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 find the potential energy (P.E.) of the mass when it momentarily comes to rest and the total energy of the system. Here are the steps to arrive at the solution:

### Step 1: Understand the System
We have a spring with a force constant \( k = 600 \, \text{N/m} \) and a mass \( m = 1.5 \, \text{kg} \) attached to it. The mass is pulled to a distance of \( 2 \, \text{cm} \) (which is \( 0.02 \, \text{m} \)) and released.

### Step 2: Identify Maximum Extension
When the mass is pulled to \( 2 \, \text{cm} \) and released, this distance represents the maximum extension (or amplitude) of the spring. At this point, the kinetic energy is zero, and all the energy is stored as potential energy.

### Step 3: Calculate Potential Energy
The potential energy stored in the spring when it is extended by a distance \( x \) is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
Substituting the values:
- \( k = 600 \, \text{N/m} \)
- \( x = 0.02 \, \text{m} \)

Calculating:
\[ PE = \frac{1}{2} \times 600 \times (0.02)^2 \]
\[ PE = \frac{1}{2} \times 600 \times 0.0004 \]
\[ PE = 300 \times 0.0004 = 0.12 \, \text{J} \]

### Step 4: Determine Total Energy
At the moment the mass comes to rest, all the energy in the system is potential energy since the kinetic energy is zero. Therefore, the total energy \( E \) of the system is equal to the potential energy:
\[ E = PE + KE \]
Since \( KE = 0 \) at this moment:
\[ E = PE = 0.12 \, \text{J} \]

### Final Results
- Potential Energy \( PE = 0.12 \, \text{J} \)
- Total Energy \( E = 0.12 \, \text{J} \)

Both the potential energy and the total energy when the mass momentarily comes to rest are \( 0.12 \, \text{J} \).

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