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308 Chapter 7 Conservation of Energy and an Introduction to Energy and Work

and so on. The only quantities that survive on the right-hand side are −K and , so

K
i
f
we are left with
net = K − i K
W
f
This is exactly the same statement of the work-energy theorem for an object as shown
.
in Equation 7-9 So the work-energy theorem is valid for any path and for any forces,
whether constant or not. This is one of the reasons why the work-energy theorem is so
important: You can apply it to situations where using forces and Newton’s laws would
be difficult or impossible.
EXAMPLE 7-6 A Swinging Spider
Figure 7-12 shows a South African kite spider ( Gasteracantha ) swinging on a

strand of spider silk. Suppose a momentary gust of wind blows on a spider of
mass 1.00 × 10 − 4 kg hanging on such a strand. As a result, the spider acquires a

horizontal velocity of 1.3 m/s when it is hanging straight down. How high will the
spider swing? Ignore air resistance. Emil von Maltitz/Getty Images
Figure 7-12 A spider swinging on silk If we know the spider’s speed at the low point of its
arc, how do we determine how high it swings?
Set Up
Because the spider follows a curved path, this would be position to be x and the vertical position to be y . Because
a very difficult problem to solve using Newton’s laws the path is a circle, we need the segments to be very
directly. Instead, we can use the work-energy theorem for tiny and to be approximated by lines. This also lets us
an object. We are given the spider’s mass and initial speed approximate the tension force as a constant over each
i v = 1.3m/s which allows us to calculate its initial kinetic segment.

,
energy. We want to find its maximum height h at the point Work-energy theorem for an
where the spider comes momentarily to rest—so its speed object:
and kinetic energy are zero—before it begins swinging

net =
back downward. W K − i K (7-9)
f
The diagram shows that only two forces are exerted Kinetic energy:
on the swinging spider: the gravitational force and the F F F T T
tension force exerted by the silk. All we have to do is K = 1 mv (7-8)
2
calculate the work done on the spider by these forces 2
along this curved path. We’ll do this by breaking the path
into a large number of segments as in Figure 7-11 We Work done by a constant force,
.
need to be able to determine the displacement (change in linear displacement: mg
position) for each segment, so let’s choose the horizontal W = Fd cos θ (7-2)

Solve
:
Rewrite the work-energy theorem for an object in terms Combine Equations 7-9 and 7-8
of the initial and final speeds of the spider and the work 1 1
done by each force. W mv − mv
net =
2
2
2 f 2 i
x x x f f
=
Net work is the sum of the y y y = h
f f f f
work done by the tension force
T and the work done by the
F
gravitational force:
x = 0
i
net =
W W grav + W T y y y = 0
i i
The final speed of the spider at the high point of its motion

v = 0, so the work-energy theorem for an object becomes
is f
1
2
grav +
W W T = − mv
i
2
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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