Page 248 - Calculus Demystified

Applications of the Integral
CHAPTER 8
You Try It: A man drags a 100 pound weight from x = 0to x = 300. He resists 235
3
a wind which at position x applies a force of magnitude F(x) = x + x + 40. How
much work does he perform?
EXAMPLE 8.15
According to Hooke’sLaw, the amount of force exerted by a spring ispropor-
tional to the distance of its displacement from the rest position.The constant
of proportionality iscalled the Hooke’s constant. A certain spring exerts a
force of 10 poundswhen stretched 1/2 foot beyond its rest state. What is
the work performed in stretching the spring from rest to 1/3 foot beyond its
rest length?

Fig. 8.25

SOLUTION
Let the x-variable denote the position of the right end of the spring (Fig. 8.25),
with x = 0 the rest position. The left end of the spring is pinned down. Imagine
that the spring is being stretched to the right. We know that the force exerted
by the spring has the form
F(x) = kx,

with k a negative constant (since the spring will pull to the left).Also F(0.5) =
−10. It follows that k =−20, so that

F(x) =−20x.
Now the work done in moving the spring from position x to position x + x
will be about (20x) · x (the sign is + since we will pull the spring to the
right—against the pull of the spring). Thus the total work done in stretching
the right end of the spring from x = 0to x = 1/3is
1/3
1/3
10
W = (20x) dx = 10x 2 = foot-pounds.
9
0 0