Problem 11 - Single Variable Calculus

Question

A force of \(20 \mathrm{~N}\) is required to stretch a spring \(3 \mathrm{~cm}\) beyond its natural length of \(24 \mathrm{~cm}\). Find the work required to stretch the spring from 30 to \(35 \mathrm{~cm}\).

Step-by-Step Solution

Answer

The work required to stretch the spring from 30 cm to 35 cm is approximately 10.833 Joules.

Step 1: Find the spring constant (k)

The provided information gives us a force (F) of 20 N when the spring is stretched 3 cm beyond its natural length of 24 cm. So, the spring is stretched at a total length of (24+3) cm = 27 cm. We can find the spring constant (k) using Hooke's Law: \(F = kx\) \(20 \mathrm{N} = k \cdot 0.03 \mathrm{m}\) Now divide by \(0.03 \mathrm{m}\) to find k: \(k = \frac{20 \mathrm{N}}{0.03 \mathrm{m}} = 666.67 \frac{\mathrm{N}}{\mathrm{m}}\)

Step 2: Calculate the work required to stretch the spring from 30 cm to 35 cm

To find the work (W) required to stretch the spring from 30 cm to 35 cm, we'll use the work-energy theorem, which states that the work done on an object is equal to the change in its potential energy. For a spring, the potential energy is given by: \(U = \frac{1}{2} kx^2\) The change in potential energy as the spring is stretched from x1 (30 cm) to x2 (35 cm) is given by: \(\Delta U = \frac{1}{2} k (x_2^2 - x_1^2)\) Substitute the values of k, x1, and x2: \(\Delta U = \frac{1}{2} (666.67 \frac{\mathrm{N}}{\mathrm{m}}) ((0.35 \mathrm{m})^2 - (0.30 \mathrm{m})^2)\)

Step 3: Calculate the change in potential energy and find the work required

Now, perform the calculation: \(\Delta U = \frac{1}{2}(666.67 \frac{\mathrm{N}}{\mathrm{m}})(0.1225 \mathrm{m^2} - 0.09 \mathrm{m^2})\) \(\Delta U = 333.335(\mathrm{m}) (0.0325 \mathrm{m})\) \(\Delta U = 10.833 \mathrm{J}\) Finally, the work required to stretch the spring from 30 cm to 35 cm is equal to the change in potential energy. The work required is approximately 10.833 J.

Key Concepts

Hooke's Law

Hooke’s Law is a fundamental principle that describes the behavior of springs. It states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is expressed as:

• \( F = kx \)

Here:

• \( F \) is the force applied to the spring, measured in newtons (N).
• \( k \) is the spring constant, which is unique to each spring and tells us how stiff it is, measured in newtons per meter (\( \frac{N}{m} \)).
• \( x \) is the displacement from the spring's natural length, in meters (m).

A higher spring constant means the spring is stiffer and requires more force to stretch or compress.
For example, if a spring requires a 20 N force to stretch 3 cm (which is 0.03 m) beyond its natural length, we can calculate the spring constant \( k \) using the formula. In this scenario, we find \( k = \frac{20 \mathrm{N}}{0.03 \mathrm{m}} = 666.67 \frac{N}{m} \). This tells us how much force is needed per meter to stretch the spring.

Potential Energy in Springs

Potential energy is the energy stored in an object due to its position or state. For springs, when a spring is stretched or compressed, it stores potential energy. The potential energy \( U \) in a spring can be calculated using the formula:

• \( U = \frac{1}{2} kx^2 \)

Here:

• \( k \) is the spring constant.
• \( x \) is the displacement from the spring’s natural length.

This formula tells us that the potential energy _quadruples_ if the displacement \( x \) is doubled. This relationship highlights how quickly the energy stored in a spring can increase with displacement.
In our example of stretching a spring, we calculated the change in potential energy when the spring length changed from 30 cm to 35 cm. The difference in the potential energy for the two positions gives the work done, which was determined to be about 10.833 Joules.

Work-Energy Theorem

The work-energy theorem is a great tool for understanding how energy changes form when work is done on an object. Simply put, the theorem states that the work done on an object is equal to the change in its kinetic or potential energy. In the case of a spring, as we stretch or compress it, we do work on the spring, which changes its potential energy. This is why work done is equal to the change in potential energy of the spring. To find work \( W \) in a spring, use the potential energy formula:

• Find the potential energy at each position.
• Subtract the initial energy from the final energy to find the work done.

In our scenario, stretching a spring from 30 cm to 35 cm required calculating potential energies at both points and finding the difference, resulting in about 10.833 Joules of work. This is the energy transferred in extending the spring.

Force and Motion

Force and motion are interconnected concepts that are crucial in understanding dynamics in physics. A force is any interaction that, when unopposed, changes the motion of an object. In the context of springs, this force changes the shape of the spring by compressing or stretching it.

• The relationship between force and motion is encapsulated in Newton's second law, which states \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration.
• However, when talking about springs, we refer to Hooke's Law, which relates force directly to the displacement of the spring: \( F = kx \).

Force affects an object's motion by either starting it, stopping it, or changing its direction. The movement created or resisted by the force does the work, leading to energy transactions like those calculated in spring systems.
In any scenario involving springs, understanding how force affects displacement helps explain how much energy is stored or released, which is vital for solving problems related to force and motion.