A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to `(-K//r^(2))`, where k is a constant. The total energy of the particle is -

Options

A. `-k/r`
B. `-k/(2r)`
C. `k/(2r)`
D. `(2k)/r`

Explanation

To solve the problem of finding the total energy of a particle moving in a horizontal circle under a given centripetal force, we can break it down into a series of steps.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Mass of the particle: \( m \)
- Radius of the circle: \( r \)
- Centripetal force: \( F = -\frac{k}{r^2} \), where \( k \) is a constant.

2. **Relate Centripetal Force to Velocity:**
The centripetal force acting on the particle can also be expressed in terms of its velocity:
\[ F = \frac{m v^2}{r} \]
Setting the two expressions for force equal gives:
\[ \frac{m v^2}{r} = -\frac{k}{r^2} \]

3. **Solve for Velocity \( v \):**
Rearranging the equation to find \( v^2 \):
\[ m v^2 = -\frac{k}{r} \implies v^2 = -\frac{k}{mr} \]
Since \( v^2 \) must be positive, we take the absolute value:
\[ v^2 = \frac{k}{mr} \]

4. **Calculate Kinetic Energy \( KE \):**
The kinetic energy of the particle is given by:
\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{k}{mr}\right) = \frac{k}{2r} \]

5. **Find Potential Energy \( PE \):**
The potential energy can be found using the relationship between force and potential energy:
\[ F = -\frac{du}{dr} \]
Thus, we can express potential energy as:
\[ PE = -\int F \, dr = -\int -\frac{k}{r^2} \, dr \]
Evaluating the integral:
\[ PE = \int \frac{k}{r^2} \, dr = \left[-\frac{k}{r}\right] = -\frac{k}{r} + C \]
Assuming the potential energy at infinity is zero, we have:
\[ PE = -\frac{k}{r} \]

6. **Calculate Total Energy \( E \):**
The total energy is the sum of kinetic and potential energy:
\[ E = KE + PE = \frac{k}{2r} - \frac{k}{r} \]
Simplifying this expression:
\[ E = \frac{k}{2r} - \frac{2k}{2r} = -\frac{k}{2r} \]

### Final Answer:
The total energy of the particle is:
\[ E = -\frac{k}{2r} \]