Problem 38 - University Physics with Modern Physics

Question

A spring of force constant 300.0 N/m and unstretched length 0.240 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 N. How long will the spring now be, and how much work was required to stretch it that distance?

Step-by-Step Solution

Answer

The spring will be 0.290 m long, and the work done is 0.375 Joules.

Step 1: Understanding Hooke's Law

The spring is stretched by two opposite forces of 15.0 N each. By Hooke's Law, the force required to stretch or compress a spring is given by the formula: \( F = k \times x \), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the displacement from the spring's natural length.

Step 2: Calculate Spring Displacement

Since the total force acting on the spring is 15.0 N per side, we use Hooke's Law: \( x = \frac{F}{k} = \frac{15.0}{300.0} = 0.05 \) meters. So, the spring extends by 0.05 meters.

Step 3: Calculate New Length of the Spring

The new length of the spring is the original length plus the extension: New length = 0.240 m + 0.050 m = 0.290 m.

Step 4: Calculating the Work Done

The work done on the spring is given by \( W = \frac{1}{2} k x^2 \). Substituting the values, \( W = \frac{1}{2} \times 300.0 \times (0.05)^2 = 0.375 \) Joules.

Key Concepts

Spring Constant

The spring constant, symbolized as \( k \), is a fundamental attribute of a spring and represents how stiff or stretchy it is. In the context of Hooke's Law, it defines the relationship between the force applied to a spring and the amount that the spring stretches or compresses. In more straightforward terms, the spring constant tells you how much force is needed to stretch or compress the spring by a certain distance.
For example, in the original exercise, the spring constant is 300.0 N/m. This means that every meter you wish to stretch the spring requires a force of 300 newtons. Larger spring constants indicate stiffer springs, while smaller ones suggest more pliable springs. The value itself is derived from the units of force (Newtons, N) divided by the units of displacement (meters, m), giving the spring constant its unit, N/m.
Understanding the spring constant is crucial for engineers and physicists. It helps design systems that involve springs, ensuring safety and functionality, whether in a child's toy or a vehicle's suspension system. Keep in mind that a spring constant remains constant for linear springs, but real-world springs may behave differently when stretched or compressed beyond their limit.

Spring Displacement

Spring displacement is the term used to describe how far a spring has been stretched or compressed from its original, unstressed position. In the exercise under discussion, a spring with an unstretched length of 0.240 m is subjected to forces that cause it to stretch.
The formula used to calculate spring displacement is central to Hooke’s Law: \( x = \frac{F}{k} \). Here, \( F \) is the force applied, \( k \) is the known spring constant, and \( x \) is the displacement. Substituting in values from the exercise, where a force of 15.0 N is applied and the spring constant is 300.0 N/m, we find \( x = \frac{15.0}{300.0} = 0.05 \) meters. This calculation shows that the spring will extend 0.05 meters beyond its natural length.
Spring displacement is important because it informs us how much energy is stored in the spring. It also reveals potential limits of the material, since excessive displacement beyond a spring's elastic limit can lead to permanent deformation.

Work Done on Spring

The work done on a spring is the energy required to stretch (or compress) the spring. This value can be calculated using the formula: \( W = \frac{1}{2} k x^2 \). In this case, \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the displacement from the resting position.
For our specific exercise, substituting the values into the formula, we have \( W = \frac{1}{2} \times 300.0 \times (0.05)^2 = 0.375 \) Joules. This result means that 0.375 Joules of energy was used or "work done" to stretch the spring by 0.05 meters.
This concept is vital in fields like engineering and physics, where energy efficiency, control, and stability are crucial. Knowing the work done on a spring helps assess how much energy can be stored and later released, which is useful in applications such as shock absorbers in vehicles or even in everyday tasks like operating mechanical pencils.