A particle of mass 4 kg moves along x axis, potential energy ( U ) varies with respect to x as `U = 20 + ( x-4)^(2)` , maximum speed of paritcle is at

Options

A. `x=4`
B. `x=2`
C. ` x=0`
D. `x=2.5`

Answer

To solve the problem, we need to determine the position \( x \) at which the potential energy \( U \) is minimized, as this will correspond to the maximum speed of the particle.

### Step-by-Step Solution:

1. **Identify the potential energy function**:
The potential energy \( U \) is given by the equation:
\[ U = 20 + (x - 4)^2 \]

2. **Find the derivative of the potential energy**:
To find the minimum potential energy, we need to take the derivative of \( U \) with respect to \( x \) and set it to zero:
\[ \frac{dU}{dx} = \frac{d}{dx}(20 + (x - 4)^2) \]
Using the chain rule, we get:
\[ \frac{dU}{dx} = 2(x - 4) \]

3. **Set the derivative equal to zero**:
To find the critical points, we set the derivative equal to zero:
\[ 2(x - 4) = 0 \]
Solving for \( x \):
\[ x - 4 = 0 \implies x = 4 \]

4. **Determine if this point is a minimum**:
To confirm that this point is a minimum, we can take the second derivative of \( U \):
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(2(x - 4)) = 2 \]
Since \( \frac{d^2U}{dx^2} = 2 > 0 \), this indicates that \( U \) has a minimum at \( x = 4 \).

5. **Conclusion**:
Since the potential energy is minimized at \( x = 4 \), the kinetic energy will be maximized at this point, and therefore, the maximum speed of the particle occurs at:
\[ x = 4 \]

### Final Answer:
The maximum speed of the particle is at \( x = 4 \).