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7-5 The work-energy theorem is also valid for curved paths and varying forces 309

To calculate the work done by each force along the f y
spider’s curved path, break the path up into segments so
short that each one can be considered as a line tangent to x

the path. On each segment the spider has a displacement d

with horizontal component x∆ and vertical component d ∆y
∆ . y These are different for each individual segment. In the

figure, we exaggerated the length of the segment to make i
it easier to draw. ∆x

F
On each segment of the path the tension force T Work done by the tension force as the spider moves
points radially inward, perpendicular to the path, and through a short segment of its path:
because of our very short segment approximation,

T,segment =
perpendicular to the displacement .d Hence θ = 90° in W F T d cos90° = 0

Equation 7-2 so the tension force does no work during
,
this displacement.

The gravitational force g F always points straight Work done by the gravitational force as the spider moves
downward and has constant magnitude g F = mg . through a short segment of its path:

We apply Equation 7-2 to the displacement and
grav,segment =
gravitational force drawn as shown. We are also going W g F d cos θ

φ
to use the identity cos(90°+ ) φ =− sin . Don’t be scared
by this, even if you haven’t seen it before. (If you look
at graphs of the cosine and sine of an angle versus the
angle you can see that if you add 90° to any angle the

cosine of the sum is the same as the negative of the sine
of the angle!)
Because the angle between the displacement and the The angle between the F F F T T

gravitational force is greater than 90°, the gravitational gravitational force and the d

force does negative work. Our trig identity gave us this displacement is θ = 90° + φ so f ∆y 5 d sin f
sign, so the equation we came up with looks correct.
W grav,segment = g F d cos(90° + ) φ
u590° 1 f
=− g Fd sin φ
F F F g g
The figure shows
sin φ =∆ y, so
d
∆
grav,segment =−
W g F y∆= − mgy
Now we can calculate the net work done on the spider by The tension force does zero work on each short segment,
each force as it moves along its curved path. This is just so when you add them all up
the sum of the individual bits of work done on it along the
T =
short segments. W 0
Work done by the gravitational force over the entire path:
+−
+
∆
∆
−
( mgy
W grav =− ∆ ) ( mg y ) ( mgy +
)
1 2 3
( y + ∆
=− mg ∆ 1 2 y + ∆ 3 y + )

The sum of all the ∆y terms is the net vertical displacement

W
y − i y = h − 0 = . h So grav =− mgh .

f

W

Substitute the expressions for T and grav into the work- Work-energy theorem for an object:
W

energy theorem for an object and solve for h Note that
.
we didn’t need the length of the silk for this problem. W W T = − mgh + 0 =− 1 mv
grav +
2
2 i
v 2 (1.3 m/s) 2
h = i = = 0.086 m
2
2g 2(9.80 m/s)
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.
08_stewart3e_33228_ch07_284_333_8pp.indd 309 20/08/22 8:45 AM