Question

For any spring obeying Hooke's Law, show that the work done in stretching a spring a distance \(d\) is given by \(W=\frac{1}{2} k d^{2}\).

Short answer

The work done in stretching a spring by distance \(d\) is \(W = \frac{1}{2} k d^{2}\).

Step-by-step solution

Understanding Hooke's Law

Hooke's Law states that the force required to stretch or compress a spring by a distance \(x\) is proportional to \(x\): \(F = kx\), where \(k\) is the spring constant. This is the starting point for understanding how work is calculated.

Define Work in Terms of Force and Distance

The work done on an object is the integral of the force over the distance. In the case of a spring, work \(W\) is the integral of the force \(F(x)=kx\) from the initial position \(x=0\) to the final position \(x=d\). So, \(W = \int_{0}^{d} kx \, dx\).

Perform the Integration

To find the work done, calculate the integral: \(W = \int_{0}^{d} kx \, dx = \left[\frac{kx^2}{2}\right]_{0}^{d}\). This simplifies to \(W = \frac{kd^2}{2} - \frac{k(0)^2}{2} = \frac{kd^2}{2}\).

Conclusion

Thus, the work done in stretching the spring by a distance \(d\) is given by \(W = \frac{1}{2} k d^2\). This formula shows that the work depends quadratically on the distance \(d\) and linearly on the spring constant \(k\).

Key concepts

Work Done on a Spring

The concept of work done on a spring boils down to the energy required to stretch or compress a spring. When you pull on a spring, you are performing work, which is stored in the form of potential energy in the spring. According to Hooke's Law, the force required to stretch a spring linearly depends on the distance it's stretched. This force is mathematically described as \( F = kx \), where

• F is the force applied,
• k is the spring constant, an indicator of the spring's stiffness,
• x is the displacement from its equilibrium position.

To calculate the work done (\( W \)) in stretching the spring, you must evaluate the integral of the force over the distance of the stretch, leading to \( W = \frac{1}{2}kd^2 \). Here,

• \( \frac{1}{2}kd^2 \) represents the area under the linear force-distance graph,
• which highlights how the energy needed grows with the square of the distance,
• emphasizing the non-linear relationship between work and stretch distance in springs.

Spring Constant

The spring constant (\( k \)) is also known as the elasticity or stiffness constant. It is a crucial element in understanding how a spring behaves under force.
In Hooke's Law, \( F = kx \), the spring constant characterizes the spring’s resistance to deformation.

In practical terms:

• A higher \( k \) value implies a stiffer spring, which requires more force for the same displacement,
• whereas a lower \( k \) value indicates a more easily deformable spring.

The unit of the spring constant is Newton per meter (N/m), which signifies how much force is needed to elongate or compress the spring by one meter.
This constant is unique to each spring and is determined by material properties and dimensions. Understanding \( k \) helps you predict how the spring will respond under various loads.

Integration in Physics

Integration is a fundamental mathematical tool used extensively in physics to handle various real-world problems. In the context of Hooke's Law and springs:

• It allows us to calculate the total work done over a distance by integrating the variable force.
• The integral of a force function gives us insights into the accumulation of work as the spring is stretched or compressed.

By defining the work done on a spring as \( W = \int_{0}^{d} kx \, dx \), we are essentially calculating the area under the force vs. displacement graph. This operation gives us a precise mathematical relationship between force, stretch distance, and work done.

In simpler cases, where force remains constant over a distance, work may often merely be the product of force and distance. However, with springs, where force changes with distance, integration becomes necessary to reflect the dynamic nature of how work is accumulated.
Mastery of integration assists in understanding this dynamic and helps solve more complex physics problems beyond just springs.