If a spring is compressed from 2 inches to 4 inches, how much does the potential energy (PE) increase?

The potential energy stored in a spring is determined by Hooke's Law, which states that the potential energy (PE) 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 from the spring's equilibrium position.

In this scenario, the spring is compressed from 2 inches to 4 inches. First, we need to calculate the potential energy at both positions.

1. When the spring is compressed to 2 inches (which is 2 inches or 0.167 feet):

• Convert inches to feet: 2 inches = 2/12 feet.
• Potential energy at this compression becomes: [ PE_1 = \frac{1}{2} k \left( \frac{2}{12} \right)^2 = \frac{1}{2} k \left( \frac{1}{6} \right)^2 = \frac{1}{2} k \cdot \frac{1}{36} = \frac{k}{72} ]

2. When the spring is compressed to 4 inches (which is