contact does not move, so the total work done by the AP ® Exam Tip
wall on the ball is zero. The wall transfers no energy to the ball. However, the center of A stationary object, such as a
mass of the ball does move while the wall is exerting a force on the ball, so the ball’s wall, exerting a normal force,
kinetic energy does change. cannot do work: It cannot add
When we discuss energy, we will be most focused on the changes in energy. In or remove energy. However,
Equation 7-9 we wrote out K and . i K It is easier to use the notation for changes we the normal force exerted by the
f
introduced in Chapter 2, K∆ = K − . i K Using this notation, we can now write the full stationary object on a system
f
work-energy theorem: allows the system to convert
energy from one type to another
Work is the amount of energy The change in energy of the without changing the system’s
transferred to the system by system (assuming work is the The change in all other total energy.
external forces only source of energy transfer) types of energy inside
the system due to its
con guration or the
W = ∆E = ∆K + ∆U + ∆E internal internal motion of its
constituent parts
EQUATION IN WORDS
The change in kinetic energy of the system, equal (7-12) Work-Energy Theorem
to the product of the magnitude of the force F The change in potential
exerted on the system and the magnitude of the energy of the system due
displacement of the center of mass of the system to reversible changes in
in the direction of the force. its con guration
AP ® Exam Tip
Both Equations 7-11 and 7-12 simplify our earlier equations if the displacement You are likely to need the area
of the point of contact of the force on the system is the same as the displacement of the rule on an AP® exam: Be ready to
center of mass, which is a requirement for the object model to be valid. This is true even estimate the area under a force
if the center of mass is not in the same place as the point of contact; it just requires the versus displacement graph,
system to be rigid so that all points in the system move the same distance and direction. but also be sure to distinguish
When things are rotating, we will need additional tools to solve such problems. Those that from a force versus time
tools will be provided later in the text. These equations also do not consider other ways graph (the area of which gives
of adding energy to a system, such as heating. You will get to the full expression if you impulse), which you will learn
take the next course. For now, just remember this equation doesn’t include energy added about in Chapter 9.
to or removed from the system by thermal processes so it doesn’t work in those contexts!
Work Done by a Varying Force Work done by a constant force along
direction of motion = Fd = area
Up until now, we have always assumed a constant force when calculating work done. under graph of force versus position
In Example 7-6 we saw how we might cope with a varying force, but because the ten-
sion force exerted on the spider did no work, it might not have seemed very important. F x
In many situations, however, a force of variable magnitude does do work on an object F
or a system. As an example, you must do work to stretch a spring. The force you exert
on the spring to stretch it is not constant: The farther you stretch the spring, the greater
the magnitude of the force you must exert on the spring. How can we calculate the
amount of work that you do while stretching the spring?
To see the answer let’s first consider a constant force F that is exerted on an object in x x + d x
the direction of its linear motion. Because we are working in one dimension, we are just Figure 7-13 The area rule for work
using one component, so positive and negative are enough to determine direction and Finding the work done by a constant
we don’t need arrows to denote vectors. Figure 7-13 shows a graph of this force versus force exerted on an object or a system
position as the object undergoes a displacement d. The area under the graph of force using a graph of force versus position.

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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