Section 4.3: Elastic Potential Energy and Conversion
Elastic potential energy is stored in objects that can be stretched or compressed, such as springs or elastic bands. Energy conversion occurs when this potential energy is transformed into kinetic or other forms of energy.
Elastic Potential Energy:
\[ PE_{elastic} = \frac{1}{2} k x^2 \]
where \(k\) is the spring constant and \(x\) is displacement from equilibrium.
Energy Conversion:
- Spring releases stored energy → kinetic energy of object
- Energy may also convert to sound or heat due to damping
Work-Energy Principle for Springs:
\[ W_{spring} = \Delta KE = \frac{1}{2} k x^2 - 0 \]
for a block initially at rest.
Example 1
A 0.5 kg block is attached to a spring with k = 200 N/m. The spring is compressed 0.1 m. Find the speed of the block when released.
Elastic PE: \( PE = \frac{1}{2} k x^2 = 0.5 \cdot 200 \cdot 0.1^2 = 1 \text{ J} \)
All converts to KE: \( \frac{1}{2} m v^2 = 1 \Rightarrow v = \sqrt{2/m} = \sqrt{2 / 0.5} = 2 \text{ m/s} \)
Example 2
A spring is stretched 0.2 m with k = 150 N/m. Find the work done on the spring.
\( W = \frac{1}{2} k x^2 = 0.5 \cdot 150 \cdot 0.2^2 = 3 \text{ J} \)
Practice Problems
- A 2 kg block attached to a spring (k = 100 N/m) is compressed 0.15 m. Find speed when released.
- Spring stretched 0.25 m with k = 200 N/m. Calculate stored elastic potential energy.
- A spring releases a block, converting 4 J of elastic energy into kinetic energy. If block mass is 0.5 kg, find speed.
- Determine work done to compress spring 0.3 m with k = 250 N/m.
- Spring-mass system: mass 0.75 kg, k = 120 N/m, displacement 0.2 m. Find kinetic energy at equilibrium.