The potential energy of a `4kg` particle free to move along the x-axis varies with x according to following relationship : `U(x) = ((x^(3))/(3)-(5x^(2))/(2)+6x+3)` Joules, where `x` is in meters. If the total mechanical energy of the particle is `25.5` Joules, then the maximum speed of the particle is `x m//s`, find `x`

Explanation

To find the maximum speed of the particle, we will follow these steps:

### Step 1: Understand the relationship between total mechanical energy, kinetic energy, and potential energy.
The total mechanical energy (E) of the particle is given by the sum of its kinetic energy (KE) and potential energy (U):
\[ E = KE + U \]
At maximum speed, the kinetic energy is at its maximum, and the potential energy is at its minimum.

### Step 2: Set up the equation for total mechanical energy.
Given that the total mechanical energy \( E = 25.5 \) Joules, we can express this as:
\[ E = KE_{max} + U_{min} \]

### Step 3: Find the potential energy function.
The potential energy \( U(x) \) is given by:
\[ U(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + 3 \]

### Step 4: Find the critical points of the potential energy function.
To find the minimum potential energy, we need to find the critical points by taking the derivative of \( U(x) \) and setting it to zero:
\[ \frac{dU}{dx} = x^2 - 5x + 6 = 0 \]
Factoring the quadratic equation:
\[ (x - 2)(x - 3) = 0 \]
Thus, \( x = 2 \) and \( x = 3 \) are the critical points.

### Step 5: Determine which critical point gives the minimum potential energy.
To find out whether these points are maxima or minima, we take the second derivative:
\[ \frac{d^2U}{dx^2} = 2x - 5 \]
- At \( x = 2 \):
\[ \frac{d^2U}{dx^2} = 2(2) - 5 = -1 \] (maximum)
- At \( x = 3 \):
\[ \frac{d^2U}{dx^2} = 2(3) - 5 = 1 \] (minimum)

Thus, \( U \) is minimum at \( x = 3 \).

### Step 6: Calculate the minimum potential energy \( U_{min} \).
Now, substitute \( x = 3 \) into the potential energy function:
\[ U(3) = \frac{3^3}{3} - \frac{5(3^2)}{2} + 6(3) + 3 \]
Calculating each term:
\[ U(3) = 9 - \frac{45}{2} + 18 + 3 \]
\[ U(3) = 9 - 22.5 + 18 + 3 \]
\[ U(3) = 9 + 18 + 3 - 22.5 = 7.5 \text{ Joules} \]

### Step 7: Substitute \( U_{min} \) back into the total mechanical energy equation.
Now we can find the maximum kinetic energy:
\[ E = KE_{max} + U_{min} \]
\[ 25.5 = KE_{max} + 7.5 \]
\[ KE_{max} = 25.5 - 7.5 = 18 \text{ Joules} \]

### Step 8: Relate kinetic energy to maximum speed.
The kinetic energy is given by:
\[ KE_{max} = \frac{1}{2}mv^2 \]
Where \( m = 4 \) kg. Thus:
\[ 18 = \frac{1}{2} \times 4 \times v_{max}^2 \]
\[ 18 = 2v_{max}^2 \]
\[ v_{max}^2 = \frac{18}{2} = 9 \]
\[ v_{max} = 3 \text{ m/s} \]

### Final Answer:
The maximum speed of the particle is \( 3 \) m/s.