A spring of force constant `600 Nm^(-1)` is mounted on a horizontal table. A mass of 1.5 kg is attached to the free end of the spring, pulled sideways to a distance of 2 cm and released .P.E. of the mass when it momentarily comes to rest and total energy are

Options

A. `0.12 J , 0`
B. ` 0,0.12 J`
C. `0,0`
D. ` 0.12 J , 0.12J`

Answer

To solve the problem, we need to calculate the potential energy (P.E.) of the mass when it momentarily comes to rest and the total energy of the spring-mass system.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Force constant of the spring, \( k = 600 \, \text{N/m} \)
- Mass attached to the spring, \( m = 1.5 \, \text{kg} \)
- Displacement from the equilibrium position (amplitude), \( x = 2 \, \text{cm} = 0.02 \, \text{m} \)

2. **Understand the Energy in the System:**
- When the mass is pulled to the maximum displacement (2 cm) and released, it will momentarily come to rest at that point. At this point, all the energy in the system is potential energy (P.E.), and the kinetic energy (K.E.) is zero.

3. **Calculate the Potential Energy:**
- The formula for potential energy stored in a spring is given by:
\[ P.E. = \frac{1}{2} k x^2 \]
- Substituting the values:
\[ P.E. = \frac{1}{2} \times 600 \, \text{N/m} \times (0.02 \, \text{m})^2 \]
\[ P.E. = \frac{1}{2} \times 600 \times 0.0004 \]
\[ P.E. = 0.12 \, \text{J} \]

4. **Calculate the Total Energy:**
- The total mechanical energy in the spring-mass system is equal to the potential energy at maximum displacement since all the energy is stored as potential energy at that point.
- Therefore, the total energy \( E \) is:
\[ E = P.E. = 0.12 \, \text{J} \]

5. **Final Answers:**
- Potential Energy when the mass comes to rest: \( 0.12 \, \text{J} \)
- Total Energy of the system: \( 0.12 \, \text{J} \)

### Summary:
- The potential energy of the mass when it momentarily comes to rest is \( 0.12 \, \text{J} \).
- The total energy of the system is also \( 0.12 \, \text{J} \).