2.8 Spring Forces

An ideal spring exerts a force proportional to how far it is stretched or compressed from its relaxed length, described by Hooke's law $\vec{F}_s = -k\Delta\vec{x}$. This force is a restoring force, always pointing back toward the equilibrium position of the object-spring system.

Spring Force Examples in AP Physics 1

For AP Physics 1, spring force examples usually use Hooke's law: $\vec{F}_s=-k\Delta\vec{x}$. The spring constant $k$ tells you stiffness, $\Delta x$ is measured from the relaxed length, and the negative sign means the force points back toward equilibrium.

Common examples include a horizontal spring pulling an object back after being stretched, a compressed spring pushing outward, and a vertical spring where a hanging mass stretches the spring until $k\Delta x=mg$ at equilibrium.

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Why This Matters for the AP Physics 1 Exam

Spring forces give you a clean example of a force that changes with position, which sets up later work on energy storage and oscillations. On the exam, you may need to translate between a verbal description of a spring, a force vs displacement graph, an equation, and a free-body diagram. That kind of translation between representations is exactly the reasoning the test rewards, and it shows up in both multiple-choice questions and free-response explanations where you justify the direction and size of a force.

You will also combine spring forces with other ideas from this unit, like setting the spring force equal to gravity for a hanging mass, or including a spring force in a free-body diagram and applying Newton's second law.

Key Takeaways

• An ideal spring has negligible mass and obeys a linear force-displacement relationship.
• Hooke's law in vector form is $\vec{F}_s = -k\Delta\vec{x}$; the magnitude is $|\vec{F}_s| = k|\Delta x|$.
• The spring constant $k$ measures stiffness and has units of N/m. A larger $k$ means a stiffer spring.
• $\Delta x$ is measured from the spring's relaxed (natural) length, not from some random starting point.
• The spring force is a restoring force that always points toward the equilibrium position of the object-spring system.
• For a vertical spring with a hanging mass, the equilibrium position is not the relaxed length, because gravity stretches the spring until the spring force balances the weight.

Force of an Ideal Spring

An ideal spring is a model that simplifies real springs so you can analyze them with one equation. Real springs are never perfect, but many behave closely enough to the model for the math to work.

Characteristics of an Ideal Spring

The mass of an ideal spring is treated as negligible compared to the objects attached to it.

• You can ignore the spring's own weight in calculations.
• You focus only on the interaction between the spring and the attached object.

An ideal spring follows a linear force-displacement relationship.

• The force is directly proportional to how far the spring is stretched or compressed.
• Stretch it twice as far and it pulls back with twice the force.
• This behavior stays consistent each time you use the spring.

The relaxed length is the spring's natural length when it is neither stretched nor compressed. The equilibrium position of an attached object is where the net force on the object is zero. These are not always the same point, especially for a vertical spring where gravity also acts.

Hooke's Law

Hooke's law describes how the spring force depends on the change in length. In vector form:

$\vec{F}_s = -k\Delta \vec{x}$

In one dimension:

$F_s = -k\Delta x$

Where:

• $F_s$ is the force exerted by the spring (in newtons, N)
• $k$ is the spring constant, a measure of stiffness (in N/m)
• $\Delta x$ is the change in length measured from the relaxed length (in meters)
• The negative sign shows the force points opposite the displacement from equilibrium

A spring with a large $k$ is stiffer and needs more force to stretch or compress by the same amount. A spring with $k = 100$ N/m exerts twice the force of a spring with $k = 50$ N/m for the same stretch.

On a force vs displacement graph for an ideal spring, the points fall on a straight line through the origin, and the slope equals the spring constant $k$. The force is zero at the relaxed length.

Direction of the Spring Force

The spring force is always a restoring force, directed toward the equilibrium position of the object-spring system.

When a spring is stretched:

• The change in length $\Delta x$ is positive.
• The spring force is negative, pulling back toward the relaxed length.

When a spring is compressed:

• The change in length $\Delta x$ is negative.
• The spring force is positive, pushing back out toward the relaxed length.

Because the force always pushes or pulls back toward equilibrium, a displaced object on a spring keeps getting pulled back, which is why spring systems oscillate. Energy storage and oscillation come up in later units, but the restoring behavior starts here.

How to Use This on the AP Physics 1 Exam

Problem Solving

• Identify $k$ and $\Delta x$ first, and confirm $\Delta x$ is measured from the relaxed length.
• Use $|\vec{F}_s| = k|\Delta x|$ for magnitude, then decide direction by asking which way points back toward equilibrium.
• For a hanging mass at rest, set the spring force equal to the weight: $k\Delta x = mg$. This lets you solve for $k$, the stretch, or the mass.
• Watch your units. $k$ in N/m times $\Delta x$ in meters gives force in N.

Free Response

• When asked to justify direction, state that the spring force is a restoring force pointing toward equilibrium, then connect that to whether the spring is stretched or compressed.
• If you draw a free-body diagram, show the spring force as a single straight arrow from the dot, pointing toward equilibrium.
• If you get a force vs displacement graph, the slope is $k$. Explain that the straight line through the origin shows the proportional relationship.

Common Trap

• Do not measure $\Delta x$ from the floor, the ceiling, or the equilibrium position of a hanging mass. Measure it from the relaxed length.

Practice Problem 1: Spring Force Calculation

Solution

Apply Hooke's law: $F_s = -k\Delta x$

Given:

• Spring constant $k = 25$ N/m
• Change in length $\Delta x = 0.15$ m (positive because the spring is stretched)

Substituting: $F_s = -k\Delta x = -(25 \text{ N/m})(0.15 \text{ m}) = -3.75 \text{ N}$

The negative sign means the force points opposite the displacement. Since the spring is stretched in the positive direction, the force acts in the negative direction, pulling the spring back toward its relaxed length.

The spring exerts a force of 3.75 N directed toward the equilibrium position.

Practice Problem 2: Finding the Spring Constant

Solution

At equilibrium, the spring force balances the weight of the mass.

• Weight: $F_g = mg = (2 \text{ kg})(9.8 \text{ m/s}^2) = 19.6 \text{ N}$
• At rest, the spring force equals the weight: $k\Delta x = F_g$

So: $19.6 \text{ N} = k(0.08 \text{ m})$

Solving for $k$: $k = \frac{19.6 \text{ N}}{0.08 \text{ m}} = 245 \text{ N/m}$

The spring constant is 245 N/m.

Common Misconceptions

• The negative sign in Hooke's law does not mean the force is always negative. It means the force always points back toward equilibrium, opposite the displacement.
• $\Delta x$ is the change from the relaxed length, not the total length of the spring.
• The relaxed length and the equilibrium position are not always the same. For a vertical hanging mass, gravity stretches the spring, so equilibrium is below the relaxed length.
• A larger spring constant means a stiffer spring, not a longer or weaker one.
• The spring force is not constant. It changes as the displacement changes, which is what makes it different from a steady force like gravity near Earth's surface.
• An ideal spring has negligible mass, so do not include the spring's own weight in your force analysis.

Related AP Physics 1 Guides

• 2.1 Systems and Center of Mass
• 2.2 Forces and Free-Body Diagrams
• 2.3 Newton's Third Law
• 2.4 Newton's First Law
• 2.5 Newton's Second Law
• 2.6 Gravitational Force