Problem 46 - Precalculus with Limits

Question

A force of 220 newtons stretches a spring 0.12 meter. What force is required to stretch the spring 0.16 meter?

Step-by-Step Solution

Answer

To stretch a spring by 0.16 meter, a force of approximately 293.33 newtons is required.

Step 1: Calculate the Spring Constant (k)

First, find the spring constant. The constant \(k\) can be found from the formula \(F=kx\), where \(F\) is the force and \(x\) is the distance. By rearranging this formula to \(k=F/x\), and substituting \(F=220\) newtons and \(x=0.12\) meter into the equation, the spring constant \(k\) can be calculated.

Step 2: Find the Required Force to Stretch the Spring 0.16 Meter

Next, find the force necessary to stretch the spring by 0.16 meter. With the previously found spring constant \(k\) and the stretch \(x=0.16\) meter, substitute these values into the same Hooke's law equation \(F=kx\). This will calculate the required force.

Key Concepts

Understanding the Spring Constant

The spring constant, often denoted by the symbol k, is a measure of a spring's stiffness. It is defined as the force required to stretch or compress a spring by a unit distance. In more scientific terms, it is the ratio of the force affecting the spring to the displacement caused by it, expressed by the equation Hooke's Law:
\[ k = \frac{F}{x} \]
where F represents the force applied to the spring, and x is the displacement of the spring from its equilibrium position. A higher spring constant indicates a stiffer spring that requires more force to stretch it. Conversely, a spring with a lower spring constant is easier to stretch or compress. This value is crucial because it determines how much force is needed to stretch or compress the spring a certain distance.
In the given exercise, to compute the spring constant, we use the provided values of 220 newtons (N) for the force and 0.12 meters (m) for the distance to find k. The spring constant is a fundamental trait of the spring that doesn't change, hence it allows us to calculate how much force is required for different amounts of stretching, provided the elasticity is linear and we stay within the limits of the spring's elastic behavior.

Calculating Force in Hooke’s Law

Force calculation in springs adheres to Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the amount of stretch or compression, as long as the elastic limit is not exceeded. This linear relationship can be expressed using the equation:
\[ F = k \cdot x \]
In this formula, F signifies the force applied to the spring, k is the spring constant, and x represents the displacement of the spring from its natural length.

Applying Force Calculation

Going back to our problem, once we have calculated the spring constant, it's straightforward to use it for finding the force necessary to stretch the spring to a new distance. For example, if we want to stretch the spring 0.16 meters, knowing the spring constant obtained from the initial conditions, we apply the same law: \[ F_{new} = k \cdot x_{new} \] Here, F_{new} is the force required for the new distance, x_{new}.

Linear Elasticity and Its Limits

Linear elasticity is an essential principle within material science and mechanics, particularly when discussing springs and Hooke's Law. It refers to a material's property to deform proportionally to the applied force, and return to its original shape when that force is removed, provided the material's elastic limit has not been exceeded. For springs and other elastic objects, linearity is assumed between the force applied and the displacement produced: they exhibit linear elastic behavior.

However, this linear relationship only holds up to the elastic limit, also known as the proportional limit. Beyond this point, materials may still deform but won't return to their original shape - this is called plastic deformation. For springs, this could result in a permanent stretch and a deviation from Hooke's Law.

When solving problems like the one in the exercise, we assume that the spring remains within its elastic range. If the spring were stretched beyond this range, the force required to stretch it further would not follow the calculated constant k linearly, which means the calculations would no longer be accurate.