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, we need to find the maximum kinetic energy of a particle acted upon by the force \( F = F_0 e^{-kx} \). The particle is released from rest at \( x = 0 \).

### Step-by-Step Solution:

1. **Understanding the Force**:
The force acting on the particle is given by \( F = F_0 e^{-kx} \). This force is a function of position \( x \).

2. **Using Newton's Second Law**:
According to Newton's second law, \( F = ma \), where \( m \) is the mass of the particle and \( a \) is its acceleration. We can express acceleration \( a \) in terms of velocity \( v \) as:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]
Thus, we can rewrite the force equation as:
\[ F = m v \frac{dv}{dx} \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ \frac{F}{m} = v \frac{dv}{dx} \]
or
\[ \frac{F}{m} dx = v dv \]

4. **Integrating Both Sides**:
We will integrate both sides. The left side will be integrated from \( 0 \) to \( x \) and the right side from \( 0 \) to \( v \):
\[ \int_0^v v \, dv = \int_0^x \frac{F_0 e^{-kx}}{m} \, dx \]

5. **Calculating the Integrals**:
The integral on the left side is:
\[ \frac{v^2}{2} \]
The integral on the right side becomes:
\[ \frac{F_0}{m} \int_0^x e^{-kx} \, dx \]
The integral of \( e^{-kx} \) is:
\[ -\frac{1}{k} e^{-kx} \]
Evaluating from \( 0 \) to \( x \):
\[ -\frac{1}{k} (e^{-kx} - 1) = \frac{1}{k} (1 - e^{-kx}) \]

6. **Setting the Two Integrals Equal**:
Now we have:
\[ \frac{v^2}{2} = \frac{F_0}{m} \cdot \frac{1}{k} (1 - e^{-kx}) \]
Rearranging gives:
\[ v^2 = \frac{2F_0}{mk} (1 - e^{-kx}) \]

7. **Finding Maximum Kinetic Energy**:
The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} mv^2 = \frac{1}{2} m \left( \frac{2F_0}{mk} (1 - e^{-kx}) \right) = \frac{F_0}{k} (1 - e^{-kx}) \]
To find the maximum kinetic energy, we take the limit as \( x \) approaches infinity:
\[ \lim_{x \to \infty} K = \frac{F_0}{k} (1 - 0) = \frac{F_0}{k} \]

### Final Answer:
The maximum kinetic energy attained by the particle is:
\[ \boxed{\frac{F_0}{k}} \]