Problem 28 - College Physics

Question

\(\bullet\) \(\bullet\) An unstretched spring has a force constant of 1200 \(\mathrm{N} / \mathrm{m}\) . How large a force and how much work are required to stretch the spring: (a) by 1.0 \(\mathrm{m}\) from its unstretched length, and (b) by 1.0 \(\mathrm{m}\) beyond the length reached in part (a)?

Step-by-Step Solution

Answer

(a) 1200 N and 600 J; (b) 2400 N and 1800 J.

Step 1: Understanding Hooke's Law

A spring force follows Hooke's Law, which states that the force needed to stretch or compress a spring by a distance \( x \) is proportional to that distance. The formula is \( F = kx \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

Step 2: Calculating Force for Part (a)

To find the force required to stretch the spring by 1.0 m from its unstretched length, use Hooke's Law: \( F = kx \). Given \( k = 1200 \, \mathrm{N/m} \) and \( x = 1.0 \, \mathrm{m} \), the calculation is \( F = 1200 \times 1.0 = 1200 \, \mathrm{N} \).

Step 3: Calculating Work for Part (a)

The work done on a spring is given by the formula \( W = \frac{1}{2} k x^2 \). With \( k = 1200 \, \mathrm{N/m} \) and \( x = 1.0 \, \mathrm{m} \), the work done is \( W = \frac{1}{2} \times 1200 \times (1.0)^2 = 600 \, \mathrm{J} \).

Step 4: Calculating Force for Part (b)

For part (b), the spring is stretched an additional 1.0 m beyond the length reached in part (a), so \( x = 2.0 \, \mathrm{m} \). Using Hooke's Law: \( F = 1200 \times 2.0 = 2400 \, \mathrm{N} \).

Step 5: Calculating Work for Part (b)

The work needed for additional 1.0 m stretch beyond part (a) can be found by subtracting the total work from the work done in part (a). The total work for 2.0 m is \( W = \frac{1}{2} \times 1200 \times (2.0)^2 = 2400 \, \mathrm{J} \). So, \( W_{\mathrm{additional}} = 2400 - 600 = 1800 \, \mathrm{J} \).

Key Concepts

Spring Force

When you stretch or compress a spring, you're engaging with the spring force, a concept rooted in Hooke's Law. This law expresses a simple yet powerful idea: the force exerted by a spring is directly proportional to the displacement from its original position. In mathematical terms, it's represented as: \( F = kx \).
Here, \( F \) is the spring force, \( k \) is the spring constant, and \( x \) is the displacement. The spring constant \( k \) is a measure of the stiffness of the spring: a higher \( k \) means a stiffer spring that needs more force to stretch or compress.
This means if you stretch a spring by 1 meter, and \( k = 1200 \, \text{N/m} \), the force required is \( 1200 \, \text{N} \). If you double the distance to 2 meters, you double the force to \( 2400 \, \text{N} \). Remember, the direction of the force always opposes the direction of the stretch or compression.

Spring Constant

The spring constant, denoted by \( k \), is a fundamental characteristic of a spring that dictates its stiffness. Quite simply, it describes how resistant a spring is to being compressed or stretched. A spring with a high \( k \) requires much more force to achieve a certain amount of stretch compared to a spring with a low \( k \).
In practical applications, like the exercise given, knowing the spring constant allows you to predict how a spring will behave under various loads. For instance, the spring constant of 1200 \( \text{N/m} \) in our problem indicates that for every meter of stretch, a 1200 Newton force is exerted by the spring. This is pivotal in designing systems that rely on springs, such as vehicle suspension systems and mechanical scales.
Understanding the spring constant helps in selecting the right spring for your needs, ensuring it neither stretches too easily nor refuses to budge under intended conditions.

Work on a Spring

Work done on a spring involves changing its length by stretching or compressing it. The work measures the energy transferred to the spring to change its position. Formulaically, the work \( W \) done on a spring is calculated as: \( W = \frac{1}{2}kx^2 \).
This equation shows that work is dependent on both the spring constant \( k \) and the square of the displacement \( x \). For example, stretching the given spring by 1 meter requires \( 600 \, \text{J} \) of work. If the displacement doubles to 2 meters, the work quadruples to \( 2400 \, \text{J} \), illustrating the quadratic relationship.
It's important to note that this work is stored as potential energy within the spring, which can later be released to perform tasks, like pushing back against the force applied. By understanding how work on a spring functions, you can better appreciate the energy dynamics involved in spring-based systems.