Problem 17 - College Physics

Question

To measure the spring constant of a certain spring, a student applies a 4.0 - \(\mathrm{N}\) force, and the spring stretches by \(5.0 \mathrm{~cm} .\) What is the spring constant?

Step-by-Step Solution

Answer

The spring constant is 80 N/m.

Step 1: Identify the given values

We are given a force, \( F = 4.0 \, \mathrm{N} \) and a spring stretch, \( \Delta x = 5.0 \, \mathrm{cm} \). Since calculations should be consistent in units, convert the stretch from centimeters to meters: \( \Delta x = 0.05 \, \mathrm{m} \).

Step 2: Understand the formula

The spring constant \( k \) can be found using Hooke's Law, which states \( F = k \Delta x \). This can be rearranged to solve for \( k \), which gives \( k = \frac{F}{\Delta x} \).

Step 3: Substitute the values into the formula

Insert the given values into the rearranged formula: \[ k = \frac{4.0 \, \mathrm{N}}{0.05 \, \mathrm{m}} \].

Step 4: Calculate the spring constant

Perform the division: \( k = \frac{4.0}{0.05} = 80 \, \mathrm{N/m} \). Thus, the spring constant is \( 80 \, \mathrm{N/m} \).

Key Concepts

Spring Constant

The spring constant, represented by the symbol \( k \), is a crucial characteristic of a spring. It measures the spring's stiffness—or, how much force is needed to stretch or compress the spring by a unit length. A higher \( k \) value indicates a stiffer spring that requires more force for the same amount of stretch. To determine \( k \), we use Hooke's Law, which relates the force applied to a spring and the distance it stretches or compresses. The law is expressed as \( F = k \Delta x \), where \( F \) is the force in newtons, and \( \Delta x \) is the change in length in meters. Rearranging this formula, we solve for the spring constant:

• \( k = \frac{F}{\Delta x} \)

By substituting the specific values for force and stretch, you can easily compute the spring constant, revealing how responsive or resistant a spring is to applied forces.

Force and Spring Stretch

When you apply a force to a spring, it stretches or compresses, depending on the direction of the force. This interaction is described by Hooke’s Law. Let's go over this concept a bit more expansively.The force you apply is measured in newtons (N). Every spring has a limit to how much it can stretch while still obeying Hooke's Law. This elastic behavior occurs within the elastic limit. Beyond this limit, the spring may deform permanently.In our example, with a 4.0 N force causing a 5.0 cm stretch, each of these values represents critical parts of Hooke’s Law formula (\( F = k \Delta x \)). You can visually imagine this by thinking of the spring as the coil of a pen or a shiny metal coil we use in gadgets. The more you pull, the longer it stretches.

• A gentle pull results in a small stretch
• A stronger force results in a larger stretch

Understanding the relationship between force and stretch is vital. It helps predict how much force you'll need for a desired amount of stretching or compressing.

Unit Conversion in Physics

In physics, maintaining consistent units throughout calculations is vital for accuracy. Sometimes you'll encounter quantities in different units and need to convert them to align with the units used in standard equations. For example, in our spring problem, the spring stretch was originally given as 5.0 cm. However, in the context of Hooke's Law, we express distance in meters to match the SI unit system. Here’s how you do it:

• Recognize that 1 meter equals 100 centimeters.
• Convert centimeters to meters by dividing by 100.
• For a 5.0 cm stretch: \( \Delta x = \frac{5.0}{100} = 0.05 \, \mathrm{m} \).

This ensures all values are compatible, and you can proceed with calculations without errors. Such conversions are not just limited to this scenario. You will often need to convert between units in many other problems to ensure the mathematical operations you perform make sense.