A particle of mass m moves along a circular path of radius r with a centripetal acceleration `a_n` changing with time t as `a_n=kt^2`, where k is a positive constant. The average power developed by all the forces acting on the particle during the first `t_0` seconds is

Options

A. (a) `mkrt_0`
B. (b) `(mkrt_0^2)/(2)`
C. (c) `(mkrt_0)/(2)`
D. (d) `(mkrt_0)/(4)`

Explanation

To solve the problem, we need to find the average power developed by all the forces acting on a particle of mass \( m \) moving along a circular path of radius \( r \) with a centripetal acceleration \( a_n = kt^2 \).

### Step-by-step Solution:

1. **Understanding Centripetal Acceleration**:
The centripetal acceleration \( a_n \) is given by:
\[ a_n = \frac{v^2}{r} \]
where \( v \) is the tangential speed of the particle and \( r \) is the radius of the circular path.

2. **Equating the Given Centripetal Acceleration**:
We set the expression for centripetal acceleration equal to the given function:
\[ \frac{v^2}{r} = kt^2 \]
Rearranging gives us:
\[ v^2 = krt^2 \]

3. **Finding Tangential Acceleration**:
The tangential acceleration \( a_t \) is the rate of change of speed, which can be expressed as:
\[ a_t = \frac{dv}{dt} \]
To find \( a_t \), we differentiate \( v \) with respect to \( t \):
\[ v = \sqrt{krt^2} = \sqrt{kr} \cdot t \]
Differentiating both sides with respect to \( t \):
\[ \frac{dv}{dt} = \sqrt{kr} \]

4. **Calculating Tangential Force**:
The tangential force \( F_t \) acting on the particle is given by:
\[ F_t = m \cdot a_t = m \cdot \frac{dv}{dt} = m \cdot \sqrt{kr} \]

5. **Calculating Power**:
The power \( P \) developed by the tangential force is given by:
\[ P = F_t \cdot v \]
Substituting the expressions for \( F_t \) and \( v \):
\[ P = m \cdot \sqrt{kr} \cdot \left(\sqrt{kr} \cdot t\right) = m \cdot kr \cdot t \]

6. **Finding Average Power Over Time \( t_0 \)**:
To find the average power developed during the first \( t_0 \) seconds, we substitute \( t = t_0 \):
\[ P_{\text{avg}} = m \cdot kr \cdot t_0 \]

### Final Answer:
The average power developed by all the forces acting on the particle during the first \( t_0 \) seconds is:
\[ P_{\text{avg}} = m \cdot kr \cdot t_0 \]