The total mechanical energy of a 2.00 kg particle moving along an x axis is 5.00 J. The potential energy is given as `U(x)=(x^(4)-2.00x^(2))J`, with x in meters. Find the maximum velocity.

Answer

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

### Step 1: Understand the Total Mechanical Energy
The total mechanical energy (E) of the particle is given by the sum of its kinetic energy (K) and potential energy (U):
\[ E = K + U \]
Given that the total mechanical energy is 5 J, we can express this as:
\[ 5 = K + U \]

### Step 2: Write the Expression for Potential Energy
The potential energy is given by the equation:
\[ U(x) = x^4 - 2x^2 \]

### Step 3: Find the Minimum Potential Energy
To find the maximum kinetic energy (and thus the maximum velocity), we need to find the minimum potential energy. We do this by differentiating the potential energy function and setting the derivative to zero:
\[ \frac{dU}{dx} = 4x^3 - 4x = 4x(x^2 - 1) \]
Setting this equal to zero gives us:
\[ 4x(x^2 - 1) = 0 \]
This results in three critical points:
\[ x = 0, \quad x = 1, \quad x = -1 \]

### Step 4: Determine the Nature of Critical Points
To determine which of these points corresponds to a minimum, we take the second derivative:
\[ \frac{d^2U}{dx^2} = 12x^2 - 4 \]
Now we evaluate this at the critical points:
- For \( x = 0 \):
\[ \frac{d^2U}{dx^2} = 12(0)^2 - 4 = -4 \quad (\text{local maximum}) \]
- For \( x = 1 \):
\[ \frac{d^2U}{dx^2} = 12(1)^2 - 4 = 8 \quad (\text{local minimum}) \]
- For \( x = -1 \):
\[ \frac{d^2U}{dx^2} = 12(-1)^2 - 4 = 8 \quad (\text{local minimum}) \]

Thus, the minimum potential energy occurs at \( x = 1 \) and \( x = -1 \).

### Step 5: Calculate Minimum Potential Energy
Now we calculate the potential energy at \( x = 1 \):
\[ U(1) = 1^4 - 2(1^2) = 1 - 2 = -1 \, \text{J} \]
And at \( x = -1 \):
\[ U(-1) = (-1)^4 - 2(-1)^2 = 1 - 2 = -1 \, \text{J} \]
So, the minimum potential energy is \( U_{min} = -1 \, \text{J} \).

### Step 6: Calculate Maximum Kinetic Energy
Using the total mechanical energy equation:
\[ K = E - U_{min} \]
Substituting the values:
\[ K = 5 - (-1) = 5 + 1 = 6 \, \text{J} \]

### Step 7: Relate Kinetic Energy to Velocity
The kinetic energy is given by:
\[ K = \frac{1}{2} mv^2 \]
Substituting \( K = 6 \, \text{J} \) and \( m = 2 \, \text{kg} \):
\[ 6 = \frac{1}{2} (2)v^2 \]
\[ 6 = v^2 \]
Thus, solving for \( v \):
\[ v = \sqrt{6} \, \text{m/s} \]

### Final Answer
The maximum velocity of the particle is:
\[ v = \sqrt{6} \, \text{m/s} \]

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