A partical of mass `2kg` is moving on the `x-`axis with a constant mechanical energy `20 J`. Its potential energy at any `x` is `U=(16-x^(2))J` where `x` is in metre. The minimum velocity of particle is `:-`

Options

A. `2m//s`
B. `4m//s`
C. `6m//s`
D. zero

Explanation

To find the minimum velocity of the particle, we can follow these steps:

### Step 1: Understand the given information
We know that:
- Mass of the particle, \( m = 2 \, \text{kg} \)
- Total mechanical energy, \( E = 20 \, \text{J} \)
- Potential energy function, \( U(x) = 16 - x^2 \)

### Step 2: Determine the potential energy at its maximum
The potential energy \( U \) is maximum when \( x = 0 \):
\[ U(0) = 16 - (0)^2 = 16 \, \text{J} \]

### Step 3: Calculate the kinetic energy at maximum potential energy
The total mechanical energy is the sum of kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]
At \( x = 0 \):
\[ 20 \, \text{J} = K + 16 \, \text{J} \]
Thus, the kinetic energy \( K \) at this point is:
\[ K = 20 \, \text{J} - 16 \, \text{J} = 4 \, \text{J} \]

### Step 4: Relate kinetic energy to velocity
The kinetic energy is given by the formula:
\[ K = \frac{1}{2} mv^2 \]
Substituting the known values:
\[ 4 \, \text{J} = \frac{1}{2} \cdot 2 \, \text{kg} \cdot v^2 \]
This simplifies to:
\[ 4 = 1 \cdot v^2 \]
Thus, we have:
\[ v^2 = 4 \]

### Step 5: Solve for velocity
Taking the square root of both sides:
\[ v = \sqrt{4} = 2 \, \text{m/s} \]

### Conclusion
The minimum velocity of the particle is:
\[ \boxed{2 \, \text{m/s}} \]

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