For a particle moving on a curved path kinectic energy is given as k=AS where S is distance moved and A is constant quantity .Net force acting on particle is

Options

A. less than A
B. equal to A
C. more than A
D. may be less than or equal to A

Explanation

To solve the problem, we need to analyze the relationship between kinetic energy, distance moved, and the net force acting on the particle.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The kinetic energy (K) of a particle is given by the equation:
\[ K = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

2. **Given Kinetic Energy Relation**:
According to the problem, the kinetic energy is also expressed as:
\[ K = A S \]
where \( A \) is a constant and \( S \) is the distance moved.

3. **Equating the Two Expressions**:
From the two expressions for kinetic energy, we can set them equal to each other:
\[ \frac{1}{2} mv^2 = A S \]

4. **Differentiating with Respect to Time**:
To find the net force, we need to differentiate both sides with respect to time \( t \):
- The left-hand side:
\[ \frac{d}{dt}\left(\frac{1}{2} mv^2\right) = \frac{1}{2} m \cdot 2v \frac{dv}{dt} = mv \frac{dv}{dt} \]
- The right-hand side:
\[ \frac{d}{dt}(A S) = A \frac{dS}{dt} \]

5. **Setting the Derivatives Equal**:
Now we have:
\[ mv \frac{dv}{dt} = A \frac{dS}{dt} \]

6. **Relating to Force**:
The left-hand side \( mv \frac{dv}{dt} \) can be expressed as \( F \), the net force acting on the particle:
\[ F = A \frac{dS}{dt} \]

7. **Analyzing \( \frac{dS}{dt} \)**:
Since \( S \) represents the distance moved, \( \frac{dS}{dt} \) (the velocity) is always positive. Thus:
\[ F = A \cdot v \]
where \( v \) is the velocity of the particle.

8. **Conclusion About the Force**:
Since \( v \) is always positive, the net force \( F \) is always greater than \( A \):
\[ F > A \]

### Final Answer:
The net force acting on the particle is **greater than A**.