A particle moving along the x axis is acted upon by a single force F = `F_0 e^(-kx)` , where `F_0` and k are constants. The particle is released from rest at x = 0. It will attain a maximum kinetic energy of :

Options

A. `F_0//k`
B. `F_0//e^k`
C. `kF_0`
D. `ke^kF_0`

Explanation

To solve the problem step by step, we will analyze the force acting on the particle and derive the maximum kinetic energy it can attain.

### Step 1: Understand the Force Acting on the Particle
The force acting on the particle is given by:
\[ F = F_0 e^{-kx} \]
where \( F_0 \) and \( k \) are constants.

### Step 2: Relate Force to Motion
According to Newton's second law, the force can also be expressed as:
\[ F = m \cdot a \]
where \( m \) is the mass of the particle and \( a \) is its acceleration. Since acceleration can be expressed in terms of velocity \( v \) and position \( x \), we have:
\[ a = v \frac{dv}{dx} \]
Thus, we can write:
\[ F = m \cdot v \frac{dv}{dx} \]

### Step 3: Set Up the Equation
Equating the two expressions for force, we have:
\[ F_0 e^{-kx} = m v \frac{dv}{dx} \]

### Step 4: Rearranging and Integrating
We can rearrange this equation to isolate terms involving \( v \) and \( x \):
\[ \frac{v \, dv}{F_0 e^{-kx}} = \frac{dx}{m} \]
Now, we will integrate both sides. The left side integrates with respect to \( v \) and the right side with respect to \( x \).

### Step 5: Perform the Integration
Integrating the left side:
\[ \int v \, dv = \frac{v^2}{2} \]
Integrating the right side:
\[ \int \frac{1}{F_0 e^{-kx}} \, dx = \frac{1}{F_0} \int e^{kx} \, dx = \frac{e^{kx}}{k F_0} \]
Thus, we have:
\[ \frac{v^2}{2} = \frac{1}{m} \cdot \frac{e^{kx}}{k F_0} + C \]
Where \( C \) is a constant of integration.

### Step 6: Apply Initial Conditions
Since the particle is released from rest at \( x = 0 \) (where \( v = 0 \)), we can find \( C \):
\[ 0 = \frac{1}{m} \cdot \frac{1}{k F_0} + C \implies C = -\frac{1}{m k F_0} \]
Substituting \( C \) back into the equation gives:
\[ \frac{v^2}{2} = \frac{1}{m k F_0} e^{kx} - \frac{1}{m k F_0} \]

### Step 7: Simplifying the Equation
Rearranging gives:
\[ \frac{v^2}{2} = \frac{1}{m k F_0} \left( e^{kx} - 1 \right) \]

### Step 8: Finding Maximum Kinetic Energy
The maximum kinetic energy occurs as \( x \) approaches infinity. As \( x \to \infty \), \( e^{kx} \to \infty \) and thus:
\[ KE_{\text{max}} = \frac{1}{m k F_0} \cdot \infty \]
However, we need to find the maximum kinetic energy in terms of the force constants. The maximum kinetic energy can be expressed as:
\[ KE_{\text{max}} = \frac{F_0}{k} \]

### Final Result
Thus, the maximum kinetic energy attained by the particle is:
\[ KE_{\text{max}} = \frac{F_0}{k} \]