A block of mass 2 kg is kept at origin at t = 0 and is having velocity `4sqrt5m//s` in positive x - direction. The only force on it is a conservative and its potential energy is defined as `U=-x^(3)+6x^(2)+15` (SI units). Its velocity when the force acting on it is minimum (after the time t = 0 ) is

Options

A. `8m//s`
B. `4m//s`
C. `10sqrt(24)m//s`
D. none of these

Answer

To solve the problem, we need to find the velocity of the block when the force acting on it is at a minimum. We will follow these steps:

### Step 1: Determine the Force from Potential Energy
The force acting on the block can be derived from the potential energy function \( U(x) \). The relationship between force and potential energy is given by:
\[ F = -\frac{dU}{dx} \]
Given the potential energy:
\[ U = -x^3 + 6x^2 + 15 \]
We will differentiate \( U \) with respect to \( x \).

### Step 2: Differentiate the Potential Energy
Calculating the derivative:
\[ \frac{dU}{dx} = -\frac{d}{dx}(x^3) + \frac{d}{dx}(6x^2) + \frac{d}{dx}(15) \]
This gives:
\[ \frac{dU}{dx} = -3x^2 + 12x \]
Thus, the force can be expressed as:
\[ F = -\left(-3x^2 + 12x\right) = 3x^2 - 12x \]

### Step 3: Find the Condition for Minimum Force
To find where the force is minimum, we need to set the force function equal to zero:
\[ 3x^2 - 12x = 0 \]
Factoring the equation:
\[ 3x(x - 4) = 0 \]
This gives us two solutions:
\[ x = 0 \quad \text{and} \quad x = 4 \]

### Step 4: Determine the Nature of the Force
To determine which of these points corresponds to a minimum force, we can check the second derivative:
\[ \frac{d^2F}{dx^2} = 6x - 12 \]
Evaluating at \( x = 4 \):
\[ \frac{d^2F}{dx^2} = 6(4) - 12 = 24 - 12 = 12 > 0 \]
This indicates that \( x = 4 \) is a minimum point for the force.

### Step 5: Calculate the Velocity at \( x = 4 \)
Now we need to find the velocity of the block when it reaches \( x = 4 \). We can use the conservation of mechanical energy:
\[ \text{Total Energy} = K.E + P.E \]
At \( t = 0 \):
- Kinetic Energy \( K.E = \frac{1}{2}mv^2 = \frac{1}{2}(2)(4\sqrt{5})^2 = 20 \times 5 = 100 \, \text{J} \)
- Potential Energy \( P.E = U(0) = 15 \)

Thus, the total energy at \( t = 0 \) is:
\[ E = K.E + P.E = 100 + 15 = 115 \, \text{J} \]

At \( x = 4 \):
- Potential Energy \( P.E = U(4) = -4^3 + 6(4^2) + 15 = -64 + 96 + 15 = 47 \, \text{J} \)

Using conservation of energy:
\[ E = K.E + P.E \implies 115 = K.E + 47 \]
Thus,
\[ K.E = 115 - 47 = 68 \, \text{J} \]

Now, we can find the velocity:
\[ K.E = \frac{1}{2}mv^2 \implies 68 = \frac{1}{2}(2)v^2 \implies 68 = v^2 \implies v = \sqrt{68} = 2\sqrt{17} \, \text{m/s} \]

### Final Answer
The velocity of the block when the force acting on it is minimum is:
\[ v = 2\sqrt{17} \, \text{m/s} \]