Question

A spring with \(k=10.0 \mathrm{~N} / \mathrm{cm}\) is initially stretched \(1.00 \mathrm{~cm}\) from its equilibrium length. a) How much more energy is needed to further stretch the spring to \(5.00 \mathrm{~cm}\) beyond its equilibrium length? b) From this new position, how much energy is needed to compress the spring to \(5.00 \mathrm{~cm}\) shorter than its equilibrium position?

Short answer

Question: Calculate the additional energy needed to (a) further stretch a spring from 1 cm to 5 cm beyond its equilibrium length, and (b) compress it from being stretched 5 cm longer to 5 cm shorter than its equilibrium position given a spring constant of 10 N/cm. Answer: (a) 120 J of additional energy is needed to further stretch the spring to 5 cm beyond its equilibrium length. (b) No additional energy is needed to compress the spring from being stretched 5 cm longer to 5 cm shorter than its equilibrium position.

Step-by-step solution

Part a: Calculate the Initial and Final Potential Energy

First, we'll calculate the potential energy stored in the spring when it is initially stretched 1 cm and when it is stretched 5 cm. For the initial potential energy when the spring is 1 cm stretched, we'll use the formula \(PE = \frac{1}{2}kx^2\), where \(k = 10 \mathrm{~N/cm}\), and \(x = 1 \mathrm{~cm}\): \(PE_{initial} = \frac{1}{2}(10 \mathrm{~N/cm})(1 \mathrm{~cm})^2 = 5 \mathrm{~J}\) For the final potential energy when the spring is 5 cm stretched, we'll use the same formula with \(x = 5 \mathrm{~cm}\): \(PE_{final} = \frac{1}{2}(10 \mathrm{~N/cm})(5 \mathrm{~cm})^2 = 125 \mathrm{~J}\)

Part a: Calculate the Additional Energy Needed to Stretch to 5 cm

Now we can calculate the additional energy needed to further stretch the spring to 5 cm using the difference between the final and initial potential energy: \(additional\_energy = PE_{final} - PE_{initial} = 125 \mathrm{~J} - 5 \mathrm{~J} = 120 \mathrm{~J}\) So, \(120 \mathrm{~J}\) of energy is needed to further stretch the spring to 5 cm beyond its equilibrium length.

Part b: Calculate the Potential Energy Change to Compress the Spring

For part b, we need the potential energy of the spring when it is 5 cm shorter than the equilibrium position. The displacement will be negative, so \(x = -5 \mathrm{~cm}\): \(PE_{compressed} = \frac{1}{2}(10 \mathrm{~N/cm})(-5 \mathrm{~cm})^2 = 125 \mathrm{~J}\) Now calculate the energy needed to compress the spring from its stretched position (5 cm longer) to 5 cm shorter than its equilibrium position, by taking the difference in potential energy: \(energy_{compression} = PE_{compressed} - PE_{final} = 125 \mathrm{~J} - 125 \mathrm{~J} = 0 \mathrm{~J}\) Surprisingly, no additional energy is needed to compress the spring to 5 cm shorter than its equilibrium position from the 5 cm stretched position, because the potential energy at both positions is the same.

Key concepts

Hooke's Law

Understanding how objects behave when force is applied is crucial in physics. This is where Hooke's Law comes into play, specifically when we're dealing with springs and elasticity. Fundamentally, it states that the force required to stretch or compress a spring is directly proportional to the displacement from its equilibrium position. The law can be mathematically expressed as

\( F = -kx \),

where \( F \) is the force applied, \( k \) is the spring constant indicating the stiffness of the spring, and \( x \) is the displacement from the equilibrium position. The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement. Hooke's Law is fundamental because it sets the stage for understanding elastic potential energy and provides a baseline for calculating how much work is done when deforming an elastic object like a spring.

Elastic Potential Energy

When we apply Hooke's Law to understand elastic potential energy, things get interesting. This type of potential energy is stored when a spring is stretched or compressed. It is the energy retained within the spring, due to its position—think of it like a coiled up snake, ready to snap back!

The formula derived from Hooke's Law for elastic potential energy is:

\(PE = \frac{1}{2} k x^2\),

indicating that the potential energy stored is directly proportional to the square of the displacement from equilibrium and to the spring constant k. As demonstrated in the textbook example, the potential energy changes significantly even with a small change in displacement due to the squared relationship with x. This quadratic relationship highlights the non-linear nature of elastic potential energy as a function of displacement, which is a key concept for students to understand.

Mechanical Energy

When we talk about mechanical energy in the context of springs and Hooke's Law, we're dealing with the sum of kinetic and potential energies within a system. Mechanical energy is conserved in an isolated system, where no non-conservative forces (like friction) are at work.

For a spring, the mechanical energy would involve both the kinetic energy of the mass attached to the spring (if any) and the elastic potential energy stored in the spring. When a spring is neither stretching nor compressing, all of its mechanical energy is potential. However, when it moves, this potential energy can be converted into kinetic energy, and vice versa, while the total mechanical energy remains constant. It is an essential principle that allows students to analyze different states of the system and calculate various properties accordingly.

Equilibrium Position

The equilibrium position of a spring is the sweet spot where there are no net forces acting upon it, meaning it is neither stretched nor compressed. It's the position where the spring would naturally rest if undisturbed.

The significance of the equilibrium position is that it serves as a reference point for measuring displacement (x). It's also the position where a spring has its minimum potential energy – any deviation from this point increases the spring's potential energy, as the exercise illustrates. Understanding this concept is critical for students, as it helps clarify why, in the textbook solution, compressing the spring from a stretched position (5 cm beyond equilibrium) to an equally compressed position (5 cm before equilibrium) doesn't require additional energy – both states are symmetrical around the equilibrium position resulting in the same amount of potential energy. It's a nuanced detail that emphasizes the symmetry and conservation within physical systems.