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7-5 The work-energy theorem is also valid for curved paths and varying forces 307

A weight lifter doing bicep An object’s curved path from an
curls makes the dumbbell initial point i to a nal point f. We
move through a curved mark a large number N of equally
path. The force that he spaced intermediate points 1, 2, 3,
exerts on the dumbbell ..., N along the path and imagine
varies during its travel. breaking the path into short
segments (from i to 1, from 1 to 2,
(a) (b) from 2 to 3, ...).

f
N
N–1

4

3
2
i 1

Figure 7-11 Motion along a curved path (a) The dumbbell travels in a curved path during the
bicep curl. (b) Analyzing motion along a general curved path.

the forces are not constant (Figure 7-11a) and can be extended to describe systems as
well as objects.
To see that the work-energy theorem for an object is valid for curved paths and
varying forces, consider Figure 7-11b, which shows an object’s curved path from an ini-
tial point i to a final point f. We mark a large number N of equally spaced intermediate
points 1, 2, 3,…, N along the path and imagine breaking the path into short segments
(from i to 1, from 1 to 2, from 2 to 3, and so on). If each segment is sufficiently short,
we can treat it as a line. Furthermore, the forces exerted on the object don’t change
very much during the brief time the object traverses one of these segments, so we can
treat the forces as constant over that segment. If we really want to be sure, we could
make the segments even shorter. It would not change this analysis, it would just get
very tedious! The forces exerted on the object may be different from one segment to
the next, but all that matters for this analysis is that the segments are small enough that
the forces exerted on the object are relatively constant over each segment. Because each
segment is a line with constant forces, we can safely apply the work- energy theorem
for an object in Equation 7-9 to every segment:

W net,ito1 = K 1 − i K
W net,1to2 = K 2 − K 1
W net,2to3 = K 3 − K 2

W net,N to f = K − K N
f
Now add all of these equations together:

W net,ito1 + W net,1to2 + W net,2to3 + + W net,N to f (7-10)
K
+
+
+
= ( 1 − i K ) (K − K 1 ) (K − K 2 ) + (K − K N )
f
3
2
The left-hand side of Equation 7-10 is the sum of the amounts of work done by the
net force on the object as it moves through each segment. This sum equals the net
work done by the net force along the entire path from i to f, which we call simply
W net . Remember that although force is a vector, work is a scalar so these quantities
simply add algebraically! On the right-hand side of the equation, −K in the second
1
term cancels K in the first term, −K in the third term cancels K in the second term,
2
1
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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