Question

A massless spring initially compressed by a displacement of two centimeters is now compressed by four centimeters. How has the potential energy of this system changed? (A) The potential energy has not changed. (B) The potential energy has doubled. (C) The potential energy has increased by two joules. (D) The potential energy has quadrupled.

Short answer

The potential energy has quadrupled. Thus, the correct answer is (D).

Step-by-step solution

Understand the relationship

The potential energy stored in a spring is given by the formula: \[ PE = \frac{1}{2} k x^2 \] where \( k \) is the spring constant, and \( x \) is the displacement.

Initial potential energy calculation

Initially, the spring is compressed by 2 cm (0.02 meters). Using the formula, \[ PE_{initial} = \frac{1}{2} k (0.02)^2 \]

New potential energy calculation

Now, the spring is compressed by 4 cm (0.04 meters). The new potential energy is: \[ PE_{new} = \frac{1}{2} k (0.04)^2 \]

Compare the results

To find how the potential energy has changed, compare \( PE_{new} \) with \( PE_{initial} \). We have: \[ PE_{initial} = \frac{1}{2} k (0.02)^2 = \frac{1}{2} k (0.0004) \] and \[ PE_{new} = \frac{1}{2} k (0.04)^2 = \frac{1}{2} k (0.0016) \] Clearly, \( PE_{new} = 4 \times PE_{initial} \).

Key concepts

potential energy formula

The potential energy in a spring is elastic potential energy. It is stored when a spring is compressed or extended.
The formula used to calculate this potential energy is: \[ PE = \frac{1}{2} k x^2 \] where:
• \( k \) is the spring constant, which measures the stiffness of the spring.
• \( x \) is the displacement of the spring, or how much the spring is compressed or stretched from its natural length.

This relationship shows that potential energy is proportional to the square of the displacement. This means that even a small change in displacement can greatly affect the potential energy stored in the spring.

spring constant

The spring constant \( k \) is essential for understanding how a spring behaves.
It measures how stiff or flexible the spring is.
For a stiff spring, the spring constant is larger, and for a flexible spring, it is smaller.

In our formula for potential energy, \( PE = \frac{1}{2} k x^2 \), \( k \) serves as a coefficient that scales the energy stored in the spring according to its stiffness.
Knowing the spring constant helps in calculating various properties of the spring. For example, it helps determine how much potential energy will be stored for a given displacement.

displacement in springs

Displacement represents how much the spring is compressed or stretched from its natural state.
In our problem, the spring is initially compressed by 2 centimeters (0.02 meters) and then by 4 centimeters (0.04 meters).
It’s important to always convert measurements to meters for consistency in calculations.

In our formula, the displacement \( x \) is squared. Thus, doubling the displacement increases the effect on potential energy by a factor of four. This is because \( x \) squared is a quadratic relationship.
This concept is critical when solving physics problems involving springs as it significantly impacts the solution.

energy transformation

The potential energy in a compressed or stretched spring converts to other forms of energy.
When a spring is released, the stored potential energy can transform into kinetic energy as the spring returns to its natural length.
In our spring problem, we examined changes in the potential energy purely due to differences in displacement.

Understanding energy transformation in springs is important because it explains how energy is conserved and transformed from one type to another. This concept of energy transformation can be extended to many different areas in physics and engineering.

physics problem solving

Solving physics problems requires understanding the core principles and how to apply them.
In our example:

• We used the potential energy formula to calculate energy for different displacements.
• We recognized the importance of squaring the displacement values.
• We compared the two potential energy values to determine the change.

The comparison showed that increasing the displacement from 2 cm to 4 cm quadruples the potential energy.
This problem-solving method helps students grasp the principles behind equations and apply them correctly to new problems.
Practice and familiarity with these steps make solving physics problems easier and intuitive over time.