We next want to examine the work done on a particle-like object by a particular type of variable force—namely, a spring force, the force from a spring. Many forces in nature have the same mathematical form as the spring force. Thus, by examining this one force, you can gain an understanding of many others.
The Spring Force
Figure 7-11a shows a spring in its relaxed state—that is, neither compressed nor extended. One end is fixed, and a particle-like object—a block, say—is attached to the other, free end. If we stretch the spring by pulling the block to the right as in Fig. 7-11b, the spring pulls on the block toward the left. (Because a spring force acts to restore the relaxed state, it is sometimes said to be a restoring force.) If we compress the spring by pushing the block to the left as in Fig. 7-11c, the spring now pushes on the block toward the right.
To a good approximation for many springs, the force 
from a spring is proportional to the displacement 
of the free end from its position when the spring is in the relaxed state. The spring force is given by
which is known as Hooke’s law after Robert Hooke, an English scientist of the late 1600s. The minus sign in Eq. 7-20 indicates that the direction of the spring force is always opposite the direction of the displacement of the spring’s free end. The constant k is called the spring constant (or force constant) and is a measure of the stiffness of the spring. The larger k is, the stiffer the spring; that is, the larger k is, the stronger the spring’s pull or push for a given displacement. The SI unit for k is the newton per meter.
In Fig. 7-11 an x axis has been placed parallel to the length of the spring, with the origin (x = 0) at the position of the free end when the spring is in its relaxed state. For this common arrangement, we can write Eq. 7-20 as
where we have changed the subscript. If x is positive (the spring is stretched toward the right on the x axis), then Fx is negative (it is a pull toward the left). If x is negative (the spring is compressed toward the left), then Fx is positive (it is a push toward the right).
Note that a spring force is a variable force because it is a function of x, the position of the free end. Thus Fx can be symbolized as F(x). Also note that Hooke’s law is a linear relationship between Fx and x.
Fig. 7-11 (a) A spring in its relaxed state. The origin of an x axis has been placed at the end of the spring that is attached to a block. (b) The block is displaced by , and the spring is stretched by a positive amount x. Note the restoring force exerted by the spring. (c) The spring is compressed by a negative amount x. Again, note the restoring force.
The Work Done by a Spring Force
To find the work done by the spring force as the block in Fig. 7-11a moves, 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 Fx does work on the block, decreasing the kinetic energy and slowing the block. However, we cannot find this work by using Eq. 7-7 (W = Fd cos 
) because that equation assumes a constant force. The spring force is a variable force.
To find the work done by the spring, we use calculus. Let the block’s initial position be xi and its later position xf. Then divide the distance between those two positions into many segments, each of tiny length Δx. 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 segment is so short that x hardly varies. Thus, we can approximate the force magnitude as being constant within the segment. Label these magnitudes as Fx1 in segment 1, Fx2 in segment 2, and so on.
With the force now constant in each segment, we can find the work done within each segment by using Eq. 7-7. Here = 180°, and so cos = −1. Then the work done is −Fx1 Δx in segment 1, −Fx2 Δx in segment 2, and so on. The net work Ws done by the spring, from xi to xf, is the sum of all these works:
where j labels the segments. In the limit as Δx goes to zero, Eq. 7-22 becomes
From Eq. 7-21, the force magnitude Fx is kx. Thus, substitution leads to
Multiplied out, this yields
This work Ws done by the spring force can have a positive or negative value, depending on whether the net transfer of energy is to or from the block as the block moves from xi to xf. Caution: The final position xf appears in the second term on the right side of Eq. 7-25. Therefore, Eq. 7-25 tells us:
Work Ws is positive if the block ends up closer to the relaxed position (x = 0) than it was initially. It is negative if the block ends up farther away from x = 0. It is zero if the block ends up at the same distance from x = 0.
If xi = 0 and if we call the final position x, then Eq. 7-25 becomes
The Work Done by an Applied Force
Now suppose that we displace the block along the x axis while continuing to apply a force 
to it. During the displacement, our applied force does work Wa on the block while the spring force does work Ws. By Eq. 7-10, the change ΔK in the kinetic energy of the block due to these two energy transfers is
in which Kf is the kinetic energy at the end of the displacement and Ki is that at the start of the displacement. If the block is stationary before and after the displacement, then Kf and Ki are both zero and Eq. 7-27 reduces to
If a block that is attached to a spring is stationary before and after a displacement, then the work done on it by the applied force displacing it is the negative of the work done on it by the spring force.
Caution: If the block is not stationary before and after the displacement, then this statement is not true.
CHECK POINT 4 For three situations, the initial and final positions, respectively, along the x axis for the block in Fig. 7-11 are (a) −3 cm, 2 cm; (b) 2 cm, 3 cm; and (c) −2 cm, 2 cm. In each situation, is the work done by the spring force on the block positive, negative, or zero?
A package of spicy Cajun pralines lies on a frictionless floor, attached to the free end of a spring in the arrangement of Fig. 7-11a. A rightward applied force of magnitude Fa = 4.9 N would be needed to hold the package at x1 = 12 mm.
(a) How much work does the spring force do on the package if the package is pulled rightward from x0 = 0 to x2 = 17 mm?
Solution: A Key Idea here is that as the package moves from one position to another, the spring force does work on it as given by Eq. 7-25 or Eq. 7-26. We know that the initial position xi is 0 and the final position xf is 17 mm, but we do not know the spring constant k.
We can probably find k with Eq. 7-21 (Hooke’s law), but we need a second Key Idea to use it: Were the package held stationary at x1 = 12 mm, the spring force would have to balance the applied force (according to Newton’s second law). Thus, the spring force F would have to be −4.9 N (toward the left in Fig. 7-11b); so Eq. 7-21 (Fx = −kx) gives us
Now, with the package at x2 = 17 mm, Eq. 7-26 yields
(b) Next, the package is moved leftward to x3 = −12 mm. How much work does the spring force do on the package during this displacement? Explain the sign of this work.
Solution: The Key Idea here is the first one we noted in part (a). Now xi = +17 mm and xf = −12 mm, and Eq. 7-25 yields
This work done on the block by the spring force is positive because the spring force does more positive work as the block moves from xi = +17 mm to the spring’s relaxed position than it does negative work as the block moves from the spring’s relaxed position to x = −12 mm.
In Fig. 7-12, a cumin canister of mass m = 0.40 kg slides across a horizontal frictionless counter with speed v = 0.50 m/s. It then runs into and compresses a spring of spring constant k = 750 N/m. When the canister is momentarily stopped by the spring, by what distance d is the spring compressed?
Solution: There are three Key Ideas here:
1. The work Ws done on the canister by the spring force is related to the requested distance d by Eq. 7-26 (Ws = −
kx2), with d replacing x.
2. The work Ws is also related to the kinetic energy of the canister by Eq. 7-10 (Kf − Ki = W).
3. The canister’s kinetic energy has an initial value of K = mv2 and a value of zero when the canister is momentarily at rest.
Putting the first two of these ideas together, we write the work–kinetic energy theorem for the canister as
Fig. 7-12 A canister of mass m moves at velocity 
toward a spring that has spring constant k.
Substituting according to the third idea makes this expression
Simplifying, solving for d, and substituting known data then give us
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