Problem 19 - Applied Physics

Question

A spring having a force constant of \(1.25 \mathrm{~N} / \mathrm{cm}\) is stretched through a distance of \(11.5 \mathrm{~cm} .\) How much work is required to stretch the spring?

Step-by-Step Solution

Answer

The work required to stretch the spring is approximately 0.828 J.

Step 1: Identify the Known Variables

We are given the force constant of the spring, \( k = 1.25 \; \text{N/cm} \), and the amount the spring is stretched, \( x = 11.5 \; \text{cm} \).

Step 2: Convert Units

To ensure consistent units, convert the force constant from \( \text{N/cm} \) to \( \text{N/m} \) and the displacement \( x \) from \( \text{cm} \) to \( \text{m} \). \[ k = 1.25 \; \text{N/cm} = 1.25 \times 100 = 125 \; \text{N/m}\] \\[x = 11.5 \; \text{cm} = 0.115 \; \text{m}\]

Step 3: Use the Work Done by a Spring Formula

The work done to stretch or compress a spring from its equilibrium position is given by the formula \[W = \frac{1}{2} k x^2\]where \( W \) is the work done in joules, \( k \) is the spring constant in N/m, and \( x \) is the displacement in meters.

Step 4: Calculate the Work Done

Substitute the known values into the work formula:\[W = \frac{1}{2} \times 125 \; \text{N/m} \times (0.115 \; \text{m})^2\]Simplify the expression:\[W = \frac{1}{2} \times 125 \times 0.013225\]\[W = 0.5 \times 125 \times 0.013225 = 0.8278125 \; J\]

Step 5: Round the Answer to Appropriate Significant Figures

Round the computed work to an appropriate number of significant figures, based on the precision of the given data. The initial data has 3 significant figures (11.5 cm), so:\[W \approx 0.828 \; \text{J}\]

Key Concepts

Work Done by a Spring

In spring mechanics, the concept of work done is vital. When a spring is either stretched or compressed from its equilibrium state, work is done on it. The work done is calculated using the formula: \[ W = \frac{1}{2} k x^2 \] where:

• \( W \) is the work done in joules (J).
• \( k \) is the spring constant, in newtons per meter (N/m).
• \( x \) is the displacement, in meters (m).

This formula arises because the force involved is not constant but varies linearly with displacement, as described by Hooke's Law. When applying this formula, remember it calculates the energy required to change the spring's length by a certain displacement. It's crucial to understand each component to apply this correctly.

Spring Constant

The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It's a key factor in determining how much force is needed to stretch or compress a spring by a unit length. - A higher \( k \) value means a stiffer spring, needing more force for the same amount of displacement.- The units of the spring constant are in newtons per meter (N/m), showing the force needed per meter of displacement.When dealing with spring problems in mechanics, knowing the spring constant helps predict how the spring will behave under specific forces or displacements. It's crucial to express \( k \) in consistent units, typically \( ext{N/m} \), for accurate calculations.

Unit Conversion in Physics

In physics problems, consistent units are vital for accuracy. When dealing with mechanics problems, ensure all measurements use the same unit system. Let's explore unit conversion:

N/cm to N/m

The given spring constant \( k \) was 1.25 N/cm in the example. Converting to N/m involves multiplying by 100 since 1 meter equals 100 centimeters. This gives: \( k = 125 \, ext{N/m} \).

cm to m

Similarly, converting displacement is straightforward: divide the centimeter value by 100 to obtain meters. Thus, 11.5 cm becomes 0.115 m. Unit conversions prevent errors in calculations, which are common when mixing different systems or scales. Always check your units!

Displacement in Mechanics

Displacement in mechanics refers to the change in an object's position. In spring mechanics, displacement is the distance through which you stretch or compress the spring from its equilibrium position.- Displacement is symbolized by \( x \) and measured in meters (m).- Positive displacement indicates a stretch, while negative indicates compression.Understanding displacement is key to solving problems involving springs, as it directly affects the work done and the potential energy stored in the spring. In our previous example, the displacement figure was given in centimeters, so it needed conversion to meters to fit the work done formula. This adjustment ensures all elements in the equations align for correct calculations.