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312 Chapter 7 Conservation of Energy and an Introduction to Energy and Work

versus position equals the area of the colored rectangle in Figure 7-13 ; that is, the height

Area under curve = work you do of the rectangle (the force F ) multiplied by its width (the displacement d ). But the area

to stretch spring from x 1 to x 2
Fd is just equal to the work done by the constant force (notice that the units for the area
F are correct for work, Nm ). So on a graph of force versus position, the work done by the

kx 2 force equals the area under the graph. If the area extends below the x axis, the product
is negative. This area rule is a mathematical fact we used in Section 2-4 to find displace-
ment from a graph of velocity versus time. It works for any product, when you plot the
kx 1
multiplicand and multiplier on the axes of a graph; work is just the newest example.
,
Let’s apply this area rule to the work that you do in stretching a spring. In Section 5-6
x we wrote the relationship for an ideal spring, Hooke’s law as s F =− kd ( Equation 5-10 ).
,

x 1 x 2 Remember that Equation 5-10 gives the force that the spring exerts on you as you stretch
Figure 7-14 Applying the area rule it. By Newton’s third law, the force that you exert on the spring has the same magnitude but
to an ideal spring We can use the is exerted in the opposite direction. We wrote this as Equation 5-11 : youonspring,x =+ kd .

F
same technique as in Figure 7-13 to find

the work required to stretch a spring. Figure 7-14 graphs the force that you exert as a function of the distance that
the spring has been stretched. Each position can be thought of as an incredibly tiny
displacement, over which the force is a constant, just like we did for the forces on the

spider in Example 7-6 . If the spring is initially stretched to 1 x from its equilibrium
position, which we call x = 0, and you stretch it further to 2 the work W that you do

x

,

WATCH OUT ! is equal to the area of the colored trapezoid in Figure 7-14
.
From geometry, this area is equal to the average height of the graph multiplied by
Who’s doing the work? the width 2 x − . 1 x Using Equation 5-11 we find
,

Note that Equation 7-13 tells  (force youexert at x + (forceyou exertat x 
)
)
us the work that you do on the W =  1 2  ( 2 x − 1 x )
spring Because this is a contact  2 
.
force, the work that the spring = (kx + kx 2 ) ( 2 x − ) = 1 kx )( 2 x − )
1
( 1 +
does on you is equal to the 2 1 x 2 2 x 1 x
negative of Equation 7-13 .
We can simplify this to
Work that must be done on a spring Spring constant of the spring
to x = x (a measure of its stiffness)
to stretch it from x = x 1 2
EQUATION IN WORDS 1 1
2
Work done to stretch a spring ( 7-13 ) W = kx − kx 2
2 2 2 1
x = initial stretch of the spring
1
x = nal stretch of the spring
2
x and x are measured from
2
1
x = 0, the location of the end
of the spring when it is relaxed
Example 7-7 shows how to use Equation 7-13 to attack a problem that would
have been impossible to solve with the force techniques from Chapters 4 and 5 .
EXAMPLE 7-7 Work Those Muscles!
An athlete stretches a set of exercise cords 47 cm from their unstretched length. The cords behave like a spring with a spring
.

constant of 86N/m (a) How much force does the athlete exert to hold the cords in this stretched position? (b) How much

work did he do to stretch them? (c) The athlete loses his grip on the cords. If the mass of the handle is 0.25 kg, how fast is
it moving when it hits the wall to which the other end of the cords is attached? You can ignore gravity and assume that the
cords themselves have a negligible mass.
Set Up relaxed
In part (a) we’ll use Equation 5-11 to find the F =+kd (5-11)

athleteoncords,x
force that the athlete exerts, and in part (b) 1 1 47 cm
2
2
Equation 7-13 will tell us the work that he does W = kx − kx (7-13) stretched

2
1
on the cords. In part (c) we’ll see how much work 2 2
net =
the cords do on the handle as they go from being W K − i K (7-9)

f
stretched to relaxed. This work goes into the 1 released
2
kinetic energy of the handle. We’re ignoring the K = mv (7-8)
=
mass of the cords and therefore assuming that 2 v v v = ?
f f f f
they have no kinetic energy of their own.
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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