The force constant of a weightless spring is `16 N m^(-1)`. A body of mass `1.0 kg` suspended from it is pulled down through `5 cm` and then released. The maximum energy of the sysytem (spring + body) will be

Options

A. `2 xx 10^(2) J`
B. `4 xx 10^(2) J`
C. `8 xx 10^(2) J`
D. `16 xx 10^(2) J`

Explanation

To solve the problem step by step, we will calculate the maximum energy of the system (spring + body) when the body is pulled down and released.

### Step 1: Identify the Given Values
- Force constant of the spring, \( k = 16 \, \text{N/m} \)
- Mass of the body, \( m = 1.0 \, \text{kg} \)
- Displacement of the spring, \( x = 5 \, \text{cm} = 0.05 \, \text{m} \) (conversion from cm to m)

### Step 2: Calculate the Maximum Potential Energy Stored in the Spring
The potential energy stored in a spring when it is compressed or stretched is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
Substituting the values we have:
\[ PE = \frac{1}{2} \times 16 \, \text{N/m} \times (0.05 \, \text{m})^2 \]

### Step 3: Calculate \( x^2 \)
First, calculate \( (0.05)^2 \):
\[ (0.05)^2 = 0.0025 \, \text{m}^2 \]

### Step 4: Substitute \( x^2 \) into the Potential Energy Formula
Now substitute \( x^2 \) back into the potential energy formula:
\[ PE = \frac{1}{2} \times 16 \times 0.0025 \]

### Step 5: Perform the Multiplication
Calculate \( 16 \times 0.0025 \):
\[ 16 \times 0.0025 = 0.04 \]

### Step 6: Calculate the Final Potential Energy
Now, calculate \( \frac{1}{2} \times 0.04 \):
\[ PE = \frac{1}{2} \times 0.04 = 0.02 \, \text{J} \]

### Conclusion
The maximum energy of the system (spring + body) will be:
\[ \text{Maximum Energy} = 0.02 \, \text{J} = 2 \times 10^{-2} \, \text{J} \]

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