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 solve the problem step by step, we need to find the maximum speed of a 2 kg particle given its potential energy function and total mechanical energy.

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

### Step 2: Write down the given potential energy function.
The potential energy function is given as:
\[ V(x) = \frac{x^3}{3} - \frac{x^2}{2} \, \text{J} \]

### Step 3: Set up the equation for total mechanical energy.
We know the total mechanical energy is 4 J. Therefore, we can write:
\[ 4 = KE + V(x) \]
This implies:
\[ KE = 4 - V(x) \]

### Step 4: Find the critical points of the potential energy function.
To find where the potential energy is minimized (which corresponds to maximum kinetic energy), we differentiate the potential energy function with respect to \(x\) and set it to zero:
\[ \frac{dV}{dx} = x^2 - x = 0 \]
Factoring gives:
\[ x(x - 1) = 0 \]
Thus, the critical points are:
\[ x = 0 \quad \text{and} \quad x = 1 \]

### Step 5: Determine the potential energy at the critical points.
Calculate \(V(0)\) and \(V(1)\):
- For \(x = 0\):
\[ V(0) = \frac{0^3}{3} - \frac{0^2}{2} = 0 \, \text{J} \]
- For \(x = 1\):
\[ V(1) = \frac{1^3}{3} - \frac{1^2}{2} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6} \, \text{J} \]

### Step 6: Identify the minimum potential energy.
The minimum potential energy occurs at \(x = 0\) since \(V(0) = 0\) is greater than \(V(1) = -\frac{1}{6}\).

### Step 7: Calculate the maximum kinetic energy.
Using the total mechanical energy equation:
\[ KE = 4 - V(0) = 4 - 0 = 4 \, \text{J} \]

### Step 8: Relate kinetic energy to speed.
The kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ 4 = \frac{1}{2} \times 2 \times v^2 \]
This simplifies to:
\[ 4 = v^2 \]
Thus,
\[ v = \sqrt{4} = 2 \, \text{m/s} \]

### Final Answer:
The maximum speed of the particle is \(2 \, \text{m/s}\).

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