Problem 19 - Calculus Single Variable

Question

A spring is stretched \(10 \mathrm{~cm}\) beyond its equilibrium position. If the force required to maintain it in its stretched position is \(240 \mathrm{~N}\), then how much work was done?

Step-by-Step Solution

Answer

The work done is 12 J.

Step 1: Understand Hooke's Law

Hooke's Law states that the force needed to stretch or compress a spring by some distance is proportional to that distance. Mathematically, it can be expressed as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.

Step 2: Calculate the Spring Constant

Given the force \( F = 240 \text{ N} \) and the displacement \( x = 10 \text{ cm} = 0.1 \text{ m} \), use Hooke's Law to find the spring constant \( k \). The equation becomes \( 240 = k \times 0.1 \). Solve for \( k \) by dividing both sides by 0.1 to find \( k = 2400 \text{ N/m} \).

Step 3: Formula for Work Done on a Spring

The work done on a spring to stretch or compress it is given by the formula \( W = \frac{1}{2} k x^2 \). Here, \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the displacement.

Step 4: Calculate the Work Done

Substitute the values for \( k \) and \( x \) into the work formula: \( W = \frac{1}{2} \times 2400 \times (0.1)^2 \). Calculate \( (0.1)^2 = 0.01 \), then \( W = \frac{1}{2} \times 2400 \times 0.01 = 12 \text{ J} \).

Step 5: Finalize the Answer

The work done to stretch the spring 10 cm beyond its equilibrium position is \( 12 \text{ J} \).

Key Concepts

Spring Constant

The spring constant, often denoted as \( k \), is a measure of the stiffness of a spring. It tells us how much force is needed to stretch or compress the spring by a certain distance. The higher the spring constant, the stiffer the spring is, meaning more force is required to achieve the same amount of displacement.

According to Hooke's Law, the relationship between the force exerted by the spring and the displacement from its equilibrium position can be expressed mathematically as:

• \( F = kx \)

where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position. To find the spring constant, you rearrange the formula to solve for \( k \):

• \( k = \frac{F}{x} \)

In the original exercise, the spring stretches by 10 cm (or 0.1 m), and a force of 240 N is needed. Using the formula, we can calculate the spring constant as \( 240 \text{ N} = k \times 0.1 \text{ m} \), leading to \( k = 2400 \text{ N/m} \). This means the spring is relatively stiff.

Work Done on a Spring

The work done in stretching or compressing a spring refers to the energy transferred to the spring during this process. Work done on a spring can be calculated using the formula:

• \( W = \frac{1}{2} k x^2 \)

where \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the displacement. This formula indicates that the work done is dependent on both the spring constant and the square of the displacement.

In the given problem, we have already calculated the spring constant \( k \) as 2400 N/m and the displacement \( x \) as 0.1 m. Plugging these values into the work done formula, we get:

• \( W = \frac{1}{2} \times 2400 \times (0.1)^2 \)

Calculating further:

• \( (0.1)^2 = 0.01 \)
• \( W = \frac{1}{2} \times 2400 \times 0.01 = 12 \text{ J} \)

This calculation shows that 12 joules of work is required to stretch the spring 10 cm from its equilibrium position.

Force and Displacement

Force and displacement are key components in understanding how springs behave according to Hooke's Law. Hooke's Law lists a proportional relationship between the force applied to a spring and how far it is stretched or compressed, known as displacement.

The force is simply the push or pull exerted on the spring. In this context, it is often expressed in newtons (N), and is directly related to the spring’s stiffness and how much the spring is displaced. Displacement, on the other hand, is the distance from the spring's natural resting or equilibrium position to where it is stretched or compressed, measured in meters (m).

Understanding this relationship is crucial as it helps in calculating other vital parameters like the spring constant and the work done. In the exercise, the displacement is given as 10 cm (converted to 0.1 m) and the force is 240 N, allowing us to determine the stiffness of the spring by finding the spring constant \( k \) and eventually calculate how much work is done using the aforementioned formula. This interconnectedness of force, displacement, and spring constant illustrates the foundational role these concepts play in mechanical physics.