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

Answer

To solve the problem step by step, we will follow the outlined approach based on the information provided in the question and the video transcript.

### Step 1: Identify the Given Information
- Mass of the block, \( m = 2 \, \text{kg} \)
- Initial velocity, \( v_0 = 4\sqrt{5} \, \text{m/s} \)
- Potential energy function, \( U(x) = -x^3 + 6x^2 + 15 \)

### Step 2: Calculate the Force from the Potential Energy
The force \( F \) acting on the block is given by the negative gradient of the potential energy:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ \frac{dU}{dx} = -3x^2 + 12x \]
Thus, the force can be expressed as:
\[ F = 3x^2 - 12x \]

### Step 3: Find the Condition for Minimum Force
To find the position where the force is minimum, we differentiate the force with respect to \( x \) and set it to zero:
\[ \frac{dF}{dx} = 6x - 12 = 0 \]
Solving for \( x \):
\[ 6x = 12 \implies x = 2 \, \text{m} \]

### Step 4: Set Up the Energy Conservation Equation
Using the work-energy principle, we can relate the initial kinetic energy and the change in potential energy to find the final velocity when the position is \( x = 2 \, \text{m} \).

Initial kinetic energy \( KE_i \):
\[ KE_i = \frac{1}{2}mv_0^2 = \frac{1}{2} \times 2 \times (4\sqrt{5})^2 = 80 \, \text{J} \]

Potential energy at \( x = 0 \):
\[ U(0) = -0^3 + 6(0)^2 + 15 = 15 \, \text{J} \]

Potential energy at \( x = 2 \):
\[ U(2) = -2^3 + 6(2)^2 + 15 = -8 + 24 + 15 = 31 \, \text{J} \]

### Step 5: Apply Conservation of Mechanical Energy
The total mechanical energy at \( x = 0 \) is equal to the total mechanical energy at \( x = 2 \):
\[ KE_i + U(0) = KE_f + U(2) \]
Substituting the values:
\[ 80 + 15 = \frac{1}{2}mv_f^2 + 31 \]
\[ 95 = \frac{1}{2}(2)v_f^2 + 31 \]
\[ 95 - 31 = v_f^2 \]
\[ 64 = v_f^2 \]
\[ v_f = \sqrt{64} = 8 \, \text{m/s} \]

### Final Answer
The velocity of the block when the force acting on it is minimum is \( \boxed{8 \, \text{m/s}} \).