Problem 235 - Calculus Volume 1

Question

A spring has a natural length of \(10 \mathrm{~cm}\). It takes 2 \(\mathrm{J}\) to stretch the spring to \(15 \mathrm{~cm}\). How much work would it take to stretch the spring from \(15 \mathrm{~cm}\) to \(20 \mathrm{~cm} ?\)

Step-by-Step Solution

Answer

It takes 6 J to stretch the spring from 15 cm to 20 cm.

Step 1: Understand Hooke's Law

Hooke's Law explains that the force needed to extend or compress a spring by a distance \(x\) is proportional to that distance, expressed by the equation \(F = kx\), where \(k\) is the spring constant. We need to find \(k\) to proceed with calculating work done.

Step 2: Calculate the Spring Constant

First, we recognize that the work done to stretch the spring from its natural length (\(10\, \text{cm}\)) to \(15\, \text{cm}\) is \(2\, \text{J}\). The work done on a spring can also be expressed as \(W = \frac{1}{2} k x^2\). Given the work done and the extension \(x = 5\, \text{cm} = 0.05\, \text{m}\), we can solve for \(k\):\[2 = \frac{1}{2} k (0.05)^2\] \[k = \frac{2}{0.00125} = 1600\; \text{N/m}\].

Step 3: Calculate the Work from 15cm to 20cm

The work done in moving from \(15\, \text{cm}\) to \(20\, \text{cm}\) is the difference in the work done to reach \(20\, \text{cm}\) from \(10\, \text{cm}\) minus the work done to reach \(15\, \text{cm}\) from \(10\, \text{cm}\). The extensions are \(0.1\, \text{m}\) and \(0.05\, \text{m}\). \[W = \frac{1}{2}k(x_2^2 - x_1^2) = \frac{1}{2} \times 1600 \times ((0.1)^2 - (0.05)^2)\]Calculate it: \[W = 800 \times (0.01 - 0.0025) = 800 \times 0.0075 = 6\, \text{J}\]

Key Concepts

Spring Constant

The spring constant, denoted as \( k \), is a fundamental part of understanding how springs work. It tells us how stiff a spring is. The larger the spring constant, the stiffer the spring, meaning more force is required to stretch it. Hooke's Law provides the equation \( F = kx \), which shows us the force \( F \) needed to stretch or compress a spring by a distance \( x \).

The spring constant is key when calculating the potential energy stored in a spring or the work needed to change its length. In our exercise, we use the work-energy formula \( W = \frac{1}{2} kx^2 \) to calculate \( k \). We find \( k = 1600 \text{ N/m} \) by solving the equation with known values: \( W = 2 \text{ J} \) and \( x = 0.05 \text{ m} \).

• The spring constant helps determine energy changes in a spring.
• It's measured in newtons per meter (N/m).
• A higher \( k \) means a tougher spring to stretch or compress.

Work-Energy Principle

The work-energy principle is a powerful tool in physics, relating the work done by forces to changes in kinetic and potential energy. When working with springs, it highlights the relationship between the energy applied to the spring and the energy stored in it.

In the exercise, the principle helps us comprehend how work done on stretching a spring stores potential energy within it. By understanding this, we learn how energy transforms from one form to another without being lost.

We use the equation \( W = \frac{1}{2} k (x_2^2 - x_1^2) \) to describe the work from \( 15\text{cm} \) to \( 20\text{cm} \) without directly measuring force. This formula derives from the work-energy principle, highlighting how changes in position affect energy. These calculations uncover insights into real-world physics applications and challenges.

• Energy in physics is stored or transferred.
• Work done results in energy changes within a system.
• Work-energy approaches simplify complex dynamics into understandable parts.

Mechanical Work

Mechanical work measures the amount of energy transferred by a force acting over a distance. In springs, it signifies the energy needed to stretch or compress the spring. The formula \( W = f \cdot d \) becomes \( W = \frac{1}{2} k x^2 \) for springs due to the non-linear nature of spring force.

In the context of our example, mechanical work is the effort required to stretch the spring further from its rest position. Initially, \( 2 \text{ J} \) was needed to stretch it from \( 10 \text{cm} \) to \( 15 \text{cm} \). To stretch it from \( 15 \text{cm} \) to \( 20 \text{cm} \), additional work of \( 6 \text{ J} \) is calculated.

Calculating mechanical work helps students see the hidden energy transformations in seemingly simple systems.

• Mechanical work quantifies energy transfer.
• Work in springs is calculated using potential energy changes.
• Understanding work helps clarify energy transformations in physics.