Spring Forces
Why this matters
Have you ever ridden in an older car down a bumpy city street, maybe in Chicago or Boston? You hit a pothole, and for a second, the car dips and then bounces back up, a little too much. That bounce is the work of the car's suspension system, which relies on big, heavy-duty springs.
Those springs are designed to absorb the shock of the road. When the wheel drops into a pothole, the spring stretches. When it hits the other side, the spring compresses. In both cases, the spring exerts a powerful force to return the car to its normal ride height. This "restoring force" isn't random; it follows a beautifully simple and predictable rule.
In this lesson, we'll unpack that rule, called Hooke's Law. You'll learn how to calculate the exact force a spring provides, a concept that's fundamental not just for passing the AP exam, but for understanding everything from screen doors to skyscrapers.
Diagram
Concept map
flowchart TD
A[Start: Analyze a spring system] --> B{Is the spring displaced from equilibrium?};
B -->|No| C[At equilibrium: Δx = 0, F_s = 0];
B -->|Yes| D{Stretched or Compressed?};
D -->|Stretched| E[Displacement Δx > 0];
E --> F[Restoring Force F_s < 0, points toward equilibrium];
D -->|Compressed| G[Displacement Δx < 0];
G --> H[Restoring Force F_s > 0, points toward equilibrium];
F --> I[End];
H --> I[End];
C --> I;
Core explanation
When we talk about forces, we've covered things like gravity and friction. Now, let's add another essential player to our toolkit: the spring force. It shows up everywhere, from the tiny springs in your retractable pen to the massive shock absorbers in a monster truck.
To make our lives easier, in AP Physics 1 we focus on ideal springs. An ideal spring has two key properties:
- It has negligible mass. We assume the spring itself is so lightweight that we don't have to worry about its own inertia or weight.
- It perfectly obeys our main rule, Hooke's Law, without getting permanently stretched out or deformed.
Equilibrium: The Spring's Happy Place
Every spring has a natural, relaxed length where it's not being stretched or compressed. We call this the equilibrium position. At this position, the spring exerts no force at all. Think of it as the spring's neutral, resting state.
When you pull the spring, you stretch it. When you push it, you compress it. In either case, you are displacing it from its equilibrium position. The spring, in turn, will exert a restoring force in an attempt to get back to that happy place.
- If you stretch a spring, it pulls back.
- If you compress a spring, it pushes back.
Notice the pattern? The force exerted by the spring is always directed back toward the equilibrium position.
Hooke's Law: Putting a Number on the Force
So, how much force does the spring exert? This is where the brilliant and simple Hooke's Law comes in. It gives us the exact relationship between the spring's properties and the force it produces.
The equation is:
F⃗_s = -kΔx⃗
This looks simple, but every single part of it is packed with meaning. Let's break it down.
-
F⃗_sis the spring force. It's a vector, meaning it has both a magnitude (how strong the force is) and a direction. This is the force the spring exerts on whatever is attached to it. -
kis the spring constant. This is a number that tells you how "stiff" or "stubborn" a spring is. A higherkvalue means the spring is very stiff and requires a lot of force to stretch or compress (like a truck's suspension). A lowerkvalue means the spring is soft and easy to stretch (like a Slinky). The units forkare newtons per meter (N/m). Akof 100 N/m means you need 100 Newtons of force to stretch that spring by 1 meter. -
Δx⃗is the displacement from equilibrium. This is the most critical part to get right. It's not the total length of the spring; it's the change in length from its relaxed, equilibrium position. It's also a vector. If the equilibrium position is atx = 0, and you stretch the spring tox = 0.1m, thenΔxis+0.1m. If you compress it tox = -0.1m, thenΔxis-0.1m. -
The Negative Sign (
-): This is where most students slip up. The negative sign is the key to the restoring force. It tells us that the spring force vectorF⃗_salways points in the opposite direction of the displacement vectorΔx⃗.
Let's visualize this with a spring on a horizontal, frictionless surface. Let's say equilibrium is at x = 0.
- 1Stretch itYou pull the attached mass to the right, to
x = +0.2m. The displacementΔx⃗is to the right. Because of the negative sign in the equation, the forceF⃗_swill be to the left, pulling the mass back towardx = 0. - 2Compress itYou push the mass to the left, to
x = -0.2m. The displacementΔx⃗is to the left. The negative sign cancels out the negative displacement (-times-is+), so the forceF⃗_swill be to the right, pushing the mass back towardx = 0.
