Problem 55 - College Physics

Question

\(\bullet\) An unstretched spring is 12.00 \(\mathrm{cm}\) long. When you hang an 875 g weight from it, it stretches to a length of 14.40 \(\mathrm{cm}\) . (a) What is the force constant (in \(\mathrm{N} / \mathrm{m}\) ) of this spring? (b) What total mass must you hang from the spring to stretch it to a total length of 17.72 \(\mathrm{cm} ?\)

Step-by-Step Solution

Answer

(a) 357.66 N/m; (b) 2.086 kg mass needed.

Step 1: Identify Given Information

We start by noting the initial and final lengths of the spring as well as the mass hanging from it. Initially, the spring is 12.00 cm long. With an 875 g weight, it stretches to 14.40 cm. Convert the mass to kilograms to work with SI units: 875 g = 0.875 kg. The change in length (extension) of the spring is calculated as 14.40 cm - 12.00 cm = 2.40 cm = 0.024 m.

Step 2: Calculate the Spring Constant (a)

The spring constant \( k \) can be calculated using Hooke's Law: \( F = kx \), where \( F \) is the force in the spring (equal to the weight of the mass), and \( x \) is the extension of the spring. The force \( F \) can be determined by \( F = mg \), where \( m = 0.875 \) kg and \( g \approx 9.81 \) m/s². Compute \( F = (0.875)(9.81) = 8.58375 \) N. Then, \( k = \frac{F}{x} = \frac{8.58375}{0.024} \approx 357.66 \) N/m.

Step 3: Determine Required Mass for New Length (b)

To find the mass that stretches the spring to 17.72 cm, first calculate the new extension: 17.72 cm - 12.00 cm = 5.72 cm = 0.0572 m. Using Hooke's Law \( F = kx \), solve for the force \( F = kx = 357.66 \times 0.0572 \approx 20.4659 \) N. The weight causing this force is \( mg = 20.4659 \) N, solve for \( m \) by rearranging to \( m = \frac{F}{g} = \frac{20.4659}{9.81} \approx 2.086 \) kg.

Key Concepts

Hooke's Law

When you place a weight at the end of a spring, it stretches proportionally to the weight applied. This relationship is defined by Hooke's Law. The formula is simple: \( F = kx \). Here, \( F \) is the force applied to the spring, \( k \) represents the spring constant, and \( x \) is the extension or compression of the spring. This law helps us express how the spring deforms when different weights are hung from it.
With Hooke's Law in mind, it is easier to calculate the spring constant. Knowing how much force causes a certain stretch lets us determine the stiffness of the spring, which brings us to the next important concept: the spring constant.

Spring Constant

The spring constant \( k \) is a measure of a spring's stiffness. A larger value of \( k \) means a stiffer spring. It's found by using the formula from Hooke's Law: \( k = \frac{F}{x} \). In our exercise, we find \( k \) by calculating the force (due to weight) applied to the spring and dividing it by the spring's extension.
We use the formula \( F = mg \) to find the force, where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. Once we have force \( F \), calculating \( k \) becomes straightforward by dividing this force by the spring's stretch \( x \). Knowing \( k \) is essential, especially for understanding how different masses affect the spring.

Force and Mass Conversion

When working with mechanics problems, converting between force and mass is crucial. Most often, we're given mass, and we need to find the force. The force exerted by a mass in Earth's gravitational field can be calculated using the equation \( F = mg \), where \( g \approx 9.81 \, \text{m/s}^2 \) is the standard gravitational acceleration on Earth.
For instance, in the exercise, the given mass is converted into force using this formula. By doing so, we're able to seamlessly integrate this force into Hooke's Law to calculate the spring constant. It is important to always consider these conversions to maintain consistency in calculations.

Unit Conversion

In physics, consistent units are a must to avoid mistakes. Often problems deal with different unit systems, so proper conversion is necessary. For instance, lengths given in centimeters should be converted to meters by dividing by 100, as the base unit for length in the International System of Units (SI) is the meter.
Similarly, masses are often provided in grams and need conversion to kilograms by dividing by 1000, which is the SI base unit for mass. In our exercise, such conversions ensure that all values align properly when calculating forces and extensions using Hooke's Law. Ensuring that all units are compatible is an absolutely critical step in problem-solving.

Problem-solving Steps

Solving problems systematically is the key to finding the correct solution efficiently. It begins with identifying all given information and understanding what is being asked. This was our first step in reaching a solution: collecting lengths and masses, then converting them into suitable SI units.
Next, leveraging fundamental formulas like Hooke's Law to connect the given information with what needs to be calculated, such as the spring constant. We can then carry forward these steps to solve for other variables; for example, determining the mass required for a new spring length. Meticulous step-by-step execution often leads to understanding and solving more complex problems.