The kinetic energy K of a particle moving along a circle of radius R depends upon the distance s as `K=as^2`. The force acting on the particle is

Options

A. (a) `2a(s^2)/(R)`
B. (b) `2as[1+(s^2)/(R)]^(1//2)`
C. (c) `2as`
D. (d) `2a`

Explanation

To find the force acting on a particle moving in a circle with kinetic energy given by \( K = as^2 \), we can follow these steps:

### Step 1: Relate Kinetic Energy to Velocity
The kinetic energy \( K \) of the particle can also be expressed in terms of its mass \( m \) and velocity \( v \):
\[ K = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ as^2 = \frac{1}{2} mv^2 \]

### Step 2: Solve for Velocity
Rearranging the equation to solve for \( v^2 \):
\[ mv^2 = 2as^2 \quad \Rightarrow \quad v^2 = \frac{2as^2}{m} \]

### Step 3: Differentiate with Respect to Time
Now, we differentiate both sides with respect to time \( t \):
\[ \frac{d}{dt}(mv^2) = \frac{d}{dt}(2as^2) \]
Using the chain rule on both sides:
\[ m \cdot 2v \frac{dv}{dt} = 2a \cdot 2s \frac{ds}{dt} \]
This simplifies to:
\[ 2mv \frac{dv}{dt} = 4as \frac{ds}{dt} \]

### Step 4: Substitute \( \frac{ds}{dt} \) with \( v \)
Since \( \frac{ds}{dt} = v \), we can substitute:
\[ 2mv \frac{dv}{dt} = 4asv \]
Dividing both sides by \( v \) (assuming \( v \neq 0 \)):
\[ 2m \frac{dv}{dt} = 4as \]

### Step 5: Express in Terms of Tangential Force
Recognizing that \( m \frac{dv}{dt} \) is the tangential force \( F_t \):
\[ F_t = 2as \]

### Step 6: Find Centripetal Force
The centripetal force \( F_c \) required to keep the particle moving in a circle is given by:
\[ F_c = \frac{mv^2}{R} \]
Substituting \( v^2 \) from earlier:
\[ F_c = \frac{m \cdot \frac{2as^2}{m}}{R} = \frac{2as^2}{R} \]

### Step 7: Calculate the Net Force
The net force \( F_{net} \) acting on the particle is the vector sum of the tangential force and the centripetal force. Since these forces are perpendicular:
\[ F_{net} = \sqrt{F_t^2 + F_c^2} \]
Substituting the values:
\[ F_{net} = \sqrt{(2as)^2 + \left(\frac{2as^2}{R}\right)^2} \]

### Step 8: Simplify the Expression
This simplifies to:
\[ F_{net} = \sqrt{4a^2s^2 + \frac{4a^2s^4}{R^2}} = 2a \sqrt{s^2 + \frac{s^4}{R^2}} \]

### Final Answer
Thus, the force acting on the particle is:
\[ F_{net} = 2a s \sqrt{1 + \frac{s^2}{R^2}} \]