Spring Force (Hooke's Law): Videos & Practice Problems

When a force is applied to a spring, it compresses or stretches, demonstrating Hooke's Law, which states that the spring force (F_s) is equal to the negative of the applied force (F_a), or $F=-kx$, where k is the spring constant and x is the deformation. The spring force acts as a restoring force, always opposing the deformation. Understanding these principles is crucial for analyzing mass-spring systems and their motion.

Spring Force (Hooke's Law)

When interacting with a spring, the forces at play can be understood through the concept of action and reaction pairs. When you apply a force to compress or extend a spring, the spring exerts an equal but opposite force back on you. This relationship is described by Hooke's Law, which states that the spring force (\(F_s\)) is equal to the negative of the applied force (\(F_a\)), mathematically represented as:

\(F_s = -F_a = -k \cdot x\)

In this equation, \(k\) is the spring constant, a measure of the spring's stiffness, and \(x\) represents the deformation of the spring from its equilibrium position. The negative sign indicates that the spring force acts in the opposite direction to the applied force. However, when calculating magnitudes, the negative sign can be omitted, leading to the simplified equation:

\(|F_s| = k \cdot |x|\

To illustrate this, consider an example where an applied force of 120 N compresses a spring with a spring constant \(k\) of 20 N/m. Setting up the equation:

\(120 = 20 \cdot x\

Solving for \(x\) gives:

\(x = \frac{120}{20} = 6 \text{ meters}\

This indicates that the spring compresses by 6 meters from its relaxed position, also known as the equilibrium position, where \(x = 0\).

In another scenario, if you pull on a spring, the same principles apply. Suppose the spring constant \(k\) is now 40 N/m, and the spring is initially 10 meters long but is pulled to 16 meters. The deformation \(x\) is the difference in length:

\(x = 16 - 10 = 6 \text{ meters}\

Using Hooke's Law again:

\(F_s = k \cdot x = 40 \cdot 6 = 240 \text{ N}\

This demonstrates that increasing the spring constant results in a greater restoring force for the same deformation. The units for the spring constant \(k\) are expressed in newtons per meter (N/m), indicating the force required to deform the spring by one meter.

In summary, the spring force acts as a restoring force, always opposing the applied force and striving to return the system to its equilibrium position. Understanding these relationships is crucial for applying Hooke's Law effectively in various physical scenarios.

A 1.0 m-long spring is laid horizontally with one of its ends fixed. When you pull on it with 50 N, it stretches to 1.2 m. (a) What is the spring's force constant (b) How much force is needed to compress it to 0.7 m?

• A. 250 N/m; 175 N
• B. 50 N/m; 15 N
• C. 250 N/m; 75 N
• D. 50 N/m; 35 N

Answer: (C)

Acceleration of Mass-Spring Systems

When a mass is attached to a spring, it forms a mass-spring system, where the spring exerts a force in response to an applied force. The spring force, denoted as \( F_s \), acts in the opposite direction of the applied force \( F_A \). This relationship can be expressed mathematically as:

\( F_s = -F_A = -k \cdot x \)

Here, \( k \) represents the spring constant, and \( x \) is the displacement from the equilibrium position. The negative sign indicates that the spring force opposes the direction of the applied force. In a static situation where the mass is held in place, the net force is zero, leading to the equation:

\( m \cdot a = 0 \

Since the mass of the spring is negligible, the acceleration \( a \) is also zero when the applied force is balanced by the spring force. However, when the applied force is removed, the spring force becomes the only force acting on the mass, which can be described by:

\( F_s = m \cdot a \

Substituting the spring force equation gives:

\( -k \cdot x = m \cdot a \

This leads to the formula for acceleration:

\( a = -\frac{k}{m} \cdot x \

The negative sign indicates that the acceleration is directed opposite to the displacement. For example, consider a block with a mass of 0.60 kg attached to a spring with a spring constant \( k = 15 \, \text{N/m} \), stretched 0.2 meters from its equilibrium position. To find the spring force acting on the block, we use:

\( F_s = -k \cdot x = -15 \cdot 0.2 = -3 \, \text{N} \

The negative value indicates that the force acts to the left, opposing the stretch. To calculate the acceleration, we apply the derived formula:

\( a = -\frac{15}{0.6} \cdot 0.2 = -5 \, \text{m/s}^2 \

Again, the negative sign signifies that the acceleration is directed to the left, consistent with the direction of the spring force. This understanding of the mass-spring system illustrates the fundamental principles of forces, motion, and equilibrium in physics.

You push a 3-kg mass against a spring and release it from rest. Its maximum acceleration is 10m/s² when pushed back 0.5m. What is the (a)spring constant and (b) restoring force at this point?

• A. 60 N/m; 30 N
• B. 120 N/m; 45 N
• C. 100 N/m; 35 N
• D. 30 N/m; 15 N

Answer: (A)

Here’s what students ask on this topic:

What is Hooke's Law and how is it mathematically represented?

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement or deformation of the spring from its equilibrium position. Mathematically, it is represented as:

$F=-kx$

where $F$ is the spring force, $k$ is the spring constant, and $x$ is the deformation. The negative sign indicates that the force exerted by the spring is in the opposite direction of the applied force.

How do you calculate the spring constant (k) in Hooke's Law?

The spring constant $k$ can be calculated using Hooke's Law formula:

$F=-kx$

Rearranging the formula to solve for $k$ gives:

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

Here, $F$ is the force applied to the spring, and $x$ is the deformation of the spring. The units of $k$ are Newtons per meter (N/m).

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

The negative sign in Hooke's Law, $F=-kx$, indicates that the spring force is a restoring force. This means that the force exerted by the spring is always in the opposite direction of the applied force. If you compress the spring, it pushes back; if you stretch the spring, it pulls back. The negative sign ensures that the spring force always acts to return the spring to its equilibrium position.

How do you determine the deformation of a spring when a force is applied?

To determine the deformation $x$ of a spring when a force $F$ is applied, you can use Hooke's Law:

$F=-kx$

Rearranging the formula to solve for $x$ gives:

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

Here, $F$ is the applied force, and $k$ is the spring constant. The deformation $x$ is measured in meters (m).

What is a mass-spring system and how does it relate to Hooke's Law?

A mass-spring system consists of a mass attached to a spring. When the mass is displaced from its equilibrium position, the spring exerts a restoring force described by Hooke's Law:

$F=-kx$

In this system, the mass experiences an acceleration due to the spring force, which can be described by Newton's second law:

$F=ma$

Combining these equations gives:

$-kx=ma$

This relationship helps analyze the motion of the mass-spring system, including oscillations and vibrations.