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160 CHAPTER 7 KINETIC ENERGY AND WORK

state. For this common arrangement, we can write Eq. 7-20 as

Fxkx (Hooke’s law), (7-21)

where we have changed the subscript. If xis positive (the spring is stretched

toward the right on the xaxis), then Fxis negative (it is a pull toward the left). If

xis negative (the spring is compressed toward the left), then Fxis positive (it is a

push toward the right). Note that a spring force is a variable forcebecause it is a

function of x, the position of the free end. Thus Fxcan be symbolized as F(x). Also

note that Hooke’s law is a linearrelationship between Fxandx.

The Work Done by a Spring Force

To find the work done by the spring force as the block in Fig. 7-10amoves, let us

make two simplifying assumptions about the spring. (1) It is massless;that is, its

mass is negligible relative to the block’s mass. (2) It is an ideal spring;that is, it

obeys Hooke’s law exactly. Let us also assume that the contact between the block

and the floor is frictionless and that the block is particle-like.

We give the block a rightward jerk to get it moving and then leave it alone.

As the block moves rightward, the spring force Fxdoes work on the block,

decreasing the kinetic energy and slowing the block. However, we cannotfind this

work by using Eq. 7-7 (WFdcosf) because there is no one value of Fto plug

into that equation—the value of Fincreases as the block stretches the spring.

There is a neat way around this problem. (1) We break up the block’s dis-

placement into tiny segments that are so small that we can neglect the variation

inFin each segment. (2) Then in each segment, the force has (approximately) a

single value and thus we canuse Eq. 7-7 to find the work in that segment. (3)

Then we add up the work results for all the segments to get the total work. Well,

that is our intent, but we don’t really want to spend the next several days adding

up a great many results and, besides, they would be only approximations. Instead,

let’s make the segments infinitesimalso that the error in each work result goes to

zero. And then let’s add up all the results by integration instead of by hand.

Through the ease of calculus, we can do all this in minutes instead of days.

Let the block’s initial position be xiand its later position be xf. Then divide

the distance between those two positions into many segments, each of tiny length

x. Label these segments, starting from xi, as segments 1, 2, and so on. As the

block moves through a segment, the spring force hardly varies because the seg-

ment is so short that xhardly varies. Thus, we can approximate the force magni-

tude as being constant within the segment. Label these magnitudes as Fx 1 in

segment 1,Fx 2 in segment 2, and so on.

With the force now constant in each segment, we canfind the work done

within each segment by using Eq. 7-7. Here f 180 , and so cos f1. Then

the work done is Fx 1 xin segment 1,Fx 2 xin segment 2, and so on. The net

workWsdone by the spring, from xitoxf, is the sum of all these works:

(7-22)

wherejlabels the segments. In the limit as xgoes to zero, Eq. 7-22 becomes

(7-23)

From Eq. 7-21, the force magnitude Fxiskx. Thus, substitution leads to

(^12 k)[x^2 ]xxif( 21 k)(xf^2 xi^2 ). (7-24)

Ws

xf

xi

kx dxk

xf

xi

xdx

Ws

xf

xi

Fxdx.

WsFxjx,