Question

A force of \(f\) N stretches a spring \(d\) m from its natural length. How much work is performed in stretching the spring?

Short answer

The work performed in stretching the spring is \( \frac{fd}{2} \) Joules.

Step-by-step solution

Understanding Hooke's Law

First, recognize that the force needed to extend or compress a spring by some distance is given by Hooke's law, which states that the force \( F \) is proportional to the displacement \( x \), or \( F = kx \), where \( k \) is the spring constant. In this problem, \( F = f \) when the spring is stretched by \( d \) meters.

Express Spring Constant

Since \( F = fd \), we can express the spring constant \( k \) using the equation of Hooke's law: \( f = kd \). Solving for \( k \), we get \( k = \frac{f}{d} \).

Setup the Work Integral

Work done in stretching the spring is found by integrating the force over the distance the spring is stretched. The work \( W \) done is given by the integral: \[ W = \int_0^d kxdx \]

Solve the Integral

Substitute \( k = \frac{f}{d} \) into the integral, giving: \[ W = \int_0^d \frac{f}{d} x \, dx \]Integrate the equation: \[ W = \frac{f}{d} \left[ \frac{x^2}{2} \right]_0^d \] This simplifies to: \[ W = \frac{f}{d} \cdot \frac{d^2}{2} = \frac{fd}{2} \]

Concluding the Calculation

Thus, using the relation above, the work performed in stretching the spring by distance \( d \) meters when a force of \( f \) Newtons is applied, is \( \frac{fd}{2} \) Joules.

Key concepts

Spring Constant

The spring constant, commonly denoted as \( k \), is a crucial factor in understanding how springs behave. Essentially, it's a measure of a spring's stiffness or elasticity. When you have a stiff spring, the spring constant \( k \) is larger. Conversely, a more flexible spring has a smaller \( k \). Understanding this helps in predicting how the spring will move when a force is applied.

According to Hooke's Law, the force \( F \) required to stretch a spring is directly proportional to the displacement \( x \) of the spring from its rest position. This is mathematically represented as:

• \( F = kx \)

In the context of our problem, we are given a force \( f \) that stretches the spring by a distance \( d \). By substituting these values into the formula, we find the spring constant using:

• \( k = \frac{f}{d} \)

This calculation tells us how strong a force is required to produce a certain stretch, allowing us to anticipate how the spring will react under different conditions.

Work Integral

The concept of work in physics refers to the amount of energy transferred when a force is applied over a distance. In the case of a spring, calculating this work involves an integral because the force varies with the position. The work done \( W \) when stretching a spring over a distance is determined by the integral of the force with respect to displacement.

For a spring obeying Hooke's Law, the force changes linearly with displacement, creating the need to integrate the expression \( kx \), where \( k \) is the spring constant. The work integral formula looks like this:

• \( W = \int_0^d kx \, dx \)

This integral calculates the cumulative work done from the initial position \( x = 0 \) to the final position \( x = d \). By evaluating this, we incorporate the varying nature of the force as the spring extends.

Force and Displacement

Force and displacement are fundamental in understanding how objects move and interact. In the context of spring mechanics, these two concepts work hand in hand. Force \( F \) refers to a push or pull that changes the object's motion, while displacement \( x \) is the change in position of the object.

When a spring is stretched, it experiences displacement. The relationship between force and displacement is linear for springs, as given by Hooke's Law \( F = kx \). This law implies that:

• The greater the displacement, the greater the force required to stretch the spring further.
• The spring gets tougher to stretch as you increase displacement because of the rising force.

Understanding this interplay helps in computing the effort required to stretch or compress springs and in designing systems that incorporate springs. The simplicity of the linear relationship makes it feasible to model and calculate the behavior of springs under various forces.