The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as K.E = KS2 where K is a constant. Find the force acting on the particle as a function of S -

Options

A. `(2as^(2))/(R)`
B. `2as(1+(s^(2))/(R^(2)))^(1//2)`
C. `as(1+(s^(2))/(R^(2)))^(1//2)`
D. None of these

Explanation

To solve the problem, we need to find the force acting on a particle moving in a circular path with a given kinetic energy expression. Let's break it down step by step.

### Step 1: Understand the Kinetic Energy Expression
The kinetic energy (K.E) of the particle is given by:
\[ K.E = K s^2 \]
where \( K \) is a constant and \( s \) is the distance covered along the circular path.

### Step 2: Relate Kinetic Energy to Velocity
We know that kinetic energy can also be expressed in terms of mass \( m \) and velocity \( v \):
\[ K.E = \frac{1}{2} m v^2 \]
Setting the two expressions for kinetic energy equal gives us:
\[ K s^2 = \frac{1}{2} m v^2 \]

### Step 3: Solve for Velocity
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{2 K s^2}{m} \]

### Step 4: Find the Tangential Acceleration
The tangential acceleration \( a_t \) is the rate of change of velocity with respect to time. Using the chain rule, we can express it as:
\[ a_t = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = \frac{dv}{ds} \cdot v \]
First, we need to find \( \frac{dv}{ds} \):
\[ v = \sqrt{\frac{2 K s^2}{m}} \]
Differentiating \( v \) with respect to \( s \):
\[ \frac{dv}{ds} = \frac{d}{ds} \left( \sqrt{\frac{2 K s^2}{m}} \right) = \frac{1}{2} \left( \frac{2 K}{m} \right)^{1/2} \cdot \frac{2s}{1} = \frac{K^{1/2} s}{m^{1/2}} \]

Now substituting back:
\[ a_t = \frac{K^{1/2} s}{m^{1/2}} \cdot \sqrt{\frac{2 K s^2}{m}} = \frac{2 K s^2}{m} \]

### Step 5: Find the Centripetal Acceleration
The centripetal acceleration \( a_c \) for circular motion is given by:
\[ a_c = \frac{v^2}{R} \]
Substituting the expression for \( v^2 \):
\[ a_c = \frac{2 K s^2}{mR} \]

### Step 6: Calculate the Net Acceleration
The net acceleration \( a_{net} \) is the vector sum of the tangential and centripetal accelerations:
\[ a_{net} = \sqrt{a_t^2 + a_c^2} \]
Substituting the expressions we found:
\[ a_{net} = \sqrt{\left(\frac{2 K s^2}{m}\right)^2 + \left(\frac{2 K s^2}{mR}\right)^2} \]
\[ = \frac{2 K s^2}{m} \sqrt{1 + \frac{1}{R^2}} \]

### Step 7: Find the Force
The net force \( F \) acting on the particle is given by:
\[ F = m a_{net} \]
Substituting \( a_{net} \):
\[ F = m \cdot \frac{2 K s^2}{m} \sqrt{1 + \frac{1}{R^2}} \]
\[ = 2 K s^2 \sqrt{1 + \frac{1}{R^2}} \]

### Final Result
The force acting on the particle as a function of \( s \) is:
\[ F(s) = 2 K s^2 \sqrt{1 + \frac{1}{R^2}} \]