Problem 5 - Thomas Calculus in SI Units

Question

It takes a force of 96,000 N to compress a coil spring assembly on a New York City Transit Authority subway car from its free height of \(20 \mathrm{cm}\) to its fully compressed height of \(12 \mathrm{cm}\). a. What is the assembly's force constant? b. How much work does it take to compress the assembly the first centimeter? the second centimeter? Answer to the nearest joule.

Step-by-Step Solution

Answer

a. 1,200,000 N/m; b. First cm: 60 J, Second cm: 180 J.

Step 1: Understand the Force Constant Equation

The force constant, also known as the spring constant (k), is determined using Hooke's Law, which is given by: \[ F = k \cdot \Delta x \] where \( F \) is the force applied, \( k \) is the spring constant, and \( \Delta x \) is the change in length of the spring from its natural length. The change in length for the spring is from 20 cm (0.2 m) to 12 cm (0.12 m), so \( \Delta x = 0.2 - 0.12 = 0.08 \) meters.

Step 2: Solve for Force Constant k

Substitute the given values into the equation from Hooke's Law to find \( k \):\[ k = \frac{F}{\Delta x} = \frac{96000}{0.08} \]Calculate:\[ k = 1,200,000 \, \text{N/m} \]

Step 3: Calculate Work for the First Centimeter

The work done on a spring is calculated using the formula:\[ W = \frac{1}{2} k (\Delta x)^2 \]For the first centimeter (0.01 m), the displacement \( \Delta x = 0.01 \) m. Substituting the calculated value of \( k \):\[ W_1 = \frac{1}{2} \times 1,200,000 \times (0.01)^2 = \frac{1}{2} \times 1,200,000 \times 0.0001 = 60 \text{ J} \]

Step 4: Calculate Work for the Second Centimeter

For the second centimeter, the total displacement is now 0.02 m, but we must exclude the work done in the first centimeter to find the additional work. Calculate total work for 2 cm:\[ W_{total} = \frac{1}{2} \times 1,200,000 \times (0.02)^2 = \frac{1}{2} \times 1,200,000 \times 0.0004 = 240 \text{ J} \]Subtract the work done for the first centimeter to get the work done for the second centimeter:\[ W_2 = W_{total} - W_1 = 240 \text{ J} - 60 \text{ J} = 180 \text{ J} \]

Key Concepts

Force Constant

In the world of springs, the force constant, often called the spring constant, is essential to understanding how a spring behaves under force. This constant, denoted as \( k \), represents the stiffness of the spring. The higher the value of \( k \), the stiffer the spring is. To calculate this, we use Hooke's Law:

• \[ F = k \cdot \Delta x \]

Here, \( F \) is the force applied to the spring and \( \Delta x \) is the displacement from its original length. In our exercise, the spring was compressed by 8 cm, or 0.08 meters, with a force of 96,000 N. By rearranging Hooke's Law to solve for \( k \), we get:

• \[ k = \frac{F}{\Delta x} \]
• \[ k = \frac{96000}{0.08} = 1,200,000 \, \text{N/m} \]

This means that every meter you compress the spring, you need a force of 1,200,000 N, highlighting how rigid the spring is.

Work Done on a Spring

When compressing or stretching a spring, work is done. The work done on a spring essentially measures how much energy is required to move it from one position to another. For calculating this, we use the formula:

• \[ W = \frac{1}{2} k (\Delta x)^2 \]

The formula considers the spring constant \( k \) and the displacement \( \Delta x \) squared. As illustrated in our problem, we calculate the work done for the first centimeter displacement (0.01 m) as follows:

• \[ W_1 = \frac{1}{2} \times 1,200,000 \times (0.01)^2 = 60 \, \text{J} \]

To understand further, let's consider the second centimeter. The cumulative work for 2 cm is:

• \[ W_\text{total} = \frac{1}{2} \times 1,200,000 \times 0.02^2 = 240 \, \text{J} \]

The additional work done to compress from 1 cm to 2 cm would be the difference:

• \[ W_2 = 240 \text{ J} - 60 \text{ J} = 180 \text{ J} \]

This incremental calculation shows how the energy requirements increase as the spring is compressed further.

Spring Compression

Spring compression is the process of reducing the spring's length by applying an external force. In our example, the spring was compressed from 20 cm to 12 cm. This displacement of 8 cm is crucial because it determines how much force we need according to Hooke's Law.
Understanding spring compression requires knowing the uniform and gradual way force is applied with each additional centimeter of compression. The force needed grows progressively with the displacement, illustrating the spring's resistance to compression.
In practical terms, this means as you squeeze the spring further, the effort required increases. This happens because the spring's resistance is proportional to how much it is already compressed. Thus, compressing a spring becomes a notable experimentation of applying gradual force and understanding mechanical resistance.

Elastic Potential Energy

Elastic potential energy is the stored energy in an elastic object when it is stretched or compressed. For our spring problem, once energy is added by compressing it, that energy becomes stored, ready to be released when the spring returns to its original state.
This form of potential energy is calculated similarly to work, as:

• \[ E_p = \frac{1}{2} k (\Delta x)^2 \]

Here, \( E_p \) is the elastic potential energy, \( k \) is the spring constant, and \( \Delta x \) is the displacement.
When you compress a spring by 1 cm, with \( \Delta x = 0.01 \) m, it stores:

• \[ E_{p1} = \frac{1}{2} \times 1,200,000 \times (0.01)^2 = 60 \, \text{J} \]

And the potential energy when compressed by 2 cm is:

• \[ E_{p2} = \frac{1}{2} \times 1,200,000 \times (0.02)^2 = 240 \, \text{J} \]

This showcases how the energy stored in the spring increases with more compression. Elastic potential energy highlights the spring's capability to return to its natural length, displaying efficiency in energy storage and release.