Problem 17 - Calculus Single Variable

Question

If a spring with spring constant 8 pounds per inch is stretched 7 inches beyond its equilibrium position, then how much work is done?

Step-by-Step Solution

Answer

196 inch-pounds

Step 1: Understand the Work Formula

The work done on a spring when it is stretched or compressed 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 from the equilibrium position.

Step 2: Identify Given Values

From the problem, we know that the spring constant \(k = 8\) pounds per inch and the displacement \(x = 7\) inches. These values will be plugged into the work formula.

Step 3: Substitute Values into the Formula

Substitute \(k = 8\) and \(x = 7\) into the formula:\[ W = \frac{1}{2} \times 8 \times 7^2 \]

Step 4: Calculate the Displacement Squared

Calculate \(x^2\):\[ x^2 = 7^2 = 49 \]

Step 5: Calculate the Work Done

Perform the multiplication and division:\[ W = \frac{1}{2} \times 8 \times 49 = 4 \times 49 = 196 \]So, the work done is 196 inch-pounds.

Key Concepts

Understanding Spring Constant

The spring constant, often represented as \( k \) in physics equations, is a crucial measure of a spring's stiffness. It tells us how much force is needed to stretch or compress the spring by a unit length. In simple terms, a higher spring constant means a stiffer spring, which requires more force to stretch or compress.Springs with high spring constants are more rigid and resistant to change. It’s a fundamental concept in Hooke's Law which states that the force needed to compress or extend a spring is proportional to its displacement, given by \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement.In our example, the spring constant is 8 pounds per inch. This means for every inch we want to stretch the spring, we must apply 8 pounds of force. Understanding this helps us know how much work is involved when a spring is manipulated.

Exploring Displacement in Physics

In physics, displacement refers to the change in position of an object. It's a vector quantity, meaning it has both magnitude and direction. Displacement is concerned with how far an object has moved from its initial position, regardless of the path taken.When talking about springs, displacement \( x \) is the distance the spring is stretched or compressed from its equilibrium or natural position. This is important because it's directly used in calculating the work done on the spring.In the exercise, the spring is stretched 7 inches from its equilibrium position. This displacement tells us how much the spring has been extended and is a key part of determining the work done using the work formula. Understanding displacement helps us see how changes in position influence the energy required to perform work.

Both Work Formula and Application

The work formula is a critical tool in physics for calculating how much energy is used in performing work. For springs, the formula used is:\[ W = \frac{1}{2} k x^2 \]Where \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the displacement. This formula essentially derives from integrating the force applied over the distance the spring is stretched or compressed.To apply the formula, you simply plug in the values for the spring constant and displacement. In our example, we substitute \( k = 8 \) and \( x = 7 \) into the formula to compute:

• First, calculate \( x^2 \): which is \( 7^2 = 49 \).
• Then, multiply by the spring constant and divide by two: \( W = \frac{1}{2} \times 8 \times 49 = 196 \).

This calculation shows that the work done on the spring when stretched 7 inches is 196 inch-pounds. Understanding the components of the work formula helps connect physical stretching or compressing actions to quantifiable energy amounts.