Problem 34 - College Physics

Question

\(\bullet\) A force of magnitude 800.0 \(\mathrm{N}\) stretches a certain spring by 0.200 \(\mathrm{m}\) from its equilibrium position. (a) What is the force constant of this spring? (b) How much elastic potential energy is stored in the spring when it is: (i) stretched 0.300 \(\mathrm{m}\) from its equilibrium position and (ii) compressed by 0.300 \(\mathrm{m}\) from its equilibrium position? (c) How much work was done in stretch- ing the spring by the original 0.200 \(\mathrm{m} ?\)

Step-by-Step Solution

Answer

The spring constant is 4000 N/m; 180 J stored at 0.300 m stretch/compression; 80 J work done for 0.200 m stretch.

Step 1: Find the Force Constant (Spring Constant)

The force constant, also known as the spring constant, is denoted as \( k \) and is calculated using Hooke's Law, which is \( F = kx \). Rearranging for \( k \), we have:\[ k = \frac{F}{x}\]Substitute the given values:\[ k = \frac{800.0 \text{ N}}{0.200 \text{ m}} = 4000 \text{ N/m} \]

Step 2: Calculate the Elastic Potential Energy When Stretched to 0.300 m

Elastic potential energy \( U \) in a spring is given by the formula:\[ U = \frac{1}{2} k x^2 \]Substitute \( k = 4000 \text{ N/m} \) and \( x = 0.300 \text{ m} \):\[ U = \frac{1}{2} (4000 \text{ N/m}) (0.300 \text{ m})^2 = 180 \text{ J} \]

Step 3: Calculate the Elastic Potential Energy When Compressed to 0.300 m

The formula for elastic potential energy is applicable for both stretching and compressing a spring. So again, use:\[ U = \frac{1}{2} k x^2 \]Substitute \( k = 4000 \text{ N/m} \) and \( x = 0.300 \text{ m} \):\[ U = \frac{1}{2} (4000 \text{ N/m}) (0.300 \text{ m})^2 = 180 \text{ J} \]Therefore, the energy is the same if the spring is compressed by the same displacement.

Step 4: Calculate the Work Done in Stretching the Spring to 0.200 m

The work done to stretch the spring is equivalent to the elastic potential energy stored when it is stretched to \( 0.200 \text{ m} \):\[ U = \frac{1}{2} k x^2 \]Substitute \( k = 4000 \text{ N/m} \) and \( x = 0.200 \text{ m} \):\[ U = \frac{1}{2} (4000 \text{ N/m}) (0.200 \text{ m})^2 = 80 \text{ J} \]

Key Concepts

Force Constant

One of the most pivotal concepts when discussing springs is the force constant, often called the spring constant. In physics, this is denoted as \(k\) and is a measure of how stiff or rigid a spring is. Think of it as the toughness of the spring; the higher the constant, the harder it is to stretch or compress that spring. This property is central to Hooke's Law, which is expressed as:

• \( F = kx \)

Here, \( F \) represents the force applied, \( k \) is the force constant, and \( x \) is the displacement from the spring's equilibrium position.
By rearranging the formula, we can solve for the constant as \( k = \frac{F}{x} \).
In our exercise, with a force of 800 N stretching the spring by 0.200 m, we calculate \( k \) to be 4000 N/m. That's quite a significant value, indicating a relatively stiff spring.

Elastic Potential Energy

When a spring is either stretched or compressed, it stores energy in the form of elastic potential energy. Understanding this energy form is crucial because it's the energy a spring "keeps" due to its position or configuration.

The formula to calculate this energy is:

• \( U = \frac{1}{2} k x^2 \)

Here, \( U \) is the elastic potential energy, \( k \) is the spring constant, and \( x \) is the extent to which the spring is stretched or compressed.
In our example, when the spring is stretched by 0.300 m, using the already determined force constant of 4000 N/m, the elastic potential energy becomes 180 J.
Whether the spring is compressed or stretched by the same distance, the energy stored stays the same (in this case, 180 J for 0.300 m). This highlights the symmetrical nature of forces and energy in springs.

Work Done on Spring

Work done on a spring is closely tied to the concept of elastic potential energy. It's defined as the effort required to stretch or compress the spring from its natural length to a certain point.
To find this work done, we use the same elastic potential energy formula:

• \( U = \frac{1}{2} k x^2 \)

This tells us the work is equivalent to the energy stored in the spring at any given position.
For instance, stretching the spring by 0.200 m with the force constant of 4000 N/m results in 80 J of work done.
This means you invested 80 Joules of energy to achieve this deformation. The concept reinforces the relationship between exertion and storage in spring dynamics.