Problem 53 - College Physics

Question

\(\bullet\) \(\bullet\) A 12.0 N package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is supported from below by a vertical spring of force con- stant 325 \(\mathrm{N} / \mathrm{m}\) . If the pan has negligible weight, find the maxi- mum distance the spring will be compressed if no energy is dissipated by friction.

Step-by-Step Solution

Answer

The spring compresses approximately 3.69 cm.

Step 1: Understand the problem

We are asked to find the maximum distance the spring will be compressed when a 12.0 N package is placed on it. The spring has a force constant of 325 N/m, and no energy is lost to friction.

Step 2: Identify the forces involved

There are two main forces: the weight of the package exerting a force of 12.0 N downwards, and the spring force acting upwards with a force constant of 325 N/m.

Step 3: Apply Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to the distance it is compressed or stretched, expressed as \( F = kx \). Here, \( F = 12.0\, \mathrm{N} \) (the weight of the flour) and \( k = 325\, \mathrm{N/m} \).

Step 4: Solve for displacement

Rearrange Hooke's Law to solve for the displacement \( x \): \( x = \frac{F}{k} \). Substitute the values into the equation: \( x = \frac{12.0}{325} \approx 0.0369\, \mathrm{m} \).

Step 5: Interpret the result

The result of 0.0369 m describes the maximum compression of the spring in meters, which is approximately 3.69 cm.

Key Concepts

Understanding the Spring Constant

The spring constant, often symbolized as \(k\), is a crucial concept in understanding how springs behave. It measures the stiffness of a spring, determining how much force is needed to compress or stretch the spring by a given distance. The larger the spring constant, the stiffer the spring is.
For example, in our exercise, the spring has a constant of 325 N/m. This means that for each meter of compression, 325 Newtons of force is required. Likewise, if you only need a smaller compression, say a fraction of a meter, proportionally less force is needed. Think of the spring constant as a way to characterize the spring's resistance to being changed from its natural length.
Understanding this will help you grasp how different springs react to forces placed on them. This allows us to calculate how much a spring will compress or extend under a given load. In practical situations, knowing the value of the spring constant allows us to predict these changes precisely.

The Force Displacement Relationship

The core of Hooke's Law is the relationship between force and displacement in springs. Hooke’s Law is expressed by the formula: \[ F = kx \]where \(F\) is the force applied onto the spring, \(k\) is the spring constant, and \(x\) is the displacement from the spring’s equilibrium position. This equation tells us that the force needed to compress or extend a spring is linearly proportional to the distance \(x\) that the spring is compressed or stretched.
In the context of our exercise, the 12.0 N force (from the package of flour) leads to a displacement. Using the spring constant, we can calculate how far the spring compresses:

• Substitute \(F = 12.0\, \mathrm{N}\)
• Substitute \(k = 325\, \mathrm{N/m}\)
• Use the formula \(x = \frac{F}{k}\)

Carrying out the calculation gives us \(x \approx 0.0369\, \mathrm{m}\), meaning the spring compresses about 3.69 cm. This illustrates how the spring force can be predicted from the given weight of the object and the spring’s properties.

Energy Conservation in Springs

Energy conservation is a fundamental concept when dealing with springs. In this context, it involves understanding how the energy stored in the spring transforms.
When a spring is compressed or stretched, it stores potential energy. This stored energy can be calculated using the formula:\[ U = \frac{1}{2}kx^2 \]where \(U\) is the potential energy stored in the spring, \(k\) is the spring constant, and \(x\) is the displacement.
In our scenario, as the package compresses the spring, energy transfers from gravitational potential energy to the spring as elastic potential energy. If there is no friction or energy loss, the energy remains within this system, conserving the total mechanical energy.
For the maximal point the spring can compress, all the potential energy from the weight of the package is transferred into the spring. By equating the gravitational energy to this spring energy, one could solve for exact displacements or forces involved, further illustrating how Hooke's Law and energy principles collaborate to describe motion in spring systems.