Problem 64 - University Physics with Modern Physics

Question

If a fish is attached to a vertical spring and slowly lowered to its equilibrium position, it is found to stretch the spring by an amount \(d\). If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum distance does it stretch the spring? (\(Hint\): Calculate the force constant of the spring in terms of the distance \(d\) and the mass \(m\) of the fish.)

Step-by-Step Solution

Answer

The spring stretches by a maximum distance of \(2d\).

Step 1: Understand Hooke's Law and Spring Constant

Hooke's Law states that the force exerted by a spring is proportional to its extension or compression: \(F = kx\), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. Here, the force is equal to the weight of the fish \(mg\), where \(m\) is the mass of the fish and \(g\) is the acceleration due to gravity. Therefore, \(kx = mg\).

Step 2: Derive the Spring Constant, k

When the fish is slowly lowered to its equilibrium position, the spring stretches by a distance \(d\), then \(kd = mg\) because the force due to the fish's weight equals the force exerted by the spring. Solving for \(k\) gives us \(k = \frac{mg}{d}\).

Step 3: Apply Energy Conservation

When the fish is allowed to fall, mechanical energy is conserved. The initial potential energy due to gravity is converted into spring potential energy at the maximum stretch of the spring. The initial potential energy is \(mgx\) and the spring potential energy is \(\frac{1}{2}kx^2\) when the spring is maximally stretched. Setting these equal, we have \(mgx = \frac{1}{2}kx^2\).

Step 4: Solve for the Maximum Distance, x

Substituting \(k = \frac{mg}{d}\) into \(mgx = \frac{1}{2}kx^2\), we get \(mgx = \frac{1}{2}\frac{mg}{d}x^2\). Simplifying, \(x = 2d\). This means the spring stretches by a maximum distance of \(2d\).

Key Concepts

Spring Constant

In the context of Hooke's Law, the spring constant, often denoted as \(k\), is a measure of a spring's stiffness. It is defined as the ratio of the force affecting the spring to the displacement caused by it. Mathematically, this relationship can be expressed using the equation \(F = kx\), where \(F\) refers to the force applied, \(k\) is the spring constant, and \(x\) is the displacement. In our scenario with the fish, the force due to the weight of the fish is equal to \(mg\), where \(m\) represents the mass and \(g\) is the gravitational acceleration.
To find the spring constant in this problem, we use the fact that when the fish reaches its equilibrium position, \(kd = mg\). From this equation, we can rearrange for \(k\):

• \(k = \frac{mg}{d}\)

This equation helps determine how much force is required to stretch the spring by a unit distance \(d\). A larger spring constant implies a stiffer spring, meaning it takes more force to stretch the spring by the same amount.

Energy Conservation

Energy conservation is a fundamental principle in physics. It states that energy cannot be created or destroyed, but it can change forms. In this problem, we apply energy conservation to the fish and spring system.
When the fish is allowed to fall freely, the gravitational potential energy it initially possesses is transformed into the spring's potential energy as the spring stretches to its maximum. This transformation is described by the equation:

• Initial gravitational potential energy: \(mgx\)
• Maximum spring potential energy: \(\frac{1}{2}kx^2\)

By setting these energies equal, we assume that no energy is lost to friction or air resistance:
\[mgx = \frac{1}{2}kx^2\]
This equivalence tells us that as the fish falls and stretches the spring, all the gravitational potential energy is converted into spring potential energy at the maximum stretch point.

Mechanical Energy

Mechanical energy in this context refers to the sum of potential and kinetic energies within the system. Here, we specifically focus on potential energies since the fish starts from rest.

• Gravitational potential energy arises from the fish's height above the lowest point and is equal to \(mgx\), where \(x\) is the distance it falls.
• Spring potential energy is the energy stored due to the deformation of the spring, expressed as \(\frac{1}{2}kx^2\).

In this exercise, mechanical energy is conserved, meaning that the total mechanical energy remains constant throughout the fish's motion. Initially, all the energy is gravitational when the fall begins. As the spring stretches, it stores this energy as spring potential energy. By knowing the initial gravitational potential energy and using energy conservation, we can calculate the maximum stretch of the spring, which is found to be \(2d\). This conservation and transformation of mechanical energy explains how the system behaves as it progresses from a state of rest to its maximum stretch.