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Answer:
Thus the work required to stretch the spring 9 in. beyond its natural length is 27
4 = 6.75 ft-lb.
Step-by-step explanation: If the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, how much work is needed to
stretch it 9 in. beyond its natural length?1
Hooke’s Law tells us that the force to stretch a spring x units beyond its natural length is
f(x) = kx
where k is a positive constant.2 The phrase “. . . 1 ft beyond its natural length . . . ” tells us x is changing
from 0 ft to 1 ft. We also know
W =
Z b
a
f(x) dx
Substituting the given information
W =
Z b
a
f(x) dx
12 = Z 1
0
kx dx
and since k is a constant
12 = k ·
Z 1
0
x dx
12 = k ·
x
2
2
1
0
12 = k ·
1
2
2
−
0
2
2
12 = k ·
1
2
− 0
12 =
1
2
k
24 = k
So our force function is f(x) = 24x, and, for this particular spring, we have W =
R b
a
24x dx.
The problem asks “. . . how much work is needed to stretch it 9 in. beyond its natural length?” Our integral
uses feet as the length unit, so we must recognize 9 in. = 3
4
ft. The work is given by
W =
Z 3/4
0
24x dx
=
24 ·
x
2
2
x=3/4
x=0
=
24 ·
(3/4)2
2
!
−
24 ·
(0)2
2
!
1Stewart, Calculus, Early Transcendentals, p. 459, #10.
2Note that the statement of Hooke’s Law is about stretching the spring, so x is a positive number and the force f(x) is a
positive number, thus k must be positive. If we were to compress the spring, then x is a negative number, the force f(x) is a
negative number, so k remains a positive constant.
Calculus II
Work
=
24 ·
9/16
2
− 0
= 24 ·
9
32
=
27
4