Intro to Springs (Hooke's Law) - Online Tutor, Practice Problems & Exam Prep

Springs exhibit restoring forces, as described by Hooke's Law, which states that the force exerted by a spring (F_S) is proportional to its deformation (x): F=-kx, where k is the spring constant. The deformation can be compression or extension, and k indicates the stiffness of the spring. In vertical spring systems, equilibrium occurs when the spring force equals the weight of the mass (mg), leading to the equation kx = mg .

Intro to Springs

Hey, guys. So in this video, I want to start talking about springs and spring forces. Let's check it out. So when you push or pull against the spring with a force, let's call this force F_a an applied force, the spring pushes back. This is just Newton's 3rd law, action reaction. You push against something, and it pushes back against you. So the force of the spring will be the negative of the force that was applied on it, again, showcasing Newton's action reaction. That's why it's the same magnitude but in opposite direction.

This negative here simply signifies the opposite direction. What is new is that this also equals to −kx and I'll explain what k and x are in just a moment. Notice that there's another negative here, and that negative just means opposite direction. That negative, as you'll learn more about later, is almost a conceptual reminder that the force is in the opposite direction. In many calculations, we might disregard the negatives to simplify matters.

So, what are k and x? I'll start with x. X represents the deformation that the spring will experience, meaning if you push or pull against the spring, it will either be compressed (becoming shorter) or it will be stretched (becoming longer). However, x is not the actual length but the change in length. In terms of springs, it's not about how long it is, but how much longer or shorter it gets. X can be thought of as the absolute value of the final length minus the initial length. The reason it's taken as an absolute value is that the sign of x (negative or positive) doesn't impact its interpretation since it only shows the magnitude of change.

For example, if we have a spring attached to a wall, and the spring is uncompressed, its original length, with x equals 0, represents the relaxed position of the spring. In another scenario, portrayed here with the first drawing, I apply a force F_a to the right, resulting in the spring pushing back to the left with a force F_s. The difference between its original length and its new length is called its deformation, labeled as x or x₁ for distinct situations.

K represents the spring's force constant, which essentially quantifies how stiff the spring is or how challenging it is to compress or stretch. The higher the k, the harder it is to deform the spring. K is typically measured in newtons per meter, as evident from the equation F = kx, where F is in newtons and x is in meters, rendering k as newtons per meter for dimensional consistency.

The force of a spring is fundamentally a restoring force, always attempting to bring the spring back to its original, unaltered length. If stretched or compressed, the spring exerts a force in the direction of returning to this equilibrium state.

Let's consider a practical example to clarify these concepts. Imagine a spring 1 meter long laid horizontally with one end fixed. If you exert a force of 50 newtons resulting in the spring stretching to a length of 1.2 meters, the change in spring length, x, is 0.2 meters. From this, if you wanted to determine the spring's k, knowing that the force and deformation are as stated, you could rearrange the formula to find k = 250 newtons per meter.

Now, addressing vertical springs, when a mass is attached to a spring and allowed to stretch by its weight, the spring will elongate until reaching a point of equilibrium where the upward spring force equals the downward gravitational force of the mass, described by the equation kx = mg. This balance signifies that the spring has reached a state where its force exactly counters the gravitational pull, maintaining the mass in a steady position.

In summary, understanding the mechanics of springs through the spring constant k and the deformation x, alongside recognizing the relationship between forces at play, forms the cornerstone of analyzing spring dynamics accurately. Let's remember this as we proceed with solving related problems in physics.

A vertical spring is originally 60 cm long. When you attach a 5 kg object to it, the spring stretches to 70 cm. 

(a) Find the force constant on the spring.
(b) You now attach an additional 10 kg to the spring. Find its new length.

(For the multiple choice selection, answer only part (b). Use g=10 m/s².)

• A. (b) L_new=80 cm
• B. (b) L_new=90 cm
• C. (b) L_new=100 cm
• D. (b) L_new=110 cm

Here’s what students ask on this topic:

What is Hooke's Law and how is it applied to springs?

Hooke's Law states that the force exerted by a spring is directly proportional to its deformation (either compression or extension). The law is mathematically expressed as:

$F=-kx$

where $F$ is the force exerted by the spring, $k$ is the spring constant (a measure of the spring's stiffness), and $x$ is the deformation of the spring from its equilibrium position. The negative sign indicates that the force exerted by the spring is in the opposite direction of the deformation. This law is fundamental in understanding how springs behave under various forces and is widely used in physics and engineering applications.

How do you calculate the spring constant (k) of a spring?

The spring constant $k$ can be calculated using Hooke's Law, which is given by:

$F=-kx$

Rearranging the equation to solve for $k$, we get:

$k=\frac{F}{x}$

Here, $F$ is the force applied to the spring, and $x$ is the deformation of the spring. By measuring the force applied and the resulting deformation, you can calculate the spring constant. The units of $k$ are typically Newtons per meter (N/m).

What is the significance of the negative sign in Hooke's Law?

The negative sign in Hooke's Law, expressed as $F=-kx$, indicates that the force exerted by the spring is in the opposite direction of the deformation. This is a reflection of Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. In the context of springs, if you compress the spring (shorten it), the spring exerts a force to expand back to its original length. Conversely, if you stretch the spring (lengthen it), the spring exerts a force to compress back to its original length. The negative sign is crucial for understanding the restoring nature of spring forces.

How do you determine the deformation (x) of a spring?

The deformation $x$ of a spring is the change in length from its relaxed (equilibrium) position. It can be calculated as:

$x=|L{f}_{}-L{i}_{}|$

where $L{f}_{}$ is the final length of the spring and $L{i}_{}$ is the initial length of the spring. The absolute value is taken to ensure that deformation is always a positive quantity, regardless of whether the spring is compressed or stretched. This change in length is what is used in Hooke's Law to determine the force exerted by the spring.

What is the equilibrium condition for a vertical spring with a mass attached?

For a vertical spring with a mass attached, equilibrium occurs when the spring force equals the gravitational force acting on the mass. This can be expressed as:

$kx=mg$

where $k$ is the spring constant, $x$ is the deformation of the spring, $m$ is the mass, and $g$ is the acceleration due to gravity. At equilibrium, the upward force exerted by the spring balances the downward gravitational force, resulting in no net force and no further movement. This condition is crucial for understanding how springs behave in vertical systems and is often used in problems involving vertical spring-mass setups.