To solve the problem, we need to find the maximum energy of the system consisting of a spring and a mass when the mass is pulled down and released. We will use the formula for the maximum potential energy stored in a spring, which is given by:
\[
E_{\text{max}} = \frac{1}{2} k A^2
\]
where:
- \( E_{\text{max}} \) is the maximum energy,
- \( k \) is the spring constant,
- \( A \) is the amplitude of the motion.
### Step-by-Step Solution:
1. **Identify the given values:**
- Spring constant, \( k = 16 \, \text{N/m} \)
- Mass of the body, \( m = 1.0 \, \text{kg} \) (not directly needed for energy calculation)
- Displacement (pulled down), \( x = 5 \, \text{cm} = 0.05 \, \text{m} \)
2. **Convert the displacement to meters:**
- Since \( 1 \, \text{cm} = 0.01 \, \text{m} \), we convert \( 5 \, \text{cm} \) to meters:
\[
A = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m}
\]
3. **Substitute the values into the energy formula:**
- Using the formula for maximum energy:
\[
E_{\text{max}} = \frac{1}{2} k A^2
\]
- Substitute \( k = 16 \, \text{N/m} \) and \( A = 5 \times 10^{-2} \, \text{m} \):
\[
E_{\text{max}} = \frac{1}{2} \times 16 \times (5 \times 10^{-2})^2
\]
4. **Calculate \( A^2 \):**
- First, calculate \( (5 \times 10^{-2})^2 \):
\[
(5 \times 10^{-2})^2 = 25 \times 10^{-4} = 2.5 \times 10^{-3}
\]
5. **Calculate the energy:**
- Now substitute back into the energy equation:
\[
E_{\text{max}} = \frac{1}{2} \times 16 \times 2.5 \times 10^{-3}
\]
- Calculate \( \frac{1}{2} \times 16 = 8 \):
\[
E_{\text{max}} = 8 \times 2.5 \times 10^{-3} = 20 \times 10^{-3} = 2 \times 10^{-2} \, \text{J}
\]
6. **Final Result:**
- The maximum energy of the system is:
\[
E_{\text{max}} = 2 \times 10^{-2} \, \text{J}
\]
### Conclusion:
The maximum energy of the system (spring + body) is \( 2 \times 10^{-2} \, \text{J} \).
