The PE of a 2 kg particle, free to move along x-axis is given by `V(x)=((x^(3))/(3)-(x^(2))/(2))J.` The total mechanical energy of the particle is 4 J. Maximum speed (in `ms^(-1)`) is

Options

A. `(1)/(sqrt2)`
B. `sqrt2`
C. `(3)/(sqrt2)`
D. `(5)/(sqrt6)`

Answer

To find the maximum speed of a 2 kg particle moving along the x-axis with a given potential energy function, we can follow these steps:

### Step 1: Understand the relationship between total mechanical energy, potential energy, and kinetic energy.
The total mechanical energy (E) of the system is the sum of its potential energy (V) and kinetic energy (K):
\[ E = V(x) + K \]

### Step 2: Set up the equation with the given total mechanical energy.
We know that the total mechanical energy is 4 J:
\[ 4 = V(x) + K \]

### Step 3: Write the expression for potential energy.
The potential energy is given by:
\[ V(x) = \frac{x^3}{3} - \frac{x^2}{2} \]

### Step 4: Find the condition for maximum kinetic energy.
Maximum kinetic energy occurs when potential energy is at a minimum. Therefore, we need to find the minimum value of \( V(x) \).

### Step 5: Differentiate the potential energy function to find critical points.
Differentiate \( V(x) \) with respect to \( x \):
\[ \frac{dV}{dx} = x^2 - x \]

### Step 6: Set the derivative equal to zero to find critical points.
Set the derivative equal to zero:
\[ x^2 - x = 0 \]
Factoring gives:
\[ x(x - 1) = 0 \]
This yields critical points at \( x = 0 \) and \( x = 1 \).

### Step 7: Determine which critical point gives the minimum potential energy.
To determine whether these points are minima or maxima, we can use the second derivative test:
\[ \frac{d^2V}{dx^2} = 2x \]
- At \( x = 0 \):
\[ \frac{d^2V}{dx^2} = 0 \] (point of inflection)
- At \( x = 1 \):
\[ \frac{d^2V}{dx^2} = 2 \] (positive, indicating a minimum)

Thus, \( x = 1 \) gives the minimum potential energy.

### Step 8: Calculate the minimum potential energy.
Substituting \( x = 1 \) into \( V(x) \):
\[ V(1) = \frac{1^3}{3} - \frac{1^2}{2} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6} \, \text{J} \]

### Step 9: Use the total mechanical energy to find maximum kinetic energy.
Substituting back into the total energy equation:
\[ 4 = -\frac{1}{6} + K_{\text{max}} \]
\[ K_{\text{max}} = 4 + \frac{1}{6} = \frac{24}{6} + \frac{1}{6} = \frac{25}{6} \, \text{J} \]

### Step 10: Relate kinetic energy to maximum speed.
The kinetic energy is given by:
\[ K = \frac{1}{2}mv^2 \]
Setting this equal to \( K_{\text{max}} \):
\[ \frac{1}{2} \cdot 2 \cdot v_{\text{max}}^2 = \frac{25}{6} \]
\[ v_{\text{max}}^2 = \frac{25}{6} \]
\[ v_{\text{max}} = \sqrt{\frac{25}{6}} = \frac{5}{\sqrt{6}} \, \text{m/s} \]

### Final Answer:
The maximum speed of the particle is \( \frac{5}{\sqrt{6}} \, \text{m/s} \).

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