Force constant of a weightless spring is 16 N/m. A body of mass 1.0 kg suspended from it is pulled down through 5 cm from its mean position and the released. The maximum kinetic energy of the system (spring+body) will be (0.0x) J. Find value of x.

Answer

To solve the problem step by step, we will follow the principles of simple harmonic motion (SHM) and the energy conservation in a spring-mass system.

### Step 1: Identify the parameters
- Spring constant (k) = 16 N/m
- Mass (m) = 1.0 kg
- Displacement from mean position (A) = 5 cm = 0.05 m (convert to meters)

### Step 2: Calculate the maximum potential energy
In a spring-mass system, the maximum potential energy (PE) when the spring is stretched or compressed to its maximum displacement (amplitude) is given by the formula:
\[ PE_{\text{max}} = \frac{1}{2} k A^2 \]
Substituting the values:
\[ PE_{\text{max}} = \frac{1}{2} \times 16 \, \text{N/m} \times (0.05 \, \text{m})^2 \]
Calculating:
\[ PE_{\text{max}} = \frac{1}{2} \times 16 \times 0.0025 = 0.02 \, \text{J} \]

### Step 3: Relate potential energy to kinetic energy
In SHM, the total mechanical energy (E) is conserved and is equal to the maximum potential energy when the spring is at maximum displacement. Therefore, the maximum kinetic energy (KE) at the mean position is equal to the maximum potential energy:
\[ KE_{\text{max}} = PE_{\text{max}} = 0.02 \, \text{J} \]

### Step 4: Express the maximum kinetic energy in the required format
The problem states that the maximum kinetic energy is given as \(0.0x\) J. We have calculated:
\[ KE_{\text{max}} = 0.02 \, \text{J} \]
This can be expressed as:
\[ 0.0x = 0.02 \implies x = 2 \]

### Final Answer
The value of \(x\) is \(2\).