The spring force is always a restoring force, trying to bring the system back to equilibrium. It's like an invisible hand that always pushes or pulls you back to the starting line.
Worked examples
Let's walk through a couple of common scenarios you'll see on the AP exam.
Horizontal Stretch
Problem: A toy car of mass 0.5 kg is attached to a horizontal spring with a spring constant k = 50 N/m. The spring is stretched from its equilibrium position by 10 cm. What is the magnitude and direction of the force exerted by the spring on the car?
Solution:
- 1Identify your knowns and unknowns
- Spring constant,
k = 50N/m - Displacement,
Δx = 10cm. - We need to find the spring force,
F_s.
- Spring constant,
- 2Convert unitsThis is a classic AP trap! Hooke's Law works with meters, not centimeters.
Δx = 10cm= 0.10m.
- 3Set up your coordinate systemLet's define the equilibrium position as
x = 0and the direction of the stretch (to the right) as the positive direction. So,Δx = +0.10m. - 4Apply Hooke's LawWe'll use the magnitude version first to find the strength of the force.
F_s = k * |Δx|F_s = (50 N/m) * (0.10 m)F_s = 5.0N
- 5Determine the directionNow, let's use the full vector equation
F⃗_s = -kΔx⃗.- Since we stretched the spring in the positive direction (
Δxis positive), the force must be in the negative direction. - The force is 5.0 N, directed back toward equilibrium (to the left).
- Since we stretched the spring in the positive direction (
Why this matters: Notice we didn't need the mass of the car at all. The AP exam will often give you extra information to see if you can pick out what's relevant. For calculating the spring force itself, mass is irrelevant.
Vertical Hang
Problem: Priya hangs a 2.0 kg bag of apples from a spring in the grocery store. The spring has a spring constant k = 400 N/m. How much does the spring stretch from its original length once the bag comes to rest?
Solution:
- 1Draw a free-body diagramThis is crucial for any problem involving multiple forces. The bag of apples has two forces acting on it when it's hanging at rest:
- Gravity (
F_g) pulling it down. - The spring force (
F_s) pulling it up.
- Gravity (
- 2Apply Newton's First LawSince the bag is "at rest" (in equilibrium), the net force is zero. This means the upward forces must balance the downward forces.
F_net = F_s - F_g = 0- Therefore,
F_s = F_g
- 3Calculate the force of gravity
F_g = mgF_g = (2.0 kg) * (9.8 m/s²) = 19.6N- (On the AP exam, you can often use
g = 10 m/s²for easier math if allowed, which would give 20 N. We'll use 9.8 here for precision.)
- 4Use the spring force to find the stretchWe now know the spring must be exerting 19.6 N of upward force to balance gravity. We can use the magnitude of Hooke's Law to find the
Δxthat produces this force.F_s = k * Δx19.6 N = (400 N/m) * Δx
- 5
Solve for
Δx.Δx = 19.6 N / 400 N/mΔx = 0.049m, or 4.9 cm.
Where students get stuck: The most common mistake here is to forget about gravity. They try to solve for Δx without realizing that the system is in equilibrium, where the spring force must equal the weight of the object. Always start with a free-body diagram!
Try it yourself
Ready to try one on your own?
Problem 1: A spring-loaded launcher on a pinball machine has a spring constant of k = 250 N/m. To launch the ball, Marcus compresses the spring by 5.0 cm from its rest position. What is the magnitude of the force the spring exerts on the pinball just before release?
Hint 1: Watch your units! The spring constant is in N/m. Hint 2: Does the direction of the force matter for the magnitude? The question asks for magnitude, but which way is the force pointing?
Problem 2: Aaliyah is testing the suspension of her mountain bike. She finds that when she sits on the bike, her full weight of 600 N compresses the rear shock's spring by 4.0 cm. What is the spring constant k of the shock absorber?
Hint: The bike is in equilibrium when she's sitting on it. What does that tell you about the forces acting on the spring?
Practice — 8 questions
In simple terms, spring forces are about how springs pull or push back toward their resting position when stretched or squeezed, following a predictable rule.
- 2.8.A: Describe the force exerted on an object by an ideal spring.
- 2.8.A.1
- An ideal spring has negligible mass and exerts a force that is proportional to the change in its length as measured from its relaxed length.
- 2.8.A.2
- The magnitude of the force exerted by an ideal spring on an object is given by Hooke's law: F⃗_s = -kΔx⃗
- 2.8.A.3
- The force exerted on an object by a spring is always directed toward the equilibrium position of the object–spring system.
flowchart TD
A[Start: Analyze a spring system] --> B{Is the spring displaced from equilibrium?};
B -->|No| C[At equilibrium: Δx = 0, F_s = 0];
B -->|Yes| D{Stretched or Compressed?};
D -->|Stretched| E[Displacement Δx > 0];
E --> F[Restoring Force F_s < 0, points toward equilibrium];
D -->|Compressed| G[Displacement Δx < 0];
G --> H[Restoring Force F_s > 0, points toward equilibrium];
F --> I[End];
H --> I[End];
C --> I;
Read what Saavi narrates
Hi everyone, it's Saavi. Let's talk about one of the most satisfyingly simple rules in all of physics: spring forces.
Have you ever ridden in an older car down a bumpy city street? You hit a pothole, and the car dips and then bounces back up. That bounce is the work of the car's suspension springs. They absorb the shock, and in doing so, they exert a powerful force to return the car to its normal height. This "restoring force" isn't random; it follows a beautifully simple rule called Hooke's Law.
Essentially, Hooke's Law describes the predictable way springs exert a restoring force to return to their natural, relaxed shape. The farther you pull or push a spring, the harder it fights back to get to its happy place—its equilibrium position.
Let's look at an example. Imagine Priya hangs a 2.0 kilogram bag of apples from a spring in the grocery store. The spring has a known stiffness, or spring constant, of 400 Newtons per meter. We want to know how much the spring stretches.
First, we need to realize this is an equilibrium problem. The bag is just hanging there, not moving. So, the net force on it is zero. There are two forces: gravity pulling the bag down, and the spring pulling it up. For the net force to be zero, those two forces must be equal.
The force of gravity is mass times g... so 2 kilograms times 9.8 meters per second squared gives us 19.6 Newtons. This means the spring must be pulling up with a force of 19.6 Newtons.
Now we use Hooke's Law, which says Force equals the spring constant k, times the stretch, delta x. So, 19.6 equals 400 times delta x. A little algebra... we divide 19.6 by 400... and we get a stretch of 0.049 meters, or about 5 centimeters.
The most common mistake I see every year is forgetting the negative sign in the full Hooke's Law equation, which is Force equals negative k times delta x. That negative sign is so important. It tells you the force from the spring is a restoring force—it always acts in the opposite direction of the displacement. If you stretch the spring right, the force is left. If you compress it left, the force is right. The spring is always trying to get back to where it started.
Remember that, and you'll be in great shape. You've got this.
The negative sign is not just a mathematical convention; it represents the physical reality that the spring force is a *restoring* force. It always opposes the displacement. Forgetting it will cause you to get the direction of the force wrong, which is a guaranteed point-loser on free-response questions.
Always write the full equation `F⃗_s = -kΔx⃗`. Verbally remind yourself: "The force opposes the displacement." If you pull right, the force is left.
Hooke's Law cares about the *change* from equilibrium (`Δx`), not the spring's current length. A 50 cm spring stretched to 60 cm has a `Δx` of 10 cm (0.1 m), not 60 cm.
Always identify the equilibrium (relaxed) position first. `Δx` is always measured from that specific point.
To hold a spring stretched by `Δx`, you must apply a force of `F_applied = +kΔx`. By Newton's Third Law, the spring pulls back on you with a force of `F_s = -kΔx`. They are equal in magnitude but opposite in direction. The question will be specific about which force to find.
Read the question carefully. Does it ask for the force *exerted by the spring* or the force *required to stretch* the spring?
When a spring is oriented vertically, gravity is always part of the picture. When an object hangs at rest from a spring, the spring force isn't zero; it's exactly balancing the force of gravity.
For any vertical spring problem, immediately draw a free-body diagram. Sum the forces (`F_s` and `F_g`) to find the net force.
The spring constant `k` is in Newtons per *meter*. To make the units cancel correctly, your displacement `Δx` must be in meters and your mass must be in kilograms to calculate `F_g`.
Before plugging any numbers into an equation, do a quick unit check. Convert all lengths to meters (m), masses to kilograms (kg), and forces to Newtons (N).