Problem 20 - College Physics

Question

If a \(10-\mathrm{N}\) force is used to compress a spring with a spring constant of \(4.0 \times 10^{2} \mathrm{~N} / \mathrm{m},\) what is the resulting spring compression?

Step-by-Step Solution

Answer

The spring is compressed by 0.025 meters.

Step 1: Understand the Problem

We have a spring with a spring constant (k) of \(4.0 \times 10^{2} \text{ N/m}\) and a force (F) of \(10 \text{ N}\) applied to compress it. We need to find the spring compression (x).

Step 2: Recall Hooke's Law

Hooke's Law states that the force needed to compress or extend a spring by some distance (x) is proportional to that distance. It is expressed as \(F = kx\), where \(F\) is the force applied, \(k\) is the spring constant, and \(x\) is the displacement/compression of the spring.

Step 3: Rearrange Hooke's Law to Solve for x

We want to find the compression \(x\). Rearrange the formula \(F = kx\) to solve for \(x\). This gives us \(x = \frac{F}{k}\).

Step 4: Substitute the Values into the Formula

Plug in the given values into the equation \(x = \frac{F}{k}\). Here, \(F = 10 \text{ N}\) and \(k = 4.0 \times 10^{2} \text{ N/m}\). The equation becomes:\[x = \frac{10}{4.0 \times 10^{2}}\]

Step 5: Calculate the Compression x

Perform the division to calculate the spring compression:\[x = \frac{10}{400} = 0.025 \, \text{m}\]This means the spring is compressed by 0.025 meters.

Key Concepts

Spring Constant

The spring constant, denoted as "k", is a crucial factor in understanding how springs work. It measures a spring's stiffness—the higher the spring constant, the stiffer the spring. In scientific terms, the spring constant tells us the force needed to stretch or compress a spring by a unit length. For example, if a spring has a constant of 400 N/m, this means you would need 400 Newtons of force to compress or stretch it by 1 meter.

Here's what you should remember about the spring constant:

• Measured in Newtons per meter (N/m).
• A higher value implies a stiffer spring, while a lower value indicates a more flexible spring.
• Helps determine how much a spring will compress or stretch under a given force.

Understanding the spring constant allows you to predict how springs will behave under different forces. This makes it an essential concept when working with spring-related problems.

Force and Displacement

Hooke's Law forms the basis of the relationship between force and displacement in springs. It states that the force exerted by a spring is directly proportional to the amount it is stretched or compressed. This is expressed through the equation: \( F = kx \).
In this formula:

• \( F \) represents the force applied to the spring.
• \( k \) is the spring constant, measuring the spring's stiffness.
• \( x \) is the displacement, which is the change in length of the spring.

When a force is applied:

• A positive displacement signifies stretching.
• A negative displacement indicates compression.

Remember, the proportional relationship means that as the force increases, the displacement increases, provided the spring's elastic limit is not exceeded. This principle is crucial in fields like mechanical engineering and physics, where springs play a vital role.

Spring Compression Calculation

Calculating the compression of a spring involves rearranging Hooke's Law to solve for the displacement or compression distance (\( x \)). This can be done using the formula: \( x = \frac{F}{k} \), where \( F \) is the applied force, and \( k \) is the spring constant.

To understand the calculation:

• Identify the force \( F \) being applied to compress the spring.
• Identify the spring constant \( k \).
• Use the formula to find the compression \( x \).

Let's look at a practical example. If we apply a force of 10 N to a spring with a spring constant of 400 N/m, the compression calculation is: \[ x = \frac{10 \text{ N}}{400 \text{ N/m}} = 0.025 \text{ m} \].

This means the spring is compressed by 0.025 meters, showing that even a small amount of force can result in a noticeable change in spring length. This understanding is key to solving problems involving springs